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A Neumann defect bounds the inverse norm

Statement

Let $E \in C^{n \times n}$ , $n \geq 1$ , and let $0 \leq q < 1$ satisfy $∥ E ∥_{2} \leq q$ . Then $I - E$ is invertible and

∥ (I - E)^{- 1} ∥_{2} \leq \frac{1}{1 - q} .

Facts & Assumptions

Given: $E, q, n$ as in the statement.

[F1]

If $∥ X ∥_{2} < 1$ , then $I + X$ is invertible (A finite matrix within unit distance of the identity is invertible).

[F2]

The operator norm is submultiplicative (The operator norm is submultiplicative and satisfies ||T^*T|| = ||T||^2).

[F3]

The operator norm is induced by the Euclidean vector norm (The operator norm is zero on the zero domain and otherwise is max_{||v||=1} ||Tv||).

Proof

1.1givenF1

Since $∥ - E ∥_{2} = ∥ E ∥_{2} \leq q < 1$ , [F1] gives that $T = (I - E)^{- 1}$ exists.

2.1givenF2F3step 1.1algebra∎

Put $S_{N} = \sum_{k = 0}^{N} E^{k}$ . By induction and [F2], $∥ E^{k} ∥_{2} \leq q^{k}$ , so $∥ S_{N} ∥_{2} \leq \sum_{k = 0}^{N} q^{k} = (1 - q^{N + 1}) / (1 - q)$ by [F3]. The finite identity $(I - E) S_{N} = I - E^{N + 1}$ , multiplied on the left by $T$ , gives $S_{N} = T - T E^{N + 1}$ and hence $T = S_{N} + T E^{N + 1}$ . By the triangle inequality from [F3] and submultiplicativity [F2], $(1 - q^{N + 1}) ∥ T ∥_{2} \leq ∥ S_{N} ∥_{2} \leq (1 - q^{N + 1}) / (1 - q)$ . Since $q^{N + 1} < 1$ , division gives $∥ T ∥_{2} \leq (1 - q)^{- 1}$ . This proves the inverse norm bound without taking a limit of matrices.

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Sources

Siegfried M. Rump, Verification methods: Rigorous results using floating-point arithmetic, Acta Numerica 19 (2010), §10.5 (standard reference, not scraped)
Nicholas J. Higham, Numerical Linear Algebra and Matrix Analysis, MIMS EPrint 2015.103, §1 (standard reference, not scraped)