TITLE: Show that in Lyapunov equation $A^TQ+QA=-I$, the matrix $Q$ is positive definite.
QUESTION [5 upvotes]: Let $A$ be matrix whose eigenvalues all have negative real parts. Define $Q=\int^{\infty}_0 B(t)dt$ where $B(t)=e^{A^Tt}e^{At}$. Prove that $Q$ is symmetric and positive definite.

This question is related to the corresponding Lyapunov equation $A^TQ+QA=-I$. 
By the above we know that $B(t)^T=B(t)$ and $\forall x \neq 0. x^TB(t)x>0$. Therefore:
\begin{align}
-I
&=\lim_{\tau \to \infty} B(\tau) -I\\
&=\lim_{\tau \to \infty} \int^{\tau}_0\frac{d B(t)}{dt} \\
&= \lim_{\tau \to \infty} \Big( A^T\int^{\tau}_0B(t)dt+\int^{\tau}_0B(t)dt\ A \Big)\\
&=A^TQ+QA\\
\end{align}
However I am confused on how to use these facts to show that $Q$ is symmetric and positive definite.

REPLY [3 votes]: The key is to note that any (pointwise) constant linear transformation commutes with integration.  For example, we can show that $Q$ is symmetric since
$$
Q^T = \left[\int B(t)\right]^T = \int[B(t)]^T = 
\int[e^{At}]^T[e^{A^Tt}]^T = 
\int e^{A^Tt}e^{At} = \int B(t) = Q
$$
similarly, show that $x^TQx > 0$ so that $Q$ is positive definite. Note in particular that
$$
x^TB(t)x = \|e^{At}x\|^2
$$
Moreover: if $x \neq 0$, $t \mapsto \|e^{At}x\|^2$ is necessarily a continuous, positive-valued function.