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This can be found at, but I don't know how to prove it:

$A$ is normal if and only if there's a unitary matrix V,s.t. $A^*=AV$

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Have you looked at Horn, pp. 109? – Hagen von Eitzen Oct 7 '12 at 9:12
I can not find the book via internet. – user39843 Oct 7 '12 at 9:15
up vote 2 down vote accepted

If $A^* = AV$, then show that $A = A^* V^* = V^* A^*$. Likewise you can show that $V$ commutes with $A$. Using this you will easily show that $A^* A = A A^*$, i.e., that $A$ is normal.

Conversely, if $A$ is normal, then $A = U \Lambda U^*$, where $U$ is unitary, and $\Lambda$ is the diagonal matrix of eigenvalues of $A$. Recognise that we can write $\Lambda = R \Phi$, where $R$ is a diagonal matrix whose entries are the magnitudes of the corresponding entries of $\Lambda$, and $\Phi$ is the diagonal matrix whose entries are complex numbers reflecting the phase of the corresponding entry in $\Lambda$. In other words if the $j^{\text{th}}$ diagonal entry of $\Lambda$ is $\lambda_j = r_j e^{i \phi_j}$, then $e^{i \phi_j}$ is the corresponding entry in $\Phi$.

Then $\Lambda^* = \Lambda \Phi^* \Phi^*$, and $$ A^* = U \Lambda^* U^* = (U \Lambda U^*)(U \Phi^* \Phi^* U^*) = AV, $$ and we observe that $V$ is unitary, since $\Phi^*$ is.

This kind of idea is reflected in the polar decomposition of a matrix.

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I wonder how to show that $A^*V^*=V^*A^*$? – user39843 Oct 7 '12 at 11:56
Perhaps you should first wonder how you would show that $A = V^* A^*$. – yasmar Oct 7 '12 at 11:59
actually,as $A^*=AV$,then $A=V^*A^*$,so $A^*A=AVV^*A^*=AA^*$. Isn't this more concise than yours? And we no need to show the commutation. – user39843 Oct 7 '12 at 12:07
Yes, that's good. – yasmar Oct 7 '12 at 12:24
But I can not understand why $\Lambda^* = \Lambda \Phi^* \Phi^*$? – user39843 Oct 7 '12 at 12:30

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