Sign of $tr(A)$ given $I_n+A+A^2+A^3=0$ Let $A$ be a real matrix such that $I_n+A+A^2+A^3=0$, what is the sign of $tr(A)$ ($tr$ being the trace) ?

What I have done : One can easily figure our the inverse of $A$ since $I_n=-A-A^2-A^3=A(-I_n-A-A^2)=AA^{-1}$, but I fail to see any way to use that ...

Any eigenvalue $\lambda$ of $A$ satisfy
$$1+\lambda+\lambda^2+\lambda^3 = 0\implies \frac{\lambda^4-1}{\lambda-1} = 0$$
so the possible eigenvalues are $\{-1,i,-i\}$. Since the matrix is real the eigenvalues $i$ and $-i$ has to come in pairs (with sum zero). Let $k$ denote the multiplicity of the eigenvalue $-1$ then
$$\text{tr}(A) = -k \implies \text{tr}(A) \leq 0$$

Note that we cannot rule out the case $k=0$ (i.e. the trace being stricktly negative) since $A = \left(\matrix{0 & -1\\1&0}\right)$ is a solution (see comment by marwalix above) of the system.
