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Nyström deterministic spectral residual bound under a full-row-rank leading sketch

Statement

Let F∈{R,C}, n≥1, s≥0, and let A∈Fn×n be Hermitian positive semidefinite with eigenvalues λ1≥⋯≥λn≥0. Let S∈Fn×s, and choose an integer 0≤k≤n and an orthonormal eigenbasis U=(U1 U2) with the first k eigenvectors in U1. Set Λ1=diag⁡(λ1,…,λk), Λ2=diag⁡(λk+1,…,λn), Ω1=U1∗S, and Ω2=U2∗S. Empty blocks have norm zero, λn+1:=0, and the 0×s matrix has full row rank vacuously. If Ω1 has full row rank, then, deterministically,

∥A−A^nys∥2≤λk+1+∥Λ21/2Ω2Ω1†∥22.

Here ∥M∥2=sup⁡∥x∥2=1∥Mx∥2 is the induced Euclidean operator norm. No probability assumption on S is made.

Facts & Assumptions

Given: A finite Hermitian positive-semidefinite A, a conformal sketch S, an eigenvalue split at k, and full row rank of Ω1.

[F1]

Nyström is a square-root orthogonal-projection compression: A^nys=A1/2PRan⁡(A1/2S)A1/2.

[F3]

The Moore-Penrose products are orthogonal projectors: if Ω1 has full row rank, then Ω1Ω1†=Ik.

[F6]

A non-negative operator has a unique non-negative square root: the positive-semidefinite square root is unique.

Proof

1.1F1F4F5algebra

Put M=A1/2, Y=MS, and let P=PRan⁡(Y). By [F1], A−A^nys=M(I−P)M=X∗X for X=(I−P)M, since P is self-adjoint and idempotent. A singular value decomposition from [F4] writes X=UXΣXVX∗ and X∗X=VXΣX2VX∗; [F5] therefore gives ∥A−A^nys∥2=∥X∥22.

1.2F2F3F6algebra

By [F2], choose an orthonormal eigenbasis U of A. The diagonal matrix whose entries are the nonnegative square roots of its eigenvalues is a positive-semidefinite square root, hence it is A1/2 by uniqueness [F6]. Thus in this basis, M=Udiag⁡(Λ11/2,Λ21/2)U∗ and U∗Y=[Λ11/2Ω1Λ21/2Ω2]. For any unit vector x=U(x1,x2), set c=Ω1†x1. Then [F3] gives Ω1c=x1, and Yc=U[Λ11/2x1Λ21/2Ω2Ω1†x1]∈Ran⁡(Y).

2.1step 1.1step 1.2algebra

Orthogonal projection gives the shortest distance to its range, so ∥(I−P)Mx∥2≤∥Mx−Yc∥2=∥Λ21/2(x2−Ω2Ω1†x1)∥2. Write G=Λ21/2Ω2Ω1† and H=Λ21/2. For unit (x1,x2), the last norm is at most ∥G∥2∥x1∥2+∥H∥2∥x2∥2≤(∥G∥22+∥H∥22)1/2 by the scalar Cauchy-Schwarz inequality. Taking the supremum over unit x yields ∥X∥22≤∥G∥22+∥H∥22.

3.1F2F5step 1.1step 2.1∎

Since H is diagonal with nonnegative entries, [F2] and [F5] give ∥H∥22=λk+1, including the declared empty-tail convention. Combine this with step 1.1 and step 2.1 to obtain ∥A−A^nys∥2≤λk+1+∥Λ21/2Ω2Ω1†∥22.

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