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This is a question from my friend.

It should be easy. But I have no knowledge about complex matrix.

Let $A,K$ be two invertible complex n-by-n matrices satisfying the following conditions: $A' = A^{-1}=\bar{A}\mbox{ and } \bar{K} = AK\bar{A},$

where $A'$ for transpose, $\bar{A}$ for the conjugate, and $A^{-1}$ for the inverse.

Prove that $K$ is a real matrix.

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up vote 5 down vote accepted

It's not true. Take $K = \begin{bmatrix} i & 0 \\\ 0 & \overline{i}\end{bmatrix}$ and let $A = \begin{bmatrix} 0 & 1 \\\ 1 & 0\end{bmatrix}$.

Then $A^{-1} = A = A' = \overline{A}$.

On the other hand, $AK\overline{A} = AKA = \begin{bmatrix} \overline{i} & 0 \\\ 0 & i\end{bmatrix} = \overline{K}$.

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Thanks for your counter-example. I need to check whether the condition $\bar{K}=AK\bar{A}$ is correct. – Jun Oct 4 '11 at 1:48

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