Q:

Is $A-B$ never normal?

Let $A,B \in M(3,\Bbb R)$, $A$ non-singular and symmetric, $B \neq 0$ such that $B^2=0$. Is $A-B$ never normal? 
So far i noticed that: 
$$(A-B)^*(A-B) = A^2 -AB-B^tA+B^tB$$
$$(A-B)(A-B)^*=A^2-BA-AB^t+BB^t$$
So if $B$ were symmetric $A-B$ would be normal but $B$ is nilpotent so it can't be symmetric (Symmetric matrices are diagonalizable for the spectral theorem and nilpotent matrices are not). $B$ is also different from $0$ so i believe that the statement is true but i can't demonstrate it. Thank you in advance for your help!

A:

As a counterexample, let
$$
A =
\pmatrix
{
0 & 1 & 1 \cr
1 & 0 & 1 \cr
1 &  1& 1
}
,\;\;
B =
\pmatrix
{
0 & 0 & 2 \cr
0 & 0 & 2\cr
0 & 0 & 0
}
$$
Then $A$ is symmetric and  non-singular, $B^2=0$, and $A-B$ is normal.

