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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 is 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.

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.

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 is 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?

My question is motivated by the fact that in the $2\times 2$ case it is possible to prove that the answer is in the affirmative.

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.

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 is 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.

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 is 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?

My question is motivated by the fact that in the $2\times 2$ case it is possible to prove that the answer is in the affirmative.

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] Kaszkurewicz, E., 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.

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 is 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?

My question is motivated by the fact that in the $2\times 2$ case it is possible to prove that the answer is in the affirmative.

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.