Trace $A$ is an integer.

Question

If $A$ is a square matrix of order $n$ with real entries and $A^3 + I =0$. Then $\operatorname{trace}(A)$ is an integer. My attempt: Here $|A| = -1$. And $\operatorname{trace}(A^3) = -n$. Then I tried to draw a contradiction assuming that $\operatorname{trace}(A)$ isn't an integer (using definition of determinant). But nowhere near the solution.

Answer 1

Hint

From the equation we conclude that the eigenvalues ($k$) will fit the equation:

$$k^3=-1$$

So, $k=-1$ or $k=\frac{1}{2}\pm \frac{\sqrt{3}}{2}i$ are the candidates for eigenvalues.

1) We know that the trace is the sum of the eigenvalues.

2) We also know that once we have a real matrix if we have a complex number as an eigenvalue then his conjugate will also be an eigenvalue.

Can you finish?