Prove or disprove: If $A^2$ is normal matrix then $A$ is normal matrix

Prove or disprove: If $A^2$ is normal matrix then $A$ is normal matrix.

I think this is wrong but simply can't build a counterexample.
Any hints on how to build a counterexample?
There are many conditions and it's hard for me to find a matrix.

Note that: $A$ is normal if $AA^* = A^*A$.

Pick $$A = \begin{pmatrix} u & v \\w & -u\end{pmatrix}$$ Now $$A^2 = (u^2+v\,w) \begin{pmatrix}1 & 0 \\0 & 1\end{pmatrix}$$ is clearly normal.

But for arbitrary $u$, $v$, $w$, $A$ would not be normal. Just pick $w \neq v$

• Also, $w\neq\pm v^*$ Dec 28, 2013 at 9:07
• I was editing my answer as you made the comment. You are right Dec 28, 2013 at 9:08
• Little thing I'm not sure about: how did you know A^2 will be (u^2+vw)*I ? is this simply matrix multiplication or other method to see it ?
– CnR
Dec 28, 2013 at 9:22
• @CnR: Yes, that's the easiest. Dec 28, 2013 at 10:47
• @CnR: These matrices show up all the time. I will edit my answer to show how you can show that all $2x2$ counter examples have to be of this form. Dec 28, 2013 at 17:59

Take the matrix $$A = \begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix}$$ Then we have $A^2 = 0$, which is normal. On the other hand, $$AA^* = \begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix}\begin{pmatrix}0 & 0 \\ 1 & 0 \end{pmatrix}=\begin{pmatrix}1&0\\0&0\end{pmatrix}$$ $$A^*A=\begin{pmatrix}0 & 0 \\ 1 & 0 \end{pmatrix}\begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix}=\begin{pmatrix}0&0\\0&1\end{pmatrix}$$ so that $A$ itself is not normal.

This is to answer OP's question in the comment section. The method I outline is a general way to attack problems of this nature.

Clearly there are no counter examples in $1\times 1$ matrices.

Try $2 \times 2$ matrices. Let $$D(A) = A^*A - AA^*$$ and let us work in the field of reals.

Let $A$ be $$A = \begin{pmatrix}u & v\\w& x\end{pmatrix}$$ Then $$D(A) = \begin{pmatrix}\left( w-v\right) \,\left( w+v\right) & \left( w-v\right) \,\left( x-u\right) \cr \left( w-v\right) \,\left( x-u\right) & -\left( w-v\right) \,\left( w+v\right) \end{pmatrix}$$ and $$D(A^2) = \begin{pmatrix}\left( w-v\right) \,\left( w+v\right) \,{\left( x+u\right) }^{2} & \left( w-v\right) \,\left( x-u\right) \,{\left( x+u\right) }^{2}\cr \left( w-v\right) \,\left( x-u\right) \,{\left( x+u\right) }^{2} & -\left( w-v\right) \,\left( w+v\right) \,{\left( x+u\right) }^{2}\end{pmatrix}$$

If you let $x = -u$ and $w \neq v$ then you get $D(A) \neq 0$ and $D(A^2) = 0$.

Note that $D(A)$ term-by-term divides $D(A^2)$. So at least for $2\times2$ matrices $D(A)=0 \Rightarrow D(A^2)=0$