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.


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.

  • $\begingroup$ I did not know that linear transformations commute with integration. This make the problem much easier. $\endgroup$ – AzJ Feb 7 '17 at 19:56
  • $\begingroup$ @AzJ If that's the case, you should consider trying to prove that this is the case. Not necessarily for this particular assignment, but at some point. $\endgroup$ – Omnomnomnom Feb 7 '17 at 20:03
  • $\begingroup$ someone might want to check transpose of exponential is exponential of the transpose: math.stackexchange.com/questions/1021900/… $\endgroup$ – Javier Jun 3 at 8:35

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