# A conjecture on the Lyapunov equation

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 $$\|P\|_2 = \frac{1}{2}=-\frac{1}{2\mathrm{tr}(A)}.$$