Introduction
Principal component analysis (PCA) is a fundamental tool for dimensionality reduction, data visualization, and low-rank approximation [2 ] . Given a data matrix A ∈ R n × d A \in\mathbb{R}^{n \times d} A ∈ R n × d , PCA finds a low-dimensional subspace that captures as much of the variation in the data as possible. If A = U Σ V T A = U\Sigma V^{\mathsf{T}} A = U Σ V T is the singular value decomposition of A A A , the optimal rank-k k k subspace is spanned by the top k k k right singular vectors V k = [ v 1 ⋯ v k ] V_k = [v_1 \ \cdots\ v_k] V k = [ v 1 ⋯ v k ] . It is also the rank k k k subspace that maximizes the total squared lengths of the projections of the data.
When the rows of A A A contain information about individuals, releasing a subspace computed from A A A can reveal information about a particular row. This motivates differentially private algorithms for PCA. We consider row-level privacy, where each row represents the data of one individual. We assume that every row has ℓ 2 \ell_2 ℓ 2 norm at most 1, and two matrices are adjacent if one row is replaced. Normalizing the lengths of the rows is natural because we would like to measure the correlation with the entire dataset and no single data point should have an outsize contribution. The goal is to estimate V k V_k V k while protecting the privacy of every individual row. In this model, Dwork et al. [ [9 ] give a private rank-k k k PCA algorithm whose subspace error scales with O ~ ( d ) \widetilde{O}(\sqrt{d}) O ( d ) . Thus, the error can be large when the number of dimensions d d d is large. This motivates utility guarantees that exploit additional structure of the input matrix. One such structural parameter is coherence, which is related to the largest absolute entries of the left singular vectors U U U . Intuitively, coherence captures the difficulty in privatizing the singular vectors because it measures the maximum correlation between the singular vectors and individual data rows. The more the singular vectors reveal about one row, the more perturbation we need to protect privacy. On the other hand, for many natural datasets, such as when the data is close to i.i.d. data generated from a generative model, the coherence can be shown to be small. Previous work has shown that coherence-dependent guarantees can substantially improve on worst-case bounds on low-coherence data [Hardt and Roth, 2012, 2013, Hardt and Price, 2014, Balcan et al., 2016, Nicolas et al., 2025, d’Orsi and Novikov, 2026, Nguyen et al., 2026].
However, most earlier coherence-dependent guarantees were obtained under privacy models different from the row-level ℓ 2 \ell_2 ℓ 2 model considered in our paper. For example, Hardt and Roth [ [12 ] consider entry-level adjacency, under which neighboring matrices differ in a single entry by at most 1. Subsequent rank-k k k results use structured adjacency models that, when restricted to a one-row change, impose a ℓ 1 \ell_1 ℓ 1 -norm restriction on the changed row [Nicolas et al., 2025, d’Orsi and Novikov, 2026]. These models do not cover all neighboring datasets in our setting, since a row a a a satisfying ∥ a ∥ 2 ≤ 1 \|a\|_2\le1 ∥ a ∥ 2 ≤ 1 can have ∥ a ∥ 1 \|a\|_1 ∥ a ∥ 1 as large as d \sqrt{d} d . Nguyen et al. [ [17 ] recently gave the first coherence-adaptive guarantee under row-level ℓ 2 \ell_2 ℓ 2 privacy. However, their algorithm and analysis are specific to estimating a single leading singular vector, leaving the analogous rank-k k k problem open.
In this work, we give a coherence-adaptive algorithm for estimating the top-k k k right singular subspace under row-level ℓ 2 \ell_2 ℓ 2 differential privacy. Our result extends the coherence-adaptive utility guarantee of Nguyen et al. [ [17 ] from the leading singular vector to the top-k k k singular subspace while retaining the same privacy model. Our algorithm is based on a preconditioned and clipped variant of block power iteration. It does not require the coherence or the ratio σ k + 1 2 / σ k 2 \sigma_{k+1}^2/\sigma_k^2 σ k + 1 2 / σ k 2 as input, and its utility improves when the input has low coherence. The prior work is also suboptimal when the ratio σ k + 1 2 / σ k 2 \sigma_{k+1}^2/\sigma_k^2 σ k + 1 2 / σ k 2 is very close to 1 whereas our algorithm remains near optimal for all regimes of the ratio.
Our contributions
Let Δ k : = σ k 2 − σ k + 1 2 \Delta_k := \sigma_k^2 - \sigma_{k+1}^2 Δ k := σ k 2 − σ k + 1 2 and let ∥ ( I − X ^ X ^ T ) V k ∥ 2 \left\lVert(I-\widehat{X}\widehat{X}^{\mathsf{T}})V_k \right\rVert_2 ( I − X X T ) V k 2 denote the error of an output X ^ ∈ R d × k \widehat{X} \in\mathbb{R}^{d \times k} X ∈ R d × k with orthonormal columns. Let a i T a_i^{\mathsf{T}} a i T denote the i i i th row of A A A , and recall that ∥ a i ∥ 2 ≤ 1 \lVert a_i \rVert_2 \le1 ∥ a i ∥ 2 ≤ 1 . We write U k = U : , 1 : k U_k = U_{:,1:k} U k = U : , 1 : k , Σ k = diag ( σ 1 , … , σ k ) \Sigma_k = \operatorname{diag}(\sigma_1,\ldots,\sigma_k) Σ k = diag ( σ 1 , … , σ k ) , ∥ U ∥ ∞ = max i , j ∣ U i j ∣ \lVert U \rVert_\infty= \max_{i,j} |U_{ij}| ∥ U ∥ ∞ = max i , j ∣ U ij ∣ . For the i i i th standard basis vector e i e_i e i , we have ∥ e i T U k ∥ 2 ≤ k ∥ U ∥ ∞ \lVert e_i^{\mathsf{T}} U_k \rVert_2 \le\sqrt{k} \lVert U \rVert_\infty ∥ e i T U k ∥ 2 ≤ k ∥ U ∥ ∞ .
Our main result is the following:
Theorem 1. Let ϵ , δ , β ∈ ( 0 , 1 ) \epsilon,\delta,\beta\in(0,1) ϵ , δ , β ∈ ( 0 , 1 ) , and suppose that every row of A A A has ℓ 2 \ell_2 ℓ 2 norm at most 1. There is an ( ϵ , δ ) (\epsilon,\delta) ( ϵ , δ ) -differentially private algorithm that, under the condition
Δ k ≥ Ω ~ ( d ϵ + ( σ k k ϵ ) 2 / 3 + ( k σ k 2 ∥ U ∥ ∞ ϵ ) 2 / 3 ) , \Delta_k \ge\widetilde{\Omega}\left(\frac{\sqrt{d}}{\epsilon}+\left(\frac{\sigma_k\sqrt{k}}{\epsilon}\right)^{2/3}+\left(\frac{k\sigma_k^2\lVert U\rVert_\infty}{\epsilon}\right)^{2/3}\right), Δ k ≥ Ω ϵ d + ( ϵ σ k k ) 2/3 + ( ϵ k σ k 2 ∥ U ∥ ∞ ) 2/3 , outputs X ^ ∈ R d × k \widehat{X} \in\mathbb{R}^{d \times k} X ∈ R d × k satisfying, with probability at least 1 − β 1-\beta 1 − β ,
∥ ( I − X ^ X ^ T ) V k ∥ 2 2 ≤ ( β δ n ) 2 + O ~ ( d k σ k 2 ∥ U ∥ ∞ 2 + 1 ϵ 2 Δ k 2 ) . \left\lVert(I-\widehat{X}\widehat{X}^{\mathsf{T}})V_k \right\rVert_2^2 \le\left(\frac{\beta\delta}{n}\right)^2+\widetilde{O}\left(\frac{dk\sigma_k^2\lVert U\rVert_\infty^2+1}{\epsilon^2\Delta_k^2}\right). ( I − X X T ) V k 2 2 ≤ ( n β δ ) 2 + O ( ϵ 2 Δ k 2 d k σ k 2 ∥ U ∥ ∞ 2 + 1 ) . The algorithm does not require ∥ U ∥ ∞ \lVert U\rVert_\infty ∥ U ∥ ∞ , Δ k \Delta_k Δ k , or σ k + 1 2 / σ k 2 \sigma_{k+1}^2/\sigma_k^2 σ k + 1 2 / σ k 2 as input.
Comparison with prior works
We additionally write ∥ V ∥ ∞ : = max i , j ∣ V i j ∣ \lVert V\rVert_\infty:= \max_{i,j}|V_{ij}| ∥ V ∥ ∞ := max i , j ∣ V ij ∣ . We summarize our results in comparison with prior work in Table 1 .
Table 1. Comparison of deterministic rank-k k k PCA guarantees. The error is ∥ ( I − X ^ X ^ ⊤ ) V k ∥ 2 2 = sin 2 ( θ max ) \|(I-\widehat{X}\widehat{X}^{\top})V_k\|_2^2=\sin^2(\theta_{\max}) ∥ ( I − X X ⊤ ) V k ∥ 2 2 = sin 2 ( θ m a x ) where θ max \theta_{\max} θ m a x is the largest principal angle between the column spaces of X ^ \widehat{X} X and V k V_k V k . Poly-logarithmic factors in n n n , d d d , 1 ϵ \frac{1}{\epsilon} ϵ 1 , 1 δ \frac{1}{\delta} δ 1 are suppressed. We let Δ k : = σ k 2 − σ k + 1 2 \Delta_k:=\sigma_k^2-\sigma_{k+1}^2 Δ k := σ k 2 − σ k + 1 2 .
Comparison with the row-ℓ 2 \ell_2 ℓ 2 baseline. The most direct general rank-k k k comparison is the Analyze Gauss algorithm of Dwork et al. [ [9 ] , since it provides privacy under the same row-ℓ 2 \ell_2 ℓ 2 model. Our result replaces the worst-case dependence on d d d by a dependence on the coherence of the input. Subject to the Δ k \Delta_k Δ k condition in Theorem 1, our bound improves on Analyze Gauss when k σ k 2 ∥ U ∥ ∞ 2 k\sigma_k^2\lVert U\rVert_\infty^2 k σ k 2 ∥ U ∥ ∞ 2 is small. Analyze Gauss does not require low coherence and can give the stronger guarantee outside this regime.
Comparison under other privacy models. The results of Hardt and Price [ [10 ] , d’Orsi and Novikov [2026] provide coherence-dependent guarantees for general rank k k k , but they use privacy models different from ours. Hardt and Price [ [10 ] consider entry-level adjacency. Any replacement between two rows of ℓ 2 \ell_2 ℓ 2 norm at most 1 can be implemented using at most 2 d 2d 2 d entry changes by passing through the zero row. Group privacy therefore converts their guarantee to our row-level model by setting ϵ e n t = Θ ( ϵ / d ) \epsilon_{\mathrm{ent}}=\Theta(\epsilon/d) ϵ ent = Θ ( ϵ / d ) and adjusting δ e n t \delta_{\mathrm{ent}} δ ent accordingly. So Hardt and Price [ [10 ] have a converted sin 2 \sin^2 sin 2 bound of:
O ~ ( k 3 σ k 4 max { n ∥ U ∥ ∞ 2 , d ∥ V ∥ ∞ 2 } ( ϵ / d ) 2 Δ k 3 ) = O ~ ( d 2 k 3 σ k 4 max { n ∥ U ∥ ∞ 2 , d ∥ V ∥ ∞ 2 } ϵ 2 Δ k 3 ) . \widetilde{O}\left(\frac{k^3\sigma_k^4\max\{n\lVert U\rVert_\infty^2,d\lVert V\rVert_\infty^2\}}{(\epsilon/d)^2\Delta_k^3}\right)
=
\widetilde{O}\left(\frac{d^2k^3\sigma_k^4\max\{n\lVert U\rVert_\infty^2,d\lVert V\rVert_\infty^2\}}{\epsilon^2\Delta_k^3}\right). O ( ( ϵ / d ) 2 Δ k 3 k 3 σ k 4 max { n ∥ U ∥ ∞ 2 , d ∥ V ∥ ∞ 2 } ) = O ( ϵ 2 Δ k 3 d 2 k 3 σ k 4 max { n ∥ U ∥ ∞ 2 , d ∥ V ∥ ∞ 2 } ) .
For d’Orsi and Novikov [2026] model, the replacement of row a a a by a ′ a' a ′ will result in: ∥ E E T ∥ 1 = ∥ a − a ′ ∥ 2 2 + ∥ a − a ′ ∥ 2 2 = O ( d ∥ a − a ′ ∥ 2 2 ) = O ( d ) \sqrt{\lVert EE^{\mathsf{T}}\rVert_1}=\sqrt{\lVert a-a'\rVert_2^2+\lVert a-a'\rVert_2^2}=O\left(\sqrt{d\lVert a-a'\rVert_2^2}\right)=O(\sqrt{d}) ∥ E E T ∥ 1 = ∥ a − a ′ ∥ 2 2 + ∥ a − a ′ ∥ 2 2 = O ( d ∥ a − a ′ ∥ 2 2 ) = O ( d ) . Thus it suffices to set Δ d n = O ( d ) \Delta_{d n}=O(\sqrt{d}) Δ d n = O ( d ) , leading to sin 2 \sin^2 sin 2 error of:
O ~ ( d k σ k 2 max { n ∥ U ∥ ∞ 2 , d ∥ V ∥ ∞ 2 } ϵ 2 Δ k 2 ) . \widetilde{O}\left(\frac{dk\sigma_k^2\max\{n\lVert U\rVert_\infty^2,d\lVert V\rVert_\infty^2\}}{\epsilon^2\Delta_k^2}\right). O ( ϵ 2 Δ k 2 d k σ k 2 max { n ∥ U ∥ ∞ 2 , d ∥ V ∥ ∞ 2 } ) .
For a replacement of row a a a by row a ′ a' a ′ , ∥ E E T ∥ 1 = ∥ a − a ′ ∥ 2 2 + ∥ a − a ′ ∥ 2 2 \sqrt{\lVert EE^{\mathsf{T}}\rVert_1}=\sqrt{\lVert a-a'\rVert_2^2+\lVert a-a'\rVert_2^2} ∥ E E T ∥ 1 = ∥ a − a ′ ∥ 2 2 + ∥ a − a ′ ∥ 2 2 .
The above bound is a factor n n n larger than ours. Note that while this bound does not require any condition, it is only meaningful if the fraction inside O ~ \widetilde{O} O is less than 1 1 1 , implying that the gap Δ k ≥ Ω ( d k σ k 2 max { n ∥ U ∥ ∞ 2 , d ∥ V ∥ ∞ 2 } / ϵ ) \Delta_k \ge\Omega\left(\sqrt{d k \sigma_k^2 \max\{n\|U\|_\infty^2,d\|V\|_\infty^2\}/\epsilon}\right) Δ k ≥ Ω ( d k σ k 2 max { n ∥ U ∥ ∞ 2 , d ∥ V ∥ ∞ 2 } / ϵ ) . This can be significant e.g. when k = 1 k=1 k = 1 , the incoherence n ∥ U ∥ ∞ 2 = O ~ ( 1 ) n\|U\|_\infty^2=\widetilde{O}(1) n ∥ U ∥ ∞ 2 = O ( 1 ) (the minimum possible value and what we obtain with i.i.d. data), we still need Δ k ≥ σ 1 d / ϵ \Delta_k \ge\sigma_1\sqrt{d}/\epsilon Δ k ≥ σ 1 d / ϵ . In contrast, the constraint for ours in the same setting is Δ k = Ω ~ ( d / ϵ + ( σ 1 / ϵ ) 2 / 3 + σ 1 4 / 3 / ( n 1 / 3 ϵ 2 / 3 ) ) \Delta_k=\widetilde{\Omega}\left(\sqrt{d/\epsilon}+(\sigma_1/\epsilon)^{2/3}+\sigma_1^{4/3}/(n^{1/3}\epsilon^{2/3})\right) Δ k = Ω ( d / ϵ + ( σ 1 / ϵ ) 2/3 + σ 1 4/3 / ( n 1/3 ϵ 2/3 ) ) , which is smaller.
Comparison for k = 1 k=1 k = 1 . The result of Nguyen et al. [2026] uses the same row-ℓ 2 \ell_2 ℓ 2 privacy model as ours and is the most relevant comparison when k = 1 k=1 k = 1 . Specializing our result to k = 1 k=1 k = 1 gives a comparable coherence-dependent guarantee. Our requirement is more stringent as we need Δ 1 = Ω ~ ( d / ϵ ) \Delta_1=\widetilde{\Omega}\left(\sqrt{d/\epsilon}\right) Δ 1 = Ω ( d / ϵ ) but our bound is better whenever the gap is small Δ 1 < σ 1 2 / d \Delta_1<\sigma_1^2/d Δ 1 < σ 1 2 / d .
Our approach and techniques
Let a i ⊤ a_i^\top a i ⊤ denote the i i i th row of A A A , and recall that ∥ a i ∥ 2 ≤ 1 \|a_i\|_2\le1 ∥ a i ∥ 2 ≤ 1 . We write U k = U : , 1 : k U_k=U_{:,1:k} U k = U : , 1 : k , Σ k = diag ( σ 1 , … , σ k ) \Sigma_k=\operatorname{diag}(\sigma_1,\ldots,\sigma_k) Σ k = diag ( σ 1 , … , σ k ) , ∥ U ∥ ∞ = max i , j ∣ U i j ∣ \|U\|_\infty=\max_{i,j}|U_{ij}| ∥ U ∥ ∞ = max i , j ∣ U ij ∣ . For the i i i th standard basis vector e i e_i e i , we have ∥ e i ⊤ U k ∥ 2 ≤ k ∥ U ∥ ∞ \|e_i^\top U_k\|_2\le\sqrt{k}\|U\|_\infty ∥ e i ⊤ U k ∥ 2 ≤ k ∥ U ∥ ∞ . Thus, when ∥ U ∥ ∞ \|U\|_\infty ∥ U ∥ ∞ is small, every row of U k U_k U k has small norm.
A standard noisy power method [10 ] maintains a matrix X ( t ) ∈ R d × k X^{(t)}\in\mathbb{R}^{d\times k} X ( t ) ∈ R d × k with orthonormal columns and performs an update of the form
Z ( t + 1 ) = A ⊤ A X ( t ) + Gaussian noise , Z^{(t+1)}=A^\top A X^{(t)}+\text{Gaussian noise}, Z ( t + 1 ) = A ⊤ A X ( t ) + Gaussian noise ,
X ( t + 1 ) R ( t + 1 ) X^{(t+1)}R^{(t+1)} X ( t + 1 ) R ( t + 1 ) be QR-decomposition of Z ( t + 1 ) Z^{(t+1)} Z ( t + 1 )
Since A ⊤ A X ( t ) = ∑ i = 1 n a i ( a i ⊤ X ( t ) ) A^\top A X^{(t)}=\sum_{i=1}^n a_i(a_i^\top X^{(t)}) A ⊤ A X ( t ) = ∑ i = 1 n a i ( a i ⊤ X ( t ) ) , the contribution of row a i a_i a i to the update has Frobenius norm ∥ a i ( a i ⊤ X ( t ) ) ∥ F = ∥ a i ∥ 2 ∥ a i ⊤ X ( t ) ∥ 2 \|a_i(a_i^\top X^{(t)})\|_F=\|a_i\|_2\|a_i^\top X^{(t)}\|_2 ∥ a i ( a i ⊤ X ( t ) ) ∥ F = ∥ a i ∥ 2 ∥ a i ⊤ X ( t ) ∥ 2 .
A direct implementation under row-level privacy would therefore calibrate the noise using bounds on these row contributions. The difficulty is that these contributions are not controlled by coherence alone. To see the difficulty, suppose that X ( t ) X^{(t)} X ( t ) has already reached the target V k V_k V k , so that X ( t ) = V k X^{(t)}=V_k X ( t ) = V k . The contribution of row a i a_i a i then has norm
∥ a i ( a i ⊤ V k ) ∥ F = ∥ a i ( e i ⊤ A V k ) ∥ F = ∥ a i e i ⊤ U k Σ k ∥ F = ∥ a i ∥ 2 ∥ e i ⊤ U k Σ k ∥ 2 \|a_i(a_i^\top V_k)\|_F=\|a_i(e_i^\top A V_k)\|_F=\|a_i e_i^\top U_k\Sigma_k\|_F=\|a_i\|_2\|e_i^\top U_k\Sigma_k\|_2 ∥ a i ( a i ⊤ V k ) ∥ F = ∥ a i ( e i ⊤ A V k ) ∥ F = ∥ a i e i ⊤ U k Σ k ∥ F = ∥ a i ∥ 2 ∥ e i ⊤ U k Σ k ∥ 2
Low coherence controls ∥ e i ⊤ U k ∥ 2 ≤ k ∥ U ∥ ∞ \|e_i^\top U_k\|_2\le\sqrt{k}\|U\|_\infty ∥ e i ⊤ U k ∥ 2 ≤ k ∥ U ∥ ∞ , but the direct update contains
∥ e i ⊤ U k Σ k ∥ 2 = ( ∑ j = 1 k σ j 2 U i j 2 ) 1 / 2 ≤ σ 1 ∥ e i ⊤ U k ∥ 2 ≤ σ 1 k ∥ U ∥ ∞ \|e_i^\top U_k\Sigma_k\|_2=\left(\sum_{j=1}^k \sigma_j^2 U_{ij}^2\right)^{1/2}\le\sigma_1\|e_i^\top U_k\|_2\le\sigma_1\sqrt{k}\|U\|_\infty ∥ e i ⊤ U k Σ k ∥ 2 = ( j = 1 ∑ k σ j 2 U ij 2 ) 1/2 ≤ σ 1 ∥ e i ⊤ U k ∥ 2 ≤ σ 1 k ∥ U ∥ ∞
This gives ∥ a i ( a i ⊤ V k ) ∥ F ≤ σ 1 k ∥ U ∥ ∞ \|a_i(a_i^\top V_k)\|_F\le\sigma_1\sqrt{k}\|U\|_\infty ∥ a i ( a i ⊤ V k ) ∥ F ≤ σ 1 k ∥ U ∥ ∞ . Thus, the Gaussian noise adjusted to the above upper bound inherits a dependence on σ 1 \sigma_1 σ 1 . On the other hand, in the normal power method, the subspace error contracts at rate σ k + 1 2 / σ k 2 \sigma_{k+1}^2/\sigma_k^2 σ k + 1 2 / σ k 2 . This creates the main difficulty in analyzing the above power method: the bound on Gaussian noise can depend on σ 1 \sigma_1 σ 1 , whereas the recurrence for the subspace error depends on σ k \sigma_k σ k and σ k + 1 \sigma_{k+1} σ k + 1 . This is not an issue for the prior work for k = 1 k=1 k = 1 but it is a major issue for our setting.
Preconditioner. Our main idea is to remove the Σ k \Sigma_k Σ k term on the contribution of row a i a_i a i , using the preconditioner method. Our new update has this form:
H ( t ) = X ( t ) ⊤ A ⊤ A X ( t ) + symmetric Gaussian noise , H^{(t)}=X^{(t)\top}A^\top A X^{(t)}+\text{symmetric Gaussian noise}, H ( t ) = X ( t ) ⊤ A ⊤ A X ( t ) + symmetric Gaussian noise ,
H ~ ( t ) = H ( t ) with every eigenvalue below γ replaced by γ , \widetilde{H}^{(t)}=H^{(t)}\text{ with every eigenvalue below }\gamma\text{ replaced by }\gamma, H ( t ) = H ( t ) with every eigenvalue below γ replaced by γ ,
B ( t ) = ( H ~ ( t ) ) − 1 / 2 , B^{(t)}=\left(\widetilde{H}^{(t)}\right)^{-1/2}, B ( t ) = ( H ( t ) ) − 1/2 ,
Z ( t + 1 ) = A ⊤ A X ( t ) B ( t ) + Gaussian noise , Z^{(t+1)}=A^\top A X^{(t)}B^{(t)}+\text{Gaussian noise}, Z ( t + 1 ) = A ⊤ A X ( t ) B ( t ) + Gaussian noise ,
X ( t + 1 ) R ( t + 1 ) be QR-decomposition of Z ( t + 1 ) X^{(t+1)}R^{(t+1)}\text{ be QR-decomposition of }Z^{(t+1)} X ( t + 1 ) R ( t + 1 ) be QR-decomposition of Z ( t + 1 )
where the second step is to ensure that the inverse of H ~ ( t ) \widetilde{H}^{(t)} H ( t ) exists and B ( t ) B^{(t)} B ( t ) is defined. The parameter γ \gamma γ is set roughly the same as the magnitude of the symmetric Gaussian noise so that we remove the artifact from the noise and the components smaller than the noise.
To explain the role of the preconditioner B ( t ) B^{(t)} B ( t ) , first suppose that the Gaussian noise is absent and H ~ ( t ) = H ( t ) = X ( t ) ⊤ A ⊤ A X ( t ) \widetilde{H}^{(t)}=H^{(t)}=X^{(t)\top}A^\top AX^{(t)} H ( t ) = H ( t ) = X ( t ) ⊤ A ⊤ A X ( t ) . Provided that X ( t ) ⊤ A ⊤ A X ( t ) X^{(t)\top}A^\top AX^{(t)} X ( t ) ⊤ A ⊤ A X ( t ) is positive definite, we have B ( t ) = ( X ( t ) ⊤ A ⊤ A X ( t ) ) − 1 / 2 B^{(t)}=(X^{(t)\top}A^\top AX^{(t)})^{-1/2} B ( t ) = ( X ( t ) ⊤ A ⊤ A X ( t ) ) − 1/2 , and the columns of A X ( t ) B ( t ) AX^{(t)}B^{(t)} A X ( t ) B ( t ) are orthonormal. Moreover, since B ( t ) B^{(t)} B ( t ) is invertible, range ( A ⊤ A X ( t ) B ( t ) ) = range ( A ⊤ A X ( t ) ) \operatorname{range}(A^\top AX^{(t)}B^{(t)})=\operatorname{range}(A^\top AX^{(t)}) range ( A ⊤ A X ( t ) B ( t ) ) = range ( A ⊤ A X ( t ) ) . Thus, in this idealized iteration, the preconditioner does not change the next subspace produced by the block power method. Instead, it changes the basis in which a i ⊤ X ( t ) B ( t ) a_i^\top X^{(t)}B^{(t)} a i ⊤ X ( t ) B ( t ) is measured.
At X ( t ) = V k X^{(t)}=V_k X ( t ) = V k , we have B ( t ) = ( V k ⊤ A ⊤ A V k ) − 1 / 2 = Σ k − 1 B^{(t)}=(V_k^\top A^\top AV_k)^{-1/2}=\Sigma_k^{-1} B ( t ) = ( V k ⊤ A ⊤ A V k ) − 1/2 = Σ k − 1 and therefore a i ⊤ V k B ( t ) = e i ⊤ U k a_i^\top V_kB^{(t)}=e_i^\top U_k a i ⊤ V k B ( t ) = e i ⊤ U k . The contribution of row a i a_i a i consequently has norm
∥ a i ( a i ⊤ V k B ( t ) ) ∥ F = ∥ a i ∥ 2 ∥ e i ⊤ U k ∥ 2 \left\|a_i\left(a_i^\top V_kB^{(t)}\right)\right\|_F=\|a_i\|_2\left\|e_i^\top U_k\right\|_2 a i ( a i ⊤ V k B ( t ) ) F = ∥ a i ∥ 2 e i ⊤ U k 2
which is controlled by the coherence and no longer contains σ 1 , … , σ k \sigma_1,\ldots,\sigma_k σ 1 , … , σ k . When the iterate is close to V k V_k V k and B ( t ) B^{(t)} B ( t ) is computed from the noisy k × k k\times k k × k matrix, this identity does not hold exactly. Our analysis instead shows that the part of a i ⊤ X ( t ) B ( t ) a_i^\top X^{(t)}B^{(t)} a i ⊤ X ( t ) B ( t ) from V k V_k V k remains bounded by a constant multiple of a coherence-related quantity. The remaining part of a i ⊤ X ( t ) B ( t ) a_i^\top X^{(t)}B^{(t)} a i ⊤ X ( t ) B ( t ) from V > k V_{>k} V > k is controlled through a separate row-wise analysis. This allows the final error bound to depend on σ k \sigma_k σ k , rather than on all of the leading singular values.
Clipping. The preceding bounds hold with high probability when the iterates remain close to V k V_k V k . Differential privacy, however, requires a sensitivity bound that holds for every input. For a row vector y ∈ R 1 × k y\in\mathbb{R}^{1\times k} y ∈ R 1 × k and a radius R > 0 R>0 R > 0 , define Clip R ( y ) = y min { 1 , R ∥ y ∥ 2 } \operatorname{Clip}_R(y)=y\min\left\{1,\frac{R}{\|y\|_2}\right\} Clip R ( y ) = y min { 1 , ∥ y ∥ 2 R } , with Clip R ( 0 ) = 0 \operatorname{Clip}_R(0)=0 Clip R ( 0 ) = 0 . Our update has the form:
Z ( t + 1 ) = ∑ i = 1 n a i Clip R ( a i ⊤ X ( t ) B ( t ) ) + Gaussian noise . Z^{(t+1)}=\sum_{i=1}^{n}a_i\operatorname{Clip}_R\left(a_i^\top X^{(t)}B^{(t)}\right)+\text{Gaussian noise}. Z ( t + 1 ) = i = 1 ∑ n a i Clip R ( a i ⊤ X ( t ) B ( t ) ) + Gaussian noise .
Since ∥ a i ∥ 2 ≤ 1 \|a_i\|_2\le1 ∥ a i ∥ 2 ≤ 1 , ∥ a i Clip R ( a i ⊤ X ( t ) B ( t ) ) ∥ F ≤ R \left\|a_i\operatorname{Clip}_R\left(a_i^\top X^{(t)}B^{(t)}\right)\right\|_F\le R a i Clip R ( a i ⊤ X ( t ) B ( t ) ) F ≤ R . Under replacement adjacency, the sensitivity of the sum is therefore at most 2 R 2R 2 R , and the Gaussian noise can be adjust to this sensitivity.
Clipping supplies a worst-case sensitivity bound but is never active with high probability. In particular, under the assumptions of our main result, our analysis gives a radius R = O ~ ( k ∥ U ∥ ∞ ) R=\widetilde{O}\left(\sqrt{k}\|U\|_\infty\right) R = O ( k ∥ U ∥ ∞ ) for which ∥ a i ⊤ X ( t ) B ( t ) ∥ 2 ≤ R \left\|a_i^\top X^{(t)}B^{(t)}\right\|_2\le R a i ⊤ X ( t ) B ( t ) 2 ≤ R for every row and every iteration, with high probability.
Private initialization. The preconditioner and row-wise bounds described above apply only when the current iterate has sufficient overlap with the target subspace. We obtain such an initialization using the Gaussian mechanism of Dwork et al. [ [9 ] . In particular, we add symmetric Gaussian noise to A ⊤ A A^\top A A ⊤ A and let X ( 0 ) X^{(0)} X ( 0 ) contain the top k k k eigenvectors of the resulting matrix. Under an initial condition, this initialization satisfies ∥ I − X ( 0 ) ( X ( 0 ) ) ⊤ ∥ 2 ≤ 1 / 4 \left\|I-X^{(0)}(X^{(0)})^\top\right\|_2\le1/4 I − X ( 0 ) ( X ( 0 ) ) ⊤ 2 ≤ 1/4 .
Private selection of R R R and T T T . The appropriate choices of R R R and T T T depend on private properties of the input: R R R depends on the coherence, while T T T depends on σ k + 1 2 / σ k 2 \sigma_{k+1}^2/\sigma_k^2 σ k + 1 2 / σ k 2 . We consider public powers-of-two grids for R R R and T T T and use the private selection method of Papernot and Steinke [ [19 ] to choose among the resulting outputs. The construction and analysis of this procedure are given in the algorithm section. Thus, our algorithm does not require the coherence or σ k + 1 2 / σ k 2 \sigma_{k+1}^2/\sigma_k^2 σ k + 1 2 / σ k 2 as input.
Analysis technique. The quality of the solution X ( t ) X^{(t)} X ( t ) is evaluated via the ratio between its projection on the top k k k subspace C = V k ⊤ X ( t ) C=V_k^\top X^{(t)} C = V k ⊤ X ( t ) and its projection on the bottom d − k d-k d − k subspace D = V > k ⊤ X ( t ) D=V_{>k}^\top X^{(t)} D = V > k ⊤ X ( t ) . Let E ( t ) = D ( t ) ( C ( t ) ) − 1 E^{(t)}=D^{(t)}(C^{(t)})^{-1} E ( t ) = D ( t ) ( C ( t ) ) − 1 . We track ∥ E ( t ) ∥ 2 \|E^{(t)}\|_2 ∥ E ( t ) ∥ 2 over the iterations, ideally showing that it decreases exponentially. If no clipping happens, we prove the recurrence E ( t + 1 ) = ( Σ > k E ( t ) Σ k − 2 + noise ) ( I + noise ) − 1 E^{(t+1)}=(\Sigma_{>k}E^{(t)}\Sigma_k^{-2}+\text{noise})(I+\text{noise})^{-1} E ( t + 1 ) = ( Σ > k E ( t ) Σ k − 2 + noise ) ( I + noise ) − 1 where the noise terms come from the Gaussian noise in the update for Z ( t + 1 ) Z^{(t+1)} Z ( t + 1 ) . Without noise, we can see that ∥ Σ > k 2 E ( t ) Σ k − 2 ∥ 2 ≤ σ k + 1 2 σ k 2 ∥ E ( t ) ∥ 2 = ( 1 − ρ k ) ∥ E ( t ) ∥ 2 \|\Sigma_{>k}^2E^{(t)}\Sigma_k^{-2}\|_2\le\frac{\sigma_{k+1}^2}{\sigma_k^2}\|E^{(t)}\|_2=(1-\rho_k)\|E^{(t)}\|_2 ∥ Σ > k 2 E ( t ) Σ k − 2 ∥ 2 ≤ σ k 2 σ k + 1 2 ∥ E ( t ) ∥ 2 = ( 1 − ρ k ) ∥ E ( t ) ∥ 2 , indeed decreasing exponentially with iterations. Naively bounding the noises per iteration would create an error term linear in the number of iterations, which is too large. In different iterations, we add independent noises so intuitively, these noises are nearly orthogonal to each others and have a lot of cancellation. To exploit the cancellation, we need to perform a global analysis involving all iterations together. However, the dependence on the noise in each iteration is complex with a term in the numerator and a term in the denominator. Using ( I + F ) − 1 = I − F ( I + F ) − 1 (I+F)^{-1}=I-F(I+F)^{-1} ( I + F ) − 1 = I − F ( I + F ) − 1 for any noise matrix F F F , the noise contribution is split into two parts: one part being added up linearly across all iterations, which is amenable to a martingale analysis, and one complicated part, which is bounded separately per iteration. The second part is small so even though the bound might be suboptimal, it is sufficiently small to not matter. This analysis is possible only if we never clip. First we show that clipping does not happen for the initial solution using Gaussian concentration. Then we use induction to show that clipping does not happen for the rest of the iterations.
Preliminaries
Let A ∈ R n × d A\in\mathbb{R}^{n\times d} A ∈ R n × d be the input data matrix whose rows a i a_i a i are datapoints in R d \mathbb{R}^d R d . We assume that n ≥ d n\ge d n ≥ d and each row a i a_i a i of A A A has length ∥ a ∥ 2 ≤ 1 \|a\|_2\le1 ∥ a ∥ 2 ≤ 1 . As noted by Dwork et al. [ [9 ] , the latter assumption is necessary for achieving non-trivial utility guarantees and it is a natural restriction for PCA, since otherwise the solution could be dominated by a single datapoint with very large norm and it could fail to capture the underlying structure shared by most of the datapoints.
Notation. Let A = U Σ V ⊤ A=U\Sigma V^{\top} A = U Σ V ⊤ be the thin singular value decomposition (SVD) of A A A , where U ∈ R n × d U\in\mathbb{R}^{n\times d} U ∈ R n × d is the matrix whose columns are the left-singular vectors u 1 , … , u d u_1,\ldots,u_d u 1 , … , u d , V ∈ R d × d V\in\mathbb{R}^{d\times d} V ∈ R d × d is the matrix whose columns are the right-singular vectors v 1 , … , v d v_1,\ldots,v_d v 1 , … , v d , and Σ = diag ( σ 1 , … , σ d ) ∈ R d × d \Sigma=\operatorname{diag}(\sigma_1,\ldots,\sigma_d)\in\mathbb{R}^{d\times d} Σ = diag ( σ 1 , … , σ d ) ∈ R d × d where σ 1 ≥ σ 2 ≥ ⋯ ≥ σ d \sigma_1\ge\sigma_2\ge\cdots\ge\sigma_d σ 1 ≥ σ 2 ≥ ⋯ ≥ σ d are the singular values. Let U k ∈ R n × k U_k\in\mathbb{R}^{n\times k} U k ∈ R n × k and V k ∈ R d × k V_k\in\mathbb{R}^{d\times k} V k ∈ R d × k be the matrices whose columns are the top-k k k left-singular and right-singular vectors, respectively. Let Σ k = diag ( σ 1 , … , σ k ) \Sigma_k=\operatorname{diag}(\sigma_1,\ldots,\sigma_k) Σ k = diag ( σ 1 , … , σ k ) . We assume that there is a positive eigengap Δ k : = σ k 2 − σ k + 1 2 > 0 \Delta_k:=\sigma_k^2-\sigma_{k+1}^2>0 Δ k := σ k 2 − σ k + 1 2 > 0 .
We let ∥ U ∥ ∞ : = max i , j ∣ U i j ∣ \|U\|_\infty:=\max_{i,j}|U_{ij}| ∥ U ∥ ∞ := max i , j ∣ U ij ∣ , and we refer to ∥ U ∥ ∞ \|U\|_\infty ∥ U ∥ ∞ as the coherence of the input matrix A A A . Note that U i j = ⟨ a i , v j ⟩ / σ j U_{ij}=\langle a_i,v_j\rangle/\sigma_j U ij = ⟨ a i , v j ⟩ / σ j and thus ∥ U ∥ ∞ \|U\|_\infty ∥ U ∥ ∞ measures the maximum correlation between the datapoints (the rows of A A A ) and the right-singular vectors. The coherence value is not known to the algorithm.
We use the O ~ / Ω ~ \widetilde{O}/\widetilde{\Omega} O / Ω notation to suppress factors that are poly-logarithmic in n n n , d d d , 1 / ϵ 1/\epsilon 1/ ϵ , 1 / δ 1/\delta 1/ δ , and 1 / β 1/\beta 1/ β , where n n n and d d d are the input dimensions, ϵ \epsilon ϵ and δ \delta δ are the privacy parameters, and β \beta β is the failure probability.
Clipping. For a row vector y ∈ R 1 × k y\in\mathbb{R}^{1\times k} y ∈ R 1 × k and a radius R > 0 R>0 R > 0 , we define the clipping operation:
Clip R ( y ) : = y ⋅ min { 1 , R ∥ y ∥ 2 } . \operatorname{Clip}_{R}(y):=y\cdot\min\left\{1,\frac{R}{\|y\|_2}\right\}. Clip R ( y ) := y ⋅ min { 1 , ∥ y ∥ 2 R } .
with the convention that Clip R ( 0 ) = 0 \operatorname{Clip}_{R}(0)=0 Clip R ( 0 ) = 0 .
Differential privacy. We consider the privacy model where each row of the input matrix A A A is the private data of a user. More precisely, two input matrices A A A and A ′ A' A ′ are neighboring inputs if A A A and A ′ A' A ′ differ in exactly one row. A randomized algorithm M \mathcal{M} M is ( ϵ , δ ) (\epsilon,\delta) ( ϵ , δ ) -differentially private (( ϵ , δ ) ( \epsilon,\delta) ( ϵ , δ ) -DP) if, for every pair of neighboring inputs A A A , A ′ A' A ′ and every measurable subset of possible outcomes O \mathcal{O} O , we have
Pr [ M ( A ) ∈ O ] ≤ e ϵ Pr [ M ( A ′ ) ∈ O ] + δ . \Pr[\mathcal{M}(A)\in\mathcal{O}]\le e^\epsilon\Pr[\mathcal{M}(A')\in\mathcal{O}]+\delta. Pr [ M ( A ) ∈ O ] ≤ e ϵ Pr [ M ( A ′ ) ∈ O ] + δ .
Our algorithm and analysis use the Gaussian mechanism and advanced composition, and we give the definitions and theorems in Section A.
Our Algorithm
In this section, we give an overview of our algorithm and analysis. Our main algorithm is shown in Algorithm 3. The algorithm uses the algorithm of Dwork et al. [2014] to construct an initial solution (Algorithm 1). Our main optimization algorithm is shown in Algorithm 2. This algorithm takes as input the
Algorithm 1 Algorithm for constructing an initial solution
Input: Matrix A ∈ R n × d A \in\mathbb{R}^{n \times d} A ∈ R n × d , privacy parameters ϵ i n i t , δ i n i t \epsilon_{\mathrm{init}}, \delta_{\mathrm{init}} ϵ init , δ init
Let Δ i n i t = 2 2 log ( 2 / δ i n i t ) / ϵ i n i t \Delta_{\mathrm{init}} = 2\sqrt{2\log(2/\delta_{\mathrm{init}})}/\epsilon_{\mathrm{init}} Δ init = 2 2 log ( 2/ δ init ) / ϵ init .
Let M ^ = A T A + Δ i n i t ⋅ sym ( G ( i n i t ) ) \widehat{M} = A^{\mathsf{T}}A + \Delta_{\mathrm{init}} \cdot\operatorname{sym}\left(G^{(\mathrm{init})}\right) M = A T A + Δ init ⋅ sym ( G ( init ) ) where G ( i n i t ) ∼ N ( 0 , 1 ) d × d G^{(\mathrm{init})} \sim\mathcal{N}(0,1)^{d \times d} G ( init ) ∼ N ( 0 , 1 ) d × d and sym ( G ) = 1 2 ( G + G T ) \operatorname{sym}(G) = \frac{1}{2}(G+G^{\mathsf{T}}) sym ( G ) = 2 1 ( G + G T )
Let X ( 0 ) X^{(0)} X ( 0 ) be a d × k d \times k d × k matrix whose columns are the top k k k eigenvectors of M ^ \widehat{M} M
Return X ( 0 ) X^{(0)} X ( 0 )
Algorithm 2 Private Preconditioned Power Method
Input: Matrix A ∈ R n × d A \in\mathbb{R}^{n \times d} A ∈ R n × d with rows a i a_i a i , initial solution X ( 0 ) X^{(0)} X ( 0 ) , number of iterations T T T , k ≤ d k \le d k ≤ d , per-iteration privacy parameters ϵ i t e r , δ i t e r \epsilon_{\mathrm{iter}}, \delta_{\mathrm{iter}} ϵ iter , δ iter , failure probability β \beta β , clipping radius R R R
Let Δ = 2 2 log ( 2 / δ i t e r ) / ϵ i t e r \Delta= 2\sqrt{2\log(2/\delta_{\mathrm{iter}})}/\epsilon_{\mathrm{iter}} Δ = 2 2 log ( 2/ δ iter ) / ϵ iter , γ = 4 Δ ( 2 k + 2 log ( 8 T / β ) ) \gamma= 4\Delta\left(2\sqrt{k}+\sqrt{2\log(8T/\beta)}\right) γ = 4Δ ( 2 k + 2 log ( 8 T / β ) )
for t = 0 … T − 1 t=0\ldots T-1 t = 0 … T − 1
Let H ( t ) = X ( t ) T A T A X ( t ) + Δ ⋅ sym ( G ( t ) ) H^{(t)} = X^{(t)\mathsf{T}}A^{\mathsf{T}}AX^{(t)} + \Delta\cdot\operatorname{sym}\left(G^{(t)}\right) H ( t ) = X ( t ) T A T A X ( t ) + Δ ⋅ sym ( G ( t ) ) where G ( t ) ∼ N ( 0 , 1 ) k × k G^{(t)} \sim\mathcal{N}(0,1)^{k \times k} G ( t ) ∼ N ( 0 , 1 ) k × k and sym ( G ) = 1 2 ( G + G T ) \operatorname{sym}(G) = \frac{1}{2}(G+G^{\mathsf{T}}) sym ( G ) = 2 1 ( G + G T )
Let H ( t ) = Q ( t ) Λ ( t ) Q ( t ) T H^{(t)} = Q^{(t)}\Lambda^{(t)}Q^{(t)\mathsf{T}} H ( t ) = Q ( t ) Λ ( t ) Q ( t ) T , H ~ ( t ) = Q ( t ) diag ( max { Λ 11 ( t ) , γ } , … , max { Λ k k ( t ) , γ } ) Q ( t ) T \widetilde{H}^{(t)} = Q^{(t)}\operatorname{diag}\left(\max\left\{\Lambda_{11}^{(t)},\gamma\right\},\ldots,\max\left\{\Lambda_{kk}^{(t)},\gamma\right\}\right)Q^{(t)\mathsf{T}} H ( t ) = Q ( t ) diag ( max { Λ 11 ( t ) , γ } , … , max { Λ k k ( t ) , γ } ) Q ( t ) T , B ( t ) = ( H ~ ( t ) ) − 1 / 2 B^{(t)} = \left(\widetilde{H}^{(t)}\right)^{-1/2} B ( t ) = ( H ( t ) ) − 1/2
Let Z ( t + 1 ) = ∑ i = 1 n a i Clip R ( a i T X ( t ) B ( t ) ) + Δ R ⋅ G ~ ( t ) Z^{(t+1)} = \sum_{i=1}^{n} a_i\operatorname{Clip}_{R}\left(a_i^{\mathsf{T}}X^{(t)}B^{(t)}\right) + \Delta R \cdot\widetilde{G}^{(t)} Z ( t + 1 ) = ∑ i = 1 n a i Clip R ( a i T X ( t ) B ( t ) ) + Δ R ⋅ G ( t ) where G ~ ( t ) ∼ N ( 0 , 1 ) d × k \widetilde{G}^{(t)} \sim\mathcal{N}(0,1)^{d \times k} G ( t ) ∼ N ( 0 , 1 ) d × k
Let X ( t + 1 ) R ( t + 1 ) = Z ( t + 1 ) X^{(t+1)}R^{(t+1)} = Z^{(t+1)} X ( t + 1 ) R ( t + 1 ) = Z ( t + 1 ) be the QR-decomposition of Z ( t + 1 ) Z^{(t+1)} Z ( t + 1 )
end for
return X ( T ) X^{(T)} X ( T )
Algorithm 3 Main Algorithm
Input: Matrix A ∈ R n × d A \in\mathbb{R}^{n \times d} A ∈ R n × d , privacy parameters ϵ , δ \epsilon,\delta ϵ , δ , failure probability β \beta β
Set ϵ i n i t = ϵ 2 \epsilon_{\mathrm{init}} = \frac{\epsilon}{2} ϵ init = 2 ϵ and δ i n i t = δ 2 \delta_{\mathrm{init}} = \frac{\delta}{2} δ init = 2 δ
Let X ( 0 ) X^{(0)} X ( 0 ) be the solution returned by Algorithm 1 when run with ϵ i n i t = ϵ 2 \epsilon_{\mathrm{init}} = \frac{\epsilon}{2} ϵ init = 2 ϵ and δ i n i t = δ 2 \delta_{\mathrm{init}} = \frac{\delta}{2} δ init = 2 δ
Set ϵ i t e r \epsilon_{\mathrm{iter}} ϵ iter and δ i t e r \delta_{\mathrm{iter}} δ iter so that Algorithm 2 is ( ϵ 2 , δ 2 ) \left(\frac{\epsilon}{2},\frac{\delta}{2}\right) ( 2 ϵ , 2 δ ) -DP
Let X ( T ) X^{(T)} X ( T ) be the solution returned by Algorithm 2 when run with ϵ i t e r \epsilon_{\mathrm{iter}} ϵ iter and δ i t e r \delta_{\mathrm{iter}} δ iter , and failure probability β 2 \frac{\beta}{2} 2 β
Return X ( T ) X^{(T)} X ( T )
initial solution constructed by Algorithm 1 together with a clipping radius R R R and an iteration count T T T , and it iteratively constructs a solution using a private variant of a preconditioned power method as discussed in the introduction. In Section C in the Appendix, we show how to privately select the hyperparameters R R R and T T T using the private hyperparameter-selection method of Papernot and Steinke [ [19 ] .
We now give an overview of our analysis. We give the complete analysis to Section B in the Appendix.
Privacy analysis. Our algorithm uses the Gaussian mechanism to privatize the updates, and we show that the algorithm is ( ϵ , δ ) (\epsilon,\delta) ( ϵ , δ ) -DP via standard applications of the Gaussian mechanism and the advanced composition theorem.
Utility analysis. We measure the utility of our algorithm via the error ∥ ( I − X X T ) V k ∥ 2 \left\|(I-XX^{\mathsf{T}})V_k\right\|_2 ( I − X X T ) V k 2 , where X X X is a solution with orthonormal columns and V k V_k V k is the matrix whose columns are the top-k k k right-singular vectors. We recall that this error is the sine of the maximum principal angle between the column spaces of X X X and V k V_k V k . We let s t : = ∥ ( I − X ( t ) X ( t ) T ) V k ∥ 2 s_t := \left\|(I-X^{(t)}X^{(t)\mathsf{T}})V_k\right\|_2 s t := ( I − X ( t ) X ( t ) T ) V k 2 be the error of solution X ( t ) X^{(t)} X ( t ) constructed in iteration t − 1 t-1 t − 1 of our algorithm.
We analyze the error s t s_t s t via the following quantity. Let C ( t ) : = V k T X ( t ) C^{(t)} := V_k^{\mathsf{T}}X^{(t)} C ( t ) := V k T X ( t ) , D ( t ) : = V > k T X ( t ) D^{(t)} := V_{>k}^{\mathsf{T}}X^{(t)} D ( t ) := V > k T X ( t ) , and E ( t ) : = D ( t ) ( C ( t ) ) − 1 E^{(t)} := D^{(t)}(C^{(t)})^{-1} E ( t ) := D ( t ) ( C ( t ) ) − 1 provided that C ( t ) C^{(t)} C ( t ) is invertible. The initial solution X ( 0 ) X^{(0)} X ( 0 ) constructed by Algorithm 1 satisfies s 0 ≤ 1 / 4 s_0 \le1/4 s 0 ≤ 1/4 , which guarantees that C ( 0 ) C^{(0)} C ( 0 ) is invertible. Given this property, we show by induction that the algorithm maintains the invariant that s t < 1 s_t < 1 s t < 1 , which ensures that C ( t ) C^{(t)} C ( t ) is invertible (Lemma 13 ). We analyze the error s t s_t s t via the spectral norm m t : = ∥ E ( t ) ∥ 2 m_t := \left\|E^{(t)}\right\|_2 m t := E ( t ) 2 of E ( t ) E^{(t)} E ( t ) , which upper bounds the error (i.e., we have s t ≤ m t s_t \le m_t s t ≤ m t ).
In an ideal update with no clipping or privacy noise, we have E ( t + 1 ) = Σ > k 2 E ( t ) Σ k − 2 E^{(t+1)} = \Sigma_{>k}^2 E^{(t)}\Sigma_k^{-2} E ( t + 1 ) = Σ > k 2 E ( t ) Σ k − 2 and thus
∥ E ( t + 1 ) ∥ 2 ≤ ∥ Σ > k 2 ∥ 2 ∥ E ( t ) ∥ 2 ∥ Σ k − 2 ∥ 2 = σ k + 1 2 σ k 2 ∥ E ( t ) ∥ 2 = ( 1 − κ k ) ∥ E ( t ) ∥ 2 , \left\|E^{(t+1)}\right\|_2 \le\left\|\Sigma_{>k}^2\right\|_2\left\|E^{(t)}\right\|_2\left\|\Sigma_k^{-2}\right\|_2 = \frac{\sigma_{k+1}^2}{\sigma_k^2}\left\|E^{(t)}\right\|_2 = (1-\kappa_k)\left\|E^{(t)}\right\|_2, E ( t + 1 ) 2 ≤ Σ > k 2 2 E ( t ) 2 Σ k − 2 2 = σ k 2 σ k + 1 2 E ( t ) 2 = ( 1 − κ k ) E ( t ) 2 ,
where κ k = σ k 2 − σ k + 1 2 σ k 2 \kappa_k = \frac{\sigma_k^2-\sigma_{k+1}^2}{\sigma_k^2} κ k = σ k 2 σ k 2 − σ k + 1 2 . Thus, in this ideal case, the error decreases exponentially with the eigengap. For our private update, we show that, if the clipping is inactive in iteration t t t (i.e., ∥ a i T X ( t ) B ( t ) ∥ 2 ≤ R \left\|a_i^{\mathsf{T}}X^{(t)}B^{(t)}\right\|_2 \le R a i T X ( t ) B ( t ) 2 ≤ R for all i ∈ [ n ] i \in[n] i ∈ [ n ] ), then the following modified recurrence holds:
E ( t + 1 ) = ( Σ > k 2 E ( t ) Σ k − 2 + L ( t ) ) ( I + F ( t ) ) − 1 , (1) E^{(t+1)} = \left(\Sigma_{>k}^2E^{(t)}\Sigma_k^{-2}+L^{(t)}\right)\left(I+F^{(t)}\right)^{-1}, \tag*{(1)} E ( t + 1 ) = ( Σ > k 2 E ( t ) Σ k − 2 + L ( t ) ) ( I + F ( t ) ) − 1 , (1)
where L ( t ) L^{(t)} L ( t ) and F ( t ) F^{(t)} F ( t ) are matrices arising from the Gaussian noise added, defined as follows: F ( t ) : = Δ R V k T G ~ ( t ) ( K ( t ) ) − 1 F^{(t)} := \Delta R V_k^{\mathsf{T}}\widetilde{G}^{(t)}(K^{(t)})^{-1} F ( t ) := Δ R V k T G ( t ) ( K ( t ) ) − 1 and L ( t ) : = Δ R V > k T G ~ ( t ) ( K ( t ) ) − 1 L^{(t)} := \Delta R V_{>k}^{\mathsf{T}}\widetilde{G}^{(t)}(K^{(t)})^{-1} L ( t ) := Δ R V > k T G ( t ) ( K ( t ) ) − 1 , where K ( t ) : = Σ k 2 C ( t ) B ( t ) K^{(t)} := \Sigma_k^2C^{(t)}B^{(t)} K ( t ) := Σ k 2 C ( t ) B ( t ) . The main progress term Σ > k 2 E ( t ) Σ k − 2 \Sigma_{>k}^2E^{(t)}\Sigma_k^{-2} Σ > k 2 E ( t ) Σ k − 2 is the same as in the ideal update, and its spectral norm decreases exponentially. Thus if we can show that the clipping remains inactive throughout the algorithm and the contribution of the error terms is sufficiently small, we can show that we approximately preserve the exponential decrease in the error.
We show that the error terms can be bounded with high probability using Gaussian concentration and the properties of the preconditioner B ( t ) B^{(t)} B ( t ) used by the main update. We have
∥ F ( t ) ∥ 2 ≤ Δ R ∥ V k T G ~ ( t ) ∥ 2 ∥ ( K ( t ) ) − 1 ∥ 2 ∥ L ( t ) ∥ 2 ≤ Δ R ∥ V > k T G ~ ( t ) ∥ 2 ∥ ( K ( t ) ) − 1 ∥ 2 . \left\|F^{(t)}\right\|_2 \le\Delta R\left\|V_k^{\mathsf{T}}\widetilde{G}^{(t)}\right\|_2\left\|(K^{(t)})^{-1}\right\|_2 \qquad\left\|L^{(t)}\right\|_2 \le\Delta R\left\|V_{>k}^{\mathsf{T}}\widetilde{G}^{(t)}\right\|_2\left\|(K^{(t)})^{-1}\right\|_2. F ( t ) 2 ≤ Δ R V k T G ( t ) 2 ( K ( t ) ) − 1 2 L ( t ) 2 ≤ Δ R V > k T G ( t ) 2 ( K ( t ) ) − 1 2 .
Since the Gaussian distribution is rotationally invariant, V k T G ~ ( t ) V_k^{\mathsf{T}}\widetilde{G}^{(t)} V k T G ( t ) is a k × k k \times k k × k Gaussian matrix and V > k T G ~ ( t ) V_{>k}^{\mathsf{T}}\widetilde{G}^{(t)} V > k T G ( t ) is a ( d − k ) × k (d-k) \times k ( d − k ) × k Gaussian matrix. Using Gaussian concentration results (Lemma 5), we can show that we have ∥ V k T G ~ ( t ) ∥ 2 ≤ O ~ ( k ) \left\|V_k^{\mathsf{T}}\widetilde{G}^{(t)}\right\|_2 \le\widetilde{O}(\sqrt{k}) V k T G ( t ) 2 ≤ O ( k ) and ∥ V > k T G ~ ( t ) ∥ 2 ≤ O ~ ( d − k + k ) \left\|V_{>k}^{\mathsf{T}}\widetilde{G}^{(t)}\right\|_2 \le\widetilde{O}(\sqrt{d-k}+\sqrt{k}) V > k T G ( t ) 2 ≤ O ( d − k + k ) with high probability. We can also show that ∥ ( K ( t ) ) − 1 ∥ 2 ≤ O ( 1 σ k ) \left\|(K^{(t)})^{-1}\right\|_2 \le O\left(\frac{1}{\sigma_k}\right) ( K ( t ) ) − 1 2 ≤ O ( σ k 1 ) (Lemma 18) as follows. We have K ( t ) : = Σ k 2 C ( t ) B ( t ) K^{(t)} := \Sigma_k^2C^{(t)}B^{(t)} K ( t ) := Σ k 2 C ( t ) B ( t ) . We can show the preconditioner B ( t ) B^{(t)} B ( t ) roughly whitens Σ k V k T X ( t ) = Σ k C ( t ) \Sigma_kV_k^{\mathsf{T}}X^{(t)}=\Sigma_kC^{(t)} Σ k V k T X ( t ) = Σ k C ( t ) , i.e., the singular values of Σ k C ( t ) B ( t ) \Sigma_kC^{(t)}B^{(t)} Σ k C ( t ) B ( t ) are bounded from above and below by absolute constants (Lemma 17 ). This allows us to bound ∥ ( K ( t ) ) − 1 ∥ 2 ≤ ∥ ( Σ k C ( t ) B ( t ) ) − 1 ∥ 2 ∥ Σ k − 1 ∥ 2 = O ( 1 / σ k ) \left\|(K^{(t)})^{-1}\right\|_2 \le\left\|(\Sigma_kC^{(t)}B^{(t)})^{-1}\right\|_2\left\|\Sigma_k^{-1}\right\|_2=O(1/\sigma_k) ( K ( t ) ) − 1 2 ≤ ( Σ k C ( t ) B ( t ) ) − 1 2 Σ k − 1 2 = O ( 1/ σ k ) since ∥ ( Σ k C ( t ) B ( t ) ) − 1 ∥ 2 = O ( 1 ) \left\|(\Sigma_kC^{(t)}B^{(t)})^{-1}\right\|_2=O(1) ( Σ k C ( t ) B ( t ) ) − 1 2 = O ( 1 ) .
If the clipping is inactive, we can use the recurrence above together with the upper bounds on the noise terms described above to analyze the error progress. In order to obtain a stronger bound, we use a martingale argument to show that the contribution of some of the error terms is smaller. To simplify notation, for a matrix M M M , let us define T ( M ) : = Σ > k 2 M Σ k − 2 \mathcal{T}(M) := \Sigma_{>k}^{2}M\Sigma_{k}^{-2} T ( M ) := Σ > k 2 M Σ k − 2 . Thus the main recurrence (1) becomes:
E ( t + 1 ) = ( T ( E ( t ) ) + L ( t ) ) ( I + F ( t ) ) − 1 . E^{(t+1)} = \left(\mathcal{T}\left(E^{(t)}\right) + L^{(t)}\right)\left(I+F^{(t)}\right)^{-1}. E ( t + 1 ) = ( T ( E ( t ) ) + L ( t ) ) ( I + F ( t ) ) − 1 .
Using the identity ( I + F ) − 1 = I − F ( I + F ) − 1 \left(I+F\right)^{-1}=I-F\left(I+F\right)^{-1} ( I + F ) − 1 = I − F ( I + F ) − 1 , we can rewrite the recurrence as follows:
E ( t + 1 ) = T ( E ( t ) ) + L ( t ) − ( T ( E ( t ) ) + L ( t ) ) F ( t ) ( I + F ( t ) ) − 1 ⏟ : = R ( t ) = T ( E ( t ) ) + L ( t ) + R ( t ) . E^{(t+1)}=\mathcal{T}\left(E^{(t)}\right)+L^{(t)}-\underbrace{\left(\mathcal{T}\left(E^{(t)}\right)+L^{(t)}\right)F^{(t)}\left(I+F^{(t)}\right)^{-1}}_{:=R^{(t)}}=\mathcal{T}\left(E^{(t)}\right)+L^{(t)}+R^{(t)}. E ( t + 1 ) = T ( E ( t ) ) + L ( t ) − := R ( t ) ( T ( E ( t ) ) + L ( t ) ) F ( t ) ( I + F ( t ) ) − 1 = T ( E ( t ) ) + L ( t ) + R ( t ) .
The reason this is beneficial is that it separates the noise terms depending on G ~ ( t ) \widetilde{G}^{(t)} G ( t ) into the term L ( t ) L^{(t)} L ( t ) that is linear in G ~ ( t ) \widetilde{G}^{(t)} G ( t ) and a more complex term R ( t ) R^{(t)} R ( t ) . The simple linear structure of the terms L ( t ) L^{(t)} L ( t ) allows us to bound their contribution via a martingale argument. By unrolling the recurrence, we obtain
E ( t ) = T t ( E ( 0 ) ) + ∑ j = 0 t − 1 T t − 1 − j ( L ( j ) ) + ∑ j = 0 t − 1 T t − 1 − j ( R ( j ) ) . E^{(t)}=\mathcal{T}^{t}\left(E^{(0)}\right)+\sum_{j=0}^{t-1}\mathcal{T}^{t-1-j}\left(L^{(j)}\right)+\sum_{j=0}^{t-1}\mathcal{T}^{t-1-j}\left(R^{(j)}\right). E ( t ) = T t ( E ( 0 ) ) + j = 0 ∑ t − 1 T t − 1 − j ( L ( j ) ) + j = 0 ∑ t − 1 T t − 1 − j ( R ( j ) ) .
We upper bound the spectral norm of the middle term ∑ j = 0 t − 1 T t − 1 − j ( L ( j ) ) \sum_{j=0}^{t-1}\mathcal{T}^{t-1-j}\left(L^{(j)}\right) ∑ j = 0 t − 1 T t − 1 − j ( L ( j ) ) via a martingale concentration argument (Lemma (30)). For the last term, we upper bound its spectral norm term by term, and we bound ∥ T r ( R ( j ) ) ∥ 2 = ∥ Σ > k 2 r R ( j ) Σ k − 2 r ∥ 2 \left\|\mathcal{T}^{r}\left(R^{(j)}\right)\right\|_{2}=\left\|\Sigma_{>k}^{2r}R^{(j)}\Sigma_{k}^{-2r}\right\|_{2} T r ( R ( j ) ) 2 = Σ > k 2 r R ( j ) Σ k − 2 r 2 using a similar approach as discussed above.
The analysis discussed above relied on the clipping being inactive in every iteration. A significant part of our analysis is devoted to showing that this event holds with high probability provided that R ≥ Ω ~ ( k ∥ U ∥ ∞ ) R\geq\widetilde{\Omega}\left(\sqrt{k}\left\|U\right\|_{\infty}\right) R ≥ Ω ( k ∥ U ∥ ∞ ) . Note that the clipping is inactive in iteration t t t if ∥ a i T X ( t ) B ( t ) ∥ 2 ≤ R \left\|a_i^{\mathsf T}X^{(t)}B^{(t)}\right\|_{2}\leq R a i T X ( t ) B ( t ) 2 ≤ R for all i ∈ [ n ] i\in[n] i ∈ [ n ] , and thus we need to upper bound the lengths ∥ a i T X ( t ) B ( t ) ∥ 2 \left\|a_i^{\mathsf T}X^{(t)}B^{(t)}\right\|_{2} a i T X ( t ) B ( t ) 2 . We can write A X ( t ) B ( t ) = ( U k + U > k S ( t ) ) P ( t ) AX^{(t)}B^{(t)}=\left(U_k+U_{>k}S^{(t)}\right)P^{(t)} A X ( t ) B ( t ) = ( U k + U > k S ( t ) ) P ( t ) where S ( t ) : = Σ > k E ( t ) Σ k − 1 S^{(t)}:=\Sigma_{>k}E^{(t)}\Sigma_k^{-1} S ( t ) := Σ > k E ( t ) Σ k − 1 and P ( t ) : = Σ k C ( t ) B ~ ( t ) P^{(t)}:=\Sigma_k C^{(t)}\widetilde{B}^{(t)} P ( t ) := Σ k C ( t ) B ( t ) . Thus, for each row i i i , we have a i T X ( t ) B ( t ) = e i T ( U k + U > k S ( t ) ) P ( t ) a_i^{\mathsf T}X^{(t)}B^{(t)}=e_i^{\mathsf T}\left(U_k+U_{>k}S^{(t)}\right)P^{(t)} a i T X ( t ) B ( t ) = e i T ( U k + U > k S ( t ) ) P ( t ) and thus
∥ a i T X ( t ) B ( t ) ∥ 2 ≤ ( ∥ e i T U k ∥ 2 + ∥ e i T U > k S ( t ) ∥ 2 ) ∥ P ( t ) ∥ 2 . \left\|a_i^{\mathsf T}X^{(t)}B^{(t)}\right\|_{2}\leq\left(\left\|e_i^{\mathsf T}U_k\right\|_{2}+\left\|e_i^{\mathsf T}U_{>k}S^{(t)}\right\|_{2}\right)\left\|P^{(t)}\right\|_{2}. a i T X ( t ) B ( t ) 2 ≤ ( e i T U k 2 + e i T U > k S ( t ) 2 ) P ( t ) 2 .
We have ∥ e i T U k ∥ 2 ≤ k ∥ U ∥ ∞ \left\|e_i^{\mathsf T}U_k\right\|_{2}\leq\sqrt{k}\left\|U\right\|_{\infty} e i T U k 2 ≤ k ∥ U ∥ ∞ and ∥ P ( t ) ∥ 2 ≤ O ( 1 ) \left\|P^{(t)}\right\|_{2}\leq O(1) P ( t ) 2 ≤ O ( 1 ) with high probability as noted earlier, and thus it only remains to upper bound ∥ e i T U > k S ( t ) ∥ 2 \left\|e_i^{\mathsf T}U_{>k}S^{(t)}\right\|_{2} e i T U > k S ( t ) 2 . To this end, we introduce the following potential for every 0 ≤ r , t ≤ T 0\leq r,t\leq T 0 ≤ r , t ≤ T :
Φ r , t : = max i ∈ [ n ] ∥ e i T U > k Σ > k 2 r S ( t ) ∥ 2 . \Phi_{r,t}:=\max_{i\in[n]}\left\|e_i^{\mathsf T}U_{>k}\Sigma_{>k}^{2r}S^{(t)}\right\|_{2}. Φ r , t := i ∈ [ n ] max e i T U > k Σ > k 2 r S ( t ) 2 .
The potential values with r = 0 r=0 r = 0 are precisely the terms that we need to upper bound, and the values with larger r r r allows us to analyze these terms. We first use a concentration argument to show that the initialization allows us to upper bound Φ r , 0 \Phi_{r,0} Φ r , 0 for all r r r with high probability (Lemma 21). We then use an inductive argument to show that we can upper bound the potentials Φ 0 , t \Phi_{0,t} Φ 0 , t using the initial potentials Φ r , 0 \Phi_{r,0} Φ r , 0 . Here we assume inductively that the clipping was inactive in all prior iterations and we use the recurrence for E ( t + 1 ) E^{(t+1)} E ( t + 1 ) discussed above, and show that the clipping is inactive in the current iteration as well (Lemma 28).
Related work
Privacy models on coherence-sensitive PCA. We study row-level adjacency, under which two matrices differ in one row and the ℓ 2 \ell_2 ℓ 2 norm of the difference between those rows is at most 1. The guarantees of Dwork et al. [2014] and Nguyen et al. [2026] apply under this model. By contrast, most earlier coherence-sensitive guarantees were established under different notions of adjacency. Earlier power-method results consider entry-level adjacency on the input matrix [12 , 10 , 1 ] . Nicolas et al. [2025] instead allow difference between neighboring input matrices after symmetrization E E E satisfying ∑ i = 1 n ∥ E i , : ∥ 1 2 ≤ 1 \sqrt{\sum_{i=1}^{n}\lVert E_{i,:}\rVert_{1}^{2}} \le1 ∑ i = 1 n ∥ E i , : ∥ 1 2 ≤ 1 , while d’Orsi and Novikov [2026] consider the same metric E E E satisfying ∑ i , j ∣ ( E E ⊤ ) i j ∣ ≤ Δ \sqrt{\sum_{i,j}\left|(EE^{\top})_{ij}\right|} \le\Delta ∑ i , j ∣ ( E E ⊤ ) ij ∣ ≤ Δ . Since these adjacency notions do not coincide with row-level ℓ 2 \ell_{2} ℓ 2 adjacency, Table 1 reports the bounds obtained after converting them to our model.
Worst-case bounds. Dwork et al. [2014] give a rank-k k k guarantee without any assumption on coherence. It provides a benchmark for coherence-independent k k k -PCA. Their guarantee gives the principal-angle error bound O ~ ( d ϵ ( σ k 2 − σ k + 1 2 ) ) \widetilde{O}\left(\frac{\sqrt{d}}{\epsilon(\sigma_{k}^{2}-\sigma_{k+1}^{2})}\right) O ( ϵ ( σ k 2 − σ k + 1 2 ) d ) . Tran et al. [2025] study the spectral-norm error ∥ M ~ k − M k ∥ 2 \left\lVert\widetilde{M}_{k}-M_{k}\right\rVert_{2} M k − M k 2 , where M k M_{k} M k and M ~ k \widetilde{M}_{k} M k are the best rank-k k k approximations before and after the Gaussian perturbation. Under their gap condition, converting this spectral-norm guarantee to principal-angle error recovers the bound of Dwork et al. [2014] up to logarithmic factors.
Beyond-worst-case bounds. Hardt and Roth [2013] is the first paper to study coherence-sensitive guarantees for private singular-vector computation. They give a private rank-1 algorithm and use repeated deflation to obtain a rank-k k k approximation. The resulting guarantee deteriorates when σ 1 / σ k \sigma_{1}/\sigma_{k} σ 1 / σ k is large. Subsequent works analyze block power methods, which update k k k directions simultaneously and avoid the dependence on σ 1 \sigma_{1} σ 1 introduced by repeated deflation [10 , 1 , 18 ] . A different approach is taken by d’Orsi and Novikov [2026], who privately estimate the spectral gap and coherence and use these estimates as input to a Gaussian mechanism for estimating the top-k k k subspace.
Private block power method. Hardt and Price [2014] give a general analysis of the noisy block power method and apply it to private rank-k k k PCA. Their algorithm uses oversampling, maintaining p ≥ k p \ge k p ≥ k vectors to approximate the subspace spanned by the top k k k singular vectors. Its coherence-sensitive principal-angle guarantee depends on the gap σ k − σ k + 1 \sigma_{k}-\sigma_{k+1} σ k − σ k + 1 . They conjecture that, with oversampling, this dependence can be replaced by dependence on the larger gap σ k − σ p + 1 \sigma_{k}-\sigma_{p+1} σ k − σ p + 1 . Balcan et al. [2016] partially address this conjecture by replacing the dependence on σ k − σ k + 1 \sigma_{k}-\sigma_{k+1} σ k − σ k + 1 with dependence on σ k − σ q + 1 \sigma_{k}-\sigma_{q+1} σ k − σ q + 1 for an intermediate rank q ∈ [ k , p / 2 ] q \in[k,p/2] q ∈ [ k , p /2 ] . Building on this framework, Nicolas et al. [2025] sharpen the sensitivity bound from p ∥ X t ∥ max \sqrt{p}\left\lVert X_{t}\right\rVert_{\max} p ∥ X t ∥ m a x to max i ∥ ( X t ) i , : ∥ 2 \max_{i}\left\lVert(X_{t})_{i,:}\right\rVert_{2} max i ∥ ( X t ) i , : ∥ 2 . By contrast, our algorithm does not use oversampling and maintains exactly k k k vectors throughout the iterations.
Statistical / i.i.d private PCA. This literature obtains stronger statistical guarantees by assuming that the data are i.i.d samples from a structured distribution. Liu et al. [2022] estimates the top singular vector with sub-Gaussian distributed data. Düngler and Sanyal [2025] generalizes this work for top k k k singular vectors.
AI use statement
In this work, we used generative AI tools for retrieving and summarizing related literature; discussing research directions; reviewing mathematical arguments and suggesting possible proofs; comparing and interpreting theoretical guarantees; and drafting, restructuring, and editing portions of the manuscript. The authors checked the cited literature against the original sources and reviewed all AI-assisted text and mathematical content. We did not use generative AI tools to generate or analyze data, conduct experiments, or implement the algorithm; these tasks are not applicable to this theoretical work. We take responsibility for the final content of this work, including all text and mathematical claims produced with the aid of generative AI.
Main Tools
In this section, we state the main theorems from prior work that we will use in our utility analysis shown in Section B.2. Our algorithm and analysis use the following tools from differential privacy.
Gaussian mechanism. Let f f f be a function that takes as input a matrix A A A and outputs a vector. We say f f f has ℓ 2 \ell_2 ℓ 2 -sensitivity Δ \Delta Δ if on any neighboring inputs A A A and A ′ A' A ′ , we have ∥ f ( A ) − f ( A ′ ) ∥ 2 ≤ Δ \lVert f(A)-f(A')\rVert_2 \le\Delta ∥ f ( A ) − f ( A ′ ) ∥ 2 ≤ Δ . The following result was shown by Dwork et al. [ [8 ] .
Theorem 2 (Gaussian mechanism, Dwork et al. [ [8 ] ). Assume 0 < ϵ ≤ 1 0 < \epsilon\le1 0 < ϵ ≤ 1 and 0 < δ < 1 0 < \delta< 1 0 < δ < 1 . Given an input matrix A A A and a function f f f with ℓ 2 \ell_2 ℓ 2 -sensitivity Δ \Delta Δ , the algorithm M M M , which outputs f ( A ) + G f(A)+G f ( A ) + G where G G G is sampled from the Gaussian distribution N ( 0 , σ 2 I ) \mathcal{N}(0,\sigma^2 I) N ( 0 , σ 2 I ) with σ ≥ Δ ϵ 2 log 1.25 δ \sigma\ge\frac{\Delta}{\epsilon}\sqrt{2\log\frac{1.25}{\delta}} σ ≥ ϵ Δ 2 log δ 1.25 , is ( ϵ , δ ) (\epsilon,\delta) ( ϵ , δ ) -DP.
Theorem 3 (Advanced composition Kairouz et al. [ [14 ] ). For all ϵ , δ \epsilon,\delta ϵ , δ , the adaptive composition of k k k algorithms, each of which is ( ϵ , δ ) (\epsilon,\delta) ( ϵ , δ ) -DP, is ( ϵ ′ , δ ′ ) (\epsilon',\delta') ( ϵ ′ , δ ′ ) -private where
ϵ ′ = ϵ 2 ln ( 1 δ ) k + ϵ e ϵ − 1 e ϵ + 1 k \epsilon'=\epsilon\sqrt{2\ln\left(\frac{1}{\delta}\right)k}+\epsilon\frac{e^\epsilon-1}{e^\epsilon+1}k ϵ ′ = ϵ 2 ln ( δ 1 ) k + ϵ e ϵ + 1 e ϵ − 1 k and
δ ′ = δ ( k + 1 ) . \delta'=\delta(k+1). δ ′ = δ ( k + 1 ) .
Gaussian concentration results. We also use the following concentration results for Gaussian random variables.
Theorem 4 (Boucheron et al. [ [3 ] ). If G ∼ N ( 0 , 1 ) d × d G \sim\mathcal{N}(0,1)^{d\times d} G ∼ N ( 0 , 1 ) d × d and f : R d × d → R f:\mathbb{R}^{d\times d}\to\mathbb{R} f : R d × d → R is L L L -Lipschitz in Euclidean norm, then, for every x ≥ 0 x\ge0 x ≥ 0 ,
Pr [ f ( G ) − E [ f ( G ) ] ≥ x ] ≤ exp ( − x 2 2 L 2 ) , Pr [ f ( G ) − E [ f ( G ) ] ≤ − x ] ≤ exp ( − x 2 2 L 2 ) . \Pr[f(G)-\mathbb{E}[f(G)]\ge x]\le\exp\left(-\frac{x^2}{2L^2}\right), \qquad\Pr[f(G)-\mathbb{E}[f(G)]\le-x]\le\exp\left(-\frac{x^2}{2L^2}\right). Pr [ f ( G ) − E [ f ( G )] ≥ x ] ≤ exp ( − 2 L 2 x 2 ) , Pr [ f ( G ) − E [ f ( G )] ≤ − x ] ≤ exp ( − 2 L 2 x 2 ) .
Lemma 5 (Davidson and Szarek [ [4 ] ). Let G ∼ N ( 0 , 1 ) q × k G\sim\mathcal{N}(0,1)^{q\times k} G ∼ N ( 0 , 1 ) q × k . For every x ≥ 0 x\ge0 x ≥ 0 , Pr [ ∥ G ∥ 2 > q + k + x ] ≤ e − x 2 / 2 \Pr\left[\lVert G\rVert_2>\sqrt{q}+\sqrt{k}+x\right]\le e^{-x^2/2} Pr [ ∥ G ∥ 2 > q + k + x ] ≤ e − x 2 /2 .
Proof. The first assertion follows from Davidson-Szarek bound Davidson and Szarek [ [4 ] . We must rescale our matrix to match their normalization. Assume first that k ≤ q k\le q k ≤ q , and define Γ : = 1 q G \Gamma:=\frac{1}{\sqrt{q}}G Γ := q 1 G . Then Γ ∈ R q × k \Gamma\in\mathbb{R}^{q\times k} Γ ∈ R q × k has independent N ( 0 , 1 / q ) \mathcal{N}(0,1/q) N ( 0 , 1/ q ) entries. Apply the Davidson-Szarek Theorem with parameters n = q n=q n = q , m = k m=k m = k , β = k q \beta=\frac{k}{q} β = q k , t = x q t=\frac{x}{\sqrt{q}} t = q x , and using the identity ∥ G ∥ 2 = q s 1 ( Γ ) \lVert G\rVert_2=\sqrt{q}s_1(\Gamma) ∥ G ∥ 2 = q s 1 ( Γ ) , we obtain:
Pr [ ∥ G ∥ 2 ≥ q + k + x ] = Pr [ s 1 ( Γ ) ≥ 1 + k q + x q ] ≤ exp ( − q 2 ( x q ) 2 ) = e − x 2 / 2 . \begin{aligned}
\Pr\left[\lVert G\rVert_2\ge\sqrt{q}+\sqrt{k}+x\right]=\Pr\left[s_1(\Gamma)\ge1+\sqrt{\frac{k}{q}}+\frac{x}{\sqrt{q}}\right] \\
&\le\exp\left(-\frac{q}{2}\left(\frac{x}{\sqrt{q}}\right)^2\right)=e^{-x^2/2}.
\end{aligned} Pr [ ∥ G ∥ 2 ≥ q + k + x ] = Pr [ s 1 ( Γ ) ≥ 1 + q k + q x ] ≤ exp ( − 2 q ( q x ) 2 ) = e − x 2 /2 .
If k > q k>q k > q , we apply the preceding argument to G T G^\mathsf{T} G T , since ∥ G T ∥ 2 = ∥ G ∥ 2 \lVert G^\mathsf{T}\rVert_2=\lVert G\rVert_2 ∥ G T ∥ 2 = ∥ G ∥ 2 .
Lemma 6. Let M ∈ R q × k M\in\mathbb{R}^{q\times k} M ∈ R q × k and R ∈ R k × k R\in\mathbb{R}^{k\times k} R ∈ R k × k be fixed matrices, and let G ∼ N ( 0 , 1 ) q × k G\sim\mathcal{N}(0,1)^{q\times k} G ∼ N ( 0 , 1 ) q × k . For every x ≥ 0 x\ge0 x ≥ 0 ,
Pr [ ∥ M T G R ∥ 2 > ∥ M ∥ 2 ∥ R ∥ 2 ( 2 k + x ) ] ≤ e − x 2 / 2 . \Pr\left[\lVert M^\mathsf{T}GR\rVert_2>\lVert M\rVert_2\lVert R\rVert_2\left(2\sqrt{k}+x\right)\right]\le e^{-x^2/2}. Pr [ ∥ M T GR ∥ 2 > ∥ M ∥ 2 ∥ R ∥ 2 ( 2 k + x ) ] ≤ e − x 2 /2 . Proof. Let r = rank ( M ) ≤ k r=\operatorname{rank}(M)\le k r = rank ( M ) ≤ k , and write a compact singular value decomposition M = P Λ Q T M=P\Lambda Q^\mathsf{T} M = P Λ Q T , where P ∈ R q × r P\in\mathbb{R}^{q\times r} P ∈ R q × r has orthonormal columns. By rotational invariance, P T G ∼ N ( 0 , 1 ) r × k P^\mathsf{T}G\sim\mathcal{N}(0,1)^{r\times k} P T G ∼ N ( 0 , 1 ) r × k . Hence Lemma 5 gives
Pr [ ∥ P T G ∥ 2 ≤ r + k + x ≤ 2 k + x ] ≥ 1 − e − x 2 / 2 \Pr\left[\lVert P^\mathsf{T}G\rVert_2\le\sqrt{r}+\sqrt{k}+x\le2\sqrt{k}+x\right]\ge1-e^{-x^2/2} Pr [ ∥ P T G ∥ 2 ≤ r + k + x ≤ 2 k + x ] ≥ 1 − e − x 2 /2
But ∥ M T G R ∥ 2 = ∥ Q Λ P T G R ∥ 2 ≤ ∥ M ∥ 2 ∥ P T G ∥ 2 ∥ R ∥ 2 \left\|M^{\mathsf T}GR\right\|_2=\left\|Q\Lambda P^{\mathsf T}GR\right\|_2\leq\left\|M\right\|_2\left\|P^{\mathsf T}G\right\|_2\left\|R\right\|_2 M T GR 2 = Q Λ P T GR 2 ≤ ∥ M ∥ 2 P T G 2 ∥ R ∥ 2 , which implies that:
Pr [ ∥ M T G R ∥ 2 > ∥ M ∥ 2 ∥ R ∥ 2 ( 2 k + x ) ] ≤ Pr [ ∥ M ∥ 2 ∥ P T G ∥ 2 ∥ R ∥ 2 > ∥ M ∥ 2 ∥ R ∥ 2 ( 2 k + x ) ] = Pr [ ∥ P T G ∥ 2 > 2 k + x ] ≤ e − x 2 / 2 \begin{aligned}
\Pr\left[\left\|M^{\mathsf T}GR\right\|_2>\left\|M\right\|_2\left\|R\right\|_2\left(2\sqrt{k}+x\right)\right]\leq\Pr\left[\left\|M\right\|_2\left\|P^{\mathsf T}G\right\|_2\left\|R\right\|_2>\left\|M\right\|_2\left\|R\right\|_2\left(2\sqrt{k}+x\right)\right] \\
&=\Pr\left[\left\|P^{\mathsf T}G\right\|_2>2\sqrt{k}+x\right] \\
&\leq e^{-x^2/2}
\end{aligned} Pr [ M T GR 2 > ∥ M ∥ 2 ∥ R ∥ 2 ( 2 k + x ) ] ≤ Pr [ ∥ M ∥ 2 P T G 2 ∥ R ∥ 2 > ∥ M ∥ 2 ∥ R ∥ 2 ( 2 k + x ) ] = Pr [ P T G 2 > 2 k + x ] ≤ e − x 2 /2
□ \square □
Spectral analysis tools. Our analysis uses Weyl’s inequality and the Davis-Kahan sin Θ \Theta Θ theorem.
Theorem 7 (Weyl’s inequality). Let M M M and W W W be two symmetric matrices and for a matrix A A A , let λ i ( A ) \lambda_i(A) λ i ( A ) denote the i i i -th largest eigenvalue of A A A . For every i i i , we have
∣ λ i ( M + W ) − λ i ( M ) ∣ ≤ ∥ W ∥ 2 . \left|\lambda_i(M+W)-\lambda_i(M)\right|\leq\left\|W\right\|_2. ∣ λ i ( M + W ) − λ i ( M ) ∣ ≤ ∥ W ∥ 2 .
Let us also recall the definition of principal angles between two subspaces. Let V , V ^ ∈ R d × k V,\widehat{V}\in\mathbb{R}^{d\times k} V , V ∈ R d × k have orthonormal columns. Let σ 1 , … , σ k \sigma_1,\ldots,\sigma_k σ 1 , … , σ k be the singular values of V ^ T V \widehat{V}^{\mathsf T}V V T V . The principal angles between the column spaces of V V V and V ^ \widehat{V} V are θ i : = arccos ( σ i ) \theta_i:=\arccos(\sigma_i) θ i := arccos ( σ i ) for 1 ≤ i ≤ k 1\leq i\leq k 1 ≤ i ≤ k . We let Θ ( V , V ^ ) = diag ( θ 1 , … , θ k ) \Theta(V,\widehat{V})=\operatorname{diag}(\theta_1,\ldots,\theta_k) Θ ( V , V ) = diag ( θ 1 , … , θ k ) and sin Θ ( V , V ^ ) = diag ( sin ( θ 1 ) , … , sin ( θ k ) ) \sin\Theta(V,\widehat{V})=\operatorname{diag}(\sin(\theta_1),\ldots,\sin(\theta_k)) sin Θ ( V , V ) = diag ( sin ( θ 1 ) , … , sin ( θ k )) .
Theorem 8 (Davis-Kahan sin Θ \Theta Θ theorem, Stewart and guang Sun [1990, Theorem V.3.6], Yu et al. [2015, Theorem 1]). Let M , M ^ ∈ R d × d M,\widehat{M}\in\mathbb{R}^{d\times d} M , M ∈ R d × d be symmetric matrices with eigenvalues λ 1 ≥ ⋯ ≥ λ d \lambda_1\geq\cdots\geq\lambda_d λ 1 ≥ ⋯ ≥ λ d and λ ^ 1 ≥ ⋯ ≥ λ ^ d \widehat{\lambda}_1\geq\cdots\geq\widehat{\lambda}_d λ 1 ≥ ⋯ ≥ λ d , respectively. Fix $1\leq r\leq s\leq d$, let k = s − r + 1 k=s-r+1 k = s − r + 1 and let V = ( v r , v r + 1 , … , v s ) ∈ R d × k V=(v_r,v_{r+1},\ldots,v_s)\in\mathbb{R}^{d\times k} V = ( v r , v r + 1 , … , v s ) ∈ R d × k and V ^ = ( v ^ r , v ^ r + 1 , … , v ^ s ) ∈ R d × k \widehat{V}=(\widehat{v}_r,\widehat{v}_{r+1},\ldots,\widehat{v}_s)\in\mathbb{R}^{d\times k} V = ( v r , v r + 1 , … , v s ) ∈ R d × k have orthonormal columns satisfying M v i = λ i v i Mv_i=\lambda_i v_i M v i = λ i v i and M ^ v ^ i = λ ^ i v ^ i \widehat{M}\widehat{v}_i=\widehat{\lambda}_i\widehat{v}_i M v i = λ i v i for all $r\leq i\leq s. Let
δ : = inf { ∣ λ − λ ^ ∣ : λ ∈ [ λ s , λ r ] , λ ^ ∈ ( − ∞ , λ ^ s + 1 ] ∪ [ λ ^ r − 1 , + ∞ ) } \delta:=\inf\left\{|\lambda-\widehat{\lambda}|:\lambda\in[\lambda_s,\lambda_r],\widehat{\lambda}\in(-\infty,\widehat{\lambda}_{s+1}]\cup[\widehat{\lambda}_{r-1},+\infty)\right\} δ := inf { ∣ λ − λ ∣ : λ ∈ [ λ s , λ r ] , λ ∈ ( − ∞ , λ s + 1 ] ∪ [ λ r − 1 , + ∞ ) } where λ ^ 0 : = + ∞ \widehat{\lambda}_0:=+\infty λ 0 := + ∞ and λ ^ d + 1 = − ∞ \widehat{\lambda}_{d+1}=-\infty λ d + 1 = − ∞ .
If δ > 0 \delta>0 δ > 0 , we have
∥ ( I − V ^ V ^ T ) V ∥ F = ∥ sin Θ ( V , V ^ ) ∥ F ≤ ∥ M − M ^ ∥ F δ , \left\|\left(I-\widehat{V}\widehat{V}^{\mathsf T}\right)V\right\|_F=\left\|\sin\Theta(V,\widehat{V})\right\|_F\leq\frac{\left\|M-\widehat{M}\right\|_F}{\delta}, ( I − V V T ) V F = sin Θ ( V , V ) F ≤ δ M − M F , and
∥ ( I − V ^ V ^ T ) V ∥ 2 = ∥ sin Θ ( V , V ^ ) ∥ 2 ≤ ∥ M − M ^ ∥ 2 δ . \left\|\left(I-\widehat{V}\widehat{V}^{\mathsf T}\right)V\right\|_2=\left\|\sin\Theta(V,\widehat{V})\right\|_2\leq\frac{\left\|M-\widehat{M}\right\|_2}{\delta}. ( I − V V T ) V 2 = sin Θ ( V , V ) 2 ≤ δ M − M 2 .
Weyl’s inequality (Theorem 7) and the Davis-Kahan theorem (Theorem 8) imply the following lemma. A similar argument appears in Dwork et al. [2014].
Lemma 9. Let M ^ = A T A + W \widehat{M}=A^{\mathsf T}A+W M = A T A + W be a symmetric matrix, and let Δ k = λ k ( A T A ) − λ k + 1 ( A T A ) > 0 \Delta_k=\lambda_k(A^{\mathsf T}A)-\lambda_{k+1}(A^{\mathsf T}A)>0 Δ k = λ k ( A T A ) − λ k + 1 ( A T A ) > 0 denote the spectral gap. If ∥ W ∥ 2 < Δ k \left\|W\right\|_2<\Delta_k ∥ W ∥ 2 < Δ k , the top-k k k eigenspace V ^ k \widehat{V}_k V k of M ^ \widehat{M} M satisfies
∥ ( I − V ^ k V ^ k T ) V k ∥ 2 ≤ ∥ W ∥ 2 Δ k − ∥ W ∥ 2 . \left\|\left(I-\widehat{V}_k\widehat{V}_k^{\mathsf T}\right)V_k\right\|_2\leq\frac{\left\|W\right\|_2}{\Delta_k-\left\|W\right\|_2}. ( I − V k V k T ) V k 2 ≤ Δ k − ∥ W ∥ 2 ∥ W ∥ 2 .
Proof. We apply the Davis-Kahan sin Θ \Theta Θ theorem (Theorem 8) with M = A T A M=A^{\mathsf T}A M = A T A and M ^ = A T A + W \widehat{M}=A^{\mathsf T}A+W M = A T A + W , and r = 1 r=1 r = 1 and s = k s=k s = k . The spectral separation appearing in the theorem is δ : = λ k ( A T A ) − λ k + 1 ( A T A + W ) \delta:=\lambda_k(A^{\mathsf T}A)-\lambda_{k+1}(A^{\mathsf T}A+W) δ := λ k ( A T A ) − λ k + 1 ( A T A + W ) . We obtain
∥ ( I − V ^ k V ^ k T ) V k ∥ 2 = ∥ sin Θ ( V ^ k , V k ) ∥ 2 ≤ ∥ W ∥ 2 δ . \left\|\left(I-\widehat{V}_k\widehat{V}_k^{\mathsf T}\right)V_k\right\|_2=\left\|\sin\Theta(\widehat{V}_k,V_k)\right\|_2\leq\frac{\left\|W\right\|_2}{\delta}. ( I − V k V k T ) V k 2 = sin Θ ( V k , V k ) 2 ≤ δ ∥ W ∥ 2 .
We can upper bound λ k + 1 ( A ⊤ A + W ) \lambda_{k+1}\left(A^{\top}A+W\right) λ k + 1 ( A ⊤ A + W ) using Weyl’s inequality (Theorem 7):
λ k + 1 ( A ⊤ A + W ) ≤ λ k + 1 ( A ⊤ A ) + ∥ W ∥ 2 . \lambda_{k+1}\left(A^{\top}A+W\right)\leq\lambda_{k+1}\left(A^{\top}A\right)+\lVert W\rVert_{2}. λ k + 1 ( A ⊤ A + W ) ≤ λ k + 1 ( A ⊤ A ) + ∥ W ∥ 2 .
and thus
δ ≥ λ k ( A ⊤ A ) − λ k + 1 ( A ⊤ A ) − ∥ W ∥ 2 = Δ k − ∥ W ∥ 2 . \delta\geq\lambda_k\left(A^{\top}A\right)-\lambda_{k+1}\left(A^{\top}A\right)-\lVert W\rVert_{2}=\Delta_k-\lVert W\rVert_{2}. δ ≥ λ k ( A ⊤ A ) − λ k + 1 ( A ⊤ A ) − ∥ W ∥ 2 = Δ k − ∥ W ∥ 2 .
Lemma 10. Let M M M be a matrix. If ∥ M ∥ 2 < 1 \lVert M\rVert_2<1 ∥ M ∥ 2 < 1 , I + M I+M I + M is invertible and ∥ ( I + M ) − 1 ∥ 2 ≤ 1 1 − ∥ M ∥ 2 \lVert(I+M)^{-1}\rVert_2\leq\frac{1}{1-\lVert M\rVert_2} ∥( I + M ) − 1 ∥ 2 ≤ 1 − ∥ M ∥ 2 1 .
Proof. We can show that I + M I+M I + M is invertible as follows. Suppose we have ( I + M ) z = 0 (I+M)z=0 ( I + M ) z = 0 for some vector z z z . Then z = − M z z=-Mz z = − M z and thus
∥ z ∥ 2 = ∥ M z ∥ 2 ≤ ∥ M ∥ 2 ∥ z ∥ 2 \lVert z\rVert_2=\lVert Mz\rVert_2\leq\lVert M\rVert_2\lVert z\rVert_2 ∥ z ∥ 2 = ∥ M z ∥ 2 ≤ ∥ M ∥ 2 ∥ z ∥ 2 Since ∥ M ∥ 2 < 1 \lVert M\rVert_2<1 ∥ M ∥ 2 < 1 , the above inequality can only hold if ∥ z ∥ 2 = 0 \lVert z\rVert_2=0 ∥ z ∥ 2 = 0 . Thus z = 0 z=0 z = 0 and hence I + M I+M I + M is invertible.
Next, we show the inequality in the lemma statement. For any vector z z z , we have
∥ ( I + M ) z ∥ 2 ≥ ∥ z ∥ 2 − ∥ M z ∥ 2 ≥ ( 1 − ∥ M ∥ 2 ) ∥ z ∥ 2 \lVert(I+M)z\rVert_2\geq\lVert z\rVert_2-\lVert Mz\rVert_2\geq(1-\lVert M\rVert_2)\lVert z\rVert_2 ∥( I + M ) z ∥ 2 ≥ ∥ z ∥ 2 − ∥ M z ∥ 2 ≥ ( 1 − ∥ M ∥ 2 ) ∥ z ∥ 2 Thus
σ min ( I + M ) ≥ 1 − ∥ M ∥ 2 \sigma_{\min}(I+M)\geq1-\lVert M\rVert_2 σ m i n ( I + M ) ≥ 1 − ∥ M ∥ 2 Combining with ∥ M ∥ 2 < 1 \lVert M\rVert_2<1 ∥ M ∥ 2 < 1 gives
∥ ( I + M ) − 1 ∥ 2 = 1 σ min ( I + M ) ≤ 1 1 − ∥ M ∥ 2 , \lVert(I+M)^{-1}\rVert_2=\frac{1}{\sigma_{\min}(I+M)}\leq\frac{1}{1-\lVert M\rVert_2}, ∥( I + M ) − 1 ∥ 2 = σ m i n ( I + M ) 1 ≤ 1 − ∥ M ∥ 2 1 , as needed.
Algorithm Analysis
Privacy Analysis
Lemma 11. Assume every row a i a_i a i of input matrix A A A satisfies ∥ a i ∥ 2 ≤ 1 \lVert a_i\rVert_2\leq1 ∥ a i ∥ 2 ≤ 1 , and let ϵ i n i t , δ i n i t ∈ ( 0 , 1 ) \epsilon_{\mathrm{init}},\delta_{\mathrm{init}}\in(0,1) ϵ init , δ init ∈ ( 0 , 1 ) . Algorithm 1 is ( ϵ i n i t , δ i n i t ) (\epsilon_{\mathrm{init}},\delta_{\mathrm{init}}) ( ϵ init , δ init ) -differentially private.
Proof. Under replacement adjacency, two neighboring datasets A A A and A ′ A' A ′ differ by exactly one row, meaning a row a a a is replaced by a ′ a' a ′ . Therefore:
A ⊤ A − A ′ ⊤ A ′ = a a ⊤ − a ′ a ′ ⊤ A^{\top}A-A'^{\top}A'=aa^{\top}-a'a'^{\top} A ⊤ A − A ′ ⊤ A ′ = a a ⊤ − a ′ a ′ ⊤ and using triangle inequality, we have:
∥ A ⊤ A − A ′ ⊤ A ′ ∥ F ≤ ∥ a a ⊤ ∥ F + ∥ a ′ a ′ ⊤ ∥ F = ∥ a ∥ 2 2 + ∥ a ′ ∥ 2 2 ≤ 2 \lVert A^{\top}A-A'^{\top}A'\rVert_F\leq\lVert aa^{\top}\rVert_F+\lVert a'a'^{\top}\rVert_F=\lVert a\rVert_2^2+\lVert a'\rVert_2^2\leq2 ∥ A ⊤ A − A ′ ⊤ A ′ ∥ F ≤ ∥ a a ⊤ ∥ F + ∥ a ′ a ′ ⊤ ∥ F = ∥ a ∥ 2 2 + ∥ a ′ ∥ 2 2 ≤ 2 Thus, Δ ∥ A ⊤ A ∥ F ≤ 2 \Delta\lVert A^{\top}A\rVert_F\leq2 Δ ∥ A ⊤ A ∥ F ≤ 2 .
In the orthonormal basis { E i i : 1 ≤ i ≤ d } ∪ { E i j + E j i 2 : 1 ≤ i < j ≤ d } \{E_{ii}:1\leq i\leq d\}\cup\left\{\frac{E_{ij}+E_{ji}}{\sqrt{2}}:1\leq i<j\leq d\right\} { E ii : 1 ≤ i ≤ d } ∪ { 2 E ij + E j i : 1 ≤ i < j ≤ d } of the symmetric-matrix space, the coordinates of the matrix sym ( G ( i n i t ) ) \operatorname{sym}(G^{(\mathrm{init})}) sym ( G ( init ) ) are independent standard Gaussian random variables. Hence M ^ = A ⊤ A + Δ i n i t sym ( G ( i n i t ) ) \widehat{M}=A^{\top}A+\Delta_{\mathrm{init}}\operatorname{sym}(G^{(\mathrm{init})}) M = A ⊤ A + Δ init sym ( G ( init ) ) is the Gaussian mechanism with sensitivity at most 2 2 2 , and is therefore ( ϵ i n i t , δ i n i t ) (\epsilon_{\mathrm{init}},\delta_{\mathrm{init}}) ( ϵ init , δ init ) -differentially private. The subsequent step of computing X ( 0 ) X^{(0)} X ( 0 ) from M ^ \widehat{M} M is a deterministic postprocessing step.
Lemma 12. Fix an initial matrix X ( 0 ) X^{(0)} X ( 0 ) with orthonormal columns, a number of iterations T ≥ 1 T \ge1 T ≥ 1 , and a clipping radius R R R . Let ϵ p o w , δ p o w ∈ ( 0 , 1 ) \epsilon_{\mathrm{pow}}, \delta_{\mathrm{pow}} \in(0,1) ϵ pow , δ pow ∈ ( 0 , 1 ) and set
ϵ i t e r = ϵ p o w 4 2 T log ( ( T + 1 ) / δ p o w ) , \epsilon_{\mathrm{iter}} = \frac{\epsilon_{\mathrm{pow}}}{4\sqrt{2T\log\left((T+1)/\delta_{\mathrm{pow}}\right)}}, ϵ iter = 4 2 T log ( ( T + 1 ) / δ pow ) ϵ pow , δ i t e r = δ p o w 2 ( T + 1 ) . \delta_{\mathrm{iter}} = \frac{\delta_{\mathrm{pow}}}{2(T+1)}. δ iter = 2 ( T + 1 ) δ pow . Then Algorithm 2 is ( ϵ p o w , δ p o w ) (\epsilon_{\mathrm{pow}},\delta_{\mathrm{pow}}) ( ϵ pow , δ pow ) -differentially private.
Proof. Since the algorithm is adaptive, we condition on the transcript preceding each release. Previously released quantities may therefore be treated as fixed when analyzing the conditional privacy guarantee. First we analyze the release of H ( t ) H^{(t)} H ( t ) . For a single row a a a , its contribution to X ( t ) T A T A X ( t ) X^{(t)\mathsf{T}}A^{\mathsf{T}}AX^{(t)} X ( t ) T A T A X ( t ) is:
X ( t ) T a a T X ( t ) = ( X ( t ) T a ) ( X ( t ) T a ) T X^{(t)\mathsf{T}}aa^{\mathsf{T}}X^{(t)} = \left(X^{(t)\mathsf{T}}a\right)\left(X^{(t)\mathsf{T}}a\right)^{\mathsf{T}} X ( t ) T a a T X ( t ) = ( X ( t ) T a ) ( X ( t ) T a ) T
Because X ( t ) X^{(t)} X ( t ) is a matrix with orthonormal columns, ∥ X ( t ) T a ∥ 2 ≤ ∥ a ∥ 2 \lVert X^{(t)\mathsf{T}}a\rVert_2 \le\lVert a\rVert_2 ∥ X ( t ) T a ∥ 2 ≤ ∥ a ∥ 2 , and thus:
∥ X ( t ) T a a T X ( t ) ∥ F = ∥ ( X ( t ) T a ) ( X ( t ) T a ) T ∥ F = ∥ X ( t ) T a ∥ 2 2 ≤ ∥ a ∥ 2 2 ≤ 1 \left\lVert X^{(t)\mathsf{T}}aa^{\mathsf{T}}X^{(t)}\right\rVert_F = \left\lVert\left(X^{(t)\mathsf{T}}a\right)\left(X^{(t)\mathsf{T}}a\right)^{\mathsf{T}}\right\rVert_F = \left\lVert X^{(t)\mathsf{T}}a\right\rVert_2^2 \le\lVert a\rVert_2^2 \le1 X ( t ) T a a T X ( t ) F = ( X ( t ) T a ) ( X ( t ) T a ) T F = X ( t ) T a 2 2 ≤ ∥ a ∥ 2 2 ≤ 1
When comparing neighboring datasets where row a a a is replaced by a ′ a' a ′ , the total change is:
∥ X ( t ) T a a T X ( t ) − X ( t ) T a ′ a ′ T X ( t ) ∥ F ≤ ∥ X ( t ) T a a T X ( t ) ∥ F + ∥ X ( t ) T a ′ a ′ T X ( t ) ∥ F ≤ 2 \left\lVert X^{(t)\mathsf{T}}aa^{\mathsf{T}}X^{(t)} - X^{(t)\mathsf{T}}a'a'^{\mathsf{T}}X^{(t)}\right\rVert_F \le\left\lVert X^{(t)\mathsf{T}}aa^{\mathsf{T}}X^{(t)}\right\rVert_F + \left\lVert X^{(t)\mathsf{T}}a'a'^{\mathsf{T}}X^{(t)}\right\rVert_F \le2 X ( t ) T a a T X ( t ) − X ( t ) T a ′ a ′ T X ( t ) F ≤ X ( t ) T a a T X ( t ) F + X ( t ) T a ′ a ′ T X ( t ) F ≤ 2
Therefore, Δ ∥ X ( t ) T A T A X ( t ) ∥ F ≤ 2 \Delta\left\lVert X^{(t)\mathsf{T}}A^{\mathsf{T}}AX^{(t)}\right\rVert_F \le2 Δ X ( t ) T A T A X ( t ) F ≤ 2 . Since G ( t ) ∼ N ( 0 , 1 ) k × k G^{(t)} \sim\mathcal{N}(0,1)^{k \times k} G ( t ) ∼ N ( 0 , 1 ) k × k , adding Δ sym ( G ( t ) ) \Delta\operatorname{sym}(G^{(t)}) Δ sym ( G ( t ) ) applies the symmetric Gaussian mechanism. Because Δ = 2 2 log ( 2 / δ i t e r ) / ϵ i t e r \Delta= 2\sqrt{2\log(2/\delta_{\mathrm{iter}})}/\epsilon_{\mathrm{iter}} Δ = 2 2 log ( 2/ δ iter ) / ϵ iter , the release H ( t ) = X ( t ) T A T A X ( t ) + Δ sym ( G ( t ) ) H^{(t)} = X^{(t)\mathsf{T}}A^{\mathsf{T}}AX^{(t)} + \Delta\operatorname{sym}(G^{(t)}) H ( t ) = X ( t ) T A T A X ( t ) + Δ sym ( G ( t ) ) is ( ϵ i t e r , δ i t e r ) (\epsilon_{\mathrm{iter}},\delta_{\mathrm{iter}}) ( ϵ iter , δ iter ) -differentially private. Construction of B ( t ) B^{(t)} B ( t ) are post-processing steps.
Second, we analyze the release of Z ( t + 1 ) Z^{(t+1)} Z ( t + 1 ) . The contribution of one row a a a to Z ( t + 1 ) Z^{(t+1)} Z ( t + 1 ) is f t ( a ) : = a Clip R ( a T X ( t ) B ( t ) ) f_t(a) := a\operatorname{Clip}_R(a^{\mathsf{T}}X^{(t)}B^{(t)}) f t ( a ) := a Clip R ( a T X ( t ) B ( t ) ) . Since the norm of an outer product is the product of the two vector norms,
∥ f t ( a i ) ∥ F = ∥ a i ∥ 2 ∥ Clip R ( a i T X ( t ) B ( t ) ) ∥ 2 ≤ R \left\lVert f_t(a_i)\right\rVert_F = \lVert a_i\rVert_2\left\lVert\operatorname{Clip}_R\left(a_i^{\mathsf{T}}X^{(t)}B^{(t)}\right)\right\rVert_2 \le R ∥ f t ( a i ) ∥ F = ∥ a i ∥ 2 Clip R ( a i T X ( t ) B ( t ) ) 2 ≤ R
Thus Δ ( ∑ i = 1 n a i Clip R ( a i T X ( t ) B ( t ) ) ) ≤ 2 R \Delta\left(\sum_{i=1}^{n}a_i\operatorname{Clip}_R(a_i^{\mathsf{T}}X^{(t)}B^{(t)})\right)\le2R Δ ( ∑ i = 1 n a i Clip R ( a i T X ( t ) B ( t ) ) ) ≤ 2 R . Consequently, Z ( t + 1 ) = ∑ i = 1 n f t ( a i ) + Δ R G ~ ( t ) Z^{(t+1)} = \sum_{i=1}^{n}f_t(a_i)+\Delta R\widetilde{G}^{(t)} Z ( t + 1 ) = ∑ i = 1 n f t ( a i ) + Δ R G ( t ) is another ( ϵ i t e r , δ i t e r ) (\epsilon_{\mathrm{iter}},\delta_{\mathrm{iter}}) ( ϵ iter , δ iter ) -differentially private Gaussian mechanism. The QR factorization is post-processing.
By sequential composition, a single iteration is ( 2 ϵ i t e r , 2 δ i t e r ) (2\epsilon_{\mathrm{iter}},2\delta_{\mathrm{iter}}) ( 2 ϵ iter , 2 δ iter ) -differentially private. Advanced composition over the T T T iterations states that the whole algorithm is ( ϵ t o t a l , δ t o t a l ) (\epsilon_{\mathrm{total}},\delta_{\mathrm{total}}) ( ϵ total , δ total ) -differentially private, where:
δ t o t a l = T ( 2 δ i t e r ) + δ p o w T + 1 = δ p o w \delta_{\mathrm{total}} = T(2\delta_{\mathrm{iter}}) + \frac{\delta_{\mathrm{pow}}}{T+1} = \delta_{\mathrm{pow}} δ total = T ( 2 δ iter ) + T + 1 δ pow = δ pow
ϵ t o t a l ≤ 2 ϵ i t e r 2 T log ( T + 1 δ p o w ) + T ( 2 ϵ i t e r ) ( e 2 ϵ i t e r − 1 ) \epsilon_{\mathrm{total}} \le2\epsilon_{\mathrm{iter}}\sqrt{2T\log\left(\frac{T+1}{\delta_{\mathrm{pow}}}\right)} + T(2\epsilon_{\mathrm{iter}})(e^{2\epsilon_{\mathrm{iter}}}-1) ϵ total ≤ 2 ϵ iter 2 T log ( δ pow T + 1 ) + T ( 2 ϵ iter ) ( e 2 ϵ iter − 1 )
We now show that ϵ t o t a l ≤ ϵ p o w \epsilon_{\mathrm{total}}\le\epsilon_{\mathrm{pow}} ϵ total ≤ ϵ pow . We have 2 ϵ i t e r < 1 2\epsilon_{\mathrm{iter}}<1 2 ϵ iter < 1 so e 2 ϵ i t e r − 1 ≤ 2 ( 2 ϵ i t e r ) e^{2\epsilon_{\mathrm{iter}}}-1\le2(2\epsilon_{\mathrm{iter}}) e 2 ϵ iter − 1 ≤ 2 ( 2 ϵ iter ) . Our choice of ϵ i t e r \epsilon_{\mathrm{iter}} ϵ iter gives
2 ϵ i t e r 2 T log ( T + 1 δ p o w ) = ϵ p o w 2 2\epsilon_{\mathrm{iter}}\sqrt{2T\log\left(\frac{T+1}{\delta_{\mathrm{pow}}}\right)} = \frac{\epsilon_{\mathrm{pow}}}{2} 2 ϵ iter 2 T log ( δ pow T + 1 ) = 2 ϵ pow
T ( 2 ϵ i t e r ) ( e 2 ϵ i t e r − 1 ) ≤ 2 T ( 2 ϵ i t e r ) 2 = 2 T ( ϵ p o w 2 2 T log ( T + 1 δ p o w ) ) 2 \begin{aligned}
T(2\epsilon_{\mathrm{iter}})(e^{2\epsilon_{\mathrm{iter}}}-1) \le2T(2\epsilon_{\mathrm{iter}})^2 \\
&= 2T\left(\frac{\epsilon_{\mathrm{pow}}}{2\sqrt{2T\log\left(\frac{T+1}{\delta_{\mathrm{pow}}}\right)}}\right)^2
\end{aligned} T ( 2 ϵ iter ) ( e 2 ϵ iter − 1 ) ≤ 2 T ( 2 ϵ iter ) 2 = 2 T 2 2 T log ( δ pow T + 1 ) ϵ pow 2
ϵ p o w 2 4 log ( T + 1 δ p o w ) ≤ ϵ p o w 2 \frac{\epsilon_{\mathrm{pow}}^{2}}{4\log\left(\frac{T+1}{\delta_{\mathrm{pow}}}\right)} \le\frac{\epsilon_{\mathrm{pow}}}{2} 4 log ( δ pow T + 1 ) ϵ pow 2 ≤ 2 ϵ pow
where the last inequality is true since ϵ p o w < 1 ≤ 2 log ( T + 1 δ p o w ) \epsilon_{\mathrm{pow}} < 1 \le2\log\left(\frac{T+1}{\delta_{\mathrm{pow}}}\right) ϵ pow < 1 ≤ 2 log ( δ pow T + 1 ) .
Hence ϵ t o t a l ≤ ϵ p o w \epsilon_{\mathrm{total}} \le\epsilon_{\mathrm{pow}} ϵ total ≤ ϵ pow , and Algorithm 2 is ( ϵ p o w , δ p o w ) (\epsilon_{\mathrm{pow}},\delta_{\mathrm{pow}}) ( ϵ pow , δ pow ) -differentially private.
Lemma 11, Lemma 12, and sequential composition imply that Algorithm 3 is ( ϵ , δ ) (\epsilon,\delta) ( ϵ , δ ) -differentially private.
Utility Analysis
We start by introducing some notation that we will use throughout the analysis. We define
κ k = σ k 2 − σ k + 1 2 σ k 2 = Δ k σ k 2 , ∥ U ∥ ∞ = max i , j ∣ U i j ∣ . \kappa_k = \frac{\sigma_k^2-\sigma_{k+1}^2}{\sigma_k^2}=\frac{\Delta_k}{\sigma_k^2}, \qquad\lVert U\rVert_\infty=\max_{i,j}|U_{ij}|. κ k = σ k 2 σ k 2 − σ k + 1 2 = σ k 2 Δ k , ∥ U ∥ ∞ = i , j max ∣ U ij ∣.
For every iteration t ≥ 0 t \ge0 t ≥ 0 of Algorithm 2, we define
C ( t ) : = V k T X ( t ) , D ( t ) : = V > k T X ( t ) , s t : = ∥ ( I − X ( t ) X ( t ) T ) V k ∥ 2 . C^{(t)} := V_k^{\mathsf T}X^{(t)}, \quad D^{(t)} := V_{>k}^{\mathsf T}X^{(t)}, \quad s_t := \left\lVert\left(I-X^{(t)}X^{(t)\mathsf T}\right)V_k\right\rVert_2. C ( t ) := V k T X ( t ) , D ( t ) := V > k T X ( t ) , s t := ( I − X ( t ) X ( t ) T ) V k 2 .
For every iteration t t t in which C ( t ) C^{(t)} C ( t ) is invertible, we define
E ( t ) : = D ( t ) ( C ( t ) ) − 1 , m t : = ∥ E ( t ) ∥ 2 . E^{(t)} := D^{(t)}\left(C^{(t)}\right)^{-1}, \quad m_t := \left\lVert E^{(t)}\right\rVert_2. E ( t ) := D ( t ) ( C ( t ) ) − 1 , m t := E ( t ) 2 .
Our goal is to analyze m t m_t m t and show that it decreases appropriately with high probability. Our analysis will show that the matrices C ( t ) C^{(t)} C ( t ) are invertible with high probability and thus the matrices E ( t ) E^{(t)} E ( t ) are well-defined. As shown in the following lemma, having s t < 1 s_t < 1 s t < 1 suffices to ensure that C ( t ) C^{(t)} C ( t ) is invertible.
Lemma 13. Let X ∈ R d × k X \in\mathbb{R}^{d\times k} X ∈ R d × k be any matrix with orthonormal columns . Let s = ∥ ( I − X X T ) V k ∥ 2 s=\left\lVert\left(I-XX^{\mathsf T}\right)V_k\right\rVert_2 s = ( I − X X T ) V k 2 and C = V k T X C=V_k^{\mathsf T}X C = V k T X . We have s 2 = 1 − σ min 2 ( C ) s^2=1-\sigma_{\min}^2(C) s 2 = 1 − σ m i n 2 ( C ) , where σ min ( C ) \sigma_{\min}(C) σ m i n ( C ) is the smallest singular value of C C C . Thus, if s < 1 s<1 s < 1 , then σ min ( C ) > 0 \sigma_{\min}(C)>0 σ m i n ( C ) > 0 and thus C C C is invertible.
Proof. Since X X X has orthonormal columns , P = I − X X T P=I-XX^{\mathsf T} P = I − X X T is an orthogonal projector and thus P T P = P 2 = P P^{\mathsf T}P=P^2=P P T P = P 2 = P . Thus we have
s 2 = ∥ ( I − X X T ) V k ∥ 2 2 = ∥ V k T ( I − X X T ) V k ∥ 2 = ∥ I − C C T ∥ 2 = 1 − σ min 2 ( C ) \begin{aligned}
s^2 &= \left\lVert\left(I-XX^{\mathsf T}\right)V_k\right\rVert_2^2 \\
&= \left\lVert V_k^{\mathsf T}\left(I-XX^{\mathsf T}\right)V_k\right\rVert_2 \\
&= \left\lVert I-CC^{\mathsf T}\right\rVert_2 \\
&= 1-\sigma_{\min}^2(C)
\end{aligned} s 2 = ( I − X X T ) V k 2 2 = V k T ( I − X X T ) V k 2 = I − C C T 2 = 1 − σ m i n 2 ( C )
as needed.
We will use the following standard identity relating s t s_t s t and m t m_t m t .
Lemma 14. Let X ∈ R d × k X \in\mathbb{R}^{d\times k} X ∈ R d × k be any matrix with orthonormal columns . Let s = ∥ ( I − X X T ) V k ∥ 2 s=\left\lVert\left(I-XX^{\mathsf T}\right)V_k\right\rVert_2 s = ( I − X X T ) V k 2 . Let C = V k T X C=V_k^{\mathsf T}X C = V k T X and D = V > k T X D=V_{>k}^{\mathsf T}X D = V > k T X . Suppose that C C C is invertible and let m = ∥ D C − 1 ∥ 2 m=\left\lVert DC^{-1}\right\rVert_2 m = D C − 1 2 . We have s = m 1 + m 2 s=\frac{m}{\sqrt{1+m^2}} s = 1 + m 2 m .
Proof. Let E = D C − 1 E=DC^{-1} E = D C − 1 . Since V = [ V k V > k ] V=[V_k\ V_{>k}] V = [ V k V > k ] is orthogonal, we can write
X = V k C + V > k D = ( V k + V > k E ) C X=V_kC+V_{>k}D=(V_k+V_{>k}E)C X = V k C + V > k D = ( V k + V > k E ) C
Since C C C is invertible, we have
range ( X ) = range ( V k + V > k E ) \operatorname{range}(X)=\operatorname{range}(V_k+V_{>k}E) range ( X ) = range ( V k + V > k E )
Note that we have
( V k + V > k E ) ⊤ ( V k + V > k E ) = I + E ⊤ E (V_k+V_{>k}E)^\top(V_k+V_{>k}E)=I+E^\top E ( V k + V > k E ) ⊤ ( V k + V > k E ) = I + E ⊤ E
Thus Y = ( V k + V > k E ) ( I + E ⊤ E ) − 1 / 2 Y=(V_k+V_{>k}E)(I+E^\top E)^{-1/2} Y = ( V k + V > k E ) ( I + E ⊤ E ) − 1/2 is an orthonormal basis for range ( X ) (X) ( X ) , and therefore Y Y ⊤ = X X ⊤ YY^\top=XX^\top Y Y ⊤ = X X ⊤ . Thus
∥ ( I − X X ⊤ ) V k ∥ 2 2 = ∥ ( I − Y Y ⊤ ) V k ∥ 2 2 = ∥ V k ⊤ ( I − Y Y ⊤ ) ⊤ ( I − Y Y ⊤ ) V k ∥ 2 \left\|(I-XX^\top)V_k\right\|_2^2=\left\|(I-YY^\top)V_k\right\|_2^2=\left\|V_k^\top(I-YY^\top)^\top(I-YY^\top)V_k\right\|_2 ( I − X X ⊤ ) V k 2 2 = ( I − Y Y ⊤ ) V k 2 2 = V k ⊤ ( I − Y Y ⊤ ) ⊤ ( I − Y Y ⊤ ) V k 2
Since Y Y Y has orthonormal columns, Y Y ⊤ YY^\top Y Y ⊤ is an orthogonal projector. Thus P = I − Y Y ⊤ P=I-YY^\top P = I − Y Y ⊤ is also an orthogonal projector, and thus P ⊤ P = P 2 = P P^\top P=P^2=P P ⊤ P = P 2 = P . Thus
∥ ( I − X ( t ) X ( t ) ⊤ ) V k ∥ 2 2 = ∥ V k ⊤ ( I − Y Y ⊤ ) V k ∥ 2 = ∥ I − ( Y ⊤ V k ) ⊤ Y ⊤ V k ∥ 2 \left\|\left(I-X^{(t)}X^{(t)\top}\right)V_k\right\|_2^2=\left\|V_k^\top(I-YY^\top)V_k\right\|_2=\left\|I-(Y^\top V_k)^\top Y^\top V_k\right\|_2 ( I − X ( t ) X ( t ) ⊤ ) V k 2 2 = V k ⊤ ( I − Y Y ⊤ ) V k 2 = I − ( Y ⊤ V k ) ⊤ Y ⊤ V k 2
Recall that Y = ( V k + V > k E ) ( I + E ⊤ E ) − 1 / 2 Y=(V_k+V_{>k}E)(I+E^\top E)^{-1/2} Y = ( V k + V > k E ) ( I + E ⊤ E ) − 1/2 . Thus we have
Y ⊤ V k = ( I + E ⊤ E ) − 1 / 2 Y^\top V_k=(I+E^\top E)^{-1/2} Y ⊤ V k = ( I + E ⊤ E ) − 1/2
Thus
∥ ( I − X X ⊤ ) V k ∥ 2 2 = ∥ I − ( I + E ⊤ E ) − 1 ∥ 2 \left\|(I-XX^\top)V_k\right\|_2^2=\left\|I-(I+E^\top E)^{-1}\right\|_2 ( I − X X ⊤ ) V k 2 2 = I − ( I + E ⊤ E ) − 1 2
If E E E has singular values s i s_i s i , I − ( I + E ⊤ E ) − 1 I-(I+E^\top E)^{-1} I − ( I + E ⊤ E ) − 1 has eigenvalues 1 − 1 1 + s i 2 = s i 2 1 + s i 2 1-\frac{1}{1+s_i^2}=\frac{s_i^2}{1+s_i^2} 1 − 1 + s i 2 1 = 1 + s i 2 s i 2 . Thus
∥ ( I − X X ⊤ ) V k ∥ 2 2 = ∥ E ∥ 2 2 1 + ∥ E ∥ 2 2 \left\|(I-XX^\top)V_k\right\|_2^2=\frac{\|E\|_2^2}{1+\|E\|_2^2} ( I − X X ⊤ ) V k 2 2 = 1 + ∥ E ∥ 2 2 ∥ E ∥ 2 2
as needed.
We now proceed with the main analysis that bounds m t m_t m t . For every t ≥ 0 t\ge0 t ≥ 0 , we also define
K ( t ) : = Σ k 2 C ( t ) B ( t ) S ( t ) : = Σ > k E ( t ) Σ k − 1 P ( t ) : = Σ k C ( t ) B ( t ) Y ( t ) : = Σ > k 2 E ( t ) Σ k − 2 F ( t ) : = Δ R V k ⊤ G ~ ( t ) ( K ( t ) ) − 1 L ( t ) : = Δ R V > k ⊤ G ~ ( t ) ( K ( t ) ) − 1 \begin{aligned}
K^{(t)}&:=\Sigma_k^2 C^{(t)}B^{(t)}\\
S^{(t)}&:=\Sigma_{>k}E^{(t)}\Sigma_k^{-1}\\
P^{(t)}&:=\Sigma_k C^{(t)}B^{(t)}\\
Y^{(t)}&:=\Sigma_{>k}^2E^{(t)}\Sigma_k^{-2}\\
F^{(t)}&:=\Delta R V_k^\top\widetilde G^{(t)}\left(K^{(t)}\right)^{-1}\\
L^{(t)}&:=\Delta R V_{>k}^\top\widetilde G^{(t)}\left(K^{(t)}\right)^{-1}
\end{aligned} K ( t ) S ( t ) P ( t ) Y ( t ) F ( t ) L ( t ) := Σ k 2 C ( t ) B ( t ) := Σ > k E ( t ) Σ k − 1 := Σ k C ( t ) B ( t ) := Σ > k 2 E ( t ) Σ k − 2 := Δ R V k ⊤ G ( t ) ( K ( t ) ) − 1 := Δ R V > k ⊤ G ( t ) ( K ( t ) ) − 1
We also define the following event.
Definition 15. Define
q k : = 2 k + 2 log ( 32 T / β ) , q_k:=2\sqrt{k}+\sqrt{2\log(32T/\beta)}, q k := 2 k + 2 log ( 32 T / β ) , q d − k : = d − k + k + 2 log ( 32 T / β ) , q_{d-k}:=\sqrt{d-k}+\sqrt{k}+\sqrt{2\log(32T/\beta)}, q d − k := d − k + k + 2 log ( 32 T / β ) , q r o w : = k + 2 log ( 16 n T ( T + 1 ) / β ) . q_{\mathrm{row}}:=\sqrt{k}+\sqrt{2\log(16nT(T+1)/\beta)}. q row := k + 2 log ( 16 n T ( T + 1 ) / β ) . Let G c l i p \mathcal{G}_{\mathrm{clip}} G clip be the event where, simultaneously for all 0 ≤ t < T 0\le t<T 0 ≤ t < T and 0 ≤ r ≤ T 0\le r\le T 0 ≤ r ≤ T , we have
∥ Δ s y m ( G ( t ) ) ∥ 2 ≤ γ 4 , ∥ V k ⊤ G ~ ( t ) ∥ 2 ≤ q k , ∥ V > k ⊤ G ~ ( t ) ∥ 2 ≤ q d − k , (2) \left\|\Delta^{\mathrm{sym}}\left(G^{(t)}\right)\right\|_2\le\frac{\gamma}{4},\quad\left\|V_k^\top\widetilde G^{(t)}\right\|_2\le q_k,\quad\left\|V_{>k}^\top\widetilde G^{(t)}\right\|_2\le q_{d-k}, \tag*{(2)} Δ sym ( G ( t ) ) 2 ≤ 4 γ , V k ⊤ G ( t ) 2 ≤ q k , V > k ⊤ G ( t ) 2 ≤ q d − k , (2) max i ∥ e i ⊤ U > k Σ > k 2 r + 1 V > k ⊤ G ~ ( t ) ∥ 2 ≤ σ k + 1 2 r q r o w . (3) \max_i\left\|e_i^\top U_{>k}\Sigma_{>k}^{2r+1}V_{>k}^\top\widetilde G^{(t)}\right\|_2\le\sigma_{k+1}^{2r}q_{\mathrm{row}}. \tag*{(3)} i max e i ⊤ U > k Σ > k 2 r + 1 V > k ⊤ G ( t ) 2 ≤ σ k + 1 2 r q row . (3)
Lemma 16. We have Pr ( G c l i p ) ≥ 1 − β 4 \Pr(\mathcal{G}_{\mathrm{clip}}) \ge1-\frac{\beta}{4} Pr ( G clip ) ≥ 1 − 4 β .
Proof. Note that ∥ sym ( G ( t ) ) ∥ 2 ≤ ∥ G ( t ) ∥ 2 \lVert\operatorname{sym}(G^{(t)})\rVert_{2} \le\lVert G^{(t)}\rVert_{2} ∥ sym ( G ( t ) ) ∥ 2 ≤ ∥ G ( t ) ∥ 2 . By Lemma 5, for a fixed t t t , we have ∥ Δ sym ( G ( t ) ) ∥ 2 ≤ γ / 4 \lVert\Delta\operatorname{sym}(G^{(t)})\rVert_{2} \le\gamma/4 ∥ Δ sym ( G ( t ) ) ∥ 2 ≤ γ /4 with probability 1 − β 8 T 1-\frac{\beta}{8T} 1 − 8 T β . Union bound over all t t t gives the total failure probability β / 8 \beta/8 β /8 .
By rotational invariance of the normal distribution, V ⊤ G ~ ( t ) V^{\top}\widetilde{G}^{(t)} V ⊤ G ( t ) is identically distributed as G ~ ( t ) \widetilde{G}^{(t)} G ( t ) . Consequently, V k ⊤ G ~ ( t ) ∼ N ( 0 , 1 ) k × k V_{k}^{\top}\widetilde{G}^{(t)}\sim\mathcal{N}(0,1)^{k\times k} V k ⊤ G ( t ) ∼ N ( 0 , 1 ) k × k and V > k ⊤ G ~ ( t ) ∼ N ( 0 , 1 ) ( d − k ) × k V_{>k}^{\top}\widetilde{G}^{(t)}\sim\mathcal{N}(0,1)^{(d-k)\times k} V > k ⊤ G ( t ) ∼ N ( 0 , 1 ) ( d − k ) × k . The two bounds in (2) then follow from Lemma 5 with failure probability β 32 T \frac{\beta}{32T} 32 T β for each bound and each t t t . Union bound over all t t t and two bounds gives the total failure probability of β / 16 \beta/16 β /16 .
For (3), the i i i th row e i ⊤ U > k Σ > k 2 r + 1 V > k ⊤ G ~ ( t ) e_i^{\top}U_{>k}\Sigma_{>k}^{2r+1}V_{>k}^{\top}\widetilde{G}^{(t)} e i ⊤ U > k Σ > k 2 r + 1 V > k ⊤ G ( t ) is a k k k -dimensional Gaussian vector whose coordinate variance is at most σ k + 1 4 r \sigma_{k+1}^{4r} σ k + 1 4 r . Indeed, each coordinate is identically distributed as a standard Gaussian scaled by
∥ e i ⊤ U > k Σ > k 2 r + 1 ∥ 2 ≤ σ k + 1 2 r ∥ e i ⊤ U > k Σ > k ∥ 2 ≤ σ k + 1 2 r + 1 ∥ a i ∥ 2 . \lVert e_i^{\top}U_{>k}\Sigma_{>k}^{2r+1}\rVert_{2}\leq\sigma_{k+1}^{2r}\lVert e_i^{\top}U_{>k}\Sigma_{>k}\rVert_{2}\leq\sigma_{k+1}^{2r+1}\lVert a_i\rVert_{2}. ∥ e i ⊤ U > k Σ > k 2 r + 1 ∥ 2 ≤ σ k + 1 2 r ∥ e i ⊤ U > k Σ > k ∥ 2 ≤ σ k + 1 2 r + 1 ∥ a i ∥ 2 . Applying Theorem 4 and a union bound over i , t , r i,t,r i , t , r give the stated row-wise bound with total failure probability β / 16 \beta/16 β /16 .
The total failure probability for all conditions is β / 8 + β / 16 + β / 16 ≤ β / 4 \beta/8+\beta/16+\beta/16\leq\beta/4 β /8 + β /16 + β /16 ≤ β /4 .
Next, we show that, under certain assumptions, the preconditioner B ( t ) B^{(t)} B ( t ) roughly whitens Σ k V k ⊤ X ( t ) = Σ k C ( t ) \Sigma_kV_k^{\top}X^{(t)}=\Sigma_kC^{(t)} Σ k V k ⊤ X ( t ) = Σ k C ( t ) , i.e., the singular values of Σ k C ( t ) B ( t ) \Sigma_kC^{(t)}B^{(t)} Σ k C ( t ) B ( t ) are bounded from above and below by absolute constants. To this end, we assume that ∥ Δ sym ( G ( t ) ) ∥ 2 ≤ γ 4 \lVert\Delta\operatorname{sym}(G^{(t)})\rVert_{2}\leq\frac{\gamma}{4} ∥ Δ sym ( G ( t ) ) ∥ 2 ≤ 4 γ , s t ≤ 1 2 s_t\leq\frac{1}{2} s t ≤ 2 1 , and γ ≤ σ k 2 4 \gamma\leq\frac{\sigma_k^2}{4} γ ≤ 4 σ k 2 . As shown in the proof of Lemma 16, the first condition holds with high probability. Later in this section, we will show that the second condition also holds with high probability and the third condition is implied by a sufficiently large eigengap Δ k \Delta_k Δ k . Recall that P ( t ) : = Σ k C ( t ) B ( t ) P^{(t)}:=\Sigma_kC^{(t)}B^{(t)} P ( t ) := Σ k C ( t ) B ( t ) .
Lemma 17. Suppose that ∥ Δ sym ( G ( t ) ) ∥ 2 ≤ γ 4 \lVert\Delta\operatorname{sym}(G^{(t)})\rVert_{2}\leq\frac{\gamma}{4} ∥ Δ sym ( G ( t ) ) ∥ 2 ≤ 4 γ , s t ≤ 1 2 s_t\leq\frac{1}{2} s t ≤ 2 1 , and γ ≤ σ k 2 4 \gamma\leq\frac{\sigma_k^2}{4} γ ≤ 4 σ k 2 . We have H ( t ) = H ~ ( t ) H^{(t)}=\widetilde{H}^{(t)} H ( t ) = H ( t ) , B ( t ) = ( H ( t ) ) − 1 / 2 B^{(t)}=(H^{(t)})^{-1/2} B ( t ) = ( H ( t ) ) − 1/2 , and σ min ( P ( t ) ) ≥ 2 3 \sigma_{\min}(P^{(t)})\geq\frac{2}{3} σ m i n ( P ( t ) ) ≥ 3 2 and ∥ P ( t ) ∥ 2 ≤ 2 \lVert P^{(t)}\rVert_{2}\leq\sqrt{2} ∥ P ( t ) ∥ 2 ≤ 2 .
Proof. We first show that H ( t ) = H ~ ( t ) H^{(t)}=\widetilde{H}^{(t)} H ( t ) = H ( t ) , i.e., we have H ( t ) ⪰ γ I k H^{(t)}\succeq\gamma I_k H ( t ) ⪰ γ I k . We have
H ( t ) = X ( t ) ⊤ A ⊤ A X ( t ) + Δ sym ( G ( t ) ) H^{(t)}=X^{(t)\top}A^{\top}AX^{(t)}+\Delta\operatorname{sym}\left(G^{(t)}\right) H ( t ) = X ( t ) ⊤ A ⊤ A X ( t ) + Δ sym ( G ( t ) ) = X ( t ) ⊤ ( V k Σ k 2 V k ⊤ + V > k Σ > k 2 V > k ⊤ ) X ( t ) + Δ sym ( G ( t ) ) =X^{(t)\top}\left(V_k\Sigma_k^2V_k^{\top}+V_{>k}\Sigma_{>k}^2V_{>k}^{\top}\right)X^{(t)}+\Delta\operatorname{sym}\left(G^{(t)}\right) = X ( t ) ⊤ ( V k Σ k 2 V k ⊤ + V > k Σ > k 2 V > k ⊤ ) X ( t ) + Δ sym ( G ( t ) ) = C ( t ) ⊤ Σ k 2 C ( t ) + D ( t ) ⊤ Σ > k 2 D ( t ) + Δ sym ( G ( t ) ) . =C^{(t)\top}\Sigma_k^2C^{(t)}+D^{(t)\top}\Sigma_{>k}^2D^{(t)}+\Delta\operatorname{sym}\left(G^{(t)}\right). = C ( t ) ⊤ Σ k 2 C ( t ) + D ( t ) ⊤ Σ > k 2 D ( t ) + Δ sym ( G ( t ) ) . We lower bound each of the terms in turn. Consider C ( t ) ⊤ Σ k 2 C ( t ) C^{(t)\top}\Sigma_k^2C^{(t)} C ( t ) ⊤ Σ k 2 C ( t ) . Since ∥ D ( t ) ∥ 2 = s t ≤ 1 2 \lVert D^{(t)}\rVert_{2}=s_t\leq\frac{1}{2} ∥ D ( t ) ∥ 2 = s t ≤ 2 1 , we have D ( t ) ⊤ D ( t ) ⪯ 1 4 I k D^{(t)\top}D^{(t)}\preceq\frac{1}{4}I_k D ( t ) ⊤ D ( t ) ⪯ 4 1 I k . Since C ( t ) ⊤ C ( t ) + D ( t ) ⊤ D ( t ) = I k C^{(t)\top}C^{(t)}+D^{(t)\top}D^{(t)}=I_k C ( t ) ⊤ C ( t ) + D ( t ) ⊤ D ( t ) = I k , it follows that C ( t ) ⊤ C ( t ) ⪰ 3 4 I k C^{(t)\top}C^{(t)}\succeq\frac{3}{4}I_k C ( t ) ⊤ C ( t ) ⪰ 4 3 I k and thus C ( t ) ⊤ Σ k 2 C ( t ) ⪰ 3 4 σ k 2 I k C^{(t)\top}\Sigma_k^2C^{(t)}\succeq\frac{3}{4}\sigma_k^2I_k C ( t ) ⊤ Σ k 2 C ( t ) ⪰ 4 3 σ k 2 I k . Next, we have D ( t ) ⊤ Σ > k 2 D ( t ) ⪰ 0 D^{(t)\top}\Sigma_{>k}^2D^{(t)}\succeq0 D ( t ) ⊤ Σ > k 2 D ( t ) ⪰ 0 . We have ∥ Δ sym ( G ( t ) ) ∥ 2 ≤ γ 4 ≤ σ k 2 16 \lVert\Delta\operatorname{sym}(G^{(t)})\rVert_{2}\leq\frac{\gamma}{4}\leq\frac{\sigma_k^2}{16} ∥ Δ sym ( G ( t ) ) ∥ 2 ≤ 4 γ ≤ 16 σ k 2 and thus Δ sym ( G ( t ) ) ⪰ − 1 16 σ k 2 I k \Delta\operatorname{sym}(G^{(t)})\succeq-\frac{1}{16}\sigma_k^2I_k Δ sym ( G ( t ) ) ⪰ − 16 1 σ k 2 I k . Putting everything together, we obtain
H ( t ) ⪰ 3 4 σ k 2 I k − 1 16 σ k 2 I k = 11 16 σ k 2 I k ⪰ γ I k . H^{(t)}\succeq\frac{3}{4}\sigma_k^2I_k-\frac{1}{16}\sigma_k^2I_k=\frac{11}{16}\sigma_k^2I_k\succeq\gamma I_k. H ( t ) ⪰ 4 3 σ k 2 I k − 16 1 σ k 2 I k = 16 11 σ k 2 I k ⪰ γ I k . Thus we have H ( t ) = H ~ ( t ) H^{(t)}=\widetilde{H}^{(t)} H ( t ) = H ( t ) and B ( t ) = ( H ( t ) ) − 1 / 2 B^{(t)}=(H^{(t)})^{-1/2} B ( t ) = ( H ( t ) ) − 1/2 .
Finally, consider P ( t ) : = Σ k C ( t ) B ( t ) P^{(t)}:=\Sigma_kC^{(t)}B^{(t)} P ( t ) := Σ k C ( t ) B ( t ) . We will lower and upper bound P ( t ) ⊤ P ( t ) P^{(t)\top}P^{(t)} P ( t ) ⊤ P ( t ) . We have
P ( t ) ⊤ P ( t ) = B ( t ) ⊤ ( C ( t ) ⊤ Σ k 2 C ( t ) ) B ( t ) . P^{(t)\top}P^{(t)}=B^{(t)\top}\left(C^{(t)\top}\Sigma_k^2C^{(t)}\right)B^{(t)}. P ( t ) ⊤ P ( t ) = B ( t ) ⊤ ( C ( t ) ⊤ Σ k 2 C ( t ) ) B ( t ) . Let M : = X ( t ) ⊤ A ⊤ A X ( t ) = C ( t ) ⊤ Σ k 2 C ( t ) + D ( t ) ⊤ Σ > k 2 D ( t ) M:=X^{(t)\top}A^{\top}AX^{(t)}=C^{(t)\top}\Sigma_k^2C^{(t)}+D^{(t)\top}\Sigma_{>k}^2D^{(t)} M := X ( t ) ⊤ A ⊤ A X ( t ) = C ( t ) ⊤ Σ k 2 C ( t ) + D ( t ) ⊤ Σ > k 2 D ( t ) and N : = Δ sym ( G ( t ) ) N:=\Delta\operatorname{sym}(G^{(t)}) N := Δ sym ( G ( t ) ) , and thus H ( t ) = M + N H^{(t)}=M+N H ( t ) = M + N . As shown above, we have M ⪰ 3 4 σ k 2 I k M\succeq\frac{3}{4}\sigma_k^2I_k M ⪰ 4 3 σ k 2 I k and N ⪯ 1 16 σ k 2 I k N\preceq\frac{1}{16}\sigma_k^2I_k N ⪯ 16 1 σ k 2 I k , and thus
− 1 12 M ⪯ N ⪯ 1 12 M , -\frac{1}{12}M\preceq N\preceq\frac{1}{12}M, − 12 1 M ⪯ N ⪯ 12 1 M , :::`
11 12 M ⪯ H ( t ) ⪯ 13 12 M . \frac{11}{12}M \preceq H^{(t)} \preceq\frac{13}{12}M. 12 11 M ⪯ H ( t ) ⪯ 12 13 M .
Thus
11 12 B ( t ) T M B ( t ) ⪯ B ( t ) T H ( t ) B ( t ) ⪯ 13 12 B ( t ) T M B ( t ) . \frac{11}{12}B^{(t)\mathsf{T}}MB^{(t)} \preceq B^{(t)\mathsf{T}}H^{(t)}B^{(t)} \preceq\frac{13}{12}B^{(t)\mathsf{T}}MB^{(t)}. 12 11 B ( t ) T M B ( t ) ⪯ B ( t ) T H ( t ) B ( t ) ⪯ 12 13 B ( t ) T M B ( t ) .
Since B ( t ) = ( H ( t ) ) − 1 / 2 B^{(t)} = \left(H^{(t)}\right)^{-1/2} B ( t ) = ( H ( t ) ) − 1/2 , we have B ( t ) T H ( t ) B ( t ) = I k B^{(t)\mathsf{T}}H^{(t)}B^{(t)} = I_k B ( t ) T H ( t ) B ( t ) = I k . Thus
12 13 I k ⪯ B ( t ) T M B ( t ) ⪯ 12 11 I k . \frac{12}{13}I_k \preceq B^{(t)\mathsf{T}}MB^{(t)} \preceq\frac{12}{11}I_k. 13 12 I k ⪯ B ( t ) T M B ( t ) ⪯ 11 12 I k .
Since D ( t ) Σ > k 2 D ( t ) ⪰ 0 D^{(t)}\Sigma^2_{>k}D^{(t)} \succeq0 D ( t ) Σ > k 2 D ( t ) ⪰ 0 , we have M ⪰ C ( t ) T Σ k 2 C ( t ) M \succeq C^{(t)\mathsf{T}}\Sigma^2_k C^{(t)} M ⪰ C ( t ) T Σ k 2 C ( t ) and thus
B ( t ) T ( C ( t ) T Σ k 2 C ( t ) ) B ( t ) ⪯ B ( t ) T M B ( t ) ⪯ 12 11 I k . B^{(t)\mathsf{T}}\left(C^{(t)\mathsf{T}}\Sigma^2_k C^{(t)}\right)B^{(t)} \preceq B^{(t)\mathsf{T}}MB^{(t)} \preceq\frac{12}{11}I_k. B ( t ) T ( C ( t ) T Σ k 2 C ( t ) ) B ( t ) ⪯ B ( t ) T M B ( t ) ⪯ 11 12 I k .
Since C ( t ) T Σ k 2 C ( t ) ⪰ 3 4 σ k 2 I k C^{(t)\mathsf{T}}\Sigma^2_k C^{(t)} \succeq\frac{3}{4}\sigma_k^2I_k C ( t ) T Σ k 2 C ( t ) ⪰ 4 3 σ k 2 I k and D ( t ) T Σ > k 2 D ( t ) ⪯ 1 4 σ k 2 I k D^{(t)\mathsf{T}}\Sigma^2_{>k}D^{(t)} \preceq\frac{1}{4}\sigma_k^2I_k D ( t ) T Σ > k 2 D ( t ) ⪯ 4 1 σ k 2 I k , we have M ⪯ 4 3 ( C ( t ) T Σ k 2 C ( t ) ) M \preceq\frac{4}{3}\left(C^{(t)\mathsf{T}}\Sigma^2_k C^{(t)}\right) M ⪯ 3 4 ( C ( t ) T Σ k 2 C ( t ) ) and thus
B ( t ) T ( C ( t ) T Σ k 2 C ( t ) ) B ( t ) ⪰ 3 4 B ( t ) T M B ( t ) ⪰ 9 13 I k . B^{(t)\mathsf{T}}\left(C^{(t)\mathsf{T}}\Sigma^2_k C^{(t)}\right)B^{(t)} \succeq\frac{3}{4}B^{(t)\mathsf{T}}MB^{(t)} \succeq\frac{9}{13}I_k. B ( t ) T ( C ( t ) T Σ k 2 C ( t ) ) B ( t ) ⪰ 4 3 B ( t ) T M B ( t ) ⪰ 13 9 I k .
Putting everything together, we have
9 13 I k ⪯ P ( t ) T P ( t ) ⪯ 12 11 I k . \frac{9}{13}I_k \preceq P^{(t)\mathsf{T}}P^{(t)} \preceq\frac{12}{11}I_k. 13 9 I k ⪯ P ( t ) T P ( t ) ⪯ 11 12 I k .
and thus
σ min ( P ( t ) ) ≥ 9 13 ≥ 2 3 , \sigma_{\min}\left(P^{(t)}\right) \ge\sqrt{\frac{9}{13}} \ge\frac{2}{3}, σ m i n ( P ( t ) ) ≥ 13 9 ≥ 3 2 ,
∥ P ( t ) ∥ 2 ≤ 12 11 ≤ 2 . \left\|P^{(t)}\right\|_2 \le\sqrt{\frac{12}{11}} \le\sqrt{2}. P ( t ) 2 ≤ 11 12 ≤ 2 .
Lemma 18. Suppose that ∥ Δ s y m ( G ( t ) ) ∥ 2 ≤ γ 4 \left\|\Delta\mathop{\mathrm{sym}}\left(G^{(t)}\right)\right\|_2 \le\frac{\gamma}{4} Δ sym ( G ( t ) ) 2 ≤ 4 γ , s t ≤ 1 2 s_t \le\frac{1}{2} s t ≤ 2 1 , and γ ≤ σ k 2 4 \gamma\le\frac{\sigma_k^2}{4} γ ≤ 4 σ k 2 . We have ∥ ( K ( t ) ) − 1 ∥ 2 ≤ 3 2 1 σ k \left\|\left(K^{(t)}\right)^{-1}\right\|_2 \le\frac{3}{2}\frac{1}{\sigma_k} ( K ( t ) ) − 1 2 ≤ 2 3 σ k 1 .
Proof. We have K ( t ) = Σ k 2 C ( t ) B ( t ) = Σ k P ( t ) K^{(t)} = \Sigma_k^2C^{(t)}B^{(t)} = \Sigma_kP^{(t)} K ( t ) = Σ k 2 C ( t ) B ( t ) = Σ k P ( t ) . By Lemma 17 , we have σ min ( P ( t ) ) ≥ 2 3 \sigma_{\min}\left(P^{(t)}\right) \ge\frac{2}{3} σ m i n ( P ( t ) ) ≥ 3 2 . Thus P ( t ) P^{(t)} P ( t ) is invertible and
∥ ( K ( t ) ) − 1 ∥ 2 = ∥ ( Σ k P ( t ) ) − 1 ∥ 2 = ∥ ( P ( t ) ) − 1 Σ k − 1 ∥ 2 ≤ ∥ ( P ( t ) ) − 1 ∥ 2 ∥ Σ k − 1 ∥ 2 \left\|\left(K^{(t)}\right)^{-1}\right\|_2 = \left\|\left(\Sigma_kP^{(t)}\right)^{-1}\right\|_2 = \left\|\left(P^{(t)}\right)^{-1}\Sigma_k^{-1}\right\|_2 \le\left\|\left(P^{(t)}\right)^{-1}\right\|_2\left\|\Sigma_k^{-1}\right\|_2 ( K ( t ) ) − 1 2 = ( Σ k P ( t ) ) − 1 2 = ( P ( t ) ) − 1 Σ k − 1 2 ≤ ( P ( t ) ) − 1 2 Σ k − 1 2 = 1 σ min ( P ( t ) ) 1 σ k ≤ 3 2 1 σ k . = \frac{1}{\sigma_{\min}\left(P^{(t)}\right)}\frac{1}{\sigma_k} \le\frac{3}{2}\frac{1}{\sigma_k}. = σ m i n ( P ( t ) ) 1 σ k 1 ≤ 2 3 σ k 1 .
Main recurrence. In the following, we say that clipping is inactive in iteration t t t if Clip R ( a i T X ( t ) B ( t ) ) = a i T X ( t ) B ( t ) \operatorname{Clip}_R\left(a_i^{\mathsf{T}}X^{(t)}B^{(t)}\right) = a_i^{\mathsf{T}}X^{(t)}B^{(t)} Clip R ( a i T X ( t ) B ( t ) ) = a i T X ( t ) B ( t ) for all i ∈ [ n ] i \in[n] i ∈ [ n ] , i.e., we have ∥ a i T X ( t ) B ( t ) ∥ 2 ≤ R \left\|a_i^{\mathsf{T}}X^{(t)}B^{(t)}\right\|_2 \le R a i T X ( t ) B ( t ) 2 ≤ R for all i ∈ [ n ] i \in[n] i ∈ [ n ] . Note that, in this case, the main update in Algorithm 2 is Z ( t + 1 ) = A T A X ( t ) B ( t ) + Δ R G ^ ( t ) Z^{(t+1)} = A^{\mathsf{T}}AX^{(t)}B^{(t)} + \Delta R\widehat{G}^{(t)} Z ( t + 1 ) = A T A X ( t ) B ( t ) + Δ R G ( t ) . First, we will show a recurrence for E ( t + 1 ) E^{(t+1)} E ( t + 1 ) assuming that clipping is inactive in iteration t t t (Lemma 19 ). This recurrence expresses E ( t + 1 ) E^{(t+1)} E ( t + 1 ) in terms of the ideal progress term Σ > k 2 E ( t ) Σ k − 2 \Sigma^2_{>k}E^{(t)}\Sigma_k^{-2} Σ > k 2 E ( t ) Σ k − 2 (note that ∥ Σ > k 2 E ( t ) Σ k − 2 ∥ 2 ≤ σ k + 1 2 σ k 2 ∥ E ( t ) ∥ 2 = ( 1 − κ k ) ∥ E ( t ) ∥ 2 \left\|\Sigma^2_{>k}E^{(t)}\Sigma_k^{-2}\right\|_2 \le\frac{\sigma_{k+1}^2}{\sigma_k^2}\left\|E^{(t)}\right\|_2 = (1-\kappa_k)\left\|E^{(t)}\right\|_2 Σ > k 2 E ( t ) Σ k − 2 2 ≤ σ k 2 σ k + 1 2 E ( t ) 2 = ( 1 − κ k ) E ( t ) 2 ), and additional error terms that we will bound in Theorem 31 .
Lemma 19. Suppose C ( t ) C^{(t)} C ( t ) is invertible and clipping is inactive in iteration t t t . If ∥ F ( t ) ∥ 2 < 1 \lVert F^{(t)}\rVert_2 < 1 ∥ F ( t ) ∥ 2 < 1 , then C ( t + 1 ) C^{(t+1)} C ( t + 1 ) is invertible and
E ( t + 1 ) = ( Σ > k 2 E ( t ) Σ k − 2 + L ( t ) ) ( I + F ( t ) ) − 1 . E^{(t+1)}=\left(\Sigma_{>k}^{2}E^{(t)}\Sigma_k^{-2}+L^{(t)}\right)\left(I+F^{(t)}\right)^{-1}. E ( t + 1 ) = ( Σ > k 2 E ( t ) Σ k − 2 + L ( t ) ) ( I + F ( t ) ) − 1 . Proof. Let us start by observing that
C ( t + 1 ) R ( t + 1 ) = V k T X ( t + 1 ) R ( t + 1 ) = V k T Z ( t + 1 ) = V k T ( A T A X ( t ) B ( t ) + Δ R G ~ ( t ) ) = V k T A T A X ( t ) B ( t ) + Δ R V k T G ~ ( t ) = Σ k 2 V k T X ( t ) B ( t ) + Δ R V k T G ~ ( t ) = Σ k 2 C ( t ) B ( t ) + Δ R V k T G ~ ( t ) = K ( t ) + Δ R V k T G ~ ( t ) . \begin{aligned}
C^{(t+1)}R^{(t+1)} &= V_k^{\mathsf T}X^{(t+1)}R^{(t+1)}\\
&=V_k^{\mathsf T}Z^{(t+1)}\\
&=V_k^{\mathsf T}\left(A^{\mathsf T}AX^{(t)}B^{(t)}+\Delta R\widetilde{G}^{(t)}\right)\\
&=V_k^{\mathsf T}A^{\mathsf T}AX^{(t)}B^{(t)}+\Delta RV_k^{\mathsf T}\widetilde{G}^{(t)}\\
&=\Sigma_k^2V_k^{\mathsf T}X^{(t)}B^{(t)}+\Delta RV_k^{\mathsf T}\widetilde{G}^{(t)}\\
&=\Sigma_k^2C^{(t)}B^{(t)}+\Delta RV_k^{\mathsf T}\widetilde{G}^{(t)}\\
&=K^{(t)}+\Delta RV_k^{\mathsf T}\widetilde{G}^{(t)}.
\end{aligned} C ( t + 1 ) R ( t + 1 ) = V k T X ( t + 1 ) R ( t + 1 ) = V k T Z ( t + 1 ) = V k T ( A T A X ( t ) B ( t ) + Δ R G ( t ) ) = V k T A T A X ( t ) B ( t ) + Δ R V k T G ( t ) = Σ k 2 V k T X ( t ) B ( t ) + Δ R V k T G ( t ) = Σ k 2 C ( t ) B ( t ) + Δ R V k T G ( t ) = K ( t ) + Δ R V k T G ( t ) . Since C ( t ) C^{(t)} C ( t ) , Σ k 2 \Sigma_k^2 Σ k 2 , and B ( t ) B^{(t)} B ( t ) are invertible, K ( t ) = Σ k 2 C ( t ) B ( t ) K^{(t)}=\Sigma_k^2C^{(t)}B^{(t)} K ( t ) = Σ k 2 C ( t ) B ( t ) is invertible. Then F ( t ) : = Δ R V k T G ~ ( t ) ( K ( t ) ) − 1 F^{(t)}:=\Delta RV_k^{\mathsf T}\widetilde{G}^{(t)}\left(K^{(t)}\right)^{-1} F ( t ) := Δ R V k T G ( t ) ( K ( t ) ) − 1 is well-defined.
Multiplying the above identity by ( K ( t ) ) − 1 \left(K^{(t)}\right)^{-1} ( K ( t ) ) − 1 then gives
C ( t + 1 ) R ( t + 1 ) ( K ( t ) ) − 1 = I + ( Δ R V k T G ~ ( t ) ) ( K ( t ) ) − 1 = I + F ( t ) . C^{(t+1)}R^{(t+1)}\left(K^{(t)}\right)^{-1}=I+\left(\Delta RV_k^{\mathsf T}\widetilde{G}^{(t)}\right)\left(K^{(t)}\right)^{-1}=I+F^{(t)}. C ( t + 1 ) R ( t + 1 ) ( K ( t ) ) − 1 = I + ( Δ R V k T G ( t ) ) ( K ( t ) ) − 1 = I + F ( t ) . Since ∥ F ( t ) ∥ 2 < 1 \lVert F^{(t)}\rVert_2<1 ∥ F ( t ) ∥ 2 < 1 , Lemma 10 implies that I + F ( t ) I+F^{(t)} I + F ( t ) is invertible. Thus C ( t + 1 ) R ( t + 1 ) C^{(t+1)}R^{(t+1)} C ( t + 1 ) R ( t + 1 ) is invertible and therefore C ( t + 1 ) C^{(t+1)} C ( t + 1 ) and R ( t + 1 ) R^{(t+1)} R ( t + 1 ) are invertible.
This identity also gives
C ( t + 1 ) = ( I + F ( t ) ) K ( t ) ( R ( t + 1 ) ) − 1 . C^{(t+1)}=\left(I+F^{(t)}\right)K^{(t)}\left(R^{(t+1)}\right)^{-1}. C ( t + 1 ) = ( I + F ( t ) ) K ( t ) ( R ( t + 1 ) ) − 1 . Now we establish the recurrence. Using the above identity, we obtain
E ( t + 1 ) = D ( t + 1 ) ( C ( t + 1 ) ) − 1 = V > k T X ( t + 1 ) ( C ( t + 1 ) ) − 1 = V > k T Z ( t + 1 ) ( R ( t + 1 ) ) − 1 ( C ( t + 1 ) ) − 1 = V > k T Z ( t + 1 ) ( C ( t + 1 ) R ( t + 1 ) ) − 1 = V > k T Z ( t + 1 ) ( K ( t ) ) − 1 ( I + F ( t ) ) − 1 . \begin{aligned}
E^{(t+1)}&=D^{(t+1)}\left(C^{(t+1)}\right)^{-1}\\
&=V_{>k}^{\mathsf T}X^{(t+1)}\left(C^{(t+1)}\right)^{-1}\\
&=V_{>k}^{\mathsf T}Z^{(t+1)}\left(R^{(t+1)}\right)^{-1}\left(C^{(t+1)}\right)^{-1}\\
&=V_{>k}^{\mathsf T}Z^{(t+1)}\left(C^{(t+1)}R^{(t+1)}\right)^{-1}\\
&=V_{>k}^{\mathsf T}Z^{(t+1)}\left(K^{(t)}\right)^{-1}\left(I+F^{(t)}\right)^{-1}.
\end{aligned} E ( t + 1 ) = D ( t + 1 ) ( C ( t + 1 ) ) − 1 = V > k T X ( t + 1 ) ( C ( t + 1 ) ) − 1 = V > k T Z ( t + 1 ) ( R ( t + 1 ) ) − 1 ( C ( t + 1 ) ) − 1 = V > k T Z ( t + 1 ) ( C ( t + 1 ) R ( t + 1 ) ) − 1 = V > k T Z ( t + 1 ) ( K ( t ) ) − 1 ( I + F ( t ) ) − 1 . Note that
V > k T Z ( t + 1 ) = V > k T ( A T A X ( t ) B ( t ) + Δ R G ~ ( t ) ) = Σ > k 2 V > k T X ( t ) B ( t ) + Δ R V > k T G ~ ( t ) = Σ > k 2 D ( t ) B ( t ) + Δ R V > k T G ~ ( t ) = Σ > k 2 E ( t ) C ( t ) B ( t ) + Δ R V > k T G ~ ( t ) = Σ > k 2 E ( t ) Σ k − 2 K ( t ) + Δ R V > k T G ~ ( t ) = ( Σ > k 2 E ( t ) Σ k − 2 + Δ R V > k T G ~ ( t ) ( K ( t ) ) − 1 ) K ( t ) . \begin{aligned}
V_{>k}^{\mathsf T}Z^{(t+1)}
&=V_{>k}^{\mathsf T}\left(A^{\mathsf T}AX^{(t)}B^{(t)}+\Delta R\widetilde{G}^{(t)}\right)\\
&=\Sigma_{>k}^2V_{>k}^{\mathsf T}X^{(t)}B^{(t)}+\Delta RV_{>k}^{\mathsf T}\widetilde{G}^{(t)}\\
&=\Sigma_{>k}^2D^{(t)}B^{(t)}+\Delta RV_{>k}^{\mathsf T}\widetilde{G}^{(t)}\\
&=\Sigma_{>k}^2E^{(t)}C^{(t)}B^{(t)}+\Delta RV_{>k}^{\mathsf T}\widetilde{G}^{(t)}\\
&=\Sigma_{>k}^2E^{(t)}\Sigma_k^{-2}K^{(t)}+\Delta RV_{>k}^{\mathsf T}\widetilde{G}^{(t)}\\
&=\left(\Sigma_{>k}^2E^{(t)}\Sigma_k^{-2}+\Delta RV_{>k}^{\mathsf T}\widetilde{G}^{(t)}\left(K^{(t)}\right)^{-1}\right)K^{(t)}.
\end{aligned} V > k T Z ( t + 1 ) = V > k T ( A T A X ( t ) B ( t ) + Δ R G ( t ) ) = Σ > k 2 V > k T X ( t ) B ( t ) + Δ R V > k T G ( t ) = Σ > k 2 D ( t ) B ( t ) + Δ R V > k T G ( t ) = Σ > k 2 E ( t ) C ( t ) B ( t ) + Δ R V > k T G ( t ) = Σ > k 2 E ( t ) Σ k − 2 K ( t ) + Δ R V > k T G ( t ) = ( Σ > k 2 E ( t ) Σ k − 2 + Δ R V > k T G ( t ) ( K ( t ) ) − 1 ) K ( t ) .
= ( Σ > k 2 E ( t ) Σ k − 2 + L ( t ) ) K ( t ) . = \left(\Sigma_{>k}^{2}E^{(t)}\Sigma_k^{-2}+L^{(t)}\right)K^{(t)}. = ( Σ > k 2 E ( t ) Σ k − 2 + L ( t ) ) K ( t ) .
Plugging this into the previous identity for E ( t + 1 ) E^{(t+1)} E ( t + 1 ) , we obtain the claimed recurrence.
Showing that clipping is inactive. Using Lemma 19, we will show that, for an appropriate choice of R R R , clipping is inactive in every iteration with high probability. Recall that the clipping is inactive in iteration t t t if ∥ a i T X ( t ) B ( t ) ∥ 2 ≤ R \lVert a_i^{\mathsf T}X^{(t)}B^{(t)}\rVert_2\le R ∥ a i T X ( t ) B ( t ) ∥ 2 ≤ R for all i ∈ [ n ] i\in[n] i ∈ [ n ] . In the following, we analyze a i T X ( t ) B ( t ) a_i^{\mathsf T}X^{(t)}B^{(t)} a i T X ( t ) B ( t ) for each i ∈ [ n ] i\in[n] i ∈ [ n ] . Recall that S ( t ) : = Σ > k E ( t ) Σ k − 1 S^{(t)}:=\Sigma_{>k}E^{(t)}\Sigma_k^{-1} S ( t ) := Σ > k E ( t ) Σ k − 1 .
Lemma 20. Suppose that \lVert\Delta^{\sym}(G^{(t)})\rVert_2\le\frac{\gamma}{4} , s t ≤ 1 2 s_t\le\frac{1}{2} s t ≤ 2 1 , and γ ≤ σ k 2 4 \gamma\le\frac{\sigma_k^2}{4} γ ≤ 4 σ k 2 . For each 1 ≤ i ≤ n 1\le i\le n 1 ≤ i ≤ n , we have
∥ a i T X ( t ) B ( t ) ∥ 2 ≤ 2 ( k ∥ U ∥ ∞ + ∥ e i T U > k S ( t ) ∥ 2 ) . \left\lVert a_i^{\mathsf T}X^{(t)}B^{(t)}\right\rVert_2
\le\sqrt{2}\left(\sqrt{k}\lVert U\rVert_\infty+\left\lVert e_i^{\mathsf T}U_{>k}S^{(t)}\right\rVert_2\right). a i T X ( t ) B ( t ) 2 ≤ 2 ( k ∥ U ∥ ∞ + e i T U > k S ( t ) 2 ) .
Proof. We have
A X ( t ) B ( t ) = ( U k Σ k V k T + U > k Σ > k V > k T ) X ( t ) B ( t ) = U k Σ k V k T X ( t ) B ( t ) + U > k Σ > k V > k T X ( t ) B ( t ) = ( U k + U > k S ( t ) ) P ( t ) . \begin{aligned}
AX^{(t)}B^{(t)}
=\left(U_k\Sigma_kV_k^{\mathsf T}+U_{>k}\Sigma_{>k}V_{>k}^{\mathsf T}\right)X^{(t)}B^{(t)} \\
&=U_k\Sigma_kV_k^{\mathsf T}X^{(t)}B^{(t)}
+U_{>k}\Sigma_{>k}V_{>k}^{\mathsf T}X^{(t)}B^{(t)} \\
&=\left(U_k+U_{>k}S^{(t)}\right)P^{(t)}.
\end{aligned} A X ( t ) B ( t ) = ( U k Σ k V k T + U > k Σ > k V > k T ) X ( t ) B ( t ) = U k Σ k V k T X ( t ) B ( t ) + U > k Σ > k V > k T X ( t ) B ( t ) = ( U k + U > k S ( t ) ) P ( t ) .
Thus, for each row i ∈ [ n ] i\in[n] i ∈ [ n ] , we have
a i T X ( t ) B ( t ) = e i T A X ( t ) B ( t ) = e i T ( U k + U > k S ( t ) ) P ( t ) a_i^{\mathsf T}X^{(t)}B^{(t)}
=e_i^{\mathsf T}AX^{(t)}B^{(t)}
=e_i^{\mathsf T}\left(U_k+U_{>k}S^{(t)}\right)P^{(t)} a i T X ( t ) B ( t ) = e i T A X ( t ) B ( t ) = e i T ( U k + U > k S ( t ) ) P ( t )
and thus
∥ a i T X ( t ) B ( t ) ∥ 2 ≤ ( ∥ e i T U k ∥ 2 + ∥ e i T U > k S ( t ) ∥ 2 ) ∥ P ( t ) ∥ 2 . \left\lVert a_i^{\mathsf T}X^{(t)}B^{(t)}\right\rVert_2
\le\left(\left\lVert e_i^{\mathsf T}U_k\right\rVert_2
+\left\lVert e_i^{\mathsf T}U_{>k}S^{(t)}\right\rVert_2\right)
\left\lVert P^{(t)}\right\rVert_2. a i T X ( t ) B ( t ) 2 ≤ ( e i T U k 2 + e i T U > k S ( t ) 2 ) P ( t ) 2 .
By Lemma 17, we have ∥ P ( t ) ∥ 2 ≤ 2 \left\lVert P^{(t)}\right\rVert_2\le\sqrt{2} P ( t ) 2 ≤ 2 . We also have ∥ e i T U k ∥ 2 ≤ k ∥ U ∥ ∞ \left\lVert e_i^{\mathsf T}U_k\right\rVert_2\le\sqrt{k}\lVert U\rVert_\infty e i T U k 2 ≤ k ∥ U ∥ ∞ .
Next, we upper bound max i ∈ [ n ] ∥ e i T U > k S ( t ) ∥ 2 \max_{i\in[n]}\left\lVert e_i^{\mathsf T}U_{>k}S^{(t)}\right\rVert_2 max i ∈ [ n ] e i T U > k S ( t ) 2 . To this end, we introduce the following potential. For every 0 ≤ r , t ≤ T 0\le r,t\le T 0 ≤ r , t ≤ T , let
Φ r , t : = max i ∈ [ n ] ∥ e i T U > k Σ > k 2 r S ( t ) ∥ 2 . \Phi_{r,t}:=\max_{i\in[n]}\left\lVert e_i^{\mathsf T}U_{>k}\Sigma_{>k}^{2r}S^{(t)}\right\rVert_2. Φ r , t := i ∈ [ n ] max e i T U > k Σ > k 2 r S ( t ) 2 .
Our goal is to upper bound Φ 0 , t = max i ∈ [ n ] ∥ e i T U > k S ( t ) ∥ 2 \Phi_{0,t}=\max_{i\in[n]}\left\lVert e_i^{\mathsf T}U_{>k}S^{(t)}\right\rVert_2 Φ 0 , t = max i ∈ [ n ] e i T U > k S ( t ) 2 , and we will use Φ r , ⋅ \Phi_{r,\cdot} Φ r , ⋅ for larger r r r to do so. First, we show that the initialization allows us to upper bound Φ r , 0 \Phi_{r,0} Φ r , 0 for all 0 ≤ r ≤ T 0\le r\le T 0 ≤ r ≤ T as follows.
Lemma 21. Suppose that Δ k ≥ 8 Δ i n i t ( 2 d + 2 log 8 β ) \Delta_k\ge8\Delta_{\mathrm{init}}\left(2\sqrt{d}+\sqrt{2\log\frac{8}{\beta}}\right) Δ k ≥ 8 Δ init ( 2 d + 2 log β 8 ) . With probability at least 1 − β 4 1-\frac{\beta}{4} 1 − 4 β , simultaneously for all 0 ≤ r ≤ T 0\le r\le T 0 ≤ r ≤ T , we have
Φ r , 0 ≤ σ k + 1 2 r ω i n i t , ω i n i t : = 3 ∥ U ∥ ∞ ( k + log ( 16 n ( T + 1 ) β ) ) . (4) \Phi_{r,0}\le\sigma_{k+1}^{2r}\omega_{\mathrm{init}},
\qquad
\omega_{\mathrm{init}}:=3\lVert U\rVert_\infty\left(\sqrt{k}+\sqrt{\log\left(\frac{16n(T+1)}{\beta}\right)}\right). \tag*{(4)} Φ r , 0 ≤ σ k + 1 2 r ω init , ω init := 3 ∥ U ∥ ∞ ( k + log ( β 16 n ( T + 1 ) ) ) . (4)
Let G i n i t \mathcal{G}_{\mathrm{init}} G init be this event.
Proof. Recall that we have Φ r , 0 = max i ∈ [ n ] ∥ e i T U > k Σ > k 2 r + 1 E ( 0 ) Σ k − 1 ∥ 2 \Phi_{r,0}=\max_{i\in[n]}\left\lVert e_i^{\mathsf T}U_{>k}\Sigma_{>k}^{2r+1}E^{(0)}\Sigma_k^{-1}\right\rVert_2 Φ r , 0 = max i ∈ [ n ] e i T U > k Σ > k 2 r + 1 E ( 0 ) Σ k − 1 2 . To analyze the norm, it is useful to view E ( 0 ) E^{(0)} E ( 0 ) as a function of the Gaussian noise matrix added by Algorithm 1. More precisely, for a Gaussian matrix G ∼ N ( 0 , 1 ) d × d G\sim\mathcal{N}(0,1)^{d\times d} G ∼ N ( 0 , 1 ) d × d , let X(\Delta_{\mathrm{init}}\sym(G)) be the top-k k k eigenvectors of the matrix A^{\mathsf T}A+\Delta_{\mathrm{init}}\sym(G) , and let
C(\Delta_{\mathrm{init}}\sym(G)):=V_k^{\mathsf T}X(\Delta_{\mathrm{init}}\sym(G))
D ( Δ i n i t s y m ( G ) ) : = V > k T X ( Δ i n i t s y m ( G ) ) D(\Delta_{\mathrm{init\ sym}}(G)) := V_{>k}^{\mathsf T}X(\Delta_{\mathrm{init\ sym}}(G)) D ( Δ init sym ( G )) := V > k T X ( Δ init sym ( G ))
E ( Δ i n i t s y m ( G ) ) : = D ( Δ i n i t s y m ( G ) ) ( C ( Δ i n i t s y m ( G ) ) ) − 1 E(\Delta_{\mathrm{init\ sym}}(G)) := D(\Delta_{\mathrm{init\ sym}}(G))\left(C(\Delta_{\mathrm{init\ sym}}(G))\right)^{-1} E ( Δ init sym ( G )) := D ( Δ init sym ( G )) ( C ( Δ init sym ( G )) ) − 1
where E ( Δ i n i t s y m ( G ) ) E(\Delta_{\mathrm{init\ sym}}(G)) E ( Δ init sym ( G )) is defined whenever C ( Δ i n i t s y m ( G ) ) C(\Delta_{\mathrm{init\ sym}}(G)) C ( Δ init sym ( G )) is invertible. If the function M → E ( M ) M \to E(M) M → E ( M ) was Lipschitz under the Frobenius norm, we could use Theorem 4 to analyze ∥ e i T U > k Σ > k 2 r + 1 E ( M ) Σ k − 1 ∥ 2 \left\lVert e_i^{\mathsf T}U_{>k}\Sigma^{2r+1}_{>k}E(M)\Sigma_{k}^{-1}\right\rVert_2 e i T U > k Σ > k 2 r + 1 E ( M ) Σ k − 1 2 . Unfortunately, the function is not globally Lipschitz, but it is Lipschitz if we restrict it to the following ball:
B : = { M : M = M T , ∥ M ∥ 2 ≤ Δ k 8 } . \mathcal{B} := \left\{M : M=M^{\mathsf T},\ \lVert M\rVert_2 \le\frac{\Delta_k}{8}\right\}. B := { M : M = M T , ∥ M ∥ 2 ≤ 8 Δ k } .
We show in the following claim that the function M → E ( M ) M \to E(M) M → E ( M ) is Lipschitz on B \mathcal{B} B . It is convenient to construct a Lipschitz extension of this function that is defined everywhere as follows. Let Π B \Pi_{\mathcal{B}} Π B be the Frobenius norm projection onto B \mathcal{B} B , i.e., for every matrix M M M , we let Π B ( M ) = arg min M ′ ∈ B ∥ M − M ′ ∥ F \Pi_{\mathcal{B}}(M)=\mathop{\operatorname{arg\,min}}_{M'\in\mathcal{B}}\lVert M-M'\rVert_F Π B ( M ) = arg min M ′ ∈ B ∥ M − M ′ ∥ F . Let
E ‾ ( G ) = E ( Π B ( Δ i n i t s y m ( G ) ) ) . \overline{E}(G)=E\left(\Pi_{\mathcal{B}}\left(\Delta_{\mathrm{init\ sym}}(G)\right)\right). E ( G ) = E ( Π B ( Δ init sym ( G ) ) ) .
Next, we show that M → E ( M ) M \to E(M) M → E ( M ) is Lipschitz on B \mathcal{B} B under the Frobenius norm, which will then imply that G → E ‾ ( G ) G \to\overline{E}(G) G → E ( G ) is globally Lipschitz under the Frobenius norm.
Claim 22. For any two matrices M , M ′ ∈ B M,M'\in\mathcal{B} M , M ′ ∈ B for which E ( M ) E(M) E ( M ) and E ( M ′ ) E(M') E ( M ′ ) are well-defined (i.e., C ( M ) C(M) C ( M ) and C ( M ′ ) C(M') C ( M ′ ) are invertible), we have
∥ E ( M ) − E ( M ′ ) ∥ F ≤ 6 Δ k ∥ M − M ′ ∥ F . \lVert E(M)-E(M')\rVert_F\le\frac{6}{\Delta_k}\lVert M-M'\rVert_F. ∥ E ( M ) − E ( M ′ ) ∥ F ≤ Δ k 6 ∥ M − M ′ ∥ F .
Proof. We start by writing E ( M ) E(M) E ( M ) and E ( M ′ ) E(M') E ( M ′ ) in a more convenient form. For W ∈ { M , M ′ } W\in\{M,M'\} W ∈ { M , M ′ } , let P ( W ) : = X ( W ) X ( W ) T P(W):=X(W)X(W)^{\mathsf T} P ( W ) := X ( W ) X ( W ) T . We can write
V T P ( W ) V = ( P 11 ( W ) P 12 ( W ) P 21 ( W ) P 22 ( W ) ) , V^{\mathsf T}P(W)V=
\begin{pmatrix}
P_{11}(W) & P_{12}(W)\\
P_{21}(W) & P_{22}(W)
\end{pmatrix}, V T P ( W ) V = ( P 11 ( W ) P 21 ( W ) P 12 ( W ) P 22 ( W ) ) ,
where
P 11 ( W ) = V k T P ( W ) V k = C ( W ) C ( W ) T ∈ R k × k P_{11}(W)=V_k^{\mathsf T}P(W)V_k=C(W)C(W)^{\mathsf T}\in\mathbb{R}^{k\times k} P 11 ( W ) = V k T P ( W ) V k = C ( W ) C ( W ) T ∈ R k × k
P 12 ( W ) = V k T P ( W ) V > k = C ( W ) D ( W ) T ∈ R k × ( d − k ) P_{12}(W)=V_k^{\mathsf T}P(W)V_{>k}=C(W)D(W)^{\mathsf T}\in\mathbb{R}^{k\times(d-k)} P 12 ( W ) = V k T P ( W ) V > k = C ( W ) D ( W ) T ∈ R k × ( d − k )
P 21 ( W ) = V > k T P ( W ) V k = D ( W ) C ( W ) T ∈ R ( d − k ) × k P_{21}(W)=V_{>k}^{\mathsf T}P(W)V_k=D(W)C(W)^{\mathsf T}\in\mathbb{R}^{(d-k)\times k} P 21 ( W ) = V > k T P ( W ) V k = D ( W ) C ( W ) T ∈ R ( d − k ) × k
P 22 ( W ) = V > k T P ( W ) V > k = D ( W ) D ( W ) T ∈ R ( d − k ) × ( d − k ) . P_{22}(W)=V_{>k}^{\mathsf T}P(W)V_{>k}=D(W)D(W)^{\mathsf T}\in\mathbb{R}^{(d-k)\times(d-k)}. P 22 ( W ) = V > k T P ( W ) V > k = D ( W ) D ( W ) T ∈ R ( d − k ) × ( d − k ) .
We have
E ( W ) = D ( W ) C ( W ) − 1 = P 21 ( W ) P 11 ( W ) − 1 . E(W)=D(W)C(W)^{-1}=P_{21}(W)P_{11}(W)^{-1}. E ( W ) = D ( W ) C ( W ) − 1 = P 21 ( W ) P 11 ( W ) − 1 .
Using the above identity, we obtain
∥ E ( M ) − E ( M ′ ) ∥ F = ∥ P 21 ( M ) P 11 ( M ) − 1 − P 21 ( M ′ ) P 11 ( M ′ ) − 1 ∥ F = ∥ ( P 21 ( M ) − P 21 ( M ′ ) ) P 11 ( M ) − 1 + P 21 ( M ′ ) ( P 11 ( M ) − 1 − P 11 ( M ′ ) − 1 ) ∥ F \begin{aligned}
\lVert E(M)-E(M')\rVert_F \\
&=\left\lVert P_{21}(M)P_{11}(M)^{-1}-P_{21}(M')P_{11}(M')^{-1}\right\rVert_F \\
&=\left\lVert\left(P_{21}(M)-P_{21}(M')\right)P_{11}(M)^{-1}+P_{21}(M')\left(P_{11}(M)^{-1}-P_{11}(M')^{-1}\right)\right\rVert_F
\end{aligned} ∥ E ( M ) − E ( M ′ ) ∥ F = P 21 ( M ) P 11 ( M ) − 1 − P 21 ( M ′ ) P 11 ( M ′ ) − 1 F = ( P 21 ( M ) − P 21 ( M ′ ) ) P 11 ( M ) − 1 + P 21 ( M ′ ) ( P 11 ( M ) − 1 − P 11 ( M ′ ) − 1 ) F
= ( a ) ∥ ( P 21 ( M ) − P 21 ( M ′ ) ) P 11 ( M ) − 1 + P 21 ( M ′ ) P 11 ( M ) − 1 ( P 11 ( M ′ ) − P 11 ( M ) ) P 11 ( M ′ ) − 1 ∥ F \overset{(a)}{=}\left\lVert\left(P_{21}(M)-P_{21}(M')\right)P_{11}(M)^{-1}+P_{21}(M')P_{11}(M)^{-1}\left(P_{11}(M')-P_{11}(M)\right)P_{11}(M')^{-1}\right\rVert_F = ( a ) ( P 21 ( M ) − P 21 ( M ′ ) ) P 11 ( M ) − 1 + P 21 ( M ′ ) P 11 ( M ) − 1 ( P 11 ( M ′ ) − P 11 ( M ) ) P 11 ( M ′ ) − 1 F
≤ ( b ) ∥ P 21 ( M ) − P 21 ( M ′ ) ∥ F ∥ P 11 ( M ) − 1 ∥ 2 + ∥ P 21 ( M ′ ) ∥ 2 ∥ P 11 ( M ) − 1 ∥ 2 ∥ P 11 ( M ′ ) − 1 ∥ 2 ∥ P ( M ) − P ( M ′ ) ∥ F \overset{(b)}{\le}\lVert P_{21}(M)-P_{21}(M')\rVert_F\lVert P_{11}(M)^{-1}\rVert_2
+\lVert P_{21}(M')\rVert_2\lVert P_{11}(M)^{-1}\rVert_2\lVert P_{11}(M')^{-1}\rVert_2\lVert P(M)-P(M')\rVert_F ≤ ( b ) ∥ P 21 ( M ) − P 21 ( M ′ ) ∥ F ∥ P 11 ( M ) − 1 ∥ 2 + ∥ P 21 ( M ′ ) ∥ 2 ∥ P 11 ( M ) − 1 ∥ 2 ∥ P 11 ( M ′ ) − 1 ∥ 2 ∥ P ( M ) − P ( M ′ ) ∥ F
≤ ( c ) ( 1 + ∥ P 21 ( M ′ ) ∥ 2 ∥ P 11 ( M ′ ) − 1 ∥ 2 ) ∥ P 11 ( M ) − 1 ∥ 2 ∥ P ( M ) − P ( M ′ ) ∥ F \overset{(c)}{\le}\left(1+\lVert P_{21}(M')\rVert_2\lVert P_{11}(M')^{-1}\rVert_2\right)\lVert P_{11}(M)^{-1}\rVert_2\lVert P(M)-P(M')\rVert_F ≤ ( c ) ( 1 + ∥ P 21 ( M ′ ) ∥ 2 ∥ P 11 ( M ′ ) − 1 ∥ 2 ) ∥ P 11 ( M ) − 1 ∥ 2 ∥ P ( M ) − P ( M ′ ) ∥ F
In ( a ) (a) ( a ) , we used the identity A − 1 − B − 1 = A − 1 ( B − A ) B − 1 A^{-1}-B^{-1}=A^{-1}(B-A)B^{-1} A − 1 − B − 1 = A − 1 ( B − A ) B − 1 . In ( b ) (b) ( b ) , we used the inequalities ∥ A + B ∥ F ≤ ∥ A ∥ F + ∥ B ∥ F \lVert A+B\rVert_F\le\lVert A\rVert_F+\lVert B\rVert_F ∥ A + B ∥ F ≤ ∥ A ∥ F + ∥ B ∥ F , ∥ A B ∥ F ≤ ∥ A ∥ F ∥ B ∥ 2 \lVert AB\rVert_F\le\lVert A\rVert_F\lVert B\rVert_2 ∥ A B ∥ F ≤ ∥ A ∥ F ∥ B ∥ 2 , and ∥ A B ∥ 2 ≤ ∥ A ∥ 2 ∥ B ∥ 2 \lVert AB\rVert_2\le\lVert A\rVert_2\lVert B\rVert_2 ∥ A B ∥ 2 ≤ ∥ A ∥ 2 ∥ B ∥ 2 . In ( c ) (c) ( c ) , we used that ∥ P 21 ( M ) − P 21 ( M ′ ) ∥ F ≤ ∥ P ( M ) − P ( M ′ ) ∥ F \lVert P_{21}(M)-P_{21}(M')\rVert_F\le\lVert P(M)-P(M')\rVert_F ∥ P 21 ( M ) − P 21 ( M ′ ) ∥ F ≤ ∥ P ( M ) − P ( M ′ ) ∥ F , which we can show as follows:
∥ P 21 ( M ) − P 21 ( M ′ ) ∥ F = ∥ V > k T ( P ( M ) − P ( M ′ ) ) V k ∥ F ≤ ∥ V > k T ∥ 2 ∥ P ( M ) − P ( M ′ ) ∥ F ∥ V k ∥ 2 = ∥ P ( M ) − P ( M ′ ) ∥ F . \begin{aligned}
\lVert P_{21}(M)-P_{21}(M')\rVert_F
&=\lVert V_{>k}^{\mathsf T}(P(M)-P(M'))V_k\rVert_F\\
&\le\lVert V_{>k}^{\mathsf T}\rVert_2\lVert P(M)-P(M')\rVert_F\lVert V_k\rVert_2\\
&=\lVert P(M)-P(M')\rVert_F.
\end{aligned} ∥ P 21 ( M ) − P 21 ( M ′ ) ∥ F = ∥ V > k T ( P ( M ) − P ( M ′ )) V k ∥ F ≤ ∥ V > k T ∥ 2 ∥ P ( M ) − P ( M ′ ) ∥ F ∥ V k ∥ 2 = ∥ P ( M ) − P ( M ′ ) ∥ F .
where the last equality follows from the fact that V k V_k V k and V > k V_{>k} V > k have orthonormal columns.
Next, we bound the terms in the above inequality. Consider ∥ P 21 ( M ′ ) ∥ 2 \lVert P_{21}(M')\rVert_2 ∥ P 21 ( M ′ ) ∥ 2 . We have
∥ P 21 ( M ′ ) ∥ 2 = ∥ V > k T X ( M ′ ) X ( M ′ ) T V k ∥ 2 ≤ ∥ V > k T ∥ 2 ∥ X ( M ′ ) ∥ 2 2 ∥ V k ∥ 2 ≤ 1. \begin{aligned}
\lVert P_{21}(M')\rVert_2
&=\lVert V_{>k}^{\mathsf T}X(M')X(M')^{\mathsf T}V_k\rVert_2\\
&\le\lVert V_{>k}^{\mathsf T}\rVert_2\lVert X(M')\rVert_2^2\lVert V_k\rVert_2\le1.
\end{aligned} ∥ P 21 ( M ′ ) ∥ 2 = ∥ V > k T X ( M ′ ) X ( M ′ ) T V k ∥ 2 ≤ ∥ V > k T ∥ 2 ∥ X ( M ′ ) ∥ 2 2 ∥ V k ∥ 2 ≤ 1.
where the last inequality follows from the fact that V k V_k V k , V > k V_{>k} V > k , and X ( M ′ ) X(M') X ( M ′ ) have orthonormal columns.
Consider ∥ P 11 ( M ) − 1 ∥ 2 \lVert P_{11}(M)^{-1}\rVert_2 ∥ P 11 ( M ) − 1 ∥ 2 and ∥ P 11 ( M ′ ) − 1 ∥ 2 \lVert P_{11}(M')^{-1}\rVert_2 ∥ P 11 ( M ′ ) − 1 ∥ 2 . For W ∈ { M , M ′ } W\in\{M,M'\} W ∈ { M , M ′ } , we have
∥ I k − P 11 ( W ) ∥ 2 = ∥ V k T ( I − P ( W ) ) V k ∥ 2 = ∥ ( I − P ( W ) ) V k ∥ 2 2 . \lVert I_k-P_{11}(W)\rVert_2
=\lVert V_k^{\mathsf T}(I-P(W))V_k\rVert_2
=\lVert(I-P(W))V_k\rVert_2^2. ∥ I k − P 11 ( W ) ∥ 2 = ∥ V k T ( I − P ( W )) V k ∥ 2 = ∥( I − P ( W )) V k ∥ 2 2 .
where we used that P 11 ( W ) = V k T P ( W ) V k P_{11}(W)=V_k^{\mathsf T}P(W)V_k P 11 ( W ) = V k T P ( W ) V k and Q : = I − P ( W ) Q:=I-P(W) Q := I − P ( W ) is an orthogonal projector and thus Q T Q = Q Q^{\mathsf T}Q=Q Q T Q = Q .
Lemma 9 together with ∥ W ∥ 2 ≤ Δ k 8 \lVert W\rVert_2\le\frac{\Delta_k}{8} ∥ W ∥ 2 ≤ 8 Δ k give
∥ ( I − P ( W ) ) V k ∥ 2 ≤ ∥ W ∥ 2 Δ k − ∥ W ∥ 2 ≤ 2 ∥ W ∥ 2 Δ k ≤ 1 4 . \lVert(I-P(W))V_k\rVert_2\le\frac{\lVert W\rVert_2}{\Delta_k-\lVert W\rVert_2}\le\frac{2\lVert W\rVert_2}{\Delta_k}\le\frac{1}{4}. ∥( I − P ( W )) V k ∥ 2 ≤ Δ k − ∥ W ∥ 2 ∥ W ∥ 2 ≤ Δ k 2 ∥ W ∥ 2 ≤ 4 1 .
Thus
∥ I k − P 11 ( W ) ∥ 2 ≤ 1 16 \lVert I_k-P_{11}(W)\rVert_2\le\frac{1}{16} ∥ I k − P 11 ( W ) ∥ 2 ≤ 16 1
and therefore P 11 ( W ) ⪰ 15 16 I k P_{11}(W)\succeq\frac{15}{16}I_k P 11 ( W ) ⪰ 16 15 I k and ∥ P 11 ( W ) − 1 ∥ 2 ≤ 16 15 \lVert P_{11}(W)^{-1}\rVert_2\le\frac{16}{15} ∥ P 11 ( W ) − 1 ∥ 2 ≤ 15 16 .
Finally, consider ∥ P ( M ) − P ( M ′ ) ∥ F \lVert P(M)-P(M')\rVert_F ∥ P ( M ) − P ( M ′ ) ∥ F . We have
∥ P ( M ) − P ( M ′ ) ∥ F = ∥ X ( M ) X ( M ) T − X ( M ′ ) X ( M ′ ) T ∥ F = 2 ∥ ( I − X ( M ′ ) X ( M ′ ) T ) X ( M ) ∥ F . \begin{aligned}
\lVert P(M)-P(M')\rVert_F
&=\lVert X(M)X(M)^{\mathsf T}-X(M')X(M')^{\mathsf T}\rVert_F\\
&=\sqrt{2}\left\lVert\left(I-X(M')X(M')^{\mathsf T}\right)X(M)\right\rVert_F.
\end{aligned} ∥ P ( M ) − P ( M ′ ) ∥ F = ∥ X ( M ) X ( M ) T − X ( M ′ ) X ( M ′ ) T ∥ F = 2 ( I − X ( M ′ ) X ( M ′ ) T ) X ( M ) F .
We can upper bound the last term using the Davis-Kahan theorem (Theorem 8 ) and Weyl's inequality (Theorem 7) as follows. Let
δ = max { λ k ( A T A + M ) − λ k + 1 ( A T A + M ′ ) , λ k ( A T A + M ′ ) − λ k + 1 ( A T A + M ) } . \delta=\max\left\{\lambda_k(A^{\mathsf T}A+M)-\lambda_{k+1}(A^{\mathsf T}A+M'),\lambda_k(A^{\mathsf T}A+M')-\lambda_{k+1}(A^{\mathsf T}A+M)\right\}. δ = max { λ k ( A T A + M ) − λ k + 1 ( A T A + M ′ ) , λ k ( A T A + M ′ ) − λ k + 1 ( A T A + M ) } .
We apply the Davis-Kahan theorem with matrices A T A + M A^{\mathsf T}A+M A T A + M and A T A + M ′ A^{\mathsf T}A+M' A T A + M ′ , and s = 1 s=1 s = 1 and r = k r=k r = k . We obtain
∥ ( I − X ( M ′ ) X ( M ′ ) T ) X ( M ) ∥ F ≤ ∥ M − M ′ ∥ F δ . \left\lVert\left(I-X(M')X(M')^{\mathsf T}\right)X(M)\right\rVert_F\le\frac{\lVert M-M'\rVert_F}{\delta}. ( I − X ( M ′ ) X ( M ′ ) T ) X ( M ) F ≤ δ ∥ M − M ′ ∥ F .
To lower bound δ \delta δ , we use Weyl's inequality and the fact that ∥ M ∥ 2 , ∥ M ′ ∥ 2 ≤ Δ k 8 \lVert M\rVert_2,\lVert M'\rVert_2\le\frac{\Delta_k}{8} ∥ M ∥ 2 , ∥ M ′ ∥ 2 ≤ 8 Δ k . By Weyl's inequality, for W ∈ { M , M ′ } W\in\{M,M'\} W ∈ { M , M ′ } and j ∈ { k , k + 1 } j\in\{k,k+1\} j ∈ { k , k + 1 } , we have
∣ λ j ( A T A + W ) − λ j ( A T A ) ∣ ≤ ∥ W ∥ 2 . \left|\lambda_j(A^{\mathsf T}A+W)-\lambda_j(A^{\mathsf T}A)\right|\le\lVert W\rVert_2. λ j ( A T A + W ) − λ j ( A T A ) ≤ ∥ W ∥ 2 .
Thus
λ k ( A ⊤ A + M ) − λ k + 1 ( A ⊤ A + M ′ ) ≥ ( λ k ( A ⊤ A ) − ∥ M ∥ 2 ) − ( λ k + 1 ( A ⊤ A ) + ∥ M ′ ∥ 2 ) = σ k 2 − σ k + 1 2 − ∥ M ∥ 2 − ∥ M ′ ∥ 2 = Δ k − ∥ M ∥ 2 − ∥ M ′ ∥ 2 ≥ 3 4 Δ k \begin{aligned}
\lambda_k(A^\top A+M)-\lambda_{k+1}(A^\top A+M') \ge\bigl(\lambda_k(A^\top A)-\lVert M\rVert_2\bigr)-\bigl(\lambda_{k+1}(A^\top A)+\lVert M'\rVert_2\bigr) \\
&= \sigma_k^2-\sigma_{k+1}^2-\lVert M\rVert_2-\lVert M'\rVert_2 \\
&= \Delta_k-\lVert M\rVert_2-\lVert M'\rVert_2 \\
&\ge\frac{3}{4}\Delta_k
\end{aligned} λ k ( A ⊤ A + M ) − λ k + 1 ( A ⊤ A + M ′ ) ≥ ( λ k ( A ⊤ A ) − ∥ M ∥ 2 ) − ( λ k + 1 ( A ⊤ A ) + ∥ M ′ ∥ 2 ) = σ k 2 − σ k + 1 2 − ∥ M ∥ 2 − ∥ M ′ ∥ 2 = Δ k − ∥ M ∥ 2 − ∥ M ′ ∥ 2 ≥ 4 3 Δ k
where in the last inequality we used that ∥ M ∥ 2 , ∥ M ′ ∥ 2 ≤ Δ k 8 \lVert M\rVert_2,\lVert M'\rVert_2\le\frac{\Delta_k}{8} ∥ M ∥ 2 , ∥ M ′ ∥ 2 ≤ 8 Δ k .
Similarly, we have
λ k ( A ⊤ A + M ′ ) − λ k + 1 ( A ⊤ A + M ) ≥ 3 4 Δ k \lambda_k(A^\top A+M')-\lambda_{k+1}(A^\top A+M)\ge\frac{3}{4}\Delta_k λ k ( A ⊤ A + M ′ ) − λ k + 1 ( A ⊤ A + M ) ≥ 4 3 Δ k
and thus δ ≥ 3 4 Δ k \delta\ge\frac{3}{4}\Delta_k δ ≥ 4 3 Δ k .
Thus we obtain
∥ ( I − X ( M ′ ) X ( M ′ ) ⊤ ) X ( M ) ∥ F ≤ ∥ M − M ′ ∥ F δ ≤ 4 3 Δ k ∥ M − M ′ ∥ F . \left\lVert\left(I-X(M')X(M')^\top\right)X(M)\right\rVert_F\le\frac{\lVert M-M'\rVert_F}{\delta}\le\frac{4}{3\Delta_k}\lVert M-M'\rVert_F. ( I − X ( M ′ ) X ( M ′ ) ⊤ ) X ( M ) F ≤ δ ∥ M − M ′ ∥ F ≤ 3 Δ k 4 ∥ M − M ′ ∥ F .
and
∥ P ( M ) − P ( M ′ ) ∥ F = 2 ∥ ( I − X ( M ′ ) X ( M ′ ) ⊤ ) X ( M ) ∥ F ≤ 8 3 Δ k ∥ M − M ′ ∥ F . \lVert P(M)-P(M')\rVert_F=\sqrt{2}\left\lVert\left(I-X(M')X(M')^\top\right)X(M)\right\rVert_F\le\frac{8}{3\Delta_k}\lVert M-M'\rVert_F. ∥ P ( M ) − P ( M ′ ) ∥ F = 2 ( I − X ( M ′ ) X ( M ′ ) ⊤ ) X ( M ) F ≤ 3 Δ k 8 ∥ M − M ′ ∥ F .
Putting everything together, we have
∥ E ( M ) − E ( M ′ ) ∥ F ≤ ( 1 + 16 15 ) 16 15 8 3 Δ k ∥ M − M ′ ∥ F ≤ 6 Δ k ∥ M − M ′ ∥ F , \lVert E(M)-E(M')\rVert_F\le\left(1+\frac{16}{15}\right)\frac{16}{15}\frac{8}{3\Delta_k}\lVert M-M'\rVert_F\le\frac{6}{\Delta_k}\lVert M-M'\rVert_F, ∥ E ( M ) − E ( M ′ ) ∥ F ≤ ( 1 + 15 16 ) 15 16 3 Δ k 8 ∥ M − M ′ ∥ F ≤ Δ k 6 ∥ M − M ′ ∥ F ,
as needed. □ \square □
Using the above claim and the fact that Π B ( ⋅ ) \Pi_{\mathcal{B}}(\cdot) Π B ( ⋅ ) and sym ( ⋅ ) \operatorname{sym}(\cdot) sym ( ⋅ ) are non-expansive in Frobenius norm, we can show that G ↦ E ‾ ( G ) G\mapsto\overline{E}(G) G ↦ E ( G ) is Lipschitz.
Claim 23. For any two matrices G , G ′ G,G' G , G ′ for which E ‾ ( G ) \overline{E}(G) E ( G ) and E ‾ ( G ′ ) \overline{E}(G') E ( G ′ ) are well-defined, we have
∥ E ‾ ( G ) − E ‾ ( G ′ ) ∥ F ≤ 6 Δ i n i t Δ k ∥ G − G ′ ∥ F . \lVert\overline{E}(G)-\overline{E}(G')\rVert_F\le\frac{6\Delta_{\mathrm{init}}}{\Delta_k}\lVert G-G'\rVert_F. ∥ E ( G ) − E ( G ′ ) ∥ F ≤ Δ k 6 Δ init ∥ G − G ′ ∥ F . where c 1 c_1 c 1 is an absolute constant.
Proof. We have
∥ E ‾ ( G ) − E ‾ ( G ′ ) ∥ F = ∥ E ‾ ( Π B ( Δ i n i t sym ( G ) ) ) − E ‾ ( Π B ( Δ i n i t sym ( G ′ ) ) ) ∥ F ≤ 6 Δ k ∥ Π B ( Δ i n i t sym ( G ) ) − Π B ( Δ i n i t sym ( G ′ ) ) ∥ F ≤ 6 Δ k ∥ Δ i n i t sym ( G ) − Δ i n i t sym ( G ′ ) ∥ F ≤ 6 Δ i n i t Δ k ∥ G − G ′ ∥ F , \begin{aligned}
\lVert\overline{E}(G)-\overline{E}(G')\rVert_F=\left\lVert\overline{E}\left(\Pi_{\mathcal{B}}\left(\Delta_{\mathrm{init}}\operatorname{sym}(G)\right)\right)-\overline{E}\left(\Pi_{\mathcal{B}}\left(\Delta_{\mathrm{init}}\operatorname{sym}(G')\right)\right)\right\rVert_F \\
&\le\frac{6}{\Delta_k}\left\lVert\Pi_{\mathcal{B}}\left(\Delta_{\mathrm{init}}\operatorname{sym}(G)\right)-\Pi_{\mathcal{B}}\left(\Delta_{\mathrm{init}}\operatorname{sym}(G')\right)\right\rVert_F \\
&\le\frac{6}{\Delta_k}\left\lVert\Delta_{\mathrm{init}}\operatorname{sym}(G)-\Delta_{\mathrm{init}}\operatorname{sym}(G')\right\rVert_F \\
&\le\frac{6\Delta_{\mathrm{init}}}{\Delta_k}\lVert G-G'\rVert_F,
\end{aligned} ∥ E ( G ) − E ( G ′ ) ∥ F = E ( Π B ( Δ init sym ( G ) ) ) − E ( Π B ( Δ init sym ( G ′ ) ) ) F ≤ Δ k 6 ∥ Π B ( Δ init sym ( G ) ) − Π B ( Δ init sym ( G ′ ) ) ∥ F ≤ Δ k 6 ∥ Δ init sym ( G ) − Δ init sym ( G ′ ) ∥ F ≤ Δ k 6 Δ init ∥ G − G ′ ∥ F ,
where we used the previous claim and the fact that Π B ( ⋅ ) \Pi_{\mathcal{B}}(\cdot) Π B ( ⋅ ) and sym ( ⋅ ) \operatorname{sym}(\cdot) sym ( ⋅ ) are non-expansive in Frobenius norm. □ \square □
Next, we use Gaussian concentration to upper bound ∥ e i ⊤ U > k Σ > k 2 r + 1 E ‾ ( G ( i n i t ) ) Σ k − 1 ∥ 2 \left\lVert e_i^\top U_{>k}\Sigma_{>k}^{2r+1}\overline{E}(G^{(\mathrm{init})})\Sigma_k^{-1}\right\rVert_2 e i ⊤ U > k Σ > k 2 r + 1 E ( G ( init ) ) Σ k − 1 2 for every row 1 ≤ i ≤ n 1\le i\le n 1 ≤ i ≤ n and every 0 ≤ r ≤ T 0\le r\le T 0 ≤ r ≤ T . To this end, for a fixed row i i i , exponent r r r , and an arbitrary unit vector z ∈ R k z\in\mathbb{R}^k z ∈ R k , we consider the following scalar-valued function:
f i , r , z ( G ) : = e i ⊤ U > k Σ > k 2 r + 1 E ‾ ( G ) Σ k − 1 z f_{i,r,z}(G):=e_i^\top U_{>k}\Sigma_{>k}^{2r+1}\overline{E}(G)\Sigma_k^{-1}z f i , r , z ( G ) := e i ⊤ U > k Σ > k 2 r + 1 E ( G ) Σ k − 1 z
Since E ‾ ( G ) \overline{E}(G) E ( G ) is Lipschitz, we can show that f i , r , z f_{i,r,z} f i , r , z is Lipschitz.
Claim 24. For any two matrices G , G ′ G,G' G , G ′ for which E ˉ ( G ) \bar{E}(G) E ˉ ( G ) and E ˉ ( G ′ ) \bar{E}(G') E ˉ ( G ′ ) are well-defined, we have
∣ f i , r , z ( G ) − f i , r , z ( G ′ ) ∣ ≤ L i , r ∥ G − G ′ ∥ F . \left|f_{i,r,z}(G)-f_{i,r,z}(G')\right|\le L_{i,r}\left\|G-G'\right\|_{F}. ∣ f i , r , z ( G ) − f i , r , z ( G ′ ) ∣ ≤ L i , r ∥ G − G ′ ∥ F . where
L i , r : = 6 Δ i n i t Δ k σ k + 1 2 r + 1 σ k 2 r + 1 ∥ e i T U > k ∥ 2 . L_{i,r}:=\frac{6\Delta_{\mathrm{init}}}{\Delta_k}\frac{\sigma_{k+1}^{2r+1}}{\sigma_k^{2r+1}}\left\|e_i^{\mathsf T}U_{>k}\right\|_2. L i , r := Δ k 6 Δ init σ k 2 r + 1 σ k + 1 2 r + 1 e i T U > k 2 . Proof. We have
∣ f i , r , z ( G ) − f i , r , z ( G ′ ) ∣ = ∣ e i T U > k Σ > k 2 r + 1 ( E ˉ ( G ) − E ˉ ( G ′ ) ) Σ k − 1 z ∣ ≤ ∥ e i T U > k ∥ 2 ∥ Σ > k 2 r + 1 ∥ 2 ∥ E ˉ ( G ) − E ˉ ( G ′ ) ∥ F ∥ Σ k − 1 ∥ 2 ∥ z ∥ 2 = ∥ e i T U > k ∥ 2 σ k + 1 2 r + 1 ∥ E ˉ ( G ) − E ˉ ( G ′ ) ∥ F 1 σ k ≤ ∥ e i T U > k ∥ 2 σ k + 1 2 r + 1 6 Δ i n i t Δ k ∥ G − G ′ ∥ F 1 σ k , \begin{aligned}
\left|f_{i,r,z}(G)-f_{i,r,z}(G')\right|
&=\left|e_i^{\mathsf T}U_{>k}\Sigma_{>k}^{2r+1}\left(\bar{E}(G)-\bar{E}(G')\right)\Sigma_k^{-1}z\right|\\
&\le\left\|e_i^{\mathsf T}U_{>k}\right\|_2\left\|\Sigma_{>k}^{2r+1}\right\|_2\left\|\bar{E}(G)-\bar{E}(G')\right\|_F\left\|\Sigma_k^{-1}\right\|_2\left\|z\right\|_2\\
&=\left\|e_i^{\mathsf T}U_{>k}\right\|_2\sigma_{k+1}^{2r+1}\left\|\bar{E}(G)-\bar{E}(G')\right\|_F\frac{1}{\sigma_k}\\
&\le\left\|e_i^{\mathsf T}U_{>k}\right\|_2\sigma_{k+1}^{2r+1}\frac{6\Delta_{\mathrm{init}}}{\Delta_k}\left\|G-G'\right\|_F\frac{1}{\sigma_k},
\end{aligned} ∣ f i , r , z ( G ) − f i , r , z ( G ′ ) ∣ = e i T U > k Σ > k 2 r + 1 ( E ˉ ( G ) − E ˉ ( G ′ ) ) Σ k − 1 z ≤ e i T U > k 2 Σ > k 2 r + 1 2 E ˉ ( G ) − E ˉ ( G ′ ) F Σ k − 1 2 ∥ z ∥ 2 = e i T U > k 2 σ k + 1 2 r + 1 E ˉ ( G ) − E ˉ ( G ′ ) F σ k 1 ≤ e i T U > k 2 σ k + 1 2 r + 1 Δ k 6 Δ init ∥ G − G ′ ∥ F σ k 1 , where in the last inequality we used the fact that E ˉ ( G ) \bar{E}(G) E ˉ ( G ) is 6 Δ i n i t Δ k \frac{6\Delta_{\mathrm{init}}}{\Delta_k} Δ k 6 Δ init -Lipschitz.
Next, we show that E [ f i , r , z ( G ) ] = 0 \mathbb{E}[f_{i,r,z}(G)]=0 E [ f i , r , z ( G )] = 0 .
Claim 25. We have E [ f i , r , z ( G ) ] = 0 \mathbb{E}[f_{i,r,z}(G)]=0 E [ f i , r , z ( G )] = 0 .
Proof. Consider a Gaussian matrix G ∼ N ( 0 , 1 ) d × d G\sim\mathcal{N}(0,1)^{d\times d} G ∼ N ( 0 , 1 ) d × d . Let
J = V ( − I k 0 0 I d − k ) V T . J=V\begin{pmatrix}-I_k&0\\0&I_{d-k}\end{pmatrix}V^{\mathsf T}. J = V ( − I k 0 0 I d − k ) V T . The matrix J J J is orthogonal and symmetric. Let G ′ = J G J T G'=JGJ^{\mathsf T} G ′ = J G J T . Since the Gaussian distribution is rotationally invariant, G ′ G' G ′ has the same distribution as G G G . Next, we will show that f i , r , z ( G ′ ) = − f i , r , z ( G ) f_{i,r,z}(G')=-f_{i,r,z}(G) f i , r , z ( G ′ ) = − f i , r , z ( G ) which will imply that E [ f i , r , z ( G ) ] = 0 \mathbb{E}[f_{i,r,z}(G)]=0 E [ f i , r , z ( G )] = 0 . Let M : = Π B ( Δ i n i t sym ( G ) ) M:=\Pi_B(\Delta_{\mathrm{init}}\operatorname{sym}(G)) M := Π B ( Δ init sym ( G )) and M ′ : = Π B ( Δ i n i t sym ( G ′ ) ) M':=\Pi_B(\Delta_{\mathrm{init}}\operatorname{sym}(G')) M ′ := Π B ( Δ init sym ( G ′ )) . Note that we have M ′ = J M J T M'=JMJ^{\mathsf T} M ′ = J M J T and J ( A T A ) J T = A T A J(A^{\mathsf T}A)J^{\mathsf T}=A^{\mathsf T}A J ( A T A ) J T = A T A , and thus
A T A + M ′ = J ( A T A + M ) J T . A^{\mathsf T}A+M'=J(A^{\mathsf T}A+M)J^{\mathsf T}. A T A + M ′ = J ( A T A + M ) J T . As in the proof of Claim 22 , for W ∈ { M , M ′ } W\in\{M,M'\} W ∈ { M , M ′ } , we let P ( W ) = X ( W ) X ( W ) T P(W)=X(W)X(W)^{\mathsf T} P ( W ) = X ( W ) X ( W ) T and we write it as follows:
V T P ( W ) V = ( P 11 ( W ) P 12 ( W ) P 21 ( W ) P 22 ( W ) ) . V^{\mathsf T}P(W)V=
\begin{pmatrix}
P_{11}(W)&P_{12}(W)\\
P_{21}(W)&P_{22}(W)
\end{pmatrix}. V T P ( W ) V = ( P 11 ( W ) P 21 ( W ) P 12 ( W ) P 22 ( W ) ) . where P 11 ( W ) = V k T P ( W ) V k P_{11}(W)=V_k^{\mathsf T}P(W)V_k P 11 ( W ) = V k T P ( W ) V k , P 12 ( W ) = V k T P ( W ) V > k P_{12}(W)=V_k^{\mathsf T}P(W)V_{>k} P 12 ( W ) = V k T P ( W ) V > k , P 21 ( W ) = V > k T P ( W ) V k P_{21}(W)=V_{>k}^{\mathsf T}P(W)V_k P 21 ( W ) = V > k T P ( W ) V k , and P 22 ( W ) = V > k T P ( W ) V > k P_{22}(W)=V_{>k}^{\mathsf T}P(W)V_{>k} P 22 ( W ) = V > k T P ( W ) V > k . Since A T A + M ′ = J ( A T A + M ) J T A^{\mathsf T}A+M'=J(A^{\mathsf T}A+M)J^{\mathsf T} A T A + M ′ = J ( A T A + M ) J T , J V k = − V k JV_k=-V_k J V k = − V k , and J V > k = V > k JV_{>k}=V_{>k} J V > k = V > k , we have P ( M ′ ) = J P ( M ) J T P(M')=JP(M)J^{\mathsf T} P ( M ′ ) = J P ( M ) J T and
V T P ( M ′ ) V = ( P 11 ( M ) − P 12 ( M ) − P 21 ( M ) P 22 ( M ) ) . V^{\mathsf T}P(M')V=
\begin{pmatrix}
P_{11}(M)&-P_{12}(M)\\
-P_{21}(M)&P_{22}(M)
\end{pmatrix}. V T P ( M ′ ) V = ( P 11 ( M ) − P 21 ( M ) − P 12 ( M ) P 22 ( M ) ) . Since E ( M ) = P 21 ( M ) P 11 ( M ) − 1 E(M)=P_{21}(M)P_{11}(M)^{-1} E ( M ) = P 21 ( M ) P 11 ( M ) − 1 , we obtain that E ˉ ( G ′ ) = − E ˉ ( G ) \bar{E}(G')=-\bar{E}(G) E ˉ ( G ′ ) = − E ˉ ( G ) . Therefore f i , r , z ( G ′ ) = − f i , r , z ( G ) f_{i,r,z}(G')=-f_{i,r,z}(G) f i , r , z ( G ′ ) = − f i , r , z ( G ) and
E [ f i , r , z ( G ) ] = E [ f i , r , z ( G ′ ) ] = − E [ f i , r , z ( G ) ] . \mathbb{E}[f_{i,r,z}(G)]=\mathbb{E}[f_{i,r,z}(G')]=-\mathbb{E}[f_{i,r,z}(G)]. E [ f i , r , z ( G )] = E [ f i , r , z ( G ′ )] = − E [ f i , r , z ( G )] . The expectation exists since f i , r , z f_{i,r,z} f i , r , z is globally Lipschitz, and thus E [ f i , r , z ( G ) ] = 0 \mathbb{E}[f_{i,r,z}(G)]=0 E [ f i , r , z ( G )] = 0 .
By applying Theorem 4 to f i , r , z f_{i,r,z} f i , r , z and using a net argument, we obtain the following guarantee.
Claim 26. With probability at least 1 − β 8 1-\frac{\beta}{8} 1 − 8 β , simultaneously for all 1 ≤ i ≤ n 1\le i\le n 1 ≤ i ≤ n and 0 ≤ r ≤ T 0\le r\le T 0 ≤ r ≤ T , we have
∥ e i T U > k Σ > k 2 r + 1 E ˉ ( G ( i n i t ) ) Σ k − 1 ∥ 2 ≤ 2 L i , r ( 2 k + 2 log ( 16 n ( T + 1 ) β ) ) . \left\|e_i^{\mathsf T}U_{>k}\Sigma_{>k}^{2r+1}\bar{E}\left(G^{(\mathrm{init})}\right)\Sigma_k^{-1}\right\|_2\le2L_{i,r}\left(2\sqrt{k}+\sqrt{2\log\left(\frac{16n(T+1)}{\beta}\right)}\right). e i T U > k Σ > k 2 r + 1 E ˉ ( G ( init ) ) Σ k − 1 2 ≤ 2 L i , r ( 2 k + 2 log ( β 16 n ( T + 1 ) ) ) .
Proof. We discretize the set of all possible unit vectors z ∈ R k z \in\mathbb{R}^k z ∈ R k (i.e., the unit sphere in R k \mathbb{R}^k R k ) using a 1 / 2 1/2 1/2 -net N \mathcal{N} N , i.e., N \mathcal{N} N is a set of unit vectors with the property that, for every unit vector z ∈ R k z \in\mathbb{R}^k z ∈ R k , there is a vector z ′ ∈ N z' \in\mathcal{N} z ′ ∈ N such that ∥ z − z ′ ∥ 2 ≤ 1 / 2 \lVert z-z'\rVert_2 \le1/2 ∥ z − z ′ ∥ 2 ≤ 1/2 . A standard volume argument shows that there exists a net of size ∣ N ∣ ≤ 5 k \lvert\mathcal{N}\rvert\le5^k ∣ N ∣ ≤ 5 k . In the following, we let N \mathcal{N} N be a 1 / 2 1/2 1/2 -net of size ∣ N ∣ ≤ 5 k \lvert\mathcal{N}\rvert\le5^k ∣ N ∣ ≤ 5 k .
Fix a row index i i i and exponent r r r . Let w : = e i T U > k Σ > k 2 r + 1 E ‾ ( G ( i n i t ) ) Σ k − 1 w := e_i^{\mathsf{T}}U_{>k}\Sigma_{>k}^{2r+1}\overline{E}(G^{(\mathrm{init})})\Sigma_k^{-1} w := e i T U > k Σ > k 2 r + 1 E ( G ( init ) ) Σ k − 1 . Note that w ∈ R 1 × k w \in\mathbb{R}^{1 \times k} w ∈ R 1 × k is a row vector. In the following, we upper bound ∥ w ∥ 2 \lVert w\rVert_2 ∥ w ∥ 2 . We may assume that w ≠ 0 w \ne0 w = 0 , since otherwise ∥ w ∥ 2 = 0 \lVert w\rVert_2=0 ∥ w ∥ 2 = 0 and we are done. Let w ‾ = w / ∥ w ∥ 2 \overline{w}=w/\lVert w\rVert_2 w = w / ∥ w ∥ 2 and thus ∥ w ∥ 2 = w w ‾ \lVert w\rVert_2=w\overline{w} ∥ w ∥ 2 = w w . Since N \mathcal{N} N is a 1 / 2 1/2 1/2 -net, there is a vector z ′ ∈ N z' \in\mathcal{N} z ′ ∈ N such that ∥ w ‾ − z ′ ∥ 2 ≤ 1 / 2 \lVert\overline{w}-z'\rVert_2 \le1/2 ∥ w − z ′ ∥ 2 ≤ 1/2 . We have
∥ w ∥ 2 = w w ‾ = w z ′ + w ( w ‾ − z ′ ) ≤ ∣ w z ′ ∣ + ∥ w ∥ 2 ∥ w ‾ − z ′ ∥ 2 ≤ max z ∈ N ∣ w z ∣ + 1 2 ∥ w ∥ 2 . \begin{aligned}
\lVert w\rVert_2=w\overline{w}=wz'+w(\overline{w}-z') \\
&\le\lvert wz'\rvert+\lVert w\rVert_2\lVert\overline{w}-z'\rVert_2 \\
&\le\max_{z\in\mathcal{N}}\lvert wz\rvert+\frac{1}{2}\lVert w\rVert_2.
\end{aligned} ∥ w ∥ 2 = w w = w z ′ + w ( w − z ′ ) ≤ ∣ w z ′ ∣ + ∥ w ∥ 2 ∥ w − z ′ ∥ 2 ≤ z ∈ N max ∣ w z ∣ + 2 1 ∥ w ∥ 2 .
Rearranging, we obtain
∥ w ∥ 2 ≤ 2 max z ∈ N ∣ w z ∣ . \lVert w\rVert_2\le2\max_{z\in\mathcal{N}}\lvert wz\rvert. ∥ w ∥ 2 ≤ 2 z ∈ N max ∣ w z ∣ .
Since w = e i T U > k Σ > k 2 r + 1 E ‾ ( G ( i n i t ) ) Σ k − 1 w=e_i^{\mathsf{T}}U_{>k}\Sigma_{>k}^{2r+1}\overline{E}(G^{(\mathrm{init})})\Sigma_k^{-1} w = e i T U > k Σ > k 2 r + 1 E ( G ( init ) ) Σ k − 1 and f i , r , z ( G ( i n i t ) ) = w z f_{i,r,z}(G^{(\mathrm{init})})=wz f i , r , z ( G ( init ) ) = w z , we have
∥ e i T U > k Σ > k 2 r + 1 E ‾ ( G ( i n i t ) ) Σ k − 1 ∥ 2 ≤ 2 max z ∈ N ∣ f i , r , z ( G ( i n i t ) ) ∣ . \left\lVert e_i^{\mathsf{T}}U_{>k}\Sigma_{>k}^{2r+1}\overline{E}(G^{(\mathrm{init})})\Sigma_k^{-1}\right\rVert_2\le2\max_{z\in\mathcal{N}}\left\lvert f_{i,r,z}(G^{(\mathrm{init})})\right\rvert. e i T U > k Σ > k 2 r + 1 E ( G ( init ) ) Σ k − 1 2 ≤ 2 z ∈ N max f i , r , z ( G ( init ) ) .
Since f i , r , z f_{i,r,z} f i , r , z is L i , r L_{i,r} L i , r -Lipschitz and has expectation 0 0 0 , Theorem 4 gives that, for every x ≥ 0 x\ge0 x ≥ 0 ,
Pr [ ∣ f i , r , z ( G ( i n i t ) ) ∣ ≥ L i , r x ] ≤ 2 exp ( − x 2 2 ) . \Pr\left[\left\lvert f_{i,r,z}\left(G^{(\mathrm{init})}\right)\right\rvert\ge L_{i,r}x\right]\le2\exp\left(-\frac{x^2}{2}\right). Pr [ f i , r , z ( G ( init ) ) ≥ L i , r x ] ≤ 2 exp ( − 2 x 2 ) .
There are n ( T + 1 ) ∣ N ∣ ≤ n ( T + 1 ) 5 k n(T+1)\lvert\mathcal{N}\rvert\le n(T+1)5^k n ( T + 1 ) ∣ N ∣ ≤ n ( T + 1 ) 5 k triples ( i , r , z ) (i,r,z) ( i , r , z ) where 1 ≤ i ≤ n 1\le i\le n 1 ≤ i ≤ n is a row index, 0 ≤ r ≤ T 0\le r\le T 0 ≤ r ≤ T is an exponent, and z z z is a vector in N \mathcal{N} N . Setting x = 2 log ( 16 n ( T + 1 ) 5 k β ) x=\sqrt{2\log\left(\frac{16n(T+1)5^k}{\beta}\right)} x = 2 log ( β 16 n ( T + 1 ) 5 k ) in the above inequality and using an union bound over all triples ( i , r , z ) (i,r,z) ( i , r , z ) , we obtain that, with probability at least 1 − β 8 1-\frac{\beta}{8} 1 − 8 β , simultaneously for all i i i and r r r , we have
∥ e i T U > k Σ > k 2 r + 1 E ‾ ( G ( i n i t ) ) Σ k − 1 ∥ 2 ≤ 2 max z ∈ N ∣ f i , r , z ( G ( i n i t ) ) ∣ \left\lVert e_i^{\mathsf{T}}U_{>k}\Sigma_{>k}^{2r+1}\overline{E}\left(G^{(\mathrm{init})}\right)\Sigma_k^{-1}\right\rVert_2\le2\max_{z\in\mathcal{N}}\left\lvert f_{i,r,z}\left(G^{(\mathrm{init})}\right)\right\rvert e i T U > k Σ > k 2 r + 1 E ( G ( init ) ) Σ k − 1 2 ≤ 2 z ∈ N max f i , r , z ( G ( init ) )
≤ 2 L i , r 2 log ( 16 n ( T + 1 ) 5 k β ) \le2L_{i,r}\sqrt{2\log\left(\frac{16n(T+1)5^k}{\beta}\right)} ≤ 2 L i , r 2 log ( β 16 n ( T + 1 ) 5 k )
≤ 2 L i , r ( 2 k + 2 log ( 16 n ( T + 1 ) β ) ) \le2L_{i,r}\left(2\sqrt{k}+\sqrt{2\log\left(\frac{16n(T+1)}{\beta}\right)}\right) ≤ 2 L i , r ( 2 k + 2 log ( β 16 n ( T + 1 ) ) )
as needed. □ \square □
Finally, we show that the assumption on Δ k \Delta_k Δ k implies that, with probability at least 1 − β 8 1-\frac{\beta}{8} 1 − 8 β , Δ i n i t β sym ( G ( i n i t ) ) ∈ B \Delta_{\mathrm{init}}^{\beta}\operatorname{sym}\left(G^{(\mathrm{init})}\right)\in\mathcal{B} Δ init β sym ( G ( init ) ) ∈ B and thus E ‾ ( G ( i n i t ) ) = E ( 0 ) \overline{E}\left(G^{(\mathrm{init})}\right)=E^{(0)} E ( G ( init ) ) = E ( 0 ) .
Claim 27. Suppose that Δ k ≥ 8 Δ i n i t ( 2 d + 2 log 8 β ) \Delta_k\ge8\Delta_{\mathrm{init}}\left(2\sqrt{d}+\sqrt{2\log\frac{8}{\beta}}\right) Δ k ≥ 8 Δ init ( 2 d + 2 log β 8 ) . With probability at least 1 − β 8 1-\frac{\beta}{8} 1 − 8 β , ∥ Δ i n i t sym ( G ( i n i t ) ) ∥ 2 ≤ 1 8 Δ k \lVert\Delta_{\mathrm{init}}\operatorname{sym}\left(G^{(\mathrm{init})}\right)\rVert_2\le\frac{1}{8}\Delta_k ∥ Δ init sym ( G ( init ) ) ∥ 2 ≤ 8 1 Δ k and thus Δ i n i t sym ( G ( i n i t ) ) ∈ B \Delta_{\mathrm{init}}\operatorname{sym}\left(G^{(\mathrm{init})}\right)\in\mathcal{B} Δ init sym ( G ( init ) ) ∈ B and E ‾ ( G ( i n i t ) ) = E ( 0 ) \overline{E}\left(G^{(\mathrm{init})}\right)=E^{(0)} E ( G ( init ) ) = E ( 0 ) .
Proof. By Lemma 5, we have
Pr [ ∥ sym ( G ( i n i t ) ) ∥ 2 > 2 d + 2 log 8 β ] ≤ exp ( − 1 2 ( 2 log 8 β ) 2 ) = β 8 . \Pr\left[\left\lVert\operatorname{sym}\left(G^{(\mathrm{init})}\right)\right\rVert_2>2\sqrt{d}+\sqrt{2\log\frac{8}{\beta}}\right]\le\exp\left(-\frac{1}{2}\left(\sqrt{2\log\frac{8}{\beta}}\right)^2\right)=\frac{\beta}{8}. Pr [ sym ( G ( init ) ) 2 > 2 d + 2 log β 8 ] ≤ exp ( − 2 1 ( 2 log β 8 ) 2 ) = 8 β .
Thus, with probability 1 − β / 8 1-\beta/8 1 − β /8 , we have
Δ i n i t ∥ sym ( G ( i n i t ) ) ∥ 2 ≤ Δ i n i t ( 2 d + 2 log 8 β ) ≤ 1 8 Δ k . \Delta_{\mathrm{init}}\left\|\operatorname{sym}\left(G^{(\mathrm{init})}\right)\right\|_2 \le\Delta_{\mathrm{init}}\left(2\sqrt{d}+\sqrt{2\log\frac{8}{\beta}}\right)\le\frac{1}{8}\Delta_k. Δ init sym ( G ( init ) ) 2 ≤ Δ init ( 2 d + 2 log β 8 ) ≤ 8 1 Δ k .
Thus, by combining the last two claims, we obtain that, with probability at least 1 − β 4 1-\frac{\beta}{4} 1 − 4 β , simultaneously for all 0 ≤ r ≤ T 0\le r\le T 0 ≤ r ≤ T , we have
Φ r , 0 : = max i ∈ [ n ] ∥ e i T U > k Σ > k 2 r + 1 E ( 0 ) Σ k − 1 ∥ 2 = max i ∈ [ n ] ∥ e i T U > k Σ > k 2 r + 1 E ( G ( i n i t ) ) Σ k − 1 ∥ 2 ≤ max i ∈ [ n ] 2 L i , r ( 2 k + 2 log ( 16 n ( T + 1 ) β ) ) = max i ∈ [ n ] 12 Δ i n i t Δ k σ k + 1 2 r + 1 σ k ∥ e i T U > k ∥ 2 ( 2 k + 2 log ( 16 n ( T + 1 ) β ) ) . \begin{aligned}
\Phi_{r,0} &:= \max_{i\in[n]}\left\|e_i^{\mathsf{T}}U_{>k}\Sigma_{>k}^{2r+1}E^{(0)}\Sigma_k^{-1}\right\|_2 \\
&= \max_{i\in[n]}\left\|e_i^{\mathsf{T}}U_{>k}\Sigma_{>k}^{2r+1}E\left(G^{(\mathrm{init})}\right)\Sigma_k^{-1}\right\|_2 \\
&\le\max_{i\in[n]}2L_{i,r}\left(2\sqrt{k}+\sqrt{2\log\left(\frac{16n(T+1)}{\beta}\right)}\right) \\
&= \max_{i\in[n]}\frac{12\Delta_{\mathrm{init}}}{\Delta_k}\frac{\sigma_{k+1}^{2r+1}}{\sigma_k}\left\|e_i^{\mathsf{T}}U_{>k}\right\|_2\left(2\sqrt{k}+\sqrt{2\log\left(\frac{16n(T+1)}{\beta}\right)}\right).
\end{aligned} Φ r , 0 := i ∈ [ n ] max e i T U > k Σ > k 2 r + 1 E ( 0 ) Σ k − 1 2 = i ∈ [ n ] max e i T U > k Σ > k 2 r + 1 E ( G ( init ) ) Σ k − 1 2 ≤ i ∈ [ n ] max 2 L i , r ( 2 k + 2 log ( β 16 n ( T + 1 ) ) ) = i ∈ [ n ] max Δ k 12 Δ init σ k σ k + 1 2 r + 1 e i T U > k 2 ( 2 k + 2 log ( β 16 n ( T + 1 ) ) ) .
Using that ∥ e i T U > k ∥ 2 ≤ d − k ∥ U ∥ ∞ \left\|e_i^{\mathsf{T}}U_{>k}\right\|_2\le\sqrt{d-k}\left\|U\right\|_\infty e i T U > k 2 ≤ d − k ∥ U ∥ ∞ and the assumption Δ k ≥ 8 Δ i n i t ( 2 d + 2 log 8 β ) \Delta_k\ge8\Delta_{\mathrm{init}}\left(2\sqrt{d}+\sqrt{2\log\frac{8}{\beta}}\right) Δ k ≥ 8 Δ init ( 2 d + 2 log β 8 ) , we can further upper bound
Φ r , 0 ≤ 12 Δ i n i t Δ k σ k + 1 2 r + 1 σ k d − k ∥ U ∥ ∞ ( 2 k + 2 log ( 16 n ( T + 1 ) β ) ) ≤ 12 8 ( 2 d + 2 log 8 β ) σ k + 1 2 r + 1 σ k d − k ∥ U ∥ ∞ ( 2 k + 2 log ( 16 n ( T + 1 ) β ) ) ≤ 3 2 σ k + 1 2 r + 1 σ k ∥ U ∥ ∞ ( 2 k + 2 log ( 16 n ( T + 1 ) β ) ) ≤ 3 2 σ k + 1 2 r + 1 ∥ U ∥ ∞ ( 2 k + 2 log ( 16 n ( T + 1 ) β ) ) \begin{aligned}
\Phi_{r,0}\le\frac{12\Delta_{\mathrm{init}}}{\Delta_k}\frac{\sigma_{k+1}^{2r+1}}{\sigma_k}\sqrt{d-k}\left\|U\right\|_\infty\left(2\sqrt{k}+\sqrt{2\log\left(\frac{16n(T+1)}{\beta}\right)}\right) \\
&\le\frac{12}{8\left(2\sqrt{d}+\sqrt{2\log\frac{8}{\beta}}\right)}\frac{\sigma_{k+1}^{2r+1}}{\sigma_k}\sqrt{d-k}\left\|U\right\|_\infty\left(2\sqrt{k}+\sqrt{2\log\left(\frac{16n(T+1)}{\beta}\right)}\right) \\
&\le\frac{3}{2}\frac{\sigma_{k+1}^{2r+1}}{\sigma_k}\left\|U\right\|_\infty\left(2\sqrt{k}+\sqrt{2\log\left(\frac{16n(T+1)}{\beta}\right)}\right) \\
&\le\frac{3}{2}\sigma_{k+1}^{2r+1}\left\|U\right\|_\infty\left(2\sqrt{k}+\sqrt{2\log\left(\frac{16n(T+1)}{\beta}\right)}\right)
\end{aligned} Φ r , 0 ≤ Δ k 12 Δ init σ k σ k + 1 2 r + 1 d − k ∥ U ∥ ∞ ( 2 k + 2 log ( β 16 n ( T + 1 ) ) ) ≤ 8 ( 2 d + 2 log β 8 ) 12 σ k σ k + 1 2 r + 1 d − k ∥ U ∥ ∞ ( 2 k + 2 log ( β 16 n ( T + 1 ) ) ) ≤ 2 3 σ k σ k + 1 2 r + 1 ∥ U ∥ ∞ ( 2 k + 2 log ( β 16 n ( T + 1 ) ) ) ≤ 2 3 σ k + 1 2 r + 1 ∥ U ∥ ∞ ( 2 k + 2 log ( β 16 n ( T + 1 ) ) )
≤ 3 σ k + 1 2 r + 1 ∥ U ∥ ∞ ( k + log ( 16 n ( T + 1 ) β ) ) \le3\sigma_{k+1}^{2r+1}\left\|U\right\|_\infty\left(\sqrt{k}+\sqrt{\log\left(\frac{16n(T+1)}{\beta}\right)}\right) ≤ 3 σ k + 1 2 r + 1 ∥ U ∥ ∞ ( k + log ( β 16 n ( T + 1 ) ) )
as needed.
Using Lemma 21 , we will show by induction that the clipping is inactive in all iterations with high probability. We will use the following lemma for the inductive step. Recall that G c l i p G_{\mathrm{clip}} G clip is the event defined in Definition 15 and G i n i t G_{\mathrm{init}} G init is the event in Lemma 21 .
Lemma 28. Suppose that Δ k ≥ max { 8 Δ i n i t ( 2 d + 2 log 8 β ) , 8 Δ q r o w } \Delta_k\ge\max\left\{8\Delta_{\mathrm{init}}\left(2\sqrt{d}+\sqrt{2\log\frac{8}{\beta}}\right),8\Delta_{q_{\mathrm{row}}}\right\} Δ k ≥ max { 8 Δ init ( 2 d + 2 log β 8 ) , 8 Δ q row } , γ ≤ σ k 2 4 \gamma\le\frac{\sigma_k^2}{4} γ ≤ 4 σ k 2 , p R : = 3 Δ R q k 2 σ k < κ k 16 p_R:=\frac{3\Delta_{Rq_k}}{2\sigma_k}<\frac{\kappa_k}{16} p R := 2 σ k 3 Δ R q k < 16 κ k , and R ≥ 15 ∥ U ∥ ∞ ( k + log ( 16 n ( T + 1 ) β ) ) R\ge15\left\|U\right\|_\infty\left(\sqrt{k}+\sqrt{\log\left(\frac{16n(T+1)}{\beta}\right)}\right) R ≥ 15 ∥ U ∥ ∞ ( k + log ( β 16 n ( T + 1 ) ) ) . Suppose that G c l i p G_{\mathrm{clip}} G clip and G i n i t G_{\mathrm{init}} G init hold. If s j ≤ 1 2 s_j\le\frac{1}{2} s j ≤ 2 1 for all 0 ≤ j ≤ t 0\le j\le t 0 ≤ j ≤ t and clipping was inactive in all iterations 0 ≤ j ≤ t − 1 0\le j\le t-1 0 ≤ j ≤ t − 1 , then clipping is also inactive in iteration t t t .
Proof. First, let us verify that the conditions of Lemma 19 are satisfied for all iterations j ≤ t − 1 j\le t-1 j ≤ t − 1 . By assumption, clipping is inactive in each of those iterations, and thus it only remains to check that we have ∥ F ( j ) ∥ 2 < 1 \lVert F^{(j)}\rVert_2<1 ∥ F ( j ) ∥ 2 < 1 for each j ≤ t − 1 j\le t-1 j ≤ t − 1 . We have
∥ F ( j ) ∥ 2 = ∥ Δ R V k ⊤ G ~ ( j ) ( K ( j ) ) − 1 ∥ 2 ≤ Δ R ∥ V k ⊤ G ~ ( j ) ∥ 2 ∥ ( K ( j ) ) − 1 ∥ 2 ≤ Δ R q k 3 2 1 σ k = p R . \begin{aligned}
\lVert F^{(j)}\rVert_2
&=\left\lVert\Delta R V_k^{\top}\widetilde{G}^{(j)}\left(K^{(j)}\right)^{-1}\right\rVert_2\\
&\le\Delta R\left\lVert V_k^{\top}\widetilde{G}^{(j)}\right\rVert_2\left\lVert\left(K^{(j)}\right)^{-1}\right\rVert_2\\
&\le\Delta R q_k\frac{3}{2}\frac{1}{\sigma_k}=p_R.
\end{aligned} ∥ F ( j ) ∥ 2 = Δ R V k ⊤ G ( j ) ( K ( j ) ) − 1 2 ≤ Δ R V k ⊤ G ( j ) 2 ( K ( j ) ) − 1 2 ≤ Δ R q k 2 3 σ k 1 = p R .
In the last inequality, we used that G c l i p \mathcal{G}_{\mathrm{clip}} G clip holds (and thus ∥ V k ⊤ G ~ ( j ) ∥ 2 ≤ q k \lVert V_k^{\top}\widetilde{G}^{(j)}\rVert_2\le q_k ∥ V k ⊤ G ( j ) ∥ 2 ≤ q k ) and Lemma 18 (note that the conditions of the lemma are satisfied by assumption). By assumption, we have p R ≤ κ k 16 ≤ 1 16 p_R\le\frac{\kappa_k}{16}\le\frac{1}{16} p R ≤ 16 κ k ≤ 16 1 , and thus ∥ F ( j ) ∥ 2 < 1 \lVert F^{(j)}\rVert_2<1 ∥ F ( j ) ∥ 2 < 1 as needed.
Thus, by Lemma 19 , for all iterations j ≤ t − 1 j\le t-1 j ≤ t − 1 , we have that C ( j + 1 ) C^{(j+1)} C ( j + 1 ) is invertible and
E ( j + 1 ) = ( Σ > k 2 Σ k E ( j ) Σ k − 2 + L ( j ) ) ( I + F ( j ) ) − 1 . E^{(j+1)}=\left(\Sigma_{>k}^{2}\Sigma_k E^{(j)}\Sigma_k^{-2}+L^{(j)}\right)\left(I+F^{(j)}\right)^{-1}. E ( j + 1 ) = ( Σ > k 2 Σ k E ( j ) Σ k − 2 + L ( j ) ) ( I + F ( j ) ) − 1 .
Recall that S ( j ) : = Σ > k E ( j ) Σ k − 1 \mathcal{S}^{(j)}:=\Sigma_{>k}E^{(j)}\Sigma_k^{-1} S ( j ) := Σ > k E ( j ) Σ k − 1 for all j j j . Multiplying the above recurrence with Σ > k \Sigma_{>k} Σ > k on the left and Σ k − 1 \Sigma_k^{-1} Σ k − 1 on the right, we obtain
S ( j + 1 ) = ( Σ > k 2 S ( j ) Σ k − 1 + Σ > k L ( j ) ) ( I + F ( j ) ) − 1 Σ k − 1 . \mathcal{S}^{(j+1)}=\left(\Sigma_{>k}^{2}\mathcal{S}^{(j)}\Sigma_k^{-1}+\Sigma_{>k}L^{(j)}\right)\left(I+F^{(j)}\right)^{-1}\Sigma_k^{-1}. S ( j + 1 ) = ( Σ > k 2 S ( j ) Σ k − 1 + Σ > k L ( j ) ) ( I + F ( j ) ) − 1 Σ k − 1 .
Using the above recurrence, we can obtain a recurrence for the potential Φ r , t \Phi_{r,t} Φ r , t . Recall that L ( j ) = Δ R V k ⊤ G ~ ( j ) ( K ( j ) ) − 1 L^{(j)}=\Delta R V_k^{\top}\widetilde{G}^{(j)}\left(K^{(j)}\right)^{-1} L ( j ) = Δ R V k ⊤ G ( j ) ( K ( j ) ) − 1 . For every r ≥ 0 r\ge0 r ≥ 0 and every j ≤ t − 1 j\le t-1 j ≤ t − 1 , we have
Φ r , j + 1 = max i ∈ [ n ] ∥ e i ⊤ U > k Σ > k 2 r S ( j + 1 ) ∥ 2 = max i ∈ [ n ] ∥ ( e i ⊤ U > k Σ > k 2 ( r + 1 ) S ( j ) Σ k − 1 + e i ⊤ U > k Σ > k 2 r Σ > k L ( j ) ) ( I + F ( j ) ) − 1 Σ k − 1 ∥ 2 ≤ max i ∈ [ n ] ( ∥ e i ⊤ U > k Σ > k 2 ( r + 1 ) S ( j ) ∥ 2 ∥ Σ k − 1 ∥ 2 + Δ R ∥ e i ⊤ U > k Σ > k 2 r + 1 V k ⊤ G ~ ( j ) ∥ 2 ∥ ( K ( j ) ) − 1 ∥ 2 ) ⋅ ∥ ( I + F ( j ) ) − 1 ∥ 2 ∥ Σ k − 1 ∥ 2 ≤ ( Φ r + 1 , j 1 σ k + Δ R q r o w σ k + 1 2 r 3 2 1 σ k ) 1 1 − p R 1 σ k = Φ r + 1 , j ( 1 − p R ) σ k 2 + 3 Δ R q r o w σ k + 1 2 r 2 ( 1 − p R ) σ k 2 . \begin{aligned}
\Phi_{r,j+1}
&=\max_{i\in[n]}\left\lVert e_i^{\top}U_{>k}\Sigma_{>k}^{2r}\mathcal{S}^{(j+1)}\right\rVert_2\\
&=\max_{i\in[n]}\left\lVert\left(e_i^{\top}U_{>k}\Sigma_{>k}^{2(r+1)}\mathcal{S}^{(j)}\Sigma_k^{-1}+e_i^{\top}U_{>k}\Sigma_{>k}^{2r}\Sigma_{>k}L^{(j)}\right)\left(I+F^{(j)}\right)^{-1}\Sigma_k^{-1}\right\rVert_2\\
&\le\max_{i\in[n]}\left(\left\lVert e_i^{\top}U_{>k}\Sigma_{>k}^{2(r+1)}\mathcal{S}^{(j)}\right\rVert_2\left\lVert\Sigma_k^{-1}\right\rVert_2+\Delta R\left\lVert e_i^{\top}U_{>k}\Sigma_{>k}^{2r+1}V_k^{\top}\widetilde{G}^{(j)}\right\rVert_2\left\lVert\left(K^{(j)}\right)^{-1}\right\rVert_2\right)\\
&\qquad\cdot\left\lVert\left(I+F^{(j)}\right)^{-1}\right\rVert_2\left\lVert\Sigma_k^{-1}\right\rVert_2\\
&\le\left(\Phi_{r+1,j}\frac{1}{\sigma_k}+\Delta R q_{\mathrm{row}}\sigma_{k+1}^{2r}\frac{3}{2}\frac{1}{\sigma_k}\right)\frac{1}{1-p_R}\frac{1}{\sigma_k}\\
&=\frac{\Phi_{r+1,j}}{(1-p_R)\sigma_k^2}+\frac{3\Delta Rq_{\mathrm{row}}\sigma_{k+1}^{2r}}{2(1-p_R)\sigma_k^2}.
\end{aligned} Φ r , j + 1 = i ∈ [ n ] max e i ⊤ U > k Σ > k 2 r S ( j + 1 ) 2 = i ∈ [ n ] max ( e i ⊤ U > k Σ > k 2 ( r + 1 ) S ( j ) Σ k − 1 + e i ⊤ U > k Σ > k 2 r Σ > k L ( j ) ) ( I + F ( j ) ) − 1 Σ k − 1 2 ≤ i ∈ [ n ] max ( e i ⊤ U > k Σ > k 2 ( r + 1 ) S ( j ) 2 Σ k − 1 2 + Δ R e i ⊤ U > k Σ > k 2 r + 1 V k ⊤ G ( j ) 2 ( K ( j ) ) − 1 2 ) ⋅ ( I + F ( j ) ) − 1 2 Σ k − 1 2 ≤ ( Φ r + 1 , j σ k 1 + Δ R q row σ k + 1 2 r 2 3 σ k 1 ) 1 − p R 1 σ k 1 = ( 1 − p R ) σ k 2 Φ r + 1 , j + 2 ( 1 − p R ) σ k 2 3Δ R q row σ k + 1 2 r .
In the last inequality, we used the following inequalities: ∥ e i ⊤ U > k Σ > k 2 r + 1 V k ⊤ G ~ ( j ) ∥ 2 ≤ σ k + 1 2 r + 1 q r o w \left\lVert e_i^{\top}U_{>k}\Sigma_{>k}^{2r+1}V_k^{\top}\widetilde{G}^{(j)}\right\rVert_2\le\sigma_{k+1}^{2r+1}q_{\mathrm{row}} e i ⊤ U > k Σ > k 2 r + 1 V k ⊤ G ( j ) 2 ≤ σ k + 1 2 r + 1 q row since G c l i p \mathcal{G}_{\mathrm{clip}} G clip holds, ∥ ( K ( j ) ) − 1 ∥ 2 ≤ 3 2 1 σ k \left\lVert\left(K^{(j)}\right)^{-1}\right\rVert_2\le\frac{3}{2}\frac{1}{\sigma_k} ( K ( j ) ) − 1 2 ≤ 2 3 σ k 1 by Lemma 18, and ∥ ( I + F ( j ) ) − 1 ∥ 2 ≤ 1 1 − p R \left\lVert\left(I+F^{(j)}\right)^{-1}\right\rVert_2\le\frac{1}{1-p_R} ( I + F ( j ) ) − 1 2 ≤ 1 − p R 1 which follows from Lemma 10 since ∥ F ( j ) ∥ 2 ≤ p R < 1 \lVert F^{(j)}\rVert_2\le p_R<1 ∥ F ( j ) ∥ 2 ≤ p R < 1 .
Unrolling the recurrence, we obtain
Φ 0 , t ≤ Φ t , 0 σ k 2 t ( 1 − p R ) t + 3 Δ R q r o w 2 σ k 2 ( 1 − p R ) ∑ j = 0 t − 1 ( 1 − κ k 1 − p R ) j = Φ t , 0 σ k 2 t ( 1 − p R ) t + 3 Δ R q r o w 2 σ k 2 ( κ k − p R ) ( 1 − ( 1 − κ k 1 − p R ) t ) . \begin{aligned}
\Phi_{0,t}\le\frac{\Phi_{t,0}}{\sigma_k^{2t}(1-p_R)^t}+\frac{3\Delta Rq_{\mathrm{row}}}{2\sigma_k^2(1-p_R)}\sum_{j=0}^{t-1}\left(\frac{1-\kappa_k}{1-p_R}\right)^j \\
&=\frac{\Phi_{t,0}}{\sigma_k^{2t}(1-p_R)^t}+\frac{3\Delta Rq_{\mathrm{row}}}{2\sigma_k^2(\kappa_k-p_R)}\left(1-\left(\frac{1-\kappa_k}{1-p_R}\right)^t\right).
\end{aligned} Φ 0 , t ≤ σ k 2 t ( 1 − p R ) t Φ t , 0 + 2 σ k 2 ( 1 − p R ) 3Δ R q row j = 0 ∑ t − 1 ( 1 − p R 1 − κ k ) j = σ k 2 t ( 1 − p R ) t Φ t , 0 + 2 σ k 2 ( κ k − p R ) 3Δ R q row ( 1 − ( 1 − p R 1 − κ k ) t ) .
Since G i n i t \mathcal{G}_{\mathrm{init}} G init holds, we have Φ t , 0 ≤ ω i n i t σ k + 1 2 t \Phi_{t,0} \le\omega_{\mathrm{init}}\sigma_{k+1}^{2t} Φ t , 0 ≤ ω init σ k + 1 2 t . Using this bound together with the assumptions Δ q r o w ≤ 1 8 Δ k \Delta_{q\mathrm{row}} \le\frac{1}{8}\Delta_k Δ q row ≤ 8 1 Δ k and p R ≤ κ k 16 p_R \le\frac{\kappa_k}{16} p R ≤ 16 κ k , we obtain
Φ 0 , t ≤ ω i n i t σ k + 1 2 t σ k 2 t ( 1 − p R ) t + 3 Δ R q r o w 2 σ k 2 ( κ k − p R ) ( 1 − ( 1 − κ k 1 − p R ) t ) = ω i n i t ( 1 − κ k 1 − p R ) t + 3 Δ R q r o w 2 σ k 2 ( κ k − p R ) ( 1 − ( 1 − κ k 1 − p R ) t ) ≤ ω i n i t + 3 Δ R q r o w 2 σ k 2 15 16 κ k = ω i n i t + 8 Δ R q r o w 5 Δ k ≤ ω i n i t + 1 5 R . \begin{aligned}
\Phi_{0,t}
&\le\frac{\omega_{\mathrm{init}}\sigma_{k+1}^{2t}}{\sigma_k^{2t}(1-p_R)^t}
+\frac{3\Delta R q_{\mathrm{row}}}{2\sigma_k^2(\kappa_k-p_R)}
\left(1-\left(\frac{1-\kappa_k}{1-p_R}\right)^t\right) \\
&= \omega_{\mathrm{init}}\left(\frac{1-\kappa_k}{1-p_R}\right)^t
+\frac{3\Delta R q_{\mathrm{row}}}{2\sigma_k^2(\kappa_k-p_R)}
\left(1-\left(\frac{1-\kappa_k}{1-p_R}\right)^t\right) \\
&\le\omega_{\mathrm{init}}+\frac{3\Delta R q_{\mathrm{row}}}{2\sigma_k^2\frac{15}{16}\kappa_k}
= \omega_{\mathrm{init}}+\frac{8\Delta R q_{\mathrm{row}}}{5\Delta_k}
\le\omega_{\mathrm{init}}+\frac{1}{5}R .
\end{aligned} Φ 0 , t ≤ σ k 2 t ( 1 − p R ) t ω init σ k + 1 2 t + 2 σ k 2 ( κ k − p R ) 3Δ R q row ( 1 − ( 1 − p R 1 − κ k ) t ) = ω init ( 1 − p R 1 − κ k ) t + 2 σ k 2 ( κ k − p R ) 3Δ R q row ( 1 − ( 1 − p R 1 − κ k ) t ) ≤ ω init + 2 σ k 2 16 15 κ k 3Δ R q row = ω init + 5 Δ k 8Δ R q row ≤ ω init + 5 1 R .
Combining with Lemma 20 and using that R ≥ max { 15 k ∥ U ∥ ∞ , 5 ω i n i t } R \ge\max\left\{15\sqrt{k}\|U\|_\infty,5\omega_{\mathrm{init}}\right\} R ≥ max { 15 k ∥ U ∥ ∞ , 5 ω init } , we obtain
max i ∈ [ n ] ∥ a i ⊤ X ( t ) B ( t ) ∥ 2 ≤ 2 ( k ∥ U ∥ ∞ + Φ 0 , t ) ≤ 2 ( k ∥ U ∥ ∞ + ω i n i t + 1 5 R ) ≤ 2 ( 1 15 R + 1 5 R + 1 5 R ) ≤ R . \begin{aligned}
\max_{i\in[n]}\left\|a_i^\top X(t)B^{(t)}\right\|_2
\le\sqrt{2}\left(\sqrt{k}\|U\|_\infty+\Phi_{0,t}\right) \\
&\le\sqrt{2}\left(\sqrt{k}\|U\|_\infty+\omega_{\mathrm{init}}+\frac{1}{5}R\right) \\
&\le\sqrt{2}\left(\frac{1}{15}R+\frac{1}{5}R+\frac{1}{5}R\right)\le R .
\end{aligned} i ∈ [ n ] max a i ⊤ X ( t ) B ( t ) 2 ≤ 2 ( k ∥ U ∥ ∞ + Φ 0 , t ) ≤ 2 ( k ∥ U ∥ ∞ + ω init + 5 1 R ) ≤ 2 ( 15 1 R + 5 1 R + 5 1 R ) ≤ R .
and thus clipping is inactive in iteration t t t as well.
Convergence analysis. Next, using Lemmas 19 and 28 , we analyze s t s_t s t and m t m_t m t for all iterations t ≥ 0 t\ge0 t ≥ 0 . We proceed by induction and show that the clipping is inactive and thus we can apply the main recurrence shown in Lemma 19 . Recall that the recurrence states that E ( t + 1 ) = ( Σ ≥ k 2 E ( t ) Σ ≥ k − 2 + L ( t ) ) ( I + F ( t ) ) − 1 E^{(t+1)}=\left(\Sigma_{\ge k}^2E^{(t)}\Sigma_{\ge k}^{-2}+L^{(t)}\right)\left(I+F^{(t)}\right)^{-1} E ( t + 1 ) = ( Σ ≥ k 2 E ( t ) Σ ≥ k − 2 + L ( t ) ) ( I + F ( t ) ) − 1 . The term Σ ≥ k 2 E ( t ) Σ ≥ k − 2 \Sigma_{\ge k}^2E^{(t)}\Sigma_{\ge k}^{-2} Σ ≥ k 2 E ( t ) Σ ≥ k − 2 captures the main progress, and we have ∥ Σ ≥ k 2 E ( t ) Σ ≥ k − 2 ∥ 2 ≤ σ k + 1 2 σ k 2 ∥ E ( t ) ∥ 2 = ( 1 − κ k ) m t \left\|\Sigma_{\ge k}^2E^{(t)}\Sigma_{\ge k}^{-2}\right\|_2\le\frac{\sigma_{k+1}^2}{\sigma_k^2}\left\|E^{(t)}\right\|_2=(1-\kappa_k)m_t Σ ≥ k 2 E ( t ) Σ ≥ k − 2 2 ≤ σ k 2 σ k + 1 2 E ( t ) 2 = ( 1 − κ k ) m t . Thus in an ideal case where L ( t ) L^{(t)} L ( t ) and F ( t ) F^{(t)} F ( t ) are zero, m t m_t m t decreases geometrically in every iteration. In our setting, L ( t ) L^{(t)} L ( t ) and F ( t ) F^{(t)} F ( t ) are non-zero and they depend on the Gaussian noise matrix G ~ ( t ) \widetilde{G}^{(t)} G ( t ) , and our analysis will upper bound their contribution.
We start by showing that the initial solution satisfies s 0 ≤ 1 4 s_0\le\frac{1}{4} s 0 ≤ 4 1 and thus m 0 ≤ 1 15 m_0\le\frac{1}{\sqrt{15}} m 0 ≤ 15 1 , which will provide the base case for the induction.
Lemma 29. Suppose that Δ k ≥ 8 Δ i n i t ( 2 d + 2 log 4 β ) \Delta_k\ge8\Delta_{\mathrm{init}}\left(2\sqrt{d}+\sqrt{2\log\frac{4}{\beta}}\right) Δ k ≥ 8 Δ init ( 2 d + 2 log β 4 ) . With probability at least 1 − β 4 1-\frac{\beta}{4} 1 − 4 β , s 0 ≤ 1 4 s_0\le\frac{1}{4} s 0 ≤ 4 1 .
Proof. Let W = Δ i n i t sym ( G ( i n i t ) ) W=\Delta_{\mathrm{init}}\operatorname{sym}\left(G^{(\mathrm{init})}\right) W = Δ init sym ( G ( init ) ) , and thus the columns of the initial solution X ( 0 ) X^{(0)} X ( 0 ) are the top-k k k eigenvectors X ( 0 ) X^{(0)} X ( 0 ) of M ~ = A ⊤ A + W \widetilde{M}=A^\top A+W M = A ⊤ A + W . By Lemma 5, we have
Pr [ ∥ sym ( G ( i n i t ) ) ∥ 2 > 2 d + 2 log 4 β ] ≤ exp ( − 1 2 ( 2 log 4 β ) 2 ) = β 4 . \Pr\left[\left\|\operatorname{sym}\left(G^{(\mathrm{init})}\right)\right\|_2>2\sqrt{d}+\sqrt{2\log\frac{4}{\beta}}\right]
\le\exp\left(-\frac{1}{2}\left(\sqrt{2\log\frac{4}{\beta}}\right)^2\right)=\frac{\beta}{4}. Pr [ sym ( G ( init ) ) 2 > 2 d + 2 log β 4 ] ≤ exp ( − 2 1 ( 2 log β 4 ) 2 ) = 4 β .
In the following, we condition on the event ∥ sym ( G ( i n i t ) ) ∥ 2 ≤ 2 d + 2 log 4 β \left\|\operatorname{sym}\left(G^{(\mathrm{init})}\right)\right\|_2\le2\sqrt{d}+\sqrt{2\log\frac{4}{\beta}} sym ( G ( init ) ) 2 ≤ 2 d + 2 log β 4 . We have
∥ W ∥ 2 = Δ i n i t ∥ sym ( G ( i n i t ) ) ∥ 2 ≤ Δ i n i t ( 2 d + 2 log 4 β ) ≤ 1 8 Δ k . \|W\|_2=\Delta_{\mathrm{init}}\left\|\operatorname{sym}\left(G^{(\mathrm{init})}\right)\right\|_2
\le\Delta_{\mathrm{init}}\left(2\sqrt{d}+\sqrt{2\log\frac{4}{\beta}}\right)
\le\frac{1}{8}\Delta_k. ∥ W ∥ 2 = Δ init sym ( G ( init ) ) 2 ≤ Δ init ( 2 d + 2 log β 4 ) ≤ 8 1 Δ k .
Thus Lemma 9 gives
s 0 = ∥ ( I − X ( 0 ) X ( 0 ) ⊤ ) V k ∥ 2 ≤ ∥ W ∥ 2 Δ k − ∥ W ∥ 2 ≤ 2 ∥ W ∥ 2 Δ k ≤ 1 4 . s_0=\left\|\left(I-X^{(0)}X^{(0)\top}\right)V_k\right\|_2
\le\frac{\|W\|_2}{\Delta_k-\|W\|_2}
\le\frac{2\|W\|_2}{\Delta_k}
\le\frac{1}{4}. s 0 = ( I − X ( 0 ) X ( 0 ) ⊤ ) V k 2 ≤ Δ k − ∥ W ∥ 2 ∥ W ∥ 2 ≤ Δ k 2∥ W ∥ 2 ≤ 4 1 .
Before giving our inductive analysis, we first show the following lemma which will be used to upper bound one of the error terms in the main recurrence. The lemma follows from a martingale argument. We first introduce some notation. We define a filtration ( F t ) 0 ≤ t ≤ T (\mathcal{F}_t)_{0\le t\le T} ( F t ) 0 ≤ t ≤ T as follows. For t < T t<T t < T , F t \mathcal{F}_t F t includes G ( i n i t ) G^{(\mathrm{init})} G ( init ) , G ( j ) G^{(j)} G ( j ) and G ~ ( j ) \widetilde{G}^{(j)} G ( j ) for all iterations 0 ≤ j < t 0\le j<t 0 ≤ j < t prior to t t t , and G ( t ) G^{(t)} G ( t ) ; G ~ ( t ) \widetilde{G}^{(t)} G ( t ) is not included in F t \mathcal{F}_t F t and is independent of F t \mathcal{F}_t F t . We let F T \mathcal{F}_T F T include the full transcript.
Let τ \tau τ be the following stopping time:
τ : = inf { 0 ≤ t < T : C ( t ) is not invertible , s t > 1 2 , or ∥ Δ sym ( G ( t ) ) ∥ 2 > γ 4 } . (5) \tau:= \inf\left\{0\le t<T:C^{(t)}\text{ is not invertible},s_t>\frac{1}{2},\text{ or }\left\|\Delta\operatorname{sym}\left(G^{(t)}\right)\right\|_2>\frac{\gamma}{4}\right\}. \tag*{(5)} τ := inf { 0 ≤ t < T : C ( t ) is not invertible , s t > 2 1 , or Δ sym ( G ( t ) ) 2 > 4 γ } . (5)
with the convention that inf ∅ = T \inf\varnothing=T inf ∅ = T .
We define
W ( t ) : = { ( K ( t ) ) − 1 t < τ 0 t ≥ τ L ^ ( t ) : = Δ R V > k T G ~ ( t ) W ( t ) . W^{(t)}:=
\begin{cases}
\left(K^{(t)}\right)^{-1} & t<\tau\\
0 & t\ge\tau
\end{cases}
\qquad
\widehat{L}^{(t)}:=\Delta R V_{>k}^{\mathsf{T}}\widetilde{G}^{(t)}W^{(t)}. W ( t ) := { ( K ( t ) ) − 1 0 t < τ t ≥ τ L ( t ) := Δ R V > k T G ( t ) W ( t ) .
Note that L ^ ( t ) = L ( t ) \widehat{L}^{(t)}=L^{(t)} L ( t ) = L ( t ) for t < τ t<\tau t < τ and L ^ ( t ) = 0 \widehat{L}^{(t)}=0 L ( t ) = 0 for t ≥ τ t\ge\tau t ≥ τ .
Finally, for a matrix M M M , we let
T ( M ) : = Σ > k 2 M Σ k − 2 . \mathcal{T}(M):=\Sigma_{>k}^{2}M\Sigma_k^{-2}. T ( M ) := Σ > k 2 M Σ k − 2 .
Lemma 30. Suppose that γ ≤ σ k 2 4 \gamma\le\frac{\sigma_k^2}{4} γ ≤ 4 σ k 2 . Let
ν R : = 3 Δ R σ k 2 d log 9 + 2 log ( 8 T / β ) κ k ( 2 − κ k ) . \nu_R:=\frac{3\Delta R}{\sigma_k}\sqrt{\frac{2d\log9+2\log(8T/\beta)}{\kappa_k(2-\kappa_k)}}. ν R := σ k 3Δ R κ k ( 2 − κ k ) 2 d log 9 + 2 log ( 8 T / β ) . With probability at least 1 − β 4 1-\frac{\beta}{4} 1 − 4 β , simultaneously for all 1 ≤ t ≤ T 1\le t\le T 1 ≤ t ≤ T , we have
∥ ∑ j = 0 t − 1 T t − 1 − j ( L ^ ( j ) ) ∥ 2 ≤ ν R . (6) \left\|\sum_{j=0}^{t-1}\mathcal{T}^{t-1-j}\left(\widehat{L}^{(j)}\right)\right\|_2\le\nu_R. \tag*{(6)} j = 0 ∑ t − 1 T t − 1 − j ( L ( j ) ) 2 ≤ ν R . (6) Let G m a r t \mathcal{G}_{\mathrm{mart}} G mart be this event.
Proof. Fix an iteration 1 ≤ t ≤ T 1\le t\le T 1 ≤ t ≤ T and two unit vectors u ∈ R d − k u\in\mathbb{R}^{d-k} u ∈ R d − k and v ∈ R k v\in\mathbb{R}^{k} v ∈ R k . For every 0 ≤ j < t 0\le j<t 0 ≤ j < t , let
ξ j : = u T T t − 1 − j ( L ^ ( j ) ) v = u T Σ > k 2 ( t − 1 − j ) L ^ ( j ) Σ k − 2 ( t − 1 − j ) v . \xi_j:=u^{\mathsf{T}}\mathcal{T}^{t-1-j}\left(\widehat{L}^{(j)}\right)v=u^{\mathsf{T}}\Sigma_{>k}^{2(t-1-j)}\widehat{L}^{(j)}\Sigma_k^{-2(t-1-j)}v. ξ j := u T T t − 1 − j ( L ( j ) ) v = u T Σ > k 2 ( t − 1 − j ) L ( j ) Σ k − 2 ( t − 1 − j ) v .
Recall that L ^ ( j ) : = Δ R V > k T G ~ ( j ) W ( j ) \widehat{L}^{(j)}:=\Delta R V_{>k}^{\mathsf{T}}\widetilde{G}^{(j)}W^{(j)} L ( j ) := Δ R V > k T G ( j ) W ( j ) and F j \mathcal{F}_j F j includes all random matrices in the iterations prior to j j j and G ( j ) G^{(j)} G ( j ) , but not G ~ ( j ) \widetilde{G}^{(j)} G ( j ) . Thus, conditioned on F j \mathcal{F}_j F j , ξ j \xi_j ξ j is a Gaussian random variable with expectation 0 0 0 and variance
E [ ξ j 2 ∣ F j ] = ( Δ R ) 2 ∥ Σ > k 2 ( t − 1 − j ) u ∥ 2 2 ∥ W ( j ) Σ k − 2 ( t − 1 − j ) v ∥ 2 2 ≤ ( Δ R ) 2 ∥ Σ > k 2 ( t − 1 − j ) ∥ 2 2 ∥ u ∥ 2 2 ∥ W ( j ) ∥ 2 2 ∥ Σ k − 2 ( t − 1 − j ) ∥ 2 2 ∥ v ∥ 2 2 = ( Δ R ) 2 ( σ k + 1 2 σ k 2 ) 2 ( t − 1 − j ) ∥ W ( j ) ∥ 2 2 ≤ ( Δ R ) 2 ( σ k + 1 2 σ k 2 ) 2 ( t − 1 − j ) ( 3 2 σ k ) 2 = ( 3 Δ R 2 σ k ) 2 ( 1 − κ k ) 2 ( t − 1 − j ) . (7) \begin{aligned}
\mathbb{E}\left[\xi_j^2\mid\mathcal{F}_j\right]
=(\Delta R)^2\left\|\Sigma_{>k}^{2(t-1-j)}u\right\|_2^2\left\|W^{(j)}\Sigma_k^{-2(t-1-j)}v\right\|_2^2 \\
&\le(\Delta R)^2\left\|\Sigma_{>k}^{2(t-1-j)}\right\|_2^2\|u\|_2^2\left\|W^{(j)}\right\|_2^2\left\|\Sigma_k^{-2(t-1-j)}\right\|_2^2\|v\|_2^2 \\
&=(\Delta R)^2\left(\frac{\sigma_{k+1}^2}{\sigma_k^2}\right)^{2(t-1-j)}\left\|W^{(j)}\right\|_2^2 \\
&\le(\Delta R)^2\left(\frac{\sigma_{k+1}^2}{\sigma_k^2}\right)^{2(t-1-j)}
\left(\frac{3}{2\sigma_k}\right)^2 \\
&=\left(\frac{3\Delta R}{2\sigma_k}\right)^2(1-\kappa_k)^{2(t-1-j)}. \tag*{(7)}
\end{aligned} E [ ξ j 2 ∣ F j ] = ( Δ R ) 2 Σ > k 2 ( t − 1 − j ) u 2 2 W ( j ) Σ k − 2 ( t − 1 − j ) v 2 2 ≤ ( Δ R ) 2 Σ > k 2 ( t − 1 − j ) 2 2 ∥ u ∥ 2 2 W ( j ) 2 2 Σ k − 2 ( t − 1 − j ) 2 2 ∥ v ∥ 2 2 = ( Δ R ) 2 ( σ k 2 σ k + 1 2 ) 2 ( t − 1 − j ) W ( j ) 2 2 ≤ ( Δ R ) 2 ( σ k 2 σ k + 1 2 ) 2 ( t − 1 − j ) ( 2 σ k 3 ) 2 = ( 2 σ k 3Δ R ) 2 ( 1 − κ k ) 2 ( t − 1 − j ) . (7)
The last inequality follows from Lemma 18: if j < τ j<\tau j < τ , we have W ( j ) = ( K ( j ) ) − 1 W^{(j)}=(K^{(j)})^{-1} W ( j ) = ( K ( j ) ) − 1 and the conditions of Lemma 18 are satisfied; otherwise, W ( j ) = 0 W^{(j)}=0 W ( j ) = 0 .
We can use the subgaussian version of Azuma’s inequality to bound ∑ j ξ j \sum_j \xi_j ∑ j ξ j (see e.g. Duchi et al. [2012, Lemma F.4] or Howard et al. [2020, Theorem 1(b)]) with ψ ( λ ) = λ 2 / 2 \psi(\lambda)=\lambda^2/2 ψ ( λ ) = λ 2 /2 . For every x ≥ 0 x\geq0 x ≥ 0 , we have
Pr [ ∣ ∑ j = 0 t − 1 ξ j ∣ ≥ x ] ≤ 2 exp ( − x 2 2 ∑ j = 0 t − 1 ( 3 Δ R 2 σ κ ) 2 ( 1 − κ k ) 2 ( t − 1 − j ) ) \Pr\left[\left\lvert\sum_{j=0}^{t-1}\xi_j\right\rvert\geq x\right]\leq2\exp\left(-\frac{x^2}{2\sum_{j=0}^{t-1}\left(\frac{3\Delta R}{2\sigma_\kappa}\right)^2(1-\kappa_k)^{2(t-1-j)}}\right) Pr [ j = 0 ∑ t − 1 ξ j ≥ x ] ≤ 2 exp − 2 ∑ j = 0 t − 1 ( 2 σ κ 3Δ R ) 2 ( 1 − κ k ) 2 ( t − 1 − j ) x 2
≤ 2 exp ( − x 2 κ k ( 2 − κ k ) 2 ( 3 Δ R 2 σ κ ) 2 ) . \leq2\exp\left(-\frac{x^2\kappa_k(2-\kappa_k)}{2\left(\frac{3\Delta R}{2\sigma_\kappa}\right)^2}\right). ≤ 2 exp − 2 ( 2 σ κ 3Δ R ) 2 x 2 κ k ( 2 − κ k ) .
We can use the above guarantee to upper bound ∥ ∑ j = 0 t − 1 T t − 1 − j ( L ^ ( j ) ) ∥ 2 \left\lVert\sum_{j=0}^{t-1}\mathcal{T}^{t-1-j}\left(\widehat{L}^{(j)}\right)\right\rVert_2 ∑ j = 0 t − 1 T t − 1 − j ( L ( j ) ) 2 using a net argument, similarly to the proof of Claim 28 and Claim 26 . Let N 1 \mathcal{N}_1 N 1 and N 2 \mathcal{N}_2 N 2 be 1 4 \frac{1}{4} 4 1 -nets of the two unit spheres in R d − k \mathbb{R}^{d-k} R d − k and R k \mathbb{R}^k R k respectively, of sizes ∣ N 1 ∣ ≤ 9 d − k |\mathcal{N}_1|\leq9^{d-k} ∣ N 1 ∣ ≤ 9 d − k and ∣ N 2 ∣ ≤ 9 k |\mathcal{N}_2|\leq9^k ∣ N 2 ∣ ≤ 9 k . For any ( d − k ) × k (d-k)\times k ( d − k ) × k matrix M M M , we have ∥ M ∥ 2 ≤ 2 max u ∈ N 1 , v ∈ N 2 ∣ u T M v ∣ \|M\|_2\leq2\max_{u\in\mathcal{N}_1,v\in\mathcal{N}_2}|u^\mathsf{T}Mv| ∥ M ∥ 2 ≤ 2 max u ∈ N 1 , v ∈ N 2 ∣ u T M v ∣ . Indeed, let u ∗ , v ∗ u^*,v^* u ∗ , v ∗ be unit left and right top singular vectors of M M M and let u , v u,v u , v be the closest vectors in the nets to u ∗ , v ∗ u^*,v^* u ∗ , v ∗ . We have
∥ M ∥ 2 = u ∗ T M v ∗ = u ∗ T M ( v ∗ − v ) + ( u ∗ T − u T ) M v + u T M v ≤ 1 4 ∥ M ∥ 2 + 1 4 ∥ M ∥ 2 + u T M v . \|M\|_2=u^{*\mathsf{T}}Mv^*=u^{*\mathsf{T}}M(v^*-v)+(u^{*\mathsf{T}}-u^\mathsf{T})Mv+u^\mathsf{T}Mv\leq\frac{1}{4}\|M\|_2+\frac{1}{4}\|M\|_2+u^\mathsf{T}Mv. ∥ M ∥ 2 = u ∗ T M v ∗ = u ∗ T M ( v ∗ − v ) + ( u ∗ T − u T ) M v + u T M v ≤ 4 1 ∥ M ∥ 2 + 4 1 ∥ M ∥ 2 + u T M v .
By rearranging, we obtain the claimed inequality.
There are ∣ N 1 ∣ ∣ N 2 ∣ T = 9 d T |\mathcal{N}_1||\mathcal{N}_2|T=9^dT ∣ N 1 ∣∣ N 2 ∣ T = 9 d T triples ( u , v , t ) (u,v,t) ( u , v , t ) with u ∈ N 1 u\in\mathcal{N}_1 u ∈ N 1 , v ∈ N 2 v\in\mathcal{N}_2 v ∈ N 2 , and 1 ≤ t ≤ T 1\leq t\leq T 1 ≤ t ≤ T . Setting x = 3 Δ R 2 σ κ 2 log ( 9 ⋅ 9 d T β ) κ k ( 2 − κ k ) x=\frac{3\Delta R}{2\sigma_\kappa}\sqrt{\frac{2\log\left(\frac{9\cdot9^dT}{\beta}\right)}{\kappa_k(2-\kappa_k)}} x = 2 σ κ 3Δ R κ k ( 2 − κ k ) 2 l o g ( β 9 ⋅ 9 d T ) in the guarantee above and taking a union bound over all 9 d T 9^dT 9 d T triples, we obtain the following guarantee. With probability at least 1 − β 4 1-\frac{\beta}{4} 1 − 4 β , simultaneously for all 1 ≤ t ≤ T 1\leq t\leq T 1 ≤ t ≤ T ,
∥ ∑ j = 0 t − 1 T t − 1 − j ( L ^ ( j ) ) ∥ 2 ≤ 2 max u ∈ N 1 , v ∈ N 2 ∣ u T M v ∣ ≤ 3 Δ R σ κ 2 log ( 8 ⋅ 9 d T β ) κ k ( 2 − κ k ) = ν R , \begin{aligned}
\left\lVert\sum_{j=0}^{t-1}\mathcal{T}^{t-1-j}\left(\widehat{L}^{(j)}\right)\right\rVert_2\leq2\max_{u\in\mathcal{N}_1,v\in\mathcal{N}_2}|u^\mathsf{T}Mv| \\
&\leq\frac{3\Delta R}{\sigma_\kappa}\sqrt{\frac{2\log\left(\frac{8\cdot9^dT}{\beta}\right)}{\kappa_k(2-\kappa_k)}}=\nu_R,
\end{aligned} j = 0 ∑ t − 1 T t − 1 − j ( L ( j ) ) 2 ≤ 2 u ∈ N 1 , v ∈ N 2 max ∣ u T M v ∣ ≤ σ κ 3Δ R κ k ( 2 − κ k ) 2 log ( β 8 ⋅ 9 d T ) = ν R ,
as needed.
We now give the main inductive argument. Recall that G c l i p \mathcal{G}_{\mathrm{clip}} G clip is the event defined in Definition 15 , and G i n i t \mathcal{G}_{\mathrm{init}} G init and G m a r t \mathcal{G}_{\mathrm{mart}} G mart are the events defined in Lemma 21 and Lemma 30 respectively. Recall also that ρ R : = 3 Δ R κ k 2 σ κ \rho_R:=\frac{3\Delta R\kappa_k}{2\sigma_\kappa} ρ R := 2 σ κ 3Δ R κ k (see Lemma 28 ) and ν R : = 3 Δ R σ κ 2 d log 9 + 2 log ( 8 T / β ) κ k ( 2 − κ k ) \nu_R:=\frac{3\Delta R}{\sigma_\kappa}\sqrt{\frac{2d\log9+2\log(8T/\beta)}{\kappa_k(2-\kappa_k)}} ν R := σ κ 3Δ R κ k ( 2 − κ k ) 2 d l o g 9 + 2 l o g ( 8 T / β ) (see Lemma 30).
Theorem 31. Suppose that we have Δ k ≥ max { 8 Δ i n i t ( 2 d + 2 log 8 β ) , 8 Δ g r o w } \Delta_k\geq\max\left\{8\Delta_{\mathrm{init}}\left(2\sqrt{d}+\sqrt{2\log\frac{8}{\beta}}\right),8\Delta_{\mathrm{grow}}\right\} Δ k ≥ max { 8 Δ init ( 2 d + 2 log β 8 ) , 8 Δ grow } , γ ≤ σ k 2 4 \gamma\leq\frac{\sigma_k^2}{4} γ ≤ 4 σ k 2 , ρ R ≤ κ k 16 \rho_R\leq\frac{\kappa_k}{16} ρ R ≤ 16 κ k , R ≥ 15 ∥ U ∥ ∞ ( k + log ( 16 n ( T + 1 ) β ) ) R\geq15\|U\|_\infty\left(\sqrt{k}+\sqrt{\log\left(\frac{16n(T+1)}{\beta}\right)}\right) R ≥ 15∥ U ∥ ∞ ( k + log ( β 16 n ( T + 1 ) ) ) , and ν R ≤ 1 8 \nu_R\leq\frac{1}{8} ν R ≤ 8 1 . Suppose that G c l i p \mathcal{G}_{\mathrm{clip}} G clip , G i n i t \mathcal{G}_{\mathrm{init}} G init , G m a r t \mathcal{G}_{\mathrm{mart}} G mart , and the event in Lemma 21 and Lemma 29 hold. Then, in every iteration, C ( t ) C^{(t)} C ( t ) is invertible, clipping is inactive, and
s t ≤ m t ≤ ( 1 − κ k 1 − ρ R ) t m 0 + 2 ν R ≤ ( 1 − κ k 2 ) t m 0 + 2 ν R . (8) s_t\leq m_t\leq\left(\frac{1-\kappa_k}{1-\rho_R}\right)^t m_0+2\nu_R\leq\left(1-\frac{\kappa_k}{2}\right)^t m_0+2\nu_R. \tag*{(8)} s t ≤ m t ≤ ( 1 − ρ R 1 − κ k ) t m 0 + 2 ν R ≤ ( 1 − 2 κ k ) t m 0 + 2 ν R . (8)
Proof. In the following, we will use the definition of the stopping time τ \tau τ (eq. (5) ) and the identity s t = m t 1 + m t 2 s_t=\frac{m_t}{\sqrt{1+m_t^2}} s t = 1 + m t 2 m t (Lemma 14 ). We will prove by induction on t ≥ 0 t\geq0 t ≥ 0 that C ( j ) C^{(j)} C ( j ) is invertible and m j ≤ 1 3 m_j\leq\frac{1}{\sqrt{3}} m j ≤ 3 1 (and thus s j ≤ 1 2 s_j \le\frac{1}{2} s j ≤ 2 1 for all j ≤ t j \le t j ≤ t , clipping is inactive in all iterations j < t j < t j < t , τ > t \tau> t τ > t (if t = T t = T t = T , τ = T \tau= T τ = T ), and (7) holds for all iterations j ≤ t j \le t j ≤ t . Consider the base case t = 0 t = 0 t = 0 . Since the event in Lemma 21 and Lemma 29 holds, we have s 0 ≤ 1 4 s_0 \le\frac{1}{4} s 0 ≤ 4 1 and thus m 0 ≤ 1 15 m_0 \le\frac{1}{\sqrt{15}} m 0 ≤ 15 1 . Since s 0 < 1 s_0 < 1 s 0 < 1 , Lemma 13 implies that C ( 0 ) C^{(0)} C ( 0 ) is invertible. Additionally, τ > 0 \tau> 0 τ > 0 and (7) holds for t = 0 t = 0 t = 0 .
Next, we consider the inductive step. Consider t ≥ 0 t \ge0 t ≥ 0 and suppose that C ( j ) C^{(j)} C ( j ) is invertible and m j ≤ 1 3 m_j \le\frac{1}{\sqrt{3}} m j ≤ 3 1 (and thus s j ≤ 1 2 s_j \le\frac{1}{2} s j ≤ 2 1 ) for all j ≤ t j \le t j ≤ t , and clipping is inactive in all iterations j < t j < t j < t , and τ ≥ t \tau\ge t τ ≥ t . These assumptions and the assumptions in the theorem statement imply that Lemma 28 applies and thus clipping is inactive in iteration t t t as well.
For each j ≤ t j \le t j ≤ t , we can show that F ( j ) : = Δ R V k ⊤ G ~ ( j ) ( K ( j ) ) − 1 F^{(j)} := \Delta R V_k^\top\widetilde{G}^{(j)}\left(K^{(j)}\right)^{-1} F ( j ) := Δ R V k ⊤ G ( j ) ( K ( j ) ) − 1 satisfies ∥ F ( j ) ∥ 2 ≤ p R \left\|F^{(j)}\right\|_2 \le p_R F ( j ) 2 ≤ p R as follows. We have
∥ F ( j ) ∥ 2 = ∥ Δ R V k ⊤ G ~ ( j ) ( K ( j ) ) − 1 ∥ 2 ≤ Δ R ∥ V k ⊤ G ~ ( j ) ∥ 2 ∥ ( K ( j ) ) − 1 ∥ 2 ≤ Δ R q k 3 2 σ k = p R . \begin{aligned}
\left\|F^{(j)}\right\|_2 = \left\|\Delta R V_k^\top\widetilde{G}^{(j)}\left(K^{(j)}\right)^{-1}\right\|_2 \\
&\le\Delta R \left\|V_k^\top\widetilde{G}^{(j)}\right\|_2 \left\|\left(K^{(j)}\right)^{-1}\right\|_2 \\
&\le\Delta R q_k \frac{3}{2\sigma_k} = p_R.
\end{aligned} F ( j ) 2 = Δ R V k ⊤ G ( j ) ( K ( j ) ) − 1 2 ≤ Δ R V k ⊤ G ( j ) 2 ( K ( j ) ) − 1 2 ≤ Δ R q k 2 σ k 3 = p R .
In the last inequality, we used that G c l i p \mathcal{G}_{\mathrm{clip}} G clip holds (and thus ∥ V k ⊤ G ~ ( j ) ∥ 2 ≤ q k \left\|V_k^\top\widetilde{G}^{(j)}\right\|_2 \le q_k V k ⊤ G ( j ) 2 ≤ q k ) and Lemma 18. Note that the conditions of Lemma 18 hold: we have ∥ Δ sym ( G ( j ) ) ∥ 2 ≤ γ 4 \left\|\Delta\operatorname{sym}\left(G^{(j)}\right)\right\|_2 \le\frac{\gamma}{4} Δ sym ( G ( j ) ) 2 ≤ 4 γ since G c l i p \mathcal{G}_{\mathrm{clip}} G clip holds, γ ≤ σ k 2 4 \gamma\le\frac{\sigma_k^2}{4} γ ≤ 4 σ k 2 , and s j ≤ 1 2 s_j \le\frac{1}{2} s j ≤ 2 1 for all j ≤ t j \le t j ≤ t .
For all 0 ≤ j ≤ t 0 \le j \le t 0 ≤ j ≤ t , we have ∥ F ( j ) ∥ 2 ≤ p R ≤ κ k 16 < 1 \left\|F^{(j)}\right\|_2 \le p_R \le\frac{\kappa_k}{16} < 1 F ( j ) 2 ≤ p R ≤ 16 κ k < 1 , C ( j ) C^{(j)} C ( j ) is invertible, and clipping is inactive in iteration j j j . Thus Lemma 19 gives that C ( t + 1 ) C^{(t+1)} C ( t + 1 ) is invertible and the following recurrence holds for all 0 ≤ j ≤ t 0 \le j \le t 0 ≤ j ≤ t :
E ( j + 1 ) = ( Σ > k 2 E ( j ) Σ k − 2 + L ( j ) ) ( I + F ( j ) ) − 1 = ( T ( E ( j ) ) + L ( j ) ) ( I + F ( j ) ) − 1 (9) E^{(j+1)} = \left(\Sigma_{>k}^2 E^{(j)} \Sigma_k^{-2} + L^{(j)}\right)\left(I+F^{(j)}\right)^{-1} = \left(\mathcal{T}\left(E^{(j)}\right)+L^{(j)}\right)\left(I+F^{(j)}\right)^{-1} \tag*{(9)} E ( j + 1 ) = ( Σ > k 2 E ( j ) Σ k − 2 + L ( j ) ) ( I + F ( j ) ) − 1 = ( T ( E ( j ) ) + L ( j ) ) ( I + F ( j ) ) − 1 (9)
where we used the notation T ( M ) = Σ > k 2 M Σ k − 2 \mathcal{T}(M)=\Sigma_{>k}^2M\Sigma_k^{-2} T ( M ) = Σ > k 2 M Σ k − 2 defined earlier. Using the above recurrence, we obtain the following.
Claim 32. Let
b R : = Δ R q d − k 3 2 σ k η R : = ν R + b R p R κ k ( 1 − p R ) h R : = ( 1 − κ k ) p R 1 − p R . b_R := \Delta Rq_{d-k}\frac{3}{2\sigma_k}
\qquad
\eta_R := \nu_R+\frac{b_Rp_R}{\kappa_k(1-p_R)}
\qquad
h_R := \frac{(1-\kappa_k)p_R}{1-p_R}. b R := Δ R q d − k 2 σ k 3 η R := ν R + κ k ( 1 − p R ) b R p R h R := 1 − p R ( 1 − κ k ) p R .
For every 1 ≤ ℓ ≤ t + 1 1 \le\ell\le t+1 1 ≤ ℓ ≤ t + 1 , we have
m ℓ ≤ ( 1 − κ k ) ℓ m 0 + η R + h R ∑ j = 0 ℓ − 1 ( 1 − κ k ) ℓ − 1 − j m j . (10) m_\ell\le(1-\kappa_k)^\ell m_0+\eta_R+h_R\sum_{j=0}^{\ell-1}(1-\kappa_k)^{\ell-1-j}m_j. \tag*{(10)} m ℓ ≤ ( 1 − κ k ) ℓ m 0 + η R + h R j = 0 ∑ ℓ − 1 ( 1 − κ k ) ℓ − 1 − j m j . (10)
Proof. Using the identity ( I + F ) − 1 = I − F ( I + F ) − 1 (I+F)^{-1}=I-F(I+F)^{-1} ( I + F ) − 1 = I − F ( I + F ) − 1 , we can write the recurrence (9) as follows:
E ( j + 1 ) = T ( E ( j ) ) + L ( j ) − ( T ( E ( j ) ) + L ( j ) ) F ( j ) ( I + F ( j ) ) − 1 ⏟ : = R ( j ) = T ( E ( j ) ) + L ( j ) + R ( j ) . (11) E^{(j+1)}=\mathcal{T}\left(E^{(j)}\right)+L^{(j)}-\underbrace{\left(\mathcal{T}\left(E^{(j)}\right)+L^{(j)}\right)F^{(j)}\left(I+F^{(j)}\right)^{-1}}_{:=R^{(j)}}=\mathcal{T}\left(E^{(j)}\right)+L^{(j)}+R^{(j)}. \tag*{(11)} E ( j + 1 ) = T ( E ( j ) ) + L ( j ) − := R ( j ) ( T ( E ( j ) ) + L ( j ) ) F ( j ) ( I + F ( j ) ) − 1 = T ( E ( j ) ) + L ( j ) + R ( j ) . (11)
We rearranged the recurrence in order to separate the noise terms depending on G ~ ( j ) \widetilde{G}^{(j)} G ( j ) into the term L ( j ) L^{(j)} L ( j ) that is linear in G ~ ( j ) \widetilde{G}^{(j)} G ( j ) and a term R ( j ) R^{(j)} R ( j ) which has a more complex dependency on G ~ ( j ) \widetilde{G}^{(j)} G ( j ) . This separation allows us to bound the contribution of the linear terms L ( j ) L^{(j)} L ( j ) via the martingale argument in Lemma 30. The terms R ( j ) R^{(j)} R ( j ) are more complex, and we will bound their contribution by bounding the norm of each R ( j ) R^{(j)} R ( j ) separately.
For each 1 ≤ ℓ ≤ t + 1 1 \le\ell\le t+1 1 ≤ ℓ ≤ t + 1 , by unrolling the recurrence (11) and using that L ( j ) = L ~ ( j ) L^{(j)}=\widetilde{L}^{(j)} L ( j ) = L ( j ) since τ > t \tau>t τ > t , we obtain
E ( ℓ ) = T ℓ ( E ( 0 ) ) + ∑ j = 0 ℓ − 1 T ℓ − 1 − j ( L ( j ) ) + ∑ j = 0 ℓ − 1 T ℓ − 1 − j ( R ( j ) ) = T ℓ ( E ( 0 ) ) + ∑ j = 0 ℓ − 1 T ℓ − 1 − j ( L ^ ( j ) ) + ∑ j = 0 ℓ − 1 T ℓ − 1 − j ( R ( j ) ) . \begin{aligned}
E^{(\ell)}=\mathcal{T}^{\ell}\left(E^{(0)}\right)+\sum_{j=0}^{\ell-1}\mathcal{T}^{\ell-1-j}\left(L^{(j)}\right)+\sum_{j=0}^{\ell-1}\mathcal{T}^{\ell-1-j}\left(R^{(j)}\right) \\
&= \mathcal{T}^{\ell}\left(E^{(0)}\right) + \sum_{j=0}^{\ell-1} \mathcal{T}^{\ell-1-j}\left(\widehat{L}^{(j)}\right) + \sum_{j=0}^{\ell-1} \mathcal{T}^{\ell-1-j}\left(R^{(j)}\right).
\end{aligned} E ( ℓ ) = T ℓ ( E ( 0 ) ) + j = 0 ∑ ℓ − 1 T ℓ − 1 − j ( L ( j ) ) + j = 0 ∑ ℓ − 1 T ℓ − 1 − j ( R ( j ) ) = T ℓ ( E ( 0 ) ) + j = 0 ∑ ℓ − 1 T ℓ − 1 − j ( L ( j ) ) + j = 0 ∑ ℓ − 1 T ℓ − 1 − j ( R ( j ) ) .
Thus, for each 1 ≤ ℓ ≤ t + 1 1 \le\ell\le t+1 1 ≤ ℓ ≤ t + 1 , we have
m ℓ = ∥ E ( ℓ ) ∥ 2 ≤ ∥ T ℓ ( E ( 0 ) ) ∥ 2 + ∥ ∑ j = 0 ℓ − 1 T ℓ − 1 − j ( L ^ ( j ) ) ∥ 2 + ∑ j = 0 ℓ − 1 ∥ T ℓ − 1 − j ( R ( j ) ) ∥ 2 . m_{\ell} = \left\|E^{(\ell)}\right\|_{2} \le\left\|\mathcal{T}^{\ell}\left(E^{(0)}\right)\right\|_{2} + \left\|\sum_{j=0}^{\ell-1}\mathcal{T}^{\ell-1-j}\left(\widehat{L}^{(j)}\right)\right\|_{2} + \sum_{j=0}^{\ell-1}\left\|\mathcal{T}^{\ell-1-j}\left(R^{(j)}\right)\right\|_{2}. m ℓ = E ( ℓ ) 2 ≤ T ℓ ( E ( 0 ) ) 2 + j = 0 ∑ ℓ − 1 T ℓ − 1 − j ( L ( j ) ) 2 + j = 0 ∑ ℓ − 1 T ℓ − 1 − j ( R ( j ) ) 2 .
We have
∥ T ℓ ( E ( 0 ) ) ∥ 2 = ∥ ∑ k 2 ℓ Σ k 2 ℓ E ( 0 ) Σ k − 2 ℓ ∥ 2 ≤ ∥ ∑ k 2 ℓ Σ k 2 ℓ ∥ 2 ∥ E ( 0 ) ∥ 2 ∥ ∑ k − 2 ℓ Σ k − 2 ℓ ∥ 2 = σ k + 1 2 ℓ σ k 2 ℓ m 0 = ( 1 − κ k ) ℓ m 0 . \left\|\mathcal{T}^{\ell}\left(E^{(0)}\right)\right\|_{2} = \left\|\sum_{k}^{2\ell} \Sigma_{k}^{2\ell} E^{(0)} \Sigma_{k}^{-2\ell}\right\|_{2}
\le\left\|\sum_{k}^{2\ell}\Sigma_{k}^{2\ell}\right\|_{2}\left\|E^{(0)}\right\|_{2}\left\|\sum_{k}^{-2\ell}\Sigma_{k}^{-2\ell}\right\|_{2}
= \frac{\sigma_{k+1}^{2\ell}}{\sigma_{k}^{2\ell}}m_{0} = (1-\kappa_{k})^{\ell}m_{0}. T ℓ ( E ( 0 ) ) 2 = k ∑ 2 ℓ Σ k 2 ℓ E ( 0 ) Σ k − 2 ℓ 2 ≤ k ∑ 2 ℓ Σ k 2 ℓ 2 E ( 0 ) 2 k ∑ − 2 ℓ Σ k − 2 ℓ 2 = σ k 2 ℓ σ k + 1 2 ℓ m 0 = ( 1 − κ k ) ℓ m 0 .
Since G m a r t \mathcal{G}_{\mathrm{mart}} G mart holds, for all 1 ≤ ℓ ≤ t + 1 1 \le\ell\le t+1 1 ≤ ℓ ≤ t + 1 , we have
∥ ∑ j = 0 ℓ − 1 T ℓ − 1 − j ( L ^ ( j ) ) ∥ 2 ≤ ν R . \left\|\sum_{j=0}^{\ell-1}\mathcal{T}^{\ell-1-j}\left(\widehat{L}^{(j)}\right)\right\|_{2} \le\nu_{R}. j = 0 ∑ ℓ − 1 T ℓ − 1 − j ( L ( j ) ) 2 ≤ ν R .
We have
∥ R ( j ) ∥ 2 = ∥ − ( T ( E ( j ) ) + L ( j ) ) F ( j ) ( I + F ( j ) ) − 1 ∥ 2 ≤ ( ∥ T ( E ( j ) ) ∥ 2 + ∥ L ( j ) ∥ 2 ) ∥ F ( j ) ∥ 2 ∥ ( I + F ( j ) ) − 1 ∥ 2 . \left\|R^{(j)}\right\|_{2} = \left\|-\left(\mathcal{T}\left(E^{(j)}\right)+L^{(j)}\right)F^{(j)}\left(I+F^{(j)}\right)^{-1}\right\|_{2}
\le\left(\left\|\mathcal{T}\left(E^{(j)}\right)\right\|_{2}+\left\|L^{(j)}\right\|_{2}\right)\left\|F^{(j)}\right\|_{2}\left\|\left(I+F^{(j)}\right)^{-1}\right\|_{2}. R ( j ) 2 = − ( T ( E ( j ) ) + L ( j ) ) F ( j ) ( I + F ( j ) ) − 1 2 ≤ ( T ( E ( j ) ) 2 + L ( j ) 2 ) F ( j ) 2 ( I + F ( j ) ) − 1 2 .
Additionally,
∥ T ( E ( j ) ) ∥ 2 = ∥ ∑ k 2 Σ k E ( j ) Σ k − 2 ∥ 2 ≤ ∥ ∑ k 2 Σ k ∥ 2 ∥ E ( j ) ∥ 2 ∥ ∑ k − 2 ∥ 2 = σ k + 1 2 σ k 2 m j = ( 1 − κ k ) m j . \left\|\mathcal{T}\left(E^{(j)}\right)\right\|_{2} = \left\|\sum_{k}^{2} \Sigma_{k} E^{(j)} \Sigma_{k}^{-2}\right\|_{2} \le\left\|\sum_{k}^{2}\Sigma_{k}\right\|_{2}\left\|E^{(j)}\right\|_{2}\left\|\sum_{k}^{-2}\right\|_{2}
= \frac{\sigma_{k+1}^{2}}{\sigma_{k}^{2}}m_{j} = (1-\kappa_{k})m_{j}. T ( E ( j ) ) 2 = k ∑ 2 Σ k E ( j ) Σ k − 2 2 ≤ k ∑ 2 Σ k 2 E ( j ) 2 k ∑ − 2 2 = σ k 2 σ k + 1 2 m j = ( 1 − κ k ) m j .
Using that ∥ V > k T G ~ ( t ) ∥ 2 ≤ q d − k \left\|V_{>k}^{\mathsf{T}}\widetilde{G}^{(t)}\right\|_{2} \le q_{d-k} V > k T G ( t ) 2 ≤ q d − k (since G c l i p \mathcal{G}_{\mathrm{clip}} G clip holds) and Lemma 18, we obtain
∥ L ( j ) ∥ 2 = Δ R ∥ V > k G ~ ( j ) ( K ( j ) ) − 1 ∥ 2 ≤ Δ R ∥ V > k G ~ ( j ) ∥ 2 ∥ ( K ( j ) ) − 1 ∥ 2 ≤ Δ R q d − k 3 2 σ k = : b R . \left\|L^{(j)}\right\|_{2} = \Delta R\left\|V_{>k}\widetilde{G}^{(j)}\left(K^{(j)}\right)^{-1}\right\|_{2}
\le\Delta R\left\|V_{>k}\widetilde{G}^{(j)}\right\|_{2}\left\|\left(K^{(j)}\right)^{-1}\right\|_{2}
\le\Delta R q_{d-k}\frac{3}{2\sigma_{k}}
=: b_{R}. L ( j ) 2 = Δ R V > k G ( j ) ( K ( j ) ) − 1 2 ≤ Δ R V > k G ( j ) 2 ( K ( j ) ) − 1 2 ≤ Δ R q d − k 2 σ k 3 =: b R .
As shown earlier, we have ∥ F ( j ) ∥ 2 ≤ p R \left\|F^{(j)}\right\|_{2} \le p_{R} F ( j ) 2 ≤ p R . Since ∥ F ( j ) ∥ 2 ≤ p R ≤ κ k 16 < 1 \left\|F^{(j)}\right\|_{2} \le p_{R} \le\frac{\kappa_{k}}{16} < 1 F ( j ) 2 ≤ p R ≤ 16 κ k < 1 , Lemma 10 gives ∥ ( I + F ( j ) ) − 1 ∥ 2 ≤ 1 1 − p R \left\|\left(I+F^{(j)}\right)^{-1}\right\|_{2} \le\frac{1}{1-p_{R}} ( I + F ( j ) ) − 1 2 ≤ 1 − p R 1 .
Putting everything together, we obtain
∥ R ( j ) ∥ 2 ≤ ( ( 1 − κ k ) m j + b R ) p R 1 − p R = : h R m j + f R , \left\|R^{(j)}\right\|_{2} \le\left((1-\kappa_{k})m_{j}+b_{R}\right)\frac{p_{R}}{1-p_{R}}=: h_{R}m_{j}+f_{R}, R ( j ) 2 ≤ ( ( 1 − κ k ) m j + b R ) 1 − p R p R =: h R m j + f R ,
where we let h R : = ( 1 − κ k ) p R 1 − p R h_R := \frac{(1-\kappa_k)p_R}{1-p_R} h R := 1 − p R ( 1 − κ k ) p R and f R : = b R p R 1 − p R f_R := \frac{b_Rp_R}{1-p_R} f R := 1 − p R b R p R .
Thus
∥ T ℓ − 1 − j ( R ( j ) ) ∥ 2 = ∥ Σ > k 2 ( ℓ − 1 − j ) R ( j ) Σ k − 2 ( ℓ − 1 − j ) ∥ 2 ≤ ∥ Σ > k 2 ( ℓ − 1 − j ) ∥ 2 ∥ R ( j ) ∥ 2 ∥ Σ k − 2 ( ℓ − 1 − j ) ∥ 2 ≤ σ k + 1 2 ( ℓ − 1 − j ) σ k 2 ( ℓ − 1 − j ) ∥ R ( j ) ∥ 2 = ( 1 − κ k ) ℓ − 1 − j ∥ R ( j ) ∥ 2 ≤ ( 1 − κ k ) ℓ − 1 − j ( h R m j + f R ) \begin{aligned}
\left\|\mathcal{T}^{\ell-1-j}\left(R^{(j)}\right)\right\|_2
= \left\|\Sigma_{>k}^{2(\ell-1-j)}R^{(j)}\Sigma_k^{-2(\ell-1-j)}\right\|_2 \\
&\le\left\|\Sigma_{>k}^{2(\ell-1-j)}\right\|_2\left\|R^{(j)}\right\|_2\left\|\Sigma_k^{-2(\ell-1-j)}\right\|_2 \\
&\le\frac{\sigma_{k+1}^{2(\ell-1-j)}}{\sigma_k^{2(\ell-1-j)}}\left\|R^{(j)}\right\|_2 \\
&= (1-\kappa_k)^{\ell-1-j}\left\|R^{(j)}\right\|_2 \\
&\le(1-\kappa_k)^{\ell-1-j}\left(h_Rm_j+f_R\right)
\end{aligned} T ℓ − 1 − j ( R ( j ) ) 2 = Σ > k 2 ( ℓ − 1 − j ) R ( j ) Σ k − 2 ( ℓ − 1 − j ) 2 ≤ Σ > k 2 ( ℓ − 1 − j ) 2 R ( j ) 2 Σ k − 2 ( ℓ − 1 − j ) 2 ≤ σ k 2 ( ℓ − 1 − j ) σ k + 1 2 ( ℓ − 1 − j ) R ( j ) 2 = ( 1 − κ k ) ℓ − 1 − j R ( j ) 2 ≤ ( 1 − κ k ) ℓ − 1 − j ( h R m j + f R )
It follows that
m ℓ ≤ ( 1 − κ k ) ℓ m 0 + ν R + ∑ j = 0 ℓ − 1 ( 1 − κ k ) ℓ − 1 − j ( h R m j + f R ) ≤ ( 1 − κ k ) ℓ m 0 + ν R + f R κ k + h R ∑ j = 0 ℓ − 1 ( 1 − κ k ) ℓ − 1 − j m j \begin{aligned}
m_\ell\le(1-\kappa_k)^\ell m_0+\nu_R+\sum_{j=0}^{\ell-1}(1-\kappa_k)^{\ell-1-j}\left(h_Rm_j+f_R\right) \\
&\le(1-\kappa_k)^\ell m_0+\nu_R+\frac{f_R}{\kappa_k}+h_R\sum_{j=0}^{\ell-1}(1-\kappa_k)^{\ell-1-j}m_j
\end{aligned} m ℓ ≤ ( 1 − κ k ) ℓ m 0 + ν R + j = 0 ∑ ℓ − 1 ( 1 − κ k ) ℓ − 1 − j ( h R m j + f R ) ≤ ( 1 − κ k ) ℓ m 0 + ν R + κ k f R + h R j = 0 ∑ ℓ − 1 ( 1 − κ k ) ℓ − 1 − j m j
as needed.
We can prove by induction that 9 implies the following.
Claim 33. For every 1 ≤ ℓ ≤ t + 1 1\le\ell\le t+1 1 ≤ ℓ ≤ t + 1 , we have
m ℓ ≤ ( 1 − κ k 1 − p R ) ℓ m 0 + κ k κ k − h R η R . (12) m_\ell\le\left(\frac{1-\kappa_k}{1-p_R}\right)^\ell m_0+\frac{\kappa_k}{\kappa_k-h_R}\eta_R. \tag*{(12)} m ℓ ≤ ( 1 − p R 1 − κ k ) ℓ m 0 + κ k − h R κ k η R . (12)
Proof. We prove by induction on ℓ \ell ℓ that
m ℓ ≤ ( 1 − κ k + h R ) ℓ m 0 + η R ( 1 + h R ∑ j = 0 ℓ − 1 ( 1 − κ k + h R ) j ) . m_\ell\le(1-\kappa_k+h_R)^\ell m_0+\eta_R\left(1+h_R\sum_{j=0}^{\ell-1}(1-\kappa_k+h_R)^j\right). m ℓ ≤ ( 1 − κ k + h R ) ℓ m 0 + η R ( 1 + h R j = 0 ∑ ℓ − 1 ( 1 − κ k + h R ) j ) .
The base case ℓ = 1 \ell=1 ℓ = 1 follows from (10) . In the inductive step, we have
m ℓ ≤ ( 1 − κ k ) ℓ m 0 + η R + h R ∑ j = 0 ℓ − 1 ( 1 − κ k ) ℓ − 1 − j m j ≤ ( 1 − κ k ) ℓ m 0 + η R + h R ∑ j = 0 ℓ − 1 ( 1 − κ k ) ℓ − 1 − j ( ( 1 − κ k + h R ) j m 0 + η R ( 1 + h R ∑ r = 0 j − 1 ( 1 − κ k + h R ) r ) ) . \begin{aligned}
m_\ell\le(1-\kappa_k)^\ell m_0+\eta_R+h_R\sum_{j=0}^{\ell-1}(1-\kappa_k)^{\ell-1-j}m_j \\
&\le(1-\kappa_k)^\ell m_0+\eta_R+h_R\sum_{j=0}^{\ell-1}(1-\kappa_k)^{\ell-1-j}\left((1-\kappa_k+h_R)^jm_0+\eta_R\left(1+h_R\sum_{r=0}^{j-1}(1-\kappa_k+h_R)^r\right)\right).
\end{aligned} m ℓ ≤ ( 1 − κ k ) ℓ m 0 + η R + h R j = 0 ∑ ℓ − 1 ( 1 − κ k ) ℓ − 1 − j m j ≤ ( 1 − κ k ) ℓ m 0 + η R + h R j = 0 ∑ ℓ − 1 ( 1 − κ k ) ℓ − 1 − j ( ( 1 − κ k + h R ) j m 0 + η R ( 1 + h R r = 0 ∑ j − 1 ( 1 − κ k + h R ) r ) ) .
First, let us consider the coefficient of m 0 m_0 m 0 . Since we have a ℓ = b ℓ + ( a − b ) ( ∑ j = 0 ℓ − 1 a j b ℓ − 1 − j ) a^\ell=b^\ell+(a-b)\left(\sum_{j=0}^{\ell-1}a^jb^{\ell-1-j}\right) a ℓ = b ℓ + ( a − b ) ( ∑ j = 0 ℓ − 1 a j b ℓ − 1 − j ) for all a , b a,b a , b , we can add the coefficients of m 0 m_0 m 0 and obtain
( 1 − κ k ) ℓ + h R ∑ j = 0 ℓ − 1 ( 1 − κ k ) ℓ − 1 − j ( 1 − κ k + h R ) j = ( 1 − κ k + h R ) ℓ . (1-\kappa_k)^\ell+h_R\sum_{j=0}^{\ell-1}(1-\kappa_k)^{\ell-1-j}(1-\kappa_k+h_R)^j=(1-\kappa_k+h_R)^\ell. ( 1 − κ k ) ℓ + h R j = 0 ∑ ℓ − 1 ( 1 − κ k ) ℓ − 1 − j ( 1 − κ k + h R ) j = ( 1 − κ k + h R ) ℓ .
Next, we consider the coefficient of η R \eta_R η R . We have
1 + h R ∑ j = 0 ℓ − 1 ( 1 − κ k ) ℓ − 1 − j ( 1 + h R ∑ r = 0 j − 1 ( 1 − κ k + h R ) r ) = 1 + h R ∑ i = 0 ℓ − 1 ( ( 1 − κ k ) i + h R ∑ j = 0 i − 1 ( 1 − κ k ) j ( 1 − κ k + h R ) i − 1 − j ) = 1 + h R ∑ i = 0 ℓ − 1 ( 1 − κ k + h R ) i 1+h_R\sum_{j=0}^{\ell-1}(1-\kappa_k)^{\ell-1-j}\left(1+h_R\sum_{r=0}^{j-1}(1-\kappa_k+h_R)^r\right)
\\
=1+h_R\sum_{i=0}^{\ell-1}\left((1-\kappa_k)^i+h_R\sum_{j=0}^{i-1}(1-\kappa_k)^j(1-\kappa_k+h_R)^{i-1-j}\right)
\\
=1+h_R\sum_{i=0}^{\ell-1}(1-\kappa_k+h_R)^i 1 + h R j = 0 ∑ ℓ − 1 ( 1 − κ k ) ℓ − 1 − j ( 1 + h R r = 0 ∑ j − 1 ( 1 − κ k + h R ) r ) = 1 + h R i = 0 ∑ ℓ − 1 ( ( 1 − κ k ) i + h R j = 0 ∑ i − 1 ( 1 − κ k ) j ( 1 − κ k + h R ) i − 1 − j ) = 1 + h R i = 0 ∑ ℓ − 1 ( 1 − κ k + h R ) i
which completes the induction.
Finally, using that h R : = ( 1 − κ k ) p R 1 − p R h_R:=\frac{(1-\kappa_k)p_R}{1-p_R} h R := 1 − p R ( 1 − κ k ) p R and the geometric sum formula, we obtain the claim. □ \square □
By (11), for every 1 ≤ ℓ ≤ t + 1 1\le\ell\le t+1 1 ≤ ℓ ≤ t + 1 , we have
m ℓ ≤ ( 1 − κ k 1 − p R ) ℓ m 0 + κ k κ k − h R η R . m_\ell\le\left(\frac{1-\kappa_k}{1-p_R}\right)^\ell m_0+\frac{\kappa_k}{\kappa_k-h_R}\eta_R. m ℓ ≤ ( 1 − p R 1 − κ k ) ℓ m 0 + κ k − h R κ k η R .
Note that b R ≤ ν R b_R\le\nu_R b R ≤ ν R , since q d − k ≤ 2 d + log ( 32 T / β ) q_{d-k}\le2\sqrt{d+\log(32T/\beta)} q d − k ≤ 2 d + log ( 32 T / β ) . The assumption p R ≤ κ k 16 p_R\le\frac{\kappa_k}{16} p R ≤ 16 κ k implies h R ≤ p R ≤ κ k 16 h_R\le p_R\le\frac{\kappa_k}{16} h R ≤ p R ≤ 16 κ k , and p R κ k b R 1 − p R ≤ b R 15 ≤ ν R 15 \frac{p_R}{\kappa_k}\frac{b_R}{1-p_R}\le\frac{b_R}{15}\le\frac{\nu_R}{15} κ k p R 1 − p R b R ≤ 15 b R ≤ 15 ν R .
η R : = ν R + b R p R κ k ( 1 − p R ) ≤ 16 15 ν R \eta_R:=\nu_R+\frac{b_Rp_R}{\kappa_k(1-p_R)}\le\frac{16}{15}\nu_R η R := ν R + κ k ( 1 − p R ) b R p R ≤ 15 16 ν R
and
κ k κ k − h R η R ≤ ( 16 15 ) 2 ν R . \frac{\kappa_k}{\kappa_k-h_R}\eta_R\le\left(\frac{16}{15}\right)^2\nu_R. κ k − h R κ k η R ≤ ( 15 16 ) 2 ν R .
Thus we obtain
m ℓ ≤ ( 1 − κ k 1 − p R ) ℓ m 0 + ( 16 15 ) 2 ν R ≤ ( 1 − κ k 2 ) ℓ m 0 + ( 16 15 ) 2 ν R , m_\ell\le\left(\frac{1-\kappa_k}{1-p_R}\right)^\ell m_0+\left(\frac{16}{15}\right)^2\nu_R\le\left(1-\frac{\kappa_k}{2}\right)^\ell m_0+\left(\frac{16}{15}\right)^2\nu_R, m ℓ ≤ ( 1 − p R 1 − κ k ) ℓ m 0 + ( 15 16 ) 2 ν R ≤ ( 1 − 2 κ k ) ℓ m 0 + ( 15 16 ) 2 ν R ,
which proves that (7) holds for all iterations ℓ ≤ t + 1 \ell\le t+1 ℓ ≤ t + 1 .
In particular, we have
m t + 1 ≤ m 0 + 2 ν R ≤ 1 15 + 1 4 < 1 3 . m_{t+1}\le m_0+2\nu_R\le\frac{1}{\sqrt{15}}+\frac{1}{4}<\frac{1}{\sqrt{3}}. m t + 1 ≤ m 0 + 2 ν R ≤ 15 1 + 4 1 < 3 1 .
This concludes the main induction, including showing that τ > t + 1 \tau>t+1 τ > t + 1 (if t + 1 = T t+1=T t + 1 = T , τ = T \tau=T τ = T ). □ \square □
Finally, we show the following corollary of the main theorem for ideal choices of T T T and R R R that depend on the data. In Section C, we show how to privately obtain an analogous guarantee via a private hyperparameter search.
Corollary 34. Suppose that R ≥ 15 ∥ U ∥ ∞ ( k + log ( 16 n ( T + 1 ) β ) ) R\ge15\|U\|_\infty\left(\sqrt{k}+\sqrt{\log\left(\frac{16n(T+1)}{\beta}\right)}\right) R ≥ 15∥ U ∥ ∞ ( k + log ( β 16 n ( T + 1 ) ) ) , T = ⌈ 2 κ k log ( n β δ ϵ ) ⌉ T=\left\lceil\frac{2}{\kappa_k}\log\left(\frac{n}{\beta\delta\epsilon}\right)\right\rceil T = ⌈ κ k 2 log ( β δ ϵ n ) ⌉ , and
Δ k ≥ Ω ~ ( d ϵ + ( σ k k ϵ ) 2 / 3 + ( R σ 2 k ϵ ) 2 / 3 ) , \Delta_k\ge\widetilde{\Omega}\left(\frac{\sqrt{d}}{\epsilon}+\left(\frac{\sigma_k\sqrt{k}}{\epsilon}\right)^{2/3}+\left(\frac{R\sigma^2\sqrt{k}}{\epsilon}\right)^{2/3}\right), Δ k ≥ Ω ϵ d + ( ϵ σ k k ) 2/3 + ( ϵ R σ 2 k ) 2/3 , where the Ω ~ ( ⋅ ) \widetilde{\Omega}(\cdot) Ω ( ⋅ ) hides sufficiently large poly-logarithmic factors. With probability at least 1 − β 1-\beta 1 − β , the solution X ( T ) X^{(T)} X ( T ) returned by Algorithm 3 satisfies
∥ ( I − X ( T ) X ( T ) ⊤ ) V k ∥ 2 2 ≤ ( β δ ϵ n ) 2 + O ~ ( d σ k 2 R 2 ϵ 2 Δ k 2 ) . \left\|\left(I-X^{(T)}X^{(T)\top}\right)V_k\right\|_2^2\le\left(\frac{\beta\delta\epsilon}{n}\right)^2+\widetilde{O}\left(\frac{d\sigma_k^2R^2}{\epsilon^2\Delta_k^2}\right). ( I − X ( T ) X ( T ) ⊤ ) V k 2 2 ≤ ( n β δ ϵ ) 2 + O ( ϵ 2 Δ k 2 d σ k 2 R 2 ) .
In particular, if R = Θ ~ ( k ∥ U ∥ ∞ ) R = \widetilde{\Theta}\left(\sqrt{k}\|U\|_{\infty}\right) R = Θ ( k ∥ U ∥ ∞ ) , we have
∥ ( I − X ( T ) X ( T ) ⊤ ) V k ∥ 2 2 ≤ ( 8 δ ε n ) 2 + O ~ ( d k σ k 2 ∥ U ∥ ∞ 2 ε 2 Δ k 2 ) . \left\|\left(I-X^{(T)}X^{(T)\top}\right)V_k\right\|_2^2 \leq\left(\frac{8\delta\varepsilon}{n}\right)^2+\widetilde{O}\left(\frac{dk\sigma_k^2\|U\|_{\infty}^2}{\varepsilon^2\Delta_k^2}\right). ( I − X ( T ) X ( T ) ⊤ ) V k 2 2 ≤ ( n 8 δ ε ) 2 + O ( ε 2 Δ k 2 d k σ k 2 ∥ U ∥ ∞ 2 ) .
Proof. Let us verify that the assumption imply the conditions in Theorem 31. The choice of T T T gives
Δ = O ~ ( 1 ε κ k ) = O ~ ( σ k ε Δ k ) . \Delta=\widetilde{O}\left(\frac{1}{\varepsilon\sqrt{\kappa_k}}\right)=\widetilde{O}\left(\frac{\sigma_k}{\varepsilon\sqrt{\Delta_k}}\right). Δ = O ( ε κ k 1 ) = O ( ε Δ k σ k ) .
The gap condition Δ k ≥ Ω ~ ( ( σ k k ε ) 2 / 3 ) \Delta_k\geq\widetilde{\Omega}\left(\left(\frac{\sigma_k\sqrt{k}}{\varepsilon}\right)^{2/3}\right) Δ k ≥ Ω ( ( ε σ k k ) 2/3 ) implies Δ q r o w = O ~ ( σ k k ε Δ k ) = O ( Δ k ) \Delta_{q_{\rm row}}=\widetilde{O}\left(\frac{\sigma_k\sqrt{k}}{\varepsilon\sqrt{\Delta_k}}\right)=O(\Delta_k) Δ q row = O ( ε Δ k σ k k ) = O ( Δ k ) and also implies γ = O ~ ( Δ k ) = O ( Δ k ) < σ k 2 4 \gamma=\widetilde{O}\left(\Delta\sqrt{k}\right)=O(\Delta_k)<\frac{\sigma_k^2}{4} γ = O ( Δ k ) = O ( Δ k ) < 4 σ k 2 .
The gap condition Δ k ≥ Ω ~ ( ( R σ k 2 k ε ) 2 / 3 ) \Delta_k\geq\widetilde{\Omega}\left(\left(\frac{R\sigma_k^2\sqrt{k}}{\varepsilon}\right)^{2/3}\right) Δ k ≥ Ω ( ( ε R σ k 2 k ) 2/3 ) implies
p R = 3 Δ R q k 2 σ k = O ~ ( R k ε Δ k ) ≤ Δ k 16 σ k 2 = κ k 16 . p_R=\frac{3\Delta Rq_k}{2\sigma_k}=\widetilde{O}\left(\frac{R\sqrt{k}}{\varepsilon\sqrt{\Delta_k}}\right)\leq\frac{\Delta_k}{16\sigma_k^2}=\frac{\kappa_k}{16}. p R = 2 σ k 3Δ R q k = O ( ε Δ k R k ) ≤ 16 σ k 2 Δ k = 16 κ k .
The gap condition Δ k ≥ Ω ~ ( d ε ) \Delta_k\geq\widetilde{\Omega}\left(\sqrt{\frac{d}{\varepsilon}}\right) Δ k ≥ Ω ( ε d ) ensures that Δ k ≥ 8 Δ i n i t ( 2 d + 2 log 8 β ) \Delta_k\geq8\Delta_{\mathrm{init}}\left(2\sqrt{d}+\sqrt{2\log\frac{8}{\beta}}\right) Δ k ≥ 8 Δ init ( 2 d + 2 log β 8 ) .
By Lemmas 16, 21, 30, and 29, the events G c l i p \mathcal{G}_{\mathrm{clip}} G clip , G i n i t \mathcal{G}_{\mathrm{init}} G init , G m a r t \mathcal{G}_{\mathrm{mart}} G mart , and the event s 0 ≤ 1 4 s_0\leq\frac{1}{4} s 0 ≤ 4 1 hold jointly with probability at least 1 − β 1-\beta 1 − β .
If ν R ≤ 1 8 \nu_R\leq\frac{1}{8} ν R ≤ 8 1 , all of the conditions of Theorem 31 are satisfied and the theorem implies the desired guarantee. Otherwise, we have s T 2 ≤ 1 ≤ 64 ν R 2 s_T^2\leq1\leq64\nu_R^2 s T 2 ≤ 1 ≤ 64 ν R 2 , and we have the same asymptotic guarantee.
Private selection of the clipping radius R R R and iteration count T T T
Both the clipping radius R R R and the iteration count T T T in Corollary 34 depend on the private data. We can try all of possible values for them from powers of two in an appropriate range and return the approximately best solution using the exponential mechanism, incurring an overhead logarithmic in the ranges (because we only consider powers of two). To be more economical, we select them jointly using the private hyperparameter-selection method of Papernot and Steinke [ [19 ] . A candidate consists of a pair ( R , T ) (R,T) ( R , T ) and the subspace produced by that run. Each run receives the same total Renyi privacy budget, so runs with fewer iterations use smaller per-iteration noise. The guarantee compares the returned subspace with the best pair satisfying the utility hypotheses in the following public grid.
Fix failure probability β ∈ ( 0 , 1 ) \beta\in(0,1) β ∈ ( 0 , 1 ) and public grids of hyperparameters
R = { R j = 2 j R min : 0 ≤ j ≤ J } , m R = J + 1 , \mathcal{R}=\{R_j=2^jR_{\min}:0\leq j\leq J\}, \qquad m_R=J+1, R = { R j = 2 j R m i n : 0 ≤ j ≤ J } , m R = J + 1 ,
T g r i d = { T ℓ = 2 ℓ T min : 0 ≤ ℓ ≤ H } , m T = H + 1 , \mathcal{T}_{\mathrm{grid}}=\{T_\ell=2^\ell T_{\min}:0\leq\ell\leq H\}, \qquad m_T=H+1, T grid = { T ℓ = 2 ℓ T m i n : 0 ≤ ℓ ≤ H } , m T = H + 1 ,
T max = 2 H T min , R max = 2 J R min m = m R m T . (13) T_{\max}=2^H T_{\min},\quad R_{\max}=2^J R_{\min}\qquad m=m_Rm_T. \tag*{(13)} T m a x = 2 H T m i n , R m a x = 2 J R m i n m = m R m T . (13)
All endpoints are fixed independently of the data. On any dataset satisfying the warm-start gap condition, Δ k ≥ 16 Δ i n i t d \Delta_k\geq16\Delta_{\mathrm{init}}\sqrt{d} Δ k ≥ 16 Δ init d , whereas σ k 2 ≤ tr ( A ⊤ A ) ≤ n \sigma_k^2\leq\operatorname{tr}(A^\top A)\leq n σ k 2 ≤ tr ( A ⊤ A ) ≤ n . Consequently κ k ≥ Δ i n i t d / n \kappa_k\geq\Delta_{\mathrm{init}}\sqrt{d}/n κ k ≥ Δ init d / n in the regime with utility guarantees. Thus we can set T min = 1 , T max = 2 n Δ i n i t d log ( n β ε ) T_{\min}=1,T_{\max}=\frac{2n}{\Delta_{\mathrm{init}}\sqrt{d}}\log\left(\frac{n}{\beta\varepsilon}\right) T m i n = 1 , T m a x = Δ init d 2 n log ( β ε n ) .
Since 1 / n ≤ ∥ U ∥ ∞ ≤ 1 \sqrt{1/n}\leq\|U\|_{\infty}\leq1 1/ n ≤ ∥ U ∥ ∞ ≤ 1 , we can set R min = k / n R_{\min}=\sqrt{k/n} R m i n = k / n . Because ∥ U ∥ ∞ ≤ 1 \|U\|_{\infty}\leq1 ∥ U ∥ ∞ ≤ 1 , we can set R max = 15 ( k + log ( 16 n ( T + 1 ) β ) ) R_{\max}=15\left(\sqrt{k}+\sqrt{\log\left(\frac{16n(T+1)}{\beta}\right)}\right) R m a x = 15 ( k + log ( β 16 n ( T + 1 ) ) ) . We run the initialization algorithm (Algorithm (1)) only once with failure probability β 0 = β / 2 \beta_0 = \beta/2 β 0 = β /2 . We use this initial solutions for all of the runs of the optimization algorithm (Algorithm 2). The optimization runs using different choices for radius and iteration count will use the following parameters. For public Δ i n i t , χ , ζ > 0 \Delta_{\mathrm{init}}, \chi, \zeta> 0 Δ init , χ , ζ > 0 , a run with t t t iterations uses
Δ ( t ) = χ t , γ ( t ) = 4 Δ ( t ) ( 2 k + 2 log ( 4 t / β 0 ) ) (14) \Delta(t) = \chi\sqrt{t}, \qquad\gamma(t) = 4\Delta(t)\left(2\sqrt{k} + \sqrt{2\log(4t/\beta_0)}\right) \tag*{(14)} Δ ( t ) = χ t , γ ( t ) = 4Δ ( t ) ( 2 k + 2 log ( 4 t / β 0 ) ) (14)
The noise parameters will be adjusted and slightly different from the even allocation between initialization and optimization in Algorithm 3. Let q k ( t ) , q r o w ( t ) , p R ( t ) , ν R ( t ) q_k(t), q_{\mathrm{row}}(t), p_R(t), \nu_R(t) q k ( t ) , q row ( t ) , p R ( t ) , ν R ( t ) be the quantities in the analysis as functions of the iteration count t t t , noise scale Δ ( t ) \Delta(t) Δ ( t ) , and failure probability β 0 \beta_0 β 0 . In particular,
p R ( t ) = 3 Δ ( t ) R q k ( t ) 2 σ k , p_R(t) = \frac{3\Delta(t)R q_k(t)}{2\sigma_k}, p R ( t ) = 2 σ k 3Δ ( t ) R q k ( t ) ,
ν R ( t ) = 3 Δ ( t ) R σ k 2 ( d log 9 + log ( 32 t / β 0 ) ) κ k ( 2 − κ k ) . \nu_R(t) = \frac{3\Delta(t)R}{\sigma_k}\sqrt{\frac{2(d\log9+\log(32t/\beta_0))}{\kappa_k(2-\kappa_k)}}. ν R ( t ) = σ k 3Δ ( t ) R κ k ( 2 − κ k ) 2 ( d log 9 + log ( 32 t / β 0 )) .
A private score for comparing candidate subspaces
In this section, we describe the score to compare the candidate subspaces. Ideally we would like to use ∥ ( I − X X T ) V k V k T ∥ 2 \lVert(I-XX^{\mathsf{T}})V_kV_k^{\mathsf{T}}\rVert_2 ∥( I − X X T ) V k V k T ∥ 2 as the score but a more robust version of this is required to reduce the sensitivity of the score when the inputs do not have a large eigenvalue gap.
Recall that the initialization algorithm (Algorithm 1) produces the matrix
M ^ = A T A + Δ i n i t sym ( G ( i n i t ) ) , ϵ w : = Δ i n i t ( 2 d + 2 log ( 8 / β ) ) . \widehat{M}=A^{\mathsf{T}}A+\Delta_{\mathrm{init}}\operatorname{sym}(G^{(\mathrm{init})}), \qquad\epsilon_w := \Delta_{\mathrm{init}}\left(2\sqrt{d}+\sqrt{2\log(8/\beta)}\right). M = A T A + Δ init sym ( G ( init ) ) , ϵ w := Δ init ( 2 d + 2 log ( 8/ β ) ) .
Write its eigenvalues as λ ^ 1 ≥ ⋯ ≥ λ ^ d \widehat{\lambda}_1\ge\cdots\ge\widehat{\lambda}_d λ 1 ≥ ⋯ ≥ λ d and let X ( 0 ) X^{(0)} X ( 0 ) be its top-k k k eigenbasis. Set
a = λ ^ k + 1 + ϵ w , b = λ ^ k − ϵ w , h = b − a . (15) a=\widehat{\lambda}_{k+1}+\epsilon_w,\qquad b=\widehat{\lambda}_k-\epsilon_w,\qquad h=b-a. \tag*{(15)} a = λ k + 1 + ϵ w , b = λ k − ϵ w , h = b − a . (15)
If h ≤ 0 h\le0 h ≤ 0 , return ( R J , 0 , X ( 0 ) ) (R_J,0,X^{(0)}) ( R J , 0 , X ( 0 ) ) immediately. Intuitively this happens when the eigenvalue gap is too small compared with the noise. If h > 0 h>0 h > 0 , consider the following scalar function
f a , b ( x ) = { 0 , x ≤ a , ( x − a ) / h , a < x < b , 1 , x ≥ b . f_{a,b}(x)=
\begin{cases}
0, & x\le a,\\
(x-a)/h, & a<x<b,\\
1, & x\ge b.
\end{cases} f a , b ( x ) = ⎩ ⎨ ⎧ 0 , ( x − a ) / h , 1 , x ≤ a , a < x < b , x ≥ b .
Let f a , b ( M ) f_{a,b}(M) f a , b ( M ) be the function that keeps all the singular vectors of M M M but applies the scalar function f a , b ( x ) f_{a,b}(x) f a , b ( x ) to all the singular values (this is called the matrix functional calculus of f a , b f_{a,b} f a , b ). For a candidate solution X ∈ R d × k X\in\mathbb{R}^{d\times k} X ∈ R d × k with orthonormal columns, its score is
s a , b ( A , X ) : = ∥ ( I − X X T ) f a , b ( A T A ) ∥ 2 ∈ [ 0 , 1 ] . (16) s_{a,b}(A,X):=\lVert(I-XX^{\mathsf{T})f_{a,b}(A^{\mathsf{T}}A)\rVert_2\in[0,1].} \tag*{(16)} s a , b ( A , X ) := ∥( I − X X T ) f a , b ( A T A ) ∥ 2 ∈ [ 0 , 1 ] . (16)
The following lemma bounds the sensitivity of f a , b ( ⋅ ) f_{a,b}(\cdot) f a , b ( ⋅ ) for all inputs regardless of the eigenvalue gap and proves that for inputs with large singular value gaps, f a , b ( ⋅ ) f_{a,b}(\cdot) f a , b ( ⋅ ) is identical to the top singular value subspace.
Lemma 35 (Sensitivity and accuracy of the comparison score). For each fixed initialization with h > 0 h>0 h > 0 , and every fixed orthonormal X X X , the replacement sensitivity of s a , b ( A , X ) s_{a,b}(A,X) s a , b ( A , X ) is at most 2 / h 2/h 2/ h . If ∥ M ^ − A T A ∥ 2 ≤ ϵ w \lVert\widehat{M}-A^{\mathsf{T}}A\rVert_2\le\epsilon_w ∥ M − A T A ∥ 2 ≤ ϵ w , Δ k ≥ 8 ϵ w \Delta_k\ge8\epsilon_w Δ k ≥ 8 ϵ w then h ≥ Δ k / 2 h\ge\Delta_k/2 h ≥ Δ k /2 , we have f a , b ( A T A ) = V k V k T f_{a,b}(A^{\mathsf{T}}A)=V_kV_k^{\mathsf{T}} f a , b ( A T A ) = V k V k T and consequently, simultaneously for all X X X , s a , b ( A , X ) = ∥ ( I − X X T ) V k ∥ 2 s_{a,b}(A,X)=\lVert(I-XX^{\mathsf{T}})V_k\rVert_2 s a , b ( A , X ) = ∥( I − X X T ) V k ∥ 2 .
Proof. The scalar function f a , b f_{a,b} f a , b is 1 / h 1/h 1/ h -Lipschitz. Consider symmetric matrices M , M ′ M,M' M , M ′ with orthonormal eigenbases u i , v j u_i,v_j u i , v j and eigenvalues λ i , λ j ′ \lambda_i,\lambda'_j λ i , λ j ′ respectively. We will show that
∑ i , j ( λ i − λ j ′ ) 2 ∣ ⟨ u i , v j ⟩ ∣ 2 = ∥ M − M ′ ∥ F 2 (17) \sum_{i,j}(\lambda_i-\lambda'_j)^2\lvert\langle u_i,v_j\rangle\rvert^2=\lVert M-M'\rVert_F^2 \tag*{(17)} i , j ∑ ( λ i − λ j ′ ) 2 ∣⟨ u i , v j ⟩ ∣ 2 = ∥ M − M ′ ∥ F 2 (17)
Because the Frobenius norm does not change under multiplication with an orthonormal matrix, ∥ M − M ′ ∥ F 2 = ∥ U ⊤ ( M − M ′ ) V ∥ F 2 = ∑ i , j ( u i ⊤ ( M − M ′ ) v j ) 2 \lVert M-M'\rVert_F^2=\lVert U^\top(M-M')V\rVert_F^2=\sum_{i,j}(u_i^\top(M-M')v_j)^2 ∥ M − M ′ ∥ F 2 = ∥ U ⊤ ( M − M ′ ) V ∥ F 2 = ∑ i , j ( u i ⊤ ( M − M ′ ) v j ) 2 . By the definition of eigenvalues, u i ⊤ M = λ i u i ⊤ u_i^\top M=\lambda_i u_i^\top u i ⊤ M = λ i u i ⊤ and M ′ v j = λ j ′ v j M'v_j=\lambda'_jv_j M ′ v j = λ j ′ v j . Thus, u i ⊤ ( M − M ′ ) v j = ( λ i − λ j ′ ) u i ⊤ v j u_i^\top(M-M')v_j=(\lambda_i-\lambda'_j)u_i^\top v_j u i ⊤ ( M − M ′ ) v j = ( λ i − λ j ′ ) u i ⊤ v j and this proves (16). Applying 16, we have
∥ f a , b ( M ) − f a , b ( M ′ ) ∥ F 2 = ∑ i , j ( f a , b ( λ i ) − f a , b ( λ j ′ ) ) 2 ∣ u i ⊤ v j ∣ 2 ≤ h − 2 ∑ i , j ( λ i − λ j ′ ) 2 ∣ u i ⊤ v j ∣ 2 = h − 2 ∥ M − M ′ ∥ F 2 . \begin{aligned}
\lVert f_{a,b}(M)-f_{a,b}(M')\rVert_F^2=\sum_{i,j}(f_{a,b}(\lambda_i)-f_{a,b}(\lambda'_j))^2\lvert u_i^\top v_j\rvert^2 \\
&\le h^{-2}\sum_{i,j}(\lambda_i-\lambda'_j)^2\lvert u_i^\top v_j\rvert^2=h^{-2}\lVert M-M'\rVert_F^2.
\end{aligned} ∥ f a , b ( M ) − f a , b ( M ′ ) ∥ F 2 = i , j ∑ ( f a , b ( λ i ) − f a , b ( λ j ′ ) ) 2 ∣ u i ⊤ v j ∣ 2 ≤ h − 2 i , j ∑ ( λ i − λ j ′ ) 2 ∣ u i ⊤ v j ∣ 2 = h − 2 ∥ M − M ′ ∥ F 2 .
Consequently, for neighboring datasets A A A , A ′ A' A ′ ,
∣ s a , b ( A , X ) − s a , b ( A ′ , X ) ∣ ≤ ∥ f a , b ( A ⊤ A ) − f a , b ( A ′ ⊤ A ′ ) ∥ F ≤ ∥ A ⊤ A − A ′ ⊤ A ′ ∥ F h ≤ 2 h . \lvert s_{a,b}(A,X)-s_{a,b}(A',X)\rvert\le\lVert f_{a,b}(A^\top A)-f_{a,b}(A'^\top A')\rVert_F\le\frac{\lVert A^\top A-A'^\top A'\rVert_F}{h}\le\frac{2}{h}. ∣ s a , b ( A , X ) − s a , b ( A ′ , X )∣ ≤ ∥ f a , b ( A ⊤ A ) − f a , b ( A ′⊤ A ′ ) ∥ F ≤ h ∥ A ⊤ A − A ′⊤ A ′ ∥ F ≤ h 2 .
For the accuracy claim, Weyl’s inequality (Theorem 7) gives σ k + 1 2 ≤ a \sigma_{k+1}^2\le a σ k + 1 2 ≤ a and b ≤ σ k 2 b\le\sigma_k^2 b ≤ σ k 2 . It also gives h ≥ Δ k − 4 ϵ w ≥ Δ k / 2 h\ge\Delta_k-4\epsilon_{\mathrm{w}}\ge\Delta_k/2 h ≥ Δ k − 4 ϵ w ≥ Δ k /2 . Thus f a , b f_{a,b} f a , b is one on the top k k k eigenvalues of A ⊤ A A^\top A A ⊤ A and zero on all remaining eigenvalues. Finally, ∥ ( I − X X ⊤ ) V k V k ⊤ ∥ 2 = ∥ ( I − X X ⊤ ) V k ∥ 2 \lVert(I-XX^\top)V_kV_k^\top\rVert_2=\lVert(I-XX^\top)V_k\rVert_2 ∥( I − X X ⊤ ) V k V k ⊤ ∥ 2 = ∥( I − X X ⊤ ) V k ∥ 2 .
Joint random search and its privacy cost
We now apply the method of Papernot and Steinke [ [19 ] to our problem with the above score function. Conditioned on the shared initialization M ‾ \overline{M} M with h > 0 h>0 h > 0 , define one base trial Q Q Q as follows. Sample ( j , ℓ ) (j,\ell) ( j , ℓ ) uniformly from { 0 , … , J } × { 0 , … , H } \{0,\ldots,J\}\times\{0,\ldots,H\} { 0 , … , J } × { 0 , … , H } , and run Algorithm 2 from X ( 0 ) X^{(0)} X ( 0 ) for T ℓ T_\ell T ℓ iterations, with radius R j R_j R j , noise parameter Δ ( T ℓ ) \Delta(T_\ell) Δ ( T ℓ ) , and floor γ ( T ℓ ) \gamma(T_\ell) γ ( T ℓ ) . Denote its output by X X X . With a fresh Z ∼ N ( 0 , 1 ) Z\sim\mathcal{N}(0,1) Z ∼ N ( 0 , 1 ) , form the noisy score
s ~ = s a , b ( A , X ) + 2 ζ h Z . (18) \widetilde{s}=s_{a,b}(A,X)+\frac{2\zeta}{h}Z. \tag*{(18)} s = s a , b ( A , X ) + h 2 ζ Z . (18)
The base output is the tuple ( R j , T ℓ , X , s ~ ) (R_j,T_\ell,X,\widetilde{s}) ( R j , T ℓ , X , s ) , ordered by smaller s ~ \widetilde{s} s being better, with a fixed total ordering to break ties.
Independently sample a number of trials N N N with
Pr [ N = r ] = θ ( 1 − θ ) r − 1 , r ≥ 1 , θ = β 8 m , E [ N ] = 1 θ . (19) \Pr[N=r]=\theta(1-\theta)^{r-1},\qquad r\ge1,\qquad\theta=\frac{\beta}{8m},\qquad\mathbb{E}[N]=\frac{1}{\theta}. \tag*{(19)} Pr [ N = r ] = θ ( 1 − θ ) r − 1 , r ≥ 1 , θ = 8 m β , E [ N ] = θ 1 . (19)
Run the base trial Q Q Q independently N N N times with the same initialization. Release the triple ( R ^ , T ^ , X ^ ) (\widehat{R},\widehat{T},\widehat{X}) ( R , T , X ) with the smallest noisy score.
Theorem 36. Define
ϱ i n i t = 2 Δ i n i t 2 , ϱ b = 4 χ 2 + 1 2 ζ 2 , ϱ t o t = ϱ i n i t + ϱ b (20) \varrho_{\mathrm{init}}=\frac{2}{\Delta_{\mathrm{init}}^2},\qquad\varrho_b=\frac{4}{\chi^2}+\frac{1}{2\zeta^2},\qquad\varrho_{\mathrm{tot}}=\varrho_{\mathrm{init}}+\varrho_b \tag*{(20)} ϱ init = Δ init 2 2 , ϱ b = χ 2 4 + 2 ζ 2 1 , ϱ tot = ϱ init + ϱ b (20) The selection procedure satisfies ( λ , ε R D P ( λ ) ) (\lambda,\varepsilon_{\mathrm{RDP}}(\lambda)) ( λ , ε RDP ( λ )) -Rényi differential privacy for every λ > 1 \lambda>1 λ > 1 , with
ε R D P ( λ ) ≤ λ ϱ t o t + 4 ϱ b log ( 1 / θ ) + log ( 1 / θ ) λ − 1 . (21) \varepsilon_{\mathrm{RDP}}(\lambda)\le\lambda\varrho_{\mathrm{tot}}+4\sqrt{\varrho_b\log(1/\theta)}+\frac{\log(1/\theta)}{\lambda-1}. \tag*{(21)} ε RDP ( λ ) ≤ λ ϱ tot + 4 ϱ b log ( 1/ θ ) + λ − 1 log ( 1/ θ ) . (21) In particular it is ( ϵ , δ ) (\epsilon,\delta) ( ϵ , δ ) -DP whenever
ϱ t o t + 2 ϱ t o t log ( 1 / ( θ δ ) ) + 4 ϱ b log ( 1 / θ ) ≤ ϵ . (22) \varrho_{\mathrm{tot}}+2\sqrt{\varrho_{\mathrm{tot}}\log(1/(\theta\delta))}+4\sqrt{\varrho_b\log(1/\theta)}\le\epsilon. \tag*{(22)} ϱ tot + 2 ϱ tot log ( 1/ ( θ δ )) + 4 ϱ b log ( 1/ θ ) ≤ ϵ . (22) For ϵ ∈ ( 0 , 1 ) \epsilon\in(0,1) ϵ ∈ ( 0 , 1 ) , an explicit setting of parameters is
ν = ϵ 2 100 log ( 1 / ( θ δ ) ) , Δ i n i t 2 = 6 ν , χ 2 = 12 ν , ζ 2 = 3 2 ν . (23) \nu=\frac{\epsilon^2}{100\log(1/(\theta\delta))},\qquad\Delta_{\mathrm{init}}^2=\frac{6}{\nu},\qquad\chi^2=\frac{12}{\nu},\qquad\zeta^2=\frac{3}{2\nu}. \tag*{(23)} ν = 100 log ( 1/ ( θ δ )) ϵ 2 , Δ init 2 = ν 6 , χ 2 = ν 12 , ζ 2 = 2 ν 3 . (23) Thus Δ ( t ) = 1200 t log ( 1 / ( θ δ ) ) ϵ \Delta(t)=\frac{\sqrt{1200t\log(1/(\theta\delta))}}{\epsilon} Δ ( t ) = ϵ 1200 t l o g ( 1/ ( θ δ )) for each candidate iteration count t t t .
Proof. A Gaussian mechanism with Euclidean sensitivity s s s and noise standard deviation σ \sigma σ has Rényi cost λ s 2 / ( 2 σ 2 ) \lambda s^2/(2\sigma^2) λ s 2 / ( 2 σ 2 ) at order λ \lambda λ ; these costs add under adaptive composition Mironov [ [16 ] . The initialization has s = 2 s=2 s = 2 and σ = Δ i n i t \sigma=\Delta_{\mathrm{init}} σ = Δ init so it has cost λ ϱ i n i t \lambda\varrho_{\mathrm{init}} λ ϱ init .
Fix any initialization output with h > 0 h>0 h > 0 . Conditional on previous noisy outputs, the preconditioner and clipped-update queries have sensitivities 2 2 2 and 2 R j 2R_j 2 R j , with noise scales Δ ( T ℓ ) \Delta(T_\ell) Δ ( T ℓ ) and Δ ( T ℓ ) R j \Delta(T_\ell)R_j Δ ( T ℓ ) R j , respectively. Their cost over T ℓ T_\ell T ℓ steps is
4 T ℓ λ Δ ( T ℓ ) 2 = 4 λ χ 2 , \frac{4T_\ell\lambda}{\Delta(T_\ell)^2}=\frac{4\lambda}{\chi^2}, Δ ( T ℓ ) 2 4 T ℓ λ = χ 2 4 λ ,
uniformly for any choice of j j j and ℓ \ell ℓ . By the sensitivity of the score shown in Lemma 35 , the noisy score additionally costs λ / ( 2 ζ 2 ) \lambda/(2\zeta^2) λ / ( 2 ζ 2 ) . Thus the entire base tuple has cost λ ϱ b \lambda\varrho_b λ ϱ b . Each trial starts from the same private initialization and uses independent iteration noises so we do not count the initialization cost toward the cost of each trial.
Apply Theorem 2 of Papernot and Steinke [ [19 ] with negative-binomial shape parameter 1 1 1 , corresponding to the geometric distribution (18). For any auxiliary order λ ^ > 1 \hat{\lambda}>1 λ ^ > 1 , the cost of returning the best base tuple is at most
λ ϱ b + 2 ( λ ^ − 1 ) ϱ b + 2 log ( 1 / θ ) λ ^ + log ( 1 / θ ) λ − 1 . \lambda\varrho_b+2(\hat{\lambda}-1)\varrho_b+\frac{2\log(1/\theta)}{\hat{\lambda}}+\frac{\log(1/\theta)}{\lambda-1}. λ ϱ b + 2 ( λ ^ − 1 ) ϱ b + λ ^ 2 log ( 1/ θ ) + λ − 1 log ( 1/ θ ) .
Taking λ ^ = 1 + log ( 1 / θ ) / ϱ b \hat{\lambda}=1+\sqrt{\log(1/\theta)/\varrho_b} λ ^ = 1 + log ( 1/ θ ) / ϱ b bounds the middle two terms by 4 ϱ b log ( 1 / θ ) 4\sqrt{\varrho_b\log(1/\theta)} 4 ϱ b log ( 1/ θ ) . This is a uniform bound for every initialization output because in the h ≤ 0 h\leq0 h ≤ 0 case, we only perform postprocessing with no privacy loss. The bound (21) then follows from adaptive composition with the initialization.
To obtain an ( ϵ , δ ) (\epsilon,\delta) ( ϵ , δ ) -DP bound, we convert the Rényi-DP bound, which adds a log ( 1 / δ ) / ( λ − 1 ) \log(1/\delta)/(\lambda-1) log ( 1/ δ ) / ( λ − 1 ) term to ϵ \epsilon ϵ Mironov [ [16 ] . Minimizing over λ > 1 \lambda>1 λ > 1 gives (22) . Under (23) , ϱ t o t = v \varrho_{\mathrm{tot}}=v ϱ tot = v and ϱ b = 2 v / 3 ≤ v \varrho_b=2v/3\leq v ϱ b = 2 v /3 ≤ v . Writing D = log ( 1 / ( θ δ ) ) > log 8 D=\log(1/(\theta\delta))>\log8 D = log ( 1/ ( θ δ )) > log 8 , the left-hand side of (22) is bounded by ϵ 2 / ( 100 D ) + ϵ / 5 + 2 ϵ / 5 < ϵ \epsilon^2/(100D)+\epsilon/5+2\epsilon/5<\epsilon ϵ 2 / ( 100 D ) + ϵ /5 + 2 ϵ /5 < ϵ .
Utility guarantee from hyperparameter search
In this section we show that the algorithm recovers a good solution if the input data satisfies the assumptions for the utility proof without prior knowledge of κ k , ∥ U ∥ ∞ \kappa_k,\lVert U\rVert_\infty κ k , ∥ U ∥ ∞ . Let T ⋆ = ⌈ 2 κ k log n β δ ϵ ⌉ T_\star=\left\lceil\frac{2}{\kappa_k}\log\frac{n}{\beta\delta\epsilon}\right\rceil T ⋆ = ⌈ κ k 2 log β δ ϵ n ⌉ and R ⋆ = 15 ∥ U ∥ ∞ ( k + log ( 16 n ( T max + 1 ) β 0 ) ) R_\star=15\lVert U\rVert_\infty\left(\sqrt{k}+\sqrt{\log\left(\frac{16n(T_{\max}+1)}{\beta_0}\right)}\right) R ⋆ = 15 ∥ U ∥ ∞ ( k + log ( β 0 16 n ( T m a x + 1 ) ) ) . As argued before, if the input satisfies the assumptions for the utility guarantee then T min ≤ T ⋆ ≤ T max T_{\min}\leq T_\star\leq T_{\max} T m i n ≤ T ⋆ ≤ T m a x and R min ≤ R ⋆ ≤ R max R_{\min}\leq R_\star\leq R_{\max} R m i n ≤ R ⋆ ≤ R m a x . Consider the smallest grid duration T † ≥ T ⋆ T^\dagger\geq T_\star T † ≥ T ⋆ and grid radius R † ≥ R ⋆ R^\dagger\geq R_\star R † ≥ R ⋆ . We have
T ⋆ ≤ T † ≤ 2 T ⋆ , R ⋆ ≤ R † ≤ 2 R ⋆ . T_\star\leq T^\dagger\leq2T_\star,\qquad R_\star\leq R^\dagger\leq2R_\star. T ⋆ ≤ T † ≤ 2 T ⋆ , R ⋆ ≤ R † ≤ 2 R ⋆ .
Using a similar proof to that of Corollary 34, we can show the following guarantee.
Theorem 37. Let ϵ , δ , β ∈ ( 0 , 1 ) \epsilon,\delta,\beta\in(0,1) ϵ , δ , β ∈ ( 0 , 1 ) and up to logarithmic factors
Δ k ≥ Ω ~ ( d ϵ + ( σ k k ϵ ) 2 / 3 + ( R ⋆ σ k 2 k ϵ ) 2 / 3 ) . \Delta_k\geq\widetilde{\Omega}\left(\frac{\sqrt{d}}{\epsilon}+\left(\frac{\sigma_k\sqrt{k}}{\epsilon}\right)^{2/3}+\left(\frac{R_\star\sigma_k^2\sqrt{k}}{\epsilon}\right)^{2/3}\right). Δ k ≥ Ω ϵ d + ( ϵ σ k k ) 2/3 + ( ϵ R ⋆ σ k 2 k ) 2/3 . With probability at least 1 − β 1-\beta 1 − β , the selected output satisfies
∥ ( I − X ^ X ^ T ) V k ∥ 2 2 ≤ ( β δ ϵ n ) 2 + O ~ ( d k σ k 2 ∥ U ∥ ∞ 2 + 1 ϵ 2 Δ k 2 ) . (24) \lVert(I-\widehat{X}\widehat{X}^{\mathsf T})V_k\rVert_2^2\leq\left(\frac{\beta\delta\epsilon}{n}\right)^2+\widetilde{O}\left(\frac{dk\sigma_k^2\lVert U\rVert_\infty^2+1}{\epsilon^2\Delta_k^2}\right). \tag*{(24)} ∥( I − X X T ) V k ∥ 2 2 ≤ ( n β δ ϵ ) 2 + O ( ϵ 2 Δ k 2 d k σ k 2 ∥ U ∥ ∞ 2 + 1 ) . (24)
Proof. Let T = T † T=T^\dagger T = T † , R = R † R=R^\dagger R = R † , and Δ = Δ ( T ) \Delta=\Delta(T) Δ = Δ ( T ) . Let us verify that the assumptions imply the conditions in Theorem 31, with failure parameter β 0 = β / 2 \beta_0=\beta/2 β 0 = β /2 . The calibration in (23) and the choice T ⋆ ≤ T ≤ 2 T ⋆ T_\star\leq T\leq2T_\star T ⋆ ≤ T ≤ 2 T ⋆ give
Δ i n i t = O ~ ( 1 ϵ ) , Δ = O ~ ( 1 ϵ κ k ) = O ~ ( σ k ϵ Δ k ) , ζ = O ~ ( 1 ϵ ) . \Delta_{\mathrm{init}}=\widetilde{O}\left(\frac{1}{\epsilon}\right),\qquad\Delta=\widetilde{O}\left(\frac{1}{\epsilon\sqrt{\kappa_k}}\right)=\widetilde{O}\left(\frac{\sigma_k}{\epsilon\sqrt{\Delta_k}}\right),\qquad\zeta=\widetilde{O}\left(\frac{1}{\epsilon}\right). Δ init = O ( ϵ 1 ) , Δ = O ( ϵ κ k 1 ) = O ( ϵ Δ k σ k ) , ζ = O ( ϵ 1 ) .
The gap condition Δ k ≥ Ω ~ ( ( σ k k ϵ ) 2 / 3 ) \Delta_k \ge\widetilde{\Omega}\left(\left(\frac{\sigma_k\sqrt{k}}{\epsilon}\right)^{2/3}\right) Δ k ≥ Ω ( ( ϵ σ k k ) 2/3 ) implies Δ q r o w ( T ) = O ~ ( σ k k ϵ Δ k ) ≤ Δ k / 8 \Delta_{\mathrm{qrow}}(T)=\widetilde{O}\left(\frac{\sigma_k\sqrt{k}}{\epsilon\sqrt{\Delta_k}}\right)\le\Delta_k/8 Δ qrow ( T ) = O ( ϵ Δ k σ k k ) ≤ Δ k /8 and also implies γ ( T ) = O ~ ( Δ k ) ≤ σ k 2 / 4 \gamma(T)=\widetilde{O}\left(\Delta\sqrt{k}\right)\le\sigma_k^2/4 γ ( T ) = O ( Δ k ) ≤ σ k 2 /4 , when the hidden factors in the gap condition are sufficiently large.
The gap condition Δ k ≥ Ω ~ ( ( R ⋆ σ k k ϵ ) 2 / 3 ) \Delta_k \ge\widetilde{\Omega}\left(\left(\frac{R_\star\sigma_k\sqrt{k}}{\epsilon}\right)^{2/3}\right) Δ k ≥ Ω ( ( ϵ R ⋆ σ k k ) 2/3 ) and R ≤ 2 R ⋆ R\le2R_\star R ≤ 2 R ⋆ imply
p R ( T ) = 3 Δ R q k ( T ) 2 σ k = O ~ ( R ⋆ k ϵ Δ k ) ≤ Δ k 16 σ k 2 = κ k 16 . p_R(T)=\frac{3\Delta Rq_k(T)}{2\sigma_k}=\widetilde{O}\left(\frac{R_\star\sqrt{k}}{\epsilon\sqrt{\Delta_k}}\right)\le\frac{\Delta_k}{16\sigma_k^2}=\frac{\kappa_k}{16}. p R ( T ) = 2 σ k 3Δ R q k ( T ) = O ( ϵ Δ k R ⋆ k ) ≤ 16 σ k 2 Δ k = 16 κ k .
The gap condition Δ k ≥ Ω ~ ( d ϵ ) \Delta_k\ge\widetilde{\Omega}\left(\frac{\sqrt{d}}{\epsilon}\right) Δ k ≥ Ω ( ϵ d ) ensures that
Δ k ≥ 8 Δ i n i t ( 2 d + 2 log 8 β 0 ) ≥ 8 ϵ w . \Delta_k\ge8\Delta_{\mathrm{init}}\left(2\sqrt{d}+\sqrt{2\log\frac{8}{\beta_0}}\right)\ge8\epsilon_w. Δ k ≥ 8 Δ init ( 2 d + 2 log β 0 8 ) ≥ 8 ϵ w .
The choice R ≥ R ⋆ R\ge R_\star R ≥ R ⋆ and T ≤ T max T\le T_{\max} T ≤ T m a x also ensure the lower bound on R R R in Theorem 31 .
By Lemmas 21 and 5, the initialization event
G i n i t ∩ { ∥ M ^ − A T A ∥ 2 ≤ ϵ w } \mathcal{G}_{\mathrm{init}}\cap\left\{\left\|\widehat{M}-A^\mathsf{T}A\right\|_2\le\epsilon_w\right\} G init ∩ { M − A T A 2 ≤ ϵ w }
holds with probability at least 1 − β 0 / 4 − β / 8 = 1 − β / 4 1-\beta_0/4-\beta/8=1-\beta/4 1 − β 0 /4 − β /8 = 1 − β /4 , where G i n i t \mathcal{G}_{\mathrm{init}} G init is defined with iteration count T T T and failure parameter β 0 \beta_0 β 0 . On this event, Lemma 9 gives s 0 ≤ 1 / 4 s_0\le1/4 s 0 ≤ 1/4 , and Lemma 35 gives h ≥ Δ k / 2 h\ge\Delta_k/2 h ≥ Δ k /2 and identifies every comparison score with the true subspace error. In the following, condition on any initialization in this event.
By the proofs of Lemmas 16 and 30, the events G c l i p \mathcal{G}_{\mathrm{clip}} G clip and G m a r t \mathcal{G}_{\mathrm{mart}} G mart for a trial with parameters ( R , T ) (R,T) ( R , T ) hold jointly with conditional probability at least 1 − 3 β 0 / 8 − β 0 / 4 ≥ 1 / 2 1-3\beta_0/8-\beta_0/4\ge1/2 1 − 3 β 0 /8 − β 0 /4 ≥ 1/2 . Here the floor in (14) gives a failure probability of at most 3 β 0 / 8 3\beta_0/8 3 β 0 /8 for G c l i p \mathcal{G}_{\mathrm{clip}} G clip , and we use the martingale event from Lemma 30 with failure parameter β 0 \beta_0 β 0 ; its threshold is no larger than ν R ( T ) \nu_R(T) ν R ( T ) .
If ν R ( T ) ≤ 1 / 8 \nu_R(T)\le1/8 ν R ( T ) ≤ 1/8 , all of the conditions of Theorem 31 are satisfied on these events, and the theorem gives
s T ≤ ( 1 − κ k 2 ) T m 0 + 2 ν R ( T ) ≤ β δ ϵ n 15 + 2 ν R ( T ) , s_T\le\left(1-\frac{\kappa_k}{2}\right)^T m_0+2\nu_R(T)\le\frac{\beta\delta\epsilon}{n\sqrt{15}}+2\nu_R(T), s T ≤ ( 1 − 2 κ k ) T m 0 + 2 ν R ( T ) ≤ n 15 β δ ϵ + 2 ν R ( T ) ,
where we used m 0 ≤ 1 / 15 m_0\le1/\sqrt{15} m 0 ≤ 1/ 15 and T ≥ T ⋆ T\ge T_\star T ≥ T ⋆ . Otherwise, we have s T ≤ 1 ≤ 8 ν R ( T ) s_T\le1\le8\nu_R(T) s T ≤ 1 ≤ 8 ν R ( T ) , and we have the same asymptotic guarantee. Thus, in both cases, the trial produces a candidate with true error at most
e † : = β δ ϵ n 15 + 8 ν R ( T ) e_\dagger:=\frac{\beta\delta\epsilon}{n\sqrt{15}}+8\nu_R(T) e † := n 15 β δ ϵ + 8 ν R ( T )
with conditional probability at least 1 / 2 1/2 1/2 . Moreover, since R ≤ 2 R ⋆ R\le2R_\star R ≤ 2 R ⋆ and R ⋆ = O ~ ( k ∥ U ∥ ∞ ) R_\star=\widetilde{O}\left(\sqrt{k}\lVert U\rVert_\infty\right) R ⋆ = O ( k ∥ U ∥ ∞ ) ,
ν R 2 ( T ) = O ~ ( Δ 2 R 2 d σ k 2 κ k ) = O ~ ( d k σ k 2 ∥ U ∥ ∞ 2 ϵ 2 Δ k 2 ) . \nu_R^2(T)=\widetilde{O}\left(\frac{\Delta^2R^2d}{\sigma_k^2\kappa_k}\right)=\widetilde{O}\left(\frac{dk\sigma_k^2\lVert U\rVert_\infty^2}{\epsilon^2\Delta_k^2}\right). ν R 2 ( T ) = O ( σ k 2 κ k Δ 2 R 2 d ) = O ( ϵ 2 Δ k 2 d k σ k 2 ∥ U ∥ ∞ 2 ) .
It remains to account for the hyperparameter search and the noisy comparison scores. Each independent base trial samples ( R , T ) (R,T) ( R , T ) with probability 1 / m 1/m 1/ m , so it produces a candidate of error at most e † e_\dagger e † with probability p ≥ 1 / ( 2 m ) p\ge1/(2m) p ≥ 1/ ( 2 m ) . The probability that no such candidate occurs before the geometric stopping time is
E ( 1 − p ) N = θ ( 1 − p ) p + θ ( 1 − p ) ≤ θ p ≤ β 4 . \mathbb{E}(1-p)^N=\frac{\theta(1-p)}{p+\theta(1-p)}\le\frac{\theta}{p}\le\frac{\beta}{4}. E ( 1 − p ) N = p + θ ( 1 − p ) θ ( 1 − p ) ≤ p θ ≤ 4 β .
Let
t v a l = 2 ζ h 2 log 8 β θ . t_{\mathrm{val}}=\frac{2\zeta}{h}\sqrt{2\log\frac{8}{\beta\theta}}. t val = h 2 ζ 2 log β θ 8 .
Each noisy score differs from its true score by more than t v a l t_{\mathrm{val}} t val with probability at most β θ / 4 \beta\theta/4 β θ /4 . Conditioning on N N N and taking a union bound over the trials, then using E N = 1 / θ \mathbb{E}N = 1/\theta E N = 1/ θ , shows that all score errors are at most t v a l t_{\mathrm{val}} t val with probability at least 1 − β / 4 1-\beta/4 1 − β /4 . If a candidate of error at most e † e_{\dagger} e † occurs and all score errors satisfy this bound, the selected candidate has true error at most e † + 2 t v a l e_{\dagger}+2t_{\mathrm{val}} e † + 2 t val .
Since h ≥ Δ k / 2 h \ge\Delta_k/2 h ≥ Δ k /2 and ζ = O ~ ( 1 / ϵ ) \zeta= \widetilde{O}(1/\epsilon) ζ = O ( 1/ ϵ ) , we have t v a l 2 = O ~ ( 1 / ( ϵ 2 Δ k 2 ) ) t_{\mathrm{val}}^2 = \widetilde{O}(1/(\epsilon^2\Delta_k^2)) t val 2 = O ( 1/ ( ϵ 2 Δ k 2 )) . Consequently,
∥ ( I − X ^ X ^ ⊤ ) V k ∥ 2 2 ≤ ( e † + 2 t v a l ) 2 ≤ 2 15 ( β δ ϵ n ) 2 + O ~ ( ν R 2 ( T ) + t v a l 2 ) ≤ ( β δ ϵ n ) 2 + O ~ ( d k σ ∞ 2 ∥ U ∥ 2 + 1 ϵ 2 Δ k 2 ) . \begin{aligned}
\left\|(I-\widehat{X}\widehat{X}^{\top})V_k\right\|_2^2
&\le(e_{\dagger}+2t_{\mathrm{val}})^2 \\
&\le\frac{2}{15}\left(\frac{\beta\delta\epsilon}{n}\right)^2+\widetilde{O}\left(\nu_R^2(T)+t_{\mathrm{val}}^2\right) \\
&\le\left(\frac{\beta\delta\epsilon}{n}\right)^2+\widetilde{O}\left(\frac{dk\sigma_\infty^2\|U\|^2+1}{\epsilon^2\Delta_k^2}\right).
\end{aligned} ( I − X X ⊤ ) V k 2 2 ≤ ( e † + 2 t val ) 2 ≤ 15 2 ( n β δ ϵ ) 2 + O ( ν R 2 ( T ) + t val 2 ) ≤ ( n β δ ϵ ) 2 + O ( ϵ 2 Δ k 2 d k σ ∞ 2 ∥ U ∥ 2 + 1 ) .
The initialization, search, and score-error events fail with total probability at most β / 4 + β / 4 + β / 4 ≤ β \beta/4+\beta/4+\beta/4\le\beta β /4 + β /4 + β /4 ≤ β , proving (24) .