11
$\begingroup$

Let $A\in\mathbb{R}^{n\times n}$ be a Hurwitz stable matrix (i.e., all the eigenvalues of $A$ have strictly negative real part). Let $X\in\mathbb{R}^{n\times n}$ be a positive semi-definite matrix of unit trace, that is $X\succeq 0$ s.t. $\mathrm{tr}(X)=1$, and let $P$ be the positive semidefinite solution of the following Lyapunov equation $$ AP+PA^\top = -X. $$

My question. Does there always exist a matrix $X\succeq 0$ with $\mathrm{tr}(X)=1$ such that the equality $$ \|P\|_2 = \frac{1}{-2\,\mathrm{tr}(A)} $$ holds true, where $\|P\|_2$ denotes the 2-norm of matrix $P$?

If $A+A^\top$ is negative semi-definite ($A+A^\top\preceq0$), then it is easy to see that the answer is in the affirmative. In fact, in this case, picking $X=\frac{1}{2\mathrm{tr}(A)}(A+A^\top)$, yields $P=\frac{1}{-2\mathrm{tr}(A)}I$, which in turn clearly implies $\|P\|_2=\frac{1}{-2\,\mathrm{tr}(A)}$.

After running an extensive amount of numerical simulations, it seems that the answer should be in the affirmative also for the case $A+A^\top\not\preceq 0$. However, proving the latter fact seems to be a daunting task. So, any help or suggestions to tackle this conjecture is very appreciated. Thanks a lot!


Remark. (Condition $A+A^\top\preceq 0$ is not necessary)

Consider the following $2\times 2$ matrix $$ A = \begin{bmatrix}-1 & \frac{\sqrt{3}+2}{2} \\ \frac{\sqrt{3}-2}{2} & 0 \end{bmatrix}. $$ Matrix $A$ has two eigenvalues at $-0.5$, whereas the eigenvalues of $A+A^\top$ are $-3$ and $1$. Let us define $$ X = \begin{bmatrix}1 & 0 \\ 0 & 0 \end{bmatrix}, \quad P = \begin{bmatrix}\frac{1}{2} & 0 \\ 0 & -\frac{\sqrt{3}-2}{2(\sqrt{3}+2)} \end{bmatrix}\succ 0. $$ It holds that $$ AP+PA^\top =-X, $$ and $$\|P\|_2 = \frac{1}{2}=-\frac{1}{2\mathrm{tr}(A)}.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.