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Nyström deterministic spectral residual bound under a full-row-rank leading sketch
Statement
Let , , , and let be Hermitian positive semidefinite with eigenvalues . Let , and choose an integer and an orthonormal eigenbasis with the first eigenvectors in . Set , , , and . Empty blocks have norm zero, , and the matrix has full row rank vacuously. If has full row rank, then, deterministically,
Here is the induced Euclidean operator norm. No probability assumption on is made.
Facts & Assumptions
Given: A finite Hermitian positive-semidefinite , a conformal sketch , an eigenvalue split at , and full row rank of .
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: has an orthonormal eigenbasis with nonnegative eigenvalues.
The Moore-Penrose products are orthogonal projectors: if has full row rank, then .
Every linear map between finite-dimensional real or complex inner product spaces admits a singular value decomposition: every finite-dimensional linear map has a singular value decomposition.
The operator norm is 0 on the zero domain and otherwise equals the largest singular value, attained at a right-singular vector: the induced operator norm equals the largest singular value.
A non-negative operator has a unique non-negative square root: the positive-semidefinite square root is unique.
Proof
Put , , and let . By [F1], for , since is self-adjoint and idempotent. A singular value decomposition from [F4] writes and ; [F5] therefore gives .
By [F2], choose an orthonormal eigenbasis of . The diagonal matrix whose entries are the nonnegative square roots of its eigenvalues is a positive-semidefinite square root, hence it is by uniqueness [F6]. Thus in this basis, and . For any unit vector , set . Then [F3] gives , and .
Orthogonal projection gives the shortest distance to its range, so . Write and . For unit , the last norm is at most by the scalar Cauchy-Schwarz inequality. Taking the supremum over unit yields .
Since is diagonal with nonnegative entries, [F2] and [F5] give , including the declared empty-tail convention. Combine this with step 1.1 and step 2.1 to obtain .
Depends on
- The positive-semidefinite Nyström approximation
- Nyström is a square-root orthogonal-projection compression
- The Moore-Penrose products are orthogonal projectors
- 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
- A non-negative operator has a unique non-negative square root
- The operator norm is 0 on the zero domain and otherwise equals the largest singular value, attained at a right-singular vector
- Every linear map between finite-dimensional real or complex inner product spaces admits a singular value decomposition
Used by
Nothing in the library uses this result yet.
Dependency tree · two levels
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Sources
- Alex Gittens and Michael W. Mahoney, Revisiting the Nyström Method, Theorem 2 and §4.1.1 (standard reference, not scraped)
- Nathan Halko, Per-Gunnar Martinsson, and Joel A. Tropp, Finding Structure with Randomness, Theorem 9.1 (standard reference, not scraped)