How to prove that A is normal if and only if there's a unitary matrix V,s.t. A*=AV? This can be found at http://en.wikipedia.org/wiki/Normal_matrix#cite_note-1, but I don't know how to prove it:
$A$ is normal if and only if there's a unitary matrix V,s.t. $A^*=AV$
 A: 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.
