Is $A-B$ never normal?

Question

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!

Answer 1

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.