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 minimal polynomial of $A$ must divide $x^3+x$.
Then the real Jordan form of $A$ can have, consequently, two kinds of blocks:

0 blocks of dimension 1
2-dimensional blocks associated to rotations of  $\pi/2$

In both cases the trace is equal to 0.
Recalling that the trace of $A$ is invariant for conjugation you have done.
Observe that this is not true over the complex numbers: $A=i I$ satisfies $A^3+A=0$ but $trA \neq 0$.
the trace of the matrix is the sum of its eigenvalues, Here  $i,-i,0$ are  the eigenvalues of the given equation $A^3+A=0$ and the sum of its eigenvalues is zero,
So the trace of the matrix is $0$.
A matrix with real entries often has non-real eigenvalues.