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Stabilizability and detectability imply the closed-region PBH rank tests

Statement

For $F \in {R, C}$ , $n, m, p \geq 1$ , $A \in F^{n \times n}$ , $B \in F^{n \times m}$ , and $C \in F^{p \times n}$ , the following necessary conditions hold.

If $(A, B)$ is continuously stabilizable, then $rank [λ I - A B] = n$ for every $λ \in C$ with $Re λ \geq 0$ . If it is discretely stabilizable, the same rank is $n$ for every $∣ λ ∣ \geq 1$ .

If $(A, C)$ is continuously detectable, then $rank [\begin{array}{c} λ I - A \\ C \end{array}] = n$ for every $Re λ \geq 0$ . If it is discretely detectable, the same rank is $n$ for every $∣ λ ∣ \geq 1$ .

This proves the necessary half of the PBH tests. The converse pole-placement implication is not asserted here.

Facts & Assumptions

Given: The matrices and continuous/discrete stabilizability and detectability definitions of Stabilizability and detectability for continuous and discrete linear systems. Every rank test and eigenvector argument is over $C$ ; when $F = R$ , the real matrices and their real feedback or injection witnesses act complex-linearly on their complexifications.

[F1]

Stabilizability and detectability for continuous and discrete linear systems: continuous stabilizability means $A - B K$ is Hurwitz for some $K$ ; discrete stabilizability means $A - B K$ is Schur stable; detectability is the dual condition with $A - L C$ .

[F2]

Row rank equals column rank, and both equal the number of pivots: for a finite matrix, row rank and column rank agree.

[F3]

Rank-nullity: $\dim_{F} V = nullity T + rank T$ : for a linear map with finite-dimensional domain, dimension of domain equals rank plus nullity.

[A1]

For $F = R$ , real matrices and feedback/injection witnesses are extended complex-linearly when tested against a complex $λ$ .

Proof

1.1A1F1F2F3contradiction

Suppose $(A, B)$ is continuously stabilizable and fix its same-field feedback $K$ from [F1]. If for some $λ$ with $Re λ \geq 0$ the horizontal matrix $H_{λ} = [λ I - A B]$ has rank less than $n$ , [F2] and [F3] applied over $C$ to $H_{λ}^{*}$ give a nonzero $y \in C^{n}$ with $y^{*} (λ I - A) = 0$ and $y^{*} B = 0$ . By [A1], the same real or complex $K$ acts on this vector, so $y^{*} (A - B K) = λ y^{*}$ . Thus $\det (λ I - (A - B K)) = 0$ , contradicting that $A - B K$ is Hurwitz.

1.2A1F1F3contradiction

Suppose $(A, C)$ is continuously detectable and fix its same-field injection $L$ from [F1]. If $V_{λ} = [\begin{array}{c} λ I - A \\ C \end{array}]$ has rank less than $n$ , rank-nullity over $C$ gives a nonzero $x \in C^{n}$ with $(λ I - A) x = 0$ and $C x = 0$ . Hence $(A - L C) x = λ x$ , which contradicts that $A - L C$ is Hurwitz because $Re λ \geq 0$ .

2.1A1F1F2F3step 1.1contradiction

Suppose $(A, B)$ is discretely stabilizable and fix its same-field Schur feedback $K$ from [F1]. If $[λ I - A B]$ has rank less than $n$ for some $∣ λ ∣ \geq 1$ , the same complex left-kernel argument as in step 1.1 gives a nonzero $y$ with $y^{*} (A - B K) = λ y^{*}$ . Thus $\det (λ I - (A - B K)) = 0$ , contradicting Schur stability.

3.1A1F1F3step 1.2contradiction∎

Suppose $(A, C)$ is discretely detectable and fix its same-field Schur injection $L$ from [F1]. If the vertical matrix has rank less than $n$ for some $∣ λ ∣ \geq 1$ , its complex kernel contains a nonzero $x$ and $C x = 0$ , so $(A - L C) x = λ x$ . This contradicts Schur stability.

Depends on

Used by

Closed-region PBH rank tests characterize continuous and discrete stabilizability and detectability Theorem

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Sources

Daniel Liberzon, Calculus of Variations and Optimal Control Theory: A Concise Introduction, Chapter 6 (standard reference, not scraped)
Peter Benner and Lihong Feng, Model Reduction for Dynamical Systems, Lecture 8 (standard reference, not scraped)
Federico Poloni, Lecture notes on Riccati-type matrix equations (standard reference, not scraped)
A.A. Stoorvogel and A. Saberi, The discrete algebraic Riccati equation and linear matrix inequality, Memorandum COSOR 94-16 (standard reference, not scraped)