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✓ 4 results · all verified · 4 also independently AI-judged
Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs; all 4 also cleared it.

Sketched kernel learning and certified approximate solves

1 · Prerequisites

2 · Logical flowchart

3 · Definitions, theorems and proofs

DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Regularized kernel prediction system

Definition

Let X be a set, k:X×X→R a real positive-semidefinite kernel, and H a real reproducing-kernel Hilbert space with kernel k. Let X=(xi)i=1n be a nonempty finite ordered training list in X, let K=(k(xi,xj))i,j=1n, let y∈Rn, and let λ>0. Set τ=nλ and define the mean-square empirical objective

J(f)=1n∑i=1n(yi−f(xi))2+λ∥f∥H2.

For any finite real symmetric positive-semidefinite matrix G∈Rn×n, define its regularized kernel prediction system by aG=(G+τIn)−1y and uG=GaG, where In is the n×n identity matrix. The original training system uses G=K. The same definition applies to any positive-semidefinite matrix approximation K~ of K, with aK~=(K~+τIn)−1y and uK~=K~aK~. In either case uG=y−τaG. The scalar τ=nλ fixes the normalization between mean-square loss and the unnormalized finite linear system.

Remarks

The definitions allow K to be singular or zero. More generally, for every finite real symmetric positive-semidefinite G and τ>0, each nonzero x satisfies xT(G+τIn)x=xTGx+τ∥x∥22≥τ∥x∥22>0. Thus each shifted matrix in the definition is positive definite and invertible, including when G=K or G=K~; no full-rank assumption is placed on the kernel Gram matrix or on a sketch.

LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

Inverse norm bound for a regularized Gram matrix

Statement

Let M,N∈Rn×n be symmetric matrices such that xTMx≥0 and xTNx≥0 for every x∈Rn, where n≥1, and let τ>0. Write In for the identity matrix, A=M+τIn, and B=N+τIn. For vectors use ∥x∥2=(∑ixi2)1/2, and for a matrix T use the induced Euclidean norm ∥T∥2=sup⁡∥x∥2=1∥Tx∥2. Positive definite means xTAx>0 for every nonzero x, and analogously for B. Then A and B are positive definite and invertible, ∥A−1∥2,∥B−1∥2≤1/τ, and

A−1−B−1=A−1(N−M)B−1,∥A−1−B−1∥2≤∥M−N∥2τ2.

Facts & Assumptions

Given: Symmetric positive-semidefinite matrices M,N∈Rn×n and τ>0.

[F1]

A non-negative operator is equivalently self-adjoint with nonnegative eigenvalues, a positive semidefinite matrix in an orthonormal basis, or an operator of the form S^*S: every finite-dimensional real symmetric positive-semidefinite operator has an orthonormal eigenbasis and nonnegative eigenvalues.

Proof

1.1givenF1algebra

Since τ>0 and M,N are positive semidefinite, every nonzero x satisfies xTAx=xTMx+τ∥x∥22≥τ∥x∥22>0 and xTBx=xTNx+τ∥x∥22≥τ∥x∥22>0. Thus A and B are positive definite and invertible. By [F1], write M=Udiag⁡(μ1,…,μn)UT with U orthogonal and each μi≥0. Then A=Udiag⁡(μi+τ)UT and A−1=Udiag⁡((μi+τ)−1)UT; applying the same diagonalization argument to N gives the corresponding formula for B−1.

2.1givenstep 1.1algebra

Orthogonality of U gives ∥A−1x∥22=∑i(UTx)i2(μi+τ)−2≤τ−2∥x∥22 for every x. Taking the supremum over unit vectors gives ∥A−1∥2≤1/τ; the identical calculation for N gives ∥B−1∥2≤1/τ.

3.1givenstep 2.1algebra∎

Since B−A=N−M, multiplying out yields A−1(B−A)B−1=A−1−B−1. For every unit vector x, the definition of the induced norm gives ∥A−1(N−M)B−1x∥2≤∥A−1∥2∥N−M∥2∥B−1∥2. Taking the supremum and using step 2.1 proves the asserted resolvent bound.

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

The kernel ridge system uniquely minimizes the mean-square RKHS objective

Statement

Let X be a set, let k:X×X→R be a real positive-semidefinite kernel, and let H be a real reproducing-kernel Hilbert space with kernel k. Take a finite ordered list X=(xi)i=1n with n≥1, a response vector y∈Rn, and λ>0. Put K=(k(xi,xj))i,j=1n and τ=nλ. Define a=(K+τIn)−1y, f0=∑i=1naik(⋅,xi), and u=Ka. Then f0 is the unique minimizer over f∈H of

J(f)=1n∑i=1n(yi−f(xi))2+λ∥f∥H2,

and its training evaluation vector is u=y−τa.

Statement references

Positive semidefinite kernels and reproducing kernel Hilbert spaces: A real RKHS contains each kernel section and satisfies ⟨f,k(⋅,x)⟩H=f(x).

Regularized kernel prediction system: The coefficient and fitted vectors are defined by a=(K+τI)−1y and u=Ka=y−τa.

Facts & Assumptions

Given: The kernel, RKHS, training list, response vector, and regularization parameter in the Statement.

[F1]

Positive semidefinite kernels and reproducing kernel Hilbert spaces: each section k(⋅,xi) belongs to H and ⟨f,k(⋅,xi)⟩H=f(xi); every finite Gram matrix of k is symmetric positive semidefinite.

[F2]

Regularized kernel prediction system: τ=nλ, the objective is the mean-square loss plus λ∥f∥H2, and a=(K+τIn)−1y.

[F3]

Inverse norm bound for a regularized Gram matrix: for K⪰0 and τ>0, K+τIn is invertible.

Proof

1.1givenF1F2F3constructalgebra

By [F1], K is a real symmetric positive-semidefinite matrix. Let a=(K+τIn)−1y, which exists by [F2]–[F3], and set f0=∑iaik(⋅,xi). The reproducing identities give f0(xj)=∑iaik(xj,xi)=(Ka)j, so if r=y−(f0(xi))i=1n, then r=y−Ka=τa by (K+τIn)a=y.

2.1givenstep 1.1F1F2algebra

Fix any h∈H. The reproducing identities yield ⟨f0,h⟩H=∑iaih(xi)=aT(h(xi))i=1n. Expanding both squared norms in J(f0+h)−J(f0) gives 1n∥(h(xi))i=1n∥22−2nrT(h(xi))i=1n+2λaT(h(xi))i=1n+λ∥h∥H2. Substituting r=τa=nλa cancels the two cross terms, leaving 1n∥(h(xi))∥22+λ∥h∥H2≥0.

3.1givenstep 1.1step 2.1algebra∎

Step 2.1 shows J(f0+h)≥J(f0) for every h, so f0 is a minimizer. Equality forces λ∥h∥H2=0, hence h=0 because λ>0; the minimizer is unique. The evaluation computation in step 1.1 gives its fitted vector u=Ka=y−τa.

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

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.

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

Kernel ridge residuals certify coefficient and training prediction errors

Statement

Let K∈Rn×n be symmetric with xTKx≥0 for every x∈Rn and n≥1, let y∈Rn, and let τ>0. Write In for the identity matrix. Define a=(K+τIn)−1y and u=Ka. For any vector b∈Rn, define the actual residual r=y−(K+τIn)b. Define ∥x∥2=(∑ixi2)1/2 and ∥M∥2=sup⁡∥x∥2=1∥Mx∥2; then

∥b−a∥2≤∥r∥2τ,∥Kb−u∥2≤∥K∥2∥r∥2τ.

Statement references

Regularized kernel prediction system: The exact coefficient and fitted vectors solve (K+τI)a=y and u=Ka.

Inverse norm bound for a regularized Gram matrix: The shifted Gram inverse has induced norm at most 1/τ.

Facts & Assumptions

Given: The positive-semidefinite matrix K, vector y, scalar τ>0, exact solution a, and arbitrary approximate coefficient vector b in the Statement.

[F1]

Regularized kernel prediction system: (K+τIn)a=y and u=Ka.

[F2]

Inverse norm bound for a regularized Gram matrix: K+τIn is invertible and ∥(K+τIn)−1∥2≤1/τ.

Proof

1.1givenF1F2algebra

By [F1], r=y−(K+τIn)b=(K+τIn)(a−b). Invertibility in [F2] therefore gives b−a=−(K+τIn)−1r, so ∥b−a∥2≤∥r∥2/τ.

2.1givenF1step 1.1algebra∎

Since u=Ka, the training fitted-vector error is Kb−u=K(b−a). For every vector x, the definition of the induced norm gives ∥Kx∥2≤∥K∥2∥x∥2. Applying this to b−a and using step 1.1 proves ∥Kb−u∥2≤∥K∥2∥r∥2/τ.

AlgorithmAlgorithm: AI-adaptedProof: Not applicableOpen item page →

Finite-budget Nyström kernel ridge fit with residual status

Definition

All matrix-vector operations below use exact real arithmetic. The inputs are a finite real symmetric matrix K∈Rn×n with n≥1 and xTKx≥0 for every x∈Rn, a vector y∈Rn, a scalar τ>0, a finite sketch S∈Rn×s with s≥0, a residual tolerance η≥0, and a finite nonnegative integer iteration budget T. Write In for the identity matrix and ∥x∥2=(∑ixi2)1/2 for the Euclidean norm. The Moore–Penrose inverse M† is the unique matrix satisfying MM†M=M, M†MM†=M†, (MM†)T=MM†, and (M†M)T=M†M.

  1. Form the Nyström matrix K~=KS(STKS)†STK. The formula is defined even when STKS is singular or s=0.

  2. Set B=K~+τIn. Initialize b0=0, r~0=y−Bb0=y, and p0=r~0.

  3. For j=0,1,…,T, recompute the actual residuals r~j=y−Bbj and rj=y−(K+τIn)bj. If both norms are at most η, return status certified, coefficient vector bj, fitted vector K~bj, and both residuals. If r~j=0 but ∥rj∥2>η, return status uncertified with the same outputs. If j=T and neither return condition holds, return status budget-exhausted with the same outputs.

  4. Otherwise compute qj=Bpj. If pjTqj=0, return status uncertified with the current outputs. If pjTqj>0, set αj=(r~jTr~j)/(pjTqj), bj+1=bj+αjpj, and r~j+1=r~j−αjqj. If r~j+1=0, recompute rj+1 and return certified when ∥rj+1∥2≤η, and otherwise return uncertified, together with the resulting outputs. If r~j+1≠0, set βj=(r~j+1Tr~j+1)/(r~jTr~j) and pj+1=r~j+1+βjpj, then continue at the next iteration.

The output always includes the actual residuals for both the Nyström system and the original system. certified asserts that both residuals meet the requested tolerance; uncertified and budget-exhausted make no solve-accuracy claim. In particular, exhausting the budget does not certify a solve. If K is a Gram matrix and τ=nλ, the original-system residual is the one used in the residual prediction certificate.

RemarkRemark: AI-adaptedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Statistical and numerical limits of finite kernel solve guarantees

Scope

The Gram-perturbation and residual bounds are deterministic finite-training-vector statements. A statistical risk here would be expected loss under a data-generating probability law; these bounds contain no such law, expected loss, or generalization term, so they do not by themselves give a statistical risk guarantee.

For a test input x∗, the kernel prediction is k∗TaK, where k∗=(k(x∗,xi))i=1n. An approximation certificate for the training Gram matrix alone does not bound the test cross-kernel vector k∗; an out-of-sample error bound requires a separate cross-kernel certificate.

Under AC, suppose the finite real PSD kernel k also has a real RKHS H with reproducing sections, use the zero mean m=0, and observe the canonical finite Gaussian process with independent noise covariance τIn. In the GP/KRR theorem take noise standard deviation σ=τ>0. Writing aK(Y)=(K+τIn)−1Y and uK(Y)=KaK(Y) for the random observation vector Y, the conditional-mean identity is

E[fX∣Y]=K(K+τIn)−1Y=uK(Y)almost surely.

The shifted observed covariance K+τIn is positive definite because τ>0. This optional interpretation uses the GP/KRR supplier and is separate from the choice-free deterministic matrix bounds.

For a fixed finite matrix A∈Rm×n or Cm×n, the cited range-finder results assume iid standard real or normalized complex Gaussian sketches, integers k,p∈N with k≥2 and p≥2, sketch width ℓ=k+p≤min⁡(m,n), exact arithmetic, and Countable Choice for the finite Gaussian probability law. They bound expected spectral error in those regimes. The theorem statements provide no per-draw bound or tail estimate for an arbitrary sketch supplied to the Nyström algorithm; a leverage-score or other sampling guarantee would require its own hypotheses and proof. The algorithm itself is specified in exact real arithmetic and includes residual statuses; no finite-precision rounding-error claim is made.

Scope references

4 · Examples, counterexamples and false statements

None yet.

Sources