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Kernel ridge predictions are stable under Gram-matrix perturbations

Statement

Let K,K~∈Rn×n be symmetric matrices such that xTKx≥0 and xTK~x≥0 for every x∈Rn, with n≥1, let y∈Rn, and let τ>0. Write In for the identity matrix. For G∈{K,K~} define aG=(G+τIn)−1y and uG=GaG. Define ∥x∥2=(∑ixi2)1/2 for vectors and ∥M∥2=sup⁡∥x∥2=1∥Mx∥2 for matrices. If ∥K−K~∥2≤δ for some δ≥0, then

∥aK−aK~∥2≤δ∥y∥2τ2,∥uK−uK~∥2≤δ∥y∥2τ.

In particular, let S∈Rn×s with s≥0 and K~=KS(STKS)†STK, where M† denotes the unique matrix satisfying the four Moore–Penrose equations. Choose 0≤r≤n and an orthogonal eigenbasis K=Udiag⁡(λ1,…,λn)UT with UTU=In and λ1≥⋯≥λn≥0. Write U=(U1 U2) with the first r eigenvectors in U1, let Λ2=diag⁡(λr+1,…,λn) and Λ21/2=diag⁡(λr+1,…,λn), and set Ω1=U1TS and Ω2=U2TS. If Ω1 has full row rank, meaning rank⁡(Ω1)=r, with the empty-row convention when r=0, then the same bounds hold with δ=λr+1+∥Λ21/2Ω2Ω1†∥22, using λn+1=0 and zero norm for empty blocks.

Facts & Assumptions

Given: Positive-semidefinite matrices K,K~, vector y, scalar τ>0, and the norm error ∥K−K~∥2≤δ in the Statement.

[F1]

Regularized kernel prediction system: for a positive-semidefinite matrix G, aG=(G+τIn)−1y and uG=GaG=y−τaG.

[F2]

Inverse norm bound for a regularized Gram matrix: G+τIn is invertible, ∥(G+τIn)−1∥2≤1/τ, and the two shifted inverses satisfy the resolvent identity.

[F3]

The positive-semidefinite Nyström approximation: for a positive-semidefinite K and finite sketch S, K~=KS(STKS)†STK is the Nyström matrix.

[F5]

Nyström deterministic spectral residual bound under a full-row-rank leading sketch: if Ω1 has full row rank, then ∥K−K~∥2≤λr+1+∥Λ21/2Ω2Ω1†∥22 under the eigenbasis and block definitions in the Statement.

Proof

1.1givenF2algebra

Put A=K+τIn and B=K~+τIn. By [F2], both inverses exist and have norm at most 1/τ.

2.1givenF1step 1.1algebra

The resolvent identity gives aK−aK~=(A−1−B−1)y=A−1(K~−K)B−1y. Taking norms and applying the bounds from step 1.1 and the assumed Gram error gives ∥aK−aK~∥2≤δ∥y∥2/τ2.

3.1F1step 2.1algebra

By [F1], uK=y−τaK and uK~=y−τaK~. Subtracting these equalities and using step 2.1 gives ∥uK−uK~∥2=τ∥aK−aK~∥2≤δ∥y∥2/τ.

4.1givenF3F4F5step 2.1step 3.1algebra∎

In the Nyström case, [F4] supplies positive semidefiniteness of K~, and [F5] supplies the stated value of δ whenever Ω1 has full row rank. Substituting that bound into steps 2.1 and 3.1 proves the conditional Nyström conclusion.

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