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I have a matrix $M$ and another $N$. $N$ is an orthogonal (orthogonal => $N^{T} = N^{-1})$ r x r matrix and $M$ is an r x r skew symmetric matrix (skew syemmtric => $M^{T} = -M$). Is $(N^{-1})$$(M^2)$$N$ symmetric? Or is it skew symmetric? Or is it neither of those two options?

My work:

$N$ is orthogonal. So $N^{-1} = N^{T}$. Then we have $(N^{T})$$(M^2)$$N$. $M$ is skew symmetric. So $M^{2} = MM = (-M)(-M) = M^{T}M^{T}$. Then we have $(N^{T})$$(M^{T}M^{T})$$N$. Now what?

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  • $\begingroup$ Then is $((N^{-1})M^{2}N)^{T} = \pm (N^{-1})M^{2}N$, or neither? $\endgroup$ – Morgan Rodgers Mar 22 at 23:31
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$M$ is skew symmetric so $M^T=-M$. Then $$(MM)^T = M^T M^T = (-M)(-M)=MM.$$ That is, $M^2$ is symmetric. From there your $N^{-1}M^2N=N^TM^2N$ is symmetric, too.

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