Sign of $tr(A)$ given $I_n+A+A^2+A^3=0$

Question

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 ...

Multiply by $(A-I_n)$ to get $A^4 = I_n$. Also notice that $A = -I_n$ is a solution so if the sign of the trace is the same for all such matrices it has to be negative. — Winther, Aug 31 '15 at 16:20
@Winther Thanks a lot, but is there a way to prove that the sign is always the same ? — Ramanewbie2, Aug 31 '15 at 16:23
Are you assuming that $A$, or at least $\operatorname{tr} A$, is real? — anomaly, Aug 31 '15 at 16:25
@anomaly I forgot to mention it (now edited), but you're right. — Ramanewbie2, Aug 31 '15 at 16:26

score 4 · Accepted Answer · answered Aug 31 '15 at 16:48

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.

Sign of $tr(A)$ given $I_n+A+A^2+A^3=0$

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