Suppose I want to prove the lines of $S, A \in \mathbb{R}^{n \times n}$ where S is a positive definite \emph{diagonal} matrix and A is a real skew symmetric matrix. I want to show these two matrices commute in such a way that the inequality $A^T S + S A = 0$ is satisfied. In general, what would I need to do to prove this?
For one case, I attempted to say that A is nonsingular such that there exists $A^{-1}$. Since A is skew symmetric, I have the property $A^T = -A$, therefore
\begin{equation}
\begin{split}
A^T S + S A = 0 \\
-A S + S A = 0 \\
– A^{-1} A S + A^{-1} S A = 0 \\
– S + A^{-1} S A = 0
\end{split}
\end{equation}
Are there any other ways that this can hold?
Best Answer
Let $A=\begin{bmatrix}0&1\\-1&0\end{bmatrix}$, and let $S=\begin{bmatrix}2&0\\0&1\end{bmatrix}$
so $A^T = \begin{bmatrix}0&-1\\1&0\end{bmatrix}$
You have then that $A^T S = \begin{bmatrix}0&-1\\1&0\end{bmatrix}\cdot \begin{bmatrix}2&0\\0&1\end{bmatrix} = \begin{bmatrix}0&-1\\2&0\end{bmatrix}$
$SA = \begin{bmatrix}2&0\\0&1\end{bmatrix}\cdot \begin{bmatrix}0&1\\-1&0\end{bmatrix}= \begin{bmatrix}0&2\\-1&0\end{bmatrix}$
Finally then, $A^TS + SA = \begin{bmatrix}0&1\\1&0\end{bmatrix}\neq 0$