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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.

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$2I_n=A^{15}+B^{15}=(A+B)(A^{14}-A^{13}B+......-AB^{13}+B^{14})$

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  • $\begingroup$ Thanks. I guess I should have kept trying... At least I was on the right track after all. :) $\endgroup$ – Dair May 1 '16 at 23:03
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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.

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