Word and normal matrix i need to prove this result
Let $A$ a square complex matrix and let $\mathcal{W}(A,A^{*})$ be a word in $A$ and $A^{*}$ with lenght $m\geq 2$, and suppose that $\mathcal{W}(A,A^{*})$ contains $A^{2}$ or $(A^{*})^{2}$. Then 
$\sigma(\mathcal{W}(A,A^{*}))=\sigma^{m}(A)$ if and only if $A$ is normal 
(where $\sigma$ is singular values of a matrix)
Thanking you in advance ! 
 A: Sketch of proof:
With the spectral theorem, it is trivial that $A$ is normal implies $\sigma(\mathcal{W}(A,A^{*}))=\sigma^{m}(A)$.
For the converse, the Schur decomposition means that we may assume that $A$ is upper triangular without loss of generality.  Suppose moreover that the eigenvalues of $A$ are on the diagonal in decreasing order of magnitude.  We can write
$$
A = \pmatrix{\lambda & x^*\\0 & B}
$$
for some $x \in \Bbb C^{n-1}$ and eigenvalue $\lambda$.
Claim: If $\sigma_{\max}^m(\mathcal W) = \sigma_{\max}^m(A)$, then $x=0$.
Proof: Let $A$ be such that $\sigma_{\max}^m(\mathcal W) = \sigma_{\max}^m(A)$.  Suppose for the purpose of contradiction that $x \neq 0$. Let $\|A\| = \sigma_{\max}(A)$, which is to say that $\|\cdot\|$ dentoes the spectral norm. First, note that for any matrix $A$ and any two words $\mathcal W_1,\mathcal W_2$, we have
$$
\|\mathcal W_1 \mathcal W_2\| \leq \|\mathcal W_1\|\cdot \|\mathcal W_2\|.
$$
Use this to conclude that for any word $\mathcal W_k$ of length $m_k$, we have $\|\mathcal W_k\| \leq \|A\|^{m_k}$.  The word $\mathcal W$ under our consideration can be decomposed into
$$
\mathcal W = \mathcal W_1 A^2 \mathcal W_3.
$$
Now, show that $\|A^2\| < \|A\|^2$. Conclude that $\|\mathcal W\| \leq \|\mathcal W_1\|\cdot \|A^2\| \cdot \|\mathcal W_3\|< \|A\|^m$, contradicting our premise.
With the above claim established, we can inductively reach the desired conclusion.

Claim: If $x \neq 0$, then $\|A^2\| < \|A\|^2$.
Ideas: Note that
$$
A^2 = \pmatrix{\lambda^2 & x^*(\lambda I + B)\\ 0 & B^2},\\
A^2A^{*2} = \pmatrix{|\lambda|^4 + \|(\lambda I + B)x\|^2
& [\lambda^2 (\lambda I + B)x]^*\\
\lambda^2 (\lambda I + B)x & B^2 B^{*2}}\\
AA^* = \pmatrix{|\lambda|^2 + x^*x & x^*B\\
Bx & BB^*}\\
(AA^*)^2 = 
\pmatrix{(|\lambda|^2 + x^*x)^2 + \|Bx\|^2 
& (|\lambda|^2 + x^*x)x^*B + x^*BBB^*\\
\cdot & B[xx^* + B^*B]B^*}
$$
