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Orthogonal-projector power error inequality
Statement
Let , let be an orthogonal projector on , and let be an integer. Then
Facts & Assumptions
Given: A finite real or complex matrix , an orthogonal projector , and an integer .
has an SVD, and its squared singular values diagonalize (Every linear map between finite-dimensional real or complex inner product spaces admits a singular value decomposition).
Positive semidefinite finite-dimensional operators have an orthonormal eigenbasis (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).
The spectral norm of a positive semidefinite operator is its largest eigenvalue (The operator norm is zero on the zero domain and otherwise is max_{||v||=1} ||Tv||, The operator norm is 0 on the zero domain and otherwise equals the largest singular value, attained at a right-singular vector).
Jensen's inequality holds for finite probability distributions (Jensen's inequality for expectation).
Proof
Choose an SVD and put and . Then is diagonal positive semidefinite and is an orthogonal projector. By [F1]–[F3], and .
If , [step 1.1] gives , so the asserted inequality follows. If , choose a unit eigenvector for the top eigenvalue of ; then and , with . The weights sum to , so [F4] for gives .
Substitute from step 1.1 into step 2.1 and take the nonnegative st root. This yields the asserted inequality, also for .
Depends on
- Every linear map between finite-dimensional real or complex inner product spaces admits a singular value decomposition
- 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
- The operator norm is zero on the zero domain and otherwise is max_{||v||=1} ||Tv||
- The operator norm is 0 on the zero domain and otherwise equals the largest singular value, attained at a right-singular vector
- Jensen's inequality for expectation
Used by
Dependency tree · two levels
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Sources
- Halko, Martinsson, and Tropp, Proposition 8.6 and Theorem 9.2 (standard reference, not scraped)
- Martinsson and Tropp, Randomized Numerical Linear Algebra: Foundations & Algorithms, Lemma 11.8 (standard reference, not scraped)