Normal matrices and the SVD I've been stuck in this problem on SVD. 

Let $A\in M_n(\mathbb{C})$, with $rank(A)=k$. Prove that A is normal if and only if exists an orthonormal set in $\mathbb{C}^n$, $ \lbrace z_1,...,z_k \rbrace $, such that $|z_i^{*}Az_i|=\sigma_i$ for all $i=1,2,3,...,k$. With $\sigma_i$ singular values of A.

My attempt:
for $\Leftarrow|) $ i've seen that if A has $rank(k)$ then exists $U,V$ unitary matrices such that: 
$A=U \Sigma V^*$, with $\Sigma$ a rectangular matrix containing the $k$ singular values. Since the singular values can be written as $|z_i^{*}Az_i|=\sigma_i$ then when you transpose $\Sigma$, the elements on the diagonal are preserved. Then I've got for instance:
$AA^* =U\Sigma V^* V\Sigma^* U^*=U\Sigma\Sigma^*U^*$. 
And
$A^*A=V\Sigma^*\Sigma V^*$. From here... can I conclude thar A is normal?
Also the other implication is bothering me a lot, so any suggestion would be appreciated. Thanks! 
My instructor tells me that i should use the fact that certain norm i the sum of the squares of the singular values of the matrix,
 A: You have solved part of the problem. Just carry through with the $\Sigma$ gymnastics.
$\Leftarrow$
$$
\begin{align}
  \mathbf{A} &=
  \mathbf{U} \, \Sigma \, \mathbf{V}^{*} \\
%
 &=
% U 
  \left[ \begin{array}{cc}
     \color{blue}{\mathbf{U}_{\mathcal{R}}} & \color{red}{\mathbf{U}_{\mathcal{N}}}
  \end{array} \right]  
% Sigma
  \left[ \begin{array}{cccccc}
     \sigma_{1} & 0 & \dots &  &   & \dots &  0 \\
     0 & \sigma_{2}  \\
     \vdots && \ddots \\
       & & & \sigma_{k} \\
       & & & & 0 & \\
     \vdots &&&&&\ddots \\
     0 & & &   &   &  & 0 \\
  \end{array} \right]
% V 
  \left[ \begin{array}{c}
     \color{blue}{\mathbf{V}_{\mathcal{R}}}^{*} \\ 
     \color{red}{\mathbf{V}_{\mathcal{N}}}^{*}
  \end{array} \right]  \\
%
  & =
% U
   \left[ \begin{array}{cccccccc}
    \color{blue}{u_{1}} & \dots & \color{blue}{u_{k}} & \color{red}{u_{k+1}} & \dots & \color{red}{u_{n}}
  \end{array} \right]
% Sigma
  \left[ \begin{array}{cc}
     \mathbf{S}_{k\times k} & \mathbf{0} \\
     \mathbf{0} & \mathbf{0} 
  \end{array} \right]
% V
   \left[ \begin{array}{c}
    \color{blue}{v_{1}^{*}} \\ 
    \vdots \\
    \color{blue}{v_{k}^{*}} \\
    \color{red}{v_{k+1}^{*}} \\
    \vdots \\ 
    \color{red}{v_{n}^{*}}
  \end{array} \right]
%
\end{align}
$$
Because the number of rows is the same as the number of columns,
$$
  \Sigma \, \Sigma^{\mathrm{T}} = 
  \Sigma ^{\mathrm{T}} \Sigma =
  \Sigma^{2} =
  \left[ \begin{array}{cc}
     \mathbf{S}^{2} & \mathbf{0} \\
     \mathbf{0} & \mathbf{0} 
  \end{array} \right]
$$
The matrix product 
$$
   \mathbf{A}^{*} \mathbf{A} = 
   \mathbf{V} \, \Sigma^{\mathrm{T}} \, \Sigma \mathbf{V}^{*}.
$$
Using the thin SVD (ignoring the nullspace terms) we can write
$$
   \mathbf{A}^{*} \mathbf{A} = 
   \color{blue}{\mathbf{V}_{\mathcal{R}}} \, \mathbf{S}^{2} \, \color{blue}{\mathbf{V}_{\mathcal{R}}}^{*}.
$$
The transpose is 
$$
   \left( \mathbf{A}^{*} \mathbf{A} \right)^{*} = 
%
   \left( \color{blue}{\mathbf{V}_{\mathcal{R}}} \, \mathbf{S}^{2} \, \color{blue}{\mathbf{V}_{\mathcal{R}}}^{*} \right)^{*} =
%
   \left( \color{blue}{\mathbf{V}_{\mathcal{R}}}^{*} \right)^{*} \, 
   \left( \mathbf{S}^{2} \right)^{\mathrm{T}} \, 
   \left( \color{blue}{\mathbf{V}_{\mathcal{R}}} \right)^{*} =
%
   \color{blue}{\mathbf{V}_{\mathcal{R}}} \, \mathbf{S}^{2} \, \color{blue}{\mathbf{V}_{\mathcal{R}}}^{*}.
$$
Therefore
$$
  \left[
   \mathbf{A}^{*} \mathbf{A}
  \right] =
%
  \mathbf{A}^{*} \mathbf{A} - \mathbf{A} \, \mathbf{A}^{*} =
%
  \mathbf{0}.
$$
