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A singular A can give a stable DARE solution and an infinite pencil eigenvalue

Example

Take scalar $A = B = 0$ , $C = R = 1$ . Then $Q = 1$ and the state matrix $A$ is singular.

Facts & Assumptions

Given: The data above and the homogeneous pencil convention of Standard CARE and DARE data, equations, and conventions.

[F1]

Standard CARE and DARE data, equations, and conventions: the DARE is $X = Q + A^{*} X A - A^{*} X B (R + B^{*} X B)^{- 1} B^{*} X A$ , its closed loop is $A_{c} = A - B (R + B^{*} X B)^{- 1} B^{*} X A$ , and its homogeneous pencil is $β N - α M$ with finite coordinate $[λ : 1]$ and infinity $[1 : 0]$ .

Proof

1.1F1givenalgebra

Here $G = 0$ and $R + B^{*} X B = 1$ . The DARE reduces to $X = 1$ , so its solution is $X = 1$ and $A_{c} = 0$ , which is Schur stable.

2.1F1step 1.1algebra∎

The pencil matrices are $N = [\begin{array}{cc} 0 & 0 \\ - 1 & 1 \end{array}]$ and $M = [\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}]$ . Therefore $\det (β N - α M) = \det [\begin{array}{cc} - α & 0 \\ - β & β \end{array}] = - α β .$ This nonzero homogeneous polynomial has roots $[0 : 1]$ (the finite eigenvalue $0$ ) and $[1 : 0]$ (an infinite eigenvalue). No inverse of $A$ is used.

Depends on

Standard CARE and DARE data, equations, and conventions

Used by

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Dependency tree · two levels

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Sources

A.A. Stoorvogel and A. Saberi, The discrete algebraic Riccati equation and linear matrix inequality, Memorandum COSOR 94-16 (standard reference, not scraped)