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A negative-real-part shift of a Hurwitz matrix is invertible

Statement

If $A$ is Hurwitz and $p \in C$ has $Re p < 0$ , then $A + p I$ is invertible over $C$ .

Facts & Assumptions

Given: A Hurwitz matrix $A$ and a scalar $p$ with negative real part.

[F1]

Hurwitz matrices and the continuous Lyapunov equation: Every eigenvalue of a Hurwitz matrix has negative real part.

[F2]

For an $m \times n$ matrix $A$ , $rank (A) + \dim N (A) = n$ : For a square matrix, its rank plus the dimension of its kernel is the dimension of its domain.

Proof

1.1F2assume-contrachoosealgebra

If $A + p I$ were singular, its kernel would be nonzero: by [F2], a zero kernel would give full rank, hence a surjective and therefore invertible endomorphism of the finite-dimensional space. Choose nonzero $v$ in this kernel. Then $(A + p I) v = 0$ , so $A v = - p v$ and $- p$ would be an eigenvalue of $A$ .

2.1F1step 1.1discharge-contradiction∎

Since $Re (- p) > 0$ , this contradicts [F1], which requires every eigenvalue of $A$ to have negative real part. Therefore $A + p I$ is invertible.

Depends on

Used by

Dependency tree · two levels

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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)