The question I am currently struggling with is
Does there exist an $\ n×n $ real matrix$\ A $ such that
$\ tr(A)= \ 0 $ and $\ A^2+A^T=\ I $?
For some reason, the solution starts with
$\ I= \ A^2−A^T $
instead of $\ I = \ A^2 + A^T $ as stated in the problem.
which leads me to think that $\ A^T =\ -A^T $
Is this true?
The solution I am looking at is this.
Well, no, that is a mistake from the solution makers. However, if you use $I=A^2+A^\top$ where they write $I=A^2-A^\top$, the argument doesn't change that much.
Edit: if the question makers intended to write $I=A^2 - A^\top$, then note that for any solution $A$, we could define $B=-A$ and we would get $I=B^2 + B^\top$. The other way around works as well. Thus, it doesn't matter what the exact sign is, the problems are equivalent, and the argument is similar.