Do $A$ and $-A^t$ have a different characteristic polynomial if $A$ is traceless has size dimension than 2.

Question

Let $A$ be a traceless $n\times n$ matrix with $n>2$. Is it always possible that -A^t$ has a different characteristic polynomial.

I solved the case that $n$ is odd, as $P_A(t)=\det(A-\lambda\cdot\text{Id})=(-1)^n\lambda ^n+\cdots+\det(A)$ where $P_{-A^t}(t)=(-1)^n\lambda^n+\cdots+(-1)^n\det(A)$, which are different. But if $n$ is even, how can we do this?

Answer 1

All you need is an example of a matrix $A$ that is traceless and for which $A$ and $-A^t$ have different characteristic polynomials. One such example is as follows: let $A$ be the $n \times n$ diagonal matrix $$ A = \pmatrix{1-n\\ & 1 \\ &&\ddots\\&&&1}. $$