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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?

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  • $\begingroup$ How did you apply $A^{-1}$ like that? I'm doubting that this statement is actually true. $\endgroup$ – JMoravitz Oct 15 '15 at 2:19
  • $\begingroup$ On line 3, you're multiplying by $A^{-1}$ on the left, so you should have $-A^{-1} A S + A^{-1} S A = 0$. Which yields $A^{-1} S A = S$. To go any further, you'd need commutativity. $\endgroup$ – Joel Cohen Oct 15 '15 at 2:21
  • $\begingroup$ When you multiply something by a matrix, you have to multiply the whole thing on the same side. The third equation is incorrect because you multiplied on the left for one term and on the right for the other. $\endgroup$ – Matt Samuel Oct 15 '15 at 2:21
  • $\begingroup$ A more serious error is that you assumed the first equation was true when it's actually what you're trying to prove. $\endgroup$ – Matt Samuel Oct 15 '15 at 2:25
  • $\begingroup$ Mistake on my part, and edited. I looked into simultaneously diagonalizing both A and S, but S is already in diagonal form. A is unitarily diagonalizable because it is normal, but what would a matrix that simultaneous diagonalization look like? $\endgroup$ – trcomet Oct 15 '15 at 2:25
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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, $A^TS + SA = \begin{bmatrix}0&1\\1&0\end{bmatrix}\neq 0$

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