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Positive Sylvester separation bounds the inverse operator norm

Statement

For the Sylvester operator SA,B of Sylvester operator and Frobenius separation number, if sep⁡(A,−B)>0, then every X∈Fm×n satisfies

∥X∥F≤∥AX+XB∥Fsep⁡(A,−B).

Facts & Assumptions

Given: F∈{R,C}, A∈Fm×m, B∈Fn×n, and sep⁡(A,−B)>0.

[F1]

Sylvester operator and Frobenius separation number: sep⁡(A,−B)=min⁡∥Y∥F=1∥AY+YB∥F.

Proof

1.1givenchoosealgebra

If X=0, the claimed inequality is equality. If X≠0, set Y=X/∥X∥F, so ∥Y∥F=1 and SA,B(Y)=SA,B(X)/∥X∥F by linearity.

2.1F1step 1.1algebra∎

By [F1], sep⁡(A,−B)≤∥SA,B(Y)∥F=∥SA,B(X)∥F/∥X∥F. Positivity of the separation permits multiplication and division by it, giving the desired bound.

Depends on

Used by

Dependency tree · two levels

4 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