Lyapunov equation with semidefinite right-hand side

Question

Consider the Lyapunov equation

$$A^TX+XA=-Q$$

with Hurwitz matrix $A$ and positive semi-definite matrix $Q\succeq0$.

When is its solution $X$ strictly positive definite?

Answer 1

As is noted on the Wikipedia page about the Lyapnov equation, the (unique) solution to this equation can be written as $$ X = \int_0^\infty e^{A^T\tau}Qe^{A\tau}\,d\tau. $$ In all cases, this matrix will be positive semidefinite. The question remains as to when this matrix is strictly positive definite, i.e. when this matrix is also invertible.

Claim: Given a Hurwitz $A$ and $Q \succeq 0$, $X = \int_0^\infty e^{A^T\tau}Qe^{A\tau}\,d\tau$ is invertible if and only if the observability matrix $$ \mathcal C = \pmatrix{Q\\ QA\\ \vdots \\ QA^{n-1}} $$ has full column-rank.

Proof:

$\implies$ (necessity of criterion): Suppose that $\mathcal C$ is not of full column rank. It follows that there exists a vector $v \neq 0$ such that $\mathcal Cv = 0$, which is to say that for $0 \leq k \leq n-1$, we have $QA^k v = 0$. By the Cayley Hamilton theorem, it follows that $QA^k v = 0$ for all $k \geq 0$. It follows that for all $t\geq 0$, we have $$ Qe^{A \tau} v = \sum_{k=0}^\infty \frac{\tau^k}{k!}QA^kv = 0. $$ It follows that $$ Xv = \left(\int_0^\infty e^{A^T \tau}Qe^{A\tau} \,d\tau\right) v = \int_0^\infty e^{A^T \tau}\left(Qe^{A\tau}v\right)\,d\tau = 0 $$ Because $Xv = 0$, $X$ fails to be invertible. By contrapositive, if $X$ is invertible, then $\mathcal C$ must have full column rank.

$\Longleftarrow$ (sufficiency): see p. 21 of these notes for a proof.