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Inverse norm bound for a regularized Gram matrix

Statement

Let $M, N \in R^{n \times n}$ be symmetric matrices such that $x^{T} M x \geq 0$ and $x^{T} N x \geq 0$ for every $x \in R^{n}$ , where $n \geq 1$ , and let $τ > 0$ . Write $I_{n}$ for the identity matrix, $A = M + τ I_{n}$ , and $B = N + τ I_{n}$ . For vectors use $∥ x ∥_{2} = (\sum_{i} x_{i}^{2})^{1 / 2}$ , and for a matrix $T$ use the induced Euclidean norm $∥ T ∥_{2} = \sup_{∥ x ∥_{2} = 1} ∥ T x ∥_{2}$ . Positive definite means $x^{T} A x > 0$ for every nonzero $x$ , and analogously for $B$ . Then $A$ and $B$ are positive definite and invertible, $∥ A^{- 1} ∥_{2}, ∥ B^{- 1} ∥_{2} \leq 1 / τ$ , and

$A^{- 1} - B^{- 1} = A^{- 1} (N - M) B^{- 1}, ∥ A^{- 1} - B^{- 1} ∥_{2} \leq \frac{∥ M - N ∥_{2}}{τ^{2}} .$

Statement references

A finite-dimensional real symmetric positive-semidefinite operator has an orthonormal eigenbasis with nonnegative eigenvalues.

Facts & Assumptions

Given: Symmetric positive-semidefinite matrices $M, N \in R^{n \times n}$ 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 $x^{T} A x = x^{T} M x + τ ∥ x ∥_{2}^{2} \geq τ ∥ x ∥_{2}^{2} > 0$ and $x^{T} B x = x^{T} N x + τ ∥ x ∥_{2}^{2} \geq τ ∥ x ∥_{2}^{2} > 0$ . Thus $A$ and $B$ are positive definite and invertible. By [F1], write $M = U diag (μ_{1}, \dots, μ_{n}) U^{T}$ with $U$ orthogonal and each $μ_{i} \geq 0$ . Then $A = U diag (μ_{i} + τ) U^{T}$ and $A^{- 1} = U diag ((μ_{i} + τ)^{- 1}) U^{T}$ ; applying the same diagonalization argument to $N$ gives the corresponding formula for $B^{- 1}$ .

2.1givenstep 1.1algebra

Orthogonality of $U$ gives $∥ A^{- 1} x ∥_{2}^{2} = \sum_{i} (U^{T} x)_{i}^{2} (μ_{i} + τ)^{- 2} \leq τ^{- 2} ∥ x ∥_{2}^{2}$ for every $x$ . Taking the supremum over unit vectors gives $∥ A^{- 1} ∥_{2} \leq 1 / τ$ ; the identical calculation for $N$ gives $∥ B^{- 1} ∥_{2} \leq 1 / τ$ .

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 $∥ A^{- 1} (N - M) B^{- 1} x ∥_{2} \leq ∥ A^{- 1} ∥_{2} ∥ N - M ∥_{2} ∥ B^{- 1} ∥_{2}$ . Taking the supremum and using step 2.1 proves the asserted resolvent bound.

Depends on

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

Used by

Dependency tree · two levels

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Sources

Sheldon Axler, Linear Algebra Done Right, 4th ed., §§7B–7C (standard reference, not scraped)