7
$\begingroup$

Let $A\in\mathbb{R}^{n\times n}$ be a matrix with eigenvalues having (strictly) negative real part. Let $X\in\mathbb{R}^{n\times n}$, $X\succ 0$, be a positive definite matrix and let $P\succ 0$ be the (unique) solution of the following Lyapunov matrix equation $$\tag{1}\label{eq1} AP+PA^\top = - X. $$

My question. Let $\{X_n\}_{n\ge 0}$, $X_n\succ 0$, be a sequence of positive definite matrices, and let $\{P_n\}_{n\ge 0}$, $P_n\succ 0$, be the corresponding sequence of solutions of \eqref{eq1}. Suppose that $\lim_{n\to \infty} X_n=\bar{X}\succeq 0$ and $\lim_{n\to \infty} P_n=\bar{P}\succeq 0$ is singular. I'm wondering whether $\lim_{n\to \infty} P_n^{-1/2}X_n P_n^{-1/2}$ always converges to a finite matrix.

A few remarks are in order.

  1. If $\bar{P}$ is singular, then, $\bar{X}$ must be singular as well.
  2. If $A$ is a scalar matrix, i.e. $A=\alpha I$, $\alpha<0$, then it is quite easy to see that $\lim_{n\to \infty} P_n^{-1/2}X_n P_n^{-1/2}$ converges to a finite matrix.
  3. By pre- and post-multiplying \eqref{eq1} by $P_n^{-1/2}$, it follows that if $\lim_{n\to \infty} P_n^{-1/2}A P_n^{1/2}$ is finite, then $\lim_{n\to \infty} P_n^{-1/2}X_n P_n^{-1/2}$ is finite as well.
  4. From point 3. it holds $\mathrm{tr}(P_n^{-1/2}X_n P_n^{-1/2})=-2\,\mathrm{tr}(A)$ for all $n$.
$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.