Let $A\in R^{n\times n}$ be a symmetric matrix, invertible. Let $X\in R^{n\times n}$ be a matrix. I was wondering is $B=A^{-1}XA-X$ always an anti-symmetric matrix? Namely, is $B^T=-B?$ I think the answer is yes, but I don't know how to show it formally.
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3Take the transpose of $B $ and see what you get. – AnyAD Aug 07 '18 at 13:13
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Have you even tried a single example ? – Aug 07 '18 at 13:26
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$$\begin{pmatrix}1&0\\0&2\end{pmatrix}\begin{pmatrix}2&2\\2&2\end{pmatrix}\begin{pmatrix}1&0\\0&\frac12\end{pmatrix}-\begin{pmatrix}2&2\\2&2\end{pmatrix}=\begin{pmatrix}0&-1\\2&0\end{pmatrix}.$$
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$$B^T = (A^{-1}XA-X)^T = (A^{-1}XA)^T - X^T = A^TX^T(A^{-1})^T-X^T = A^TX^T(A^T)^{-1}-X^T$$
If $A$ is orthogonal and $X$ is anti-symmetric, then $B$ would also be anti-symmetric.
Shirish Kulhari
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