Can we say that $\text {tr}\ (A) = 0\ $? Let $A$ be an $n \times n$ real matrix with $A^3 + A = 0.$ Can we say that $\text {tr}\ (A) = 0\ $?
I think it's true but can't prove it. Any help will be highly appreciated.
Thanks in advance.

The most straightforward way to approach the proof is to use the minimal polynomial of $A$. If $K$ is the minimal polynomial of $A$, then any other polynomial $Q$ with $Q(A)=0$ is a multiple of $K$. Hence, the eigenvalues of $A$ are either $0$, $i$ or $-i$. Moreover, as $A$ is a real matrix, the sum of its eigenvalues must be a real number. In order to clarify this, consider a characteristic equation as below:
$$ a_1 x^n+a_2x^{n-1}+ ...+a_{n+1}=0$$
Then, the sum of the eigenvalues is $\frac{-a_2}{a_1}$.
To conclude the given fact, it's enough to observe that the only real number made by adding $0,-i,i$ together is $0$. So, we are done since $tr(A)$ is the sum of its eigenvalues.
PS: the link below may be useful.
https://en.wikipedia.org/wiki/Minimal_polynomial_(linear_algebra)
