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.
 A: 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 $trA \neq 0$.
A: 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)
A: 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.
