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Sylvester separation converts a residual into a forward error bound

Statement

Let $X$ be the unique solution of $A X + X B = C$ , and let $\hat{X} \in F^{m \times n}$ be any approximation. Define its equation residual by $R = C - (A \hat{X} + \hat{X} B)$ . If $sep (A, - B) > 0$ , then

∥ X - \hat{X} ∥_{F} \leq \frac{∥ R ∥_{F}}{sep (A, - B)} .

Facts & Assumptions

Given: Compatible matrices $A, B, C$ , the unique solution $X$ , the approximation $\hat{X}$ , and positive $sep (A, - B)$ .

[F1]

Sylvester operator and Frobenius separation number: $S_{A, B} (Y) = A Y + Y B$ and $sep (A, - B) = σ_{\min} (S_{A, B})$ .

[F2]

Positive Sylvester separation bounds the inverse operator norm: $∥ Y ∥_{F} \leq ∥ S_{A, B} (Y) ∥_{F} / sep (A, - B)$ .

[F3]

A Sylvester equation is uniquely solvable exactly when the spectra are disjoint: Positive separation implies that every right-hand side has a unique solution.

Proof

1.1givenF1algebra

Put $E = X - \hat{X}$ . By linearity of $S_{A, B}$ and the two equations, $S_{A, B} (E) = S_{A, B} (X) - S_{A, B} (\hat{X}) = C - (A \hat{X} + \hat{X} B) = R$ .

2.1F2F3step 1.1algebra∎

Apply [F2] to $E$ and substitute step 1.1 to obtain $∥ X - \hat{X} ∥_{F} = ∥ E ∥_{F} \leq ∥ R ∥_{F} / sep (A, - B)$ . The unique solution used here is guaranteed by [F3].

Depends on

Used by

Dependency tree · two levels

18 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Valeria Simoncini, Computational Methods for Linear Matrix Equations, SIAM Review 58(3), 2016 (standard reference, not scraped)
Elias Jarlebring, Numerical Methods for Lyapunov Equations, lecture notes in numerical linear algebra, 2017 (standard reference, not scraped)