# Is every stable matrix orthogonally similar to a $D$-skew-symmetric matrix?

Let $$\mathrm{diag}(A)$$ denote the diagonal matrix with diagonal entries of $$A\in\mathbb{R}^{n\times n}$$ and let $$\succeq$$ denote the standard partial order in the cone of (symmetric) positive definite matrices. Let me start by recalling a definition from [1].

Definition ($$D$$-skew-symmetric matrix). A matrix $$A\in\mathbb{R}^{n\times n}$$ is said to be $$D$$-skew-symmetric if there exists a diagonal matrix $$D\succ 0$$ such that $$(A-\mathrm{diag}(A))D$$ is skew-symmetric.

Now, let $$A\in\mathbb{R}^{n\times n}$$ be a matrix with eigenvalues $$\{\lambda_i\}_{i=1}^n$$ such that $$\mathrm{Re}(\lambda_i)<0$$.

My Question: Is $$A$$ orthogonally similar to a $$D$$-skew symmetric matrix with non-positive diagonal?

Of course, if $$A+A^\top\preceq 0$$, then it is easy to see that the answer is in the affirmative (just pick the orthogonal similarity transformation that diagonalizes $$A+A^\top$$ and $$D=I$$).

My question is further motivated by the fact that in the $$2\times 2$$ case it is possible to prove that the answer is always in the affirmative, as I show below.

Proof for $$n=2$$. After an orthogonal diagonalization of the symmetric part of $$A$$, we can assume that $$A$$ is in the form $$A=\begin{bmatrix}a_{11} & a_{12} \\ -a_{12} & a_{22} \end{bmatrix},$$ where $$a_{ij}\in\mathbb{R}$$, $$a_{11}<0$$, $$a_{11}+a_{22}<0$$. Further, since $$A$$ has eigenvalues with negative real part, $$\det(A)=a_{11}a_{22}+a_{12}^{2}>0$$. We distinguish two cases:

1. If $$a_{22}\le 0$$, then $$A+A^{\top}\preceq 0$$. Hence, by picking $$D=I$$, we have that $$A$$ is $$D$$-skew-symmetric with non-positive diagonal.
2. If $$a_{22}> 0$$, then $$A+A^{\top}\not \preceq 0$$. Define $$T:=\frac{1}{\sqrt{a_{22}-a_{11}}}\begin{bmatrix} \sqrt{-a_{11}} & -\sqrt{a_{22}}\\ \sqrt{a_{22}}& \sqrt{-a_{11}} \end{bmatrix}$$ and note that $$T$$ is an orthogonal matrix. It holds $$T^{\top} A T = \begin{bmatrix} a_{11}+a_{22} & a_{12}+\sqrt{-a_{11} a_{22}}\\ -a_{12}+\sqrt{-a_{11} a_{22}}& 0 \end{bmatrix}.$$ Define $$D:=\begin{bmatrix} 1 & 0\\ 0 & \frac{a_{12}-\sqrt{-a_{11} a_{22}}}{a_{12}+\sqrt{-a_{11} a_{22}}} \end{bmatrix},$$ where $$D\succ 0$$ since $$a_{11}a_{22}+a_{12}^{2}>0$$. Thus, $$T^\top AT$$ is $$D$$-skew-symmetric with non-positive diagonal.

[1] E. Kaszkurewicz, and L. Hsu. "On two classes of matrices with positive diagonal solutions to the Lyapunov equation." Linear algebra and its applications, 59 (1984): 19-27.