If $A^n$ is normal, is $A$ normal? My question is : Given an invertible matrix $A$ ( with complex entries ) , if $A^n$ is normal,is $A$ normal?
This is related to the question : If $A$ is an invertible $n\times n$ complex matrix and some power of $A$ is diagonal, then $A$ can be diagonalized .   Since we know that such a matrix is unitarily diagonalizable iff it is normal, I thought of formulating it this way.  My Attempt so far :  I thought of proving first that if $A^n$ is normal, then $A^{n-1}$ is normal. Since $A^n$ is normal, $A^n A^{n*}=A^{n*}A^n$. We write this as : $P=A A^{n-1}A^{(n-1)*}A^*=A^*A^{(n-1)*}A^{n-1}A$. Let's keep $B=A^{n-1}A^{(n-1)*}$. Then we have, $P=A BA^*=A^*B^*A$. I was thinking of taking it this way, but I'm not sure from here. Can you help  me proceed ? 
 A: No, that's not true. Take a symmetry $\sigma$ which isn't unitary, then $\sigma^2 = I$ is normal but $\sigma$ isn't. For a more explicit example, take:
$$A = PSP^{-1} = \begin{pmatrix}
1 & 0 \\
1 & -1
\end{pmatrix},$$
where $P = \begin{pmatrix}1&1\\0&1\end{pmatrix}$ and $S = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$. Then $A^2 = I_2$, but $A^* A \neq A A^*$.
A: Consider a triangularization $A=UTU^*$ of $A$. Saying that $A^n$ is normal means that $T^n$ is normal, so diagonal. Divide $T$ in blocks,
$$
T=\begin{bmatrix}
T_1 & X \\
0 & T_2
\end{bmatrix}
$$
where $T_1$ is $2\times2$ (the assertion is trivial for $1\times1$ matrices. Then, using block multiplication, we have
$$
T^n=\begin{bmatrix}
T_1^n & X_n \\
0 & T_2^n
\end{bmatrix}
$$
so we can look to $2\times2$ matrices. It's easy to prove by induction that, assuming $a\ne c$,
$$
\begin{bmatrix}
a & b\\
0 & c
\end{bmatrix}^n=
\begin{bmatrix}
a^n & b(a^n-c^n)/(a-c) \\
0 & c^n
\end{bmatrix}
$$
so saying $T_1^n$ isn't diagonal unless $b=0$. Thus we have infinitely many counterexamples.
