TITLE: Showing $A+B$ is invertible?
QUESTION [8 upvotes]: Question number two of this released exam asks:

Let $A$, $B$ be two $n \times n$ matrices with real elements such that $A^3 = B^5 = I_n$ and $AB = BA$. Prove that $A+B$ is invertible.

I am not exactly sure what to do. I have observed that:
$$A(A^2) = I_n$$
$$B(B^4) = I_n$$
So:
$$A^{-1} = A^2$$
$$B^{-1} = B^4$$
My general approach to the problem has been to consider the binomial expansion of $(A+B)^n$ and see if things cancel out to be $I_n$ or some multiple of $I_n$, and with that construct an explicit inverse of $A+B$.
For $n \in \{1,2,3,4,5\}$ nothing seems to reduce down to the identity, leading me to believe this is the wrong approach. Any hints as how to approach this problem would be greatly appreciated thanks.

REPLY [5 votes]: Since $A^3 = B^5 = I_n$, eigenvalue of $A$ is the root of $x^3-1=0$ and eigenvalue of $B$ is the root of $x^5-1=0$. Since both equations have only simple roots, $A$ and $B$ are diagonalizable over $\Bbb{C}$. In addition $AB=BA$, $A$ and $B$ are simultaneously diagonalizable. Thus the eigenvalue of $A+B$ is $e^{2ki\pi/3}+e^{2ji\pi/5}$, where $0\leqslant k \leqslant 2, \:0\leqslant j \leqslant 4$. But none of them is zero. So $A+B$ is invertible.