TITLE: Show $A$ and $B$ have a common eigenvector if $A^2=B^2=I$
QUESTION [8 upvotes]: Let $n$ be a positive odd integer and let $A,B\in M_n(R)$ such that $A^2=B^2=I$. Prove that $A$ and $B$ have a common eigenvector.

REPLY [3 votes]: By the given condition, $(AB)(BA) = I$. Therefore $AB$ and $BA$ have reciprocal eigenvalues. Yet, in general, $AB$ and $BA$ always have identical spectra. As $n$ is odd, it follows that some eigenvalue $\lambda$ of $AB$ belongs to $\{-1,1\}$. Let $x$ be a corresponding eigenvector, i.e. $ABx=\lambda x$. Multiply both sides by $A$, we get $y:=Bx=\lambda Ax$ and in turn $By=\lambda Ay=x$. Hence $B(x\pm y)=y\pm x$ and $A(x\pm y)=\lambda(y\pm x)$. Since at least one of $x+y$ and $x-y$ is nonzero, the assertion follows.
Edit. In general, if $n$ is odd, $F$ is a field and $A,B\in M_n(F)$ are annihilated by some monic polynomial $(z-a)(z-b)=0$ with $a\ne b$, by observing $\big(z-\frac{a+b}{2}\big)^2=\big(\frac{a-b}{2}\big)^2$ and thus defining $\widetilde{A}=\frac2{a-b}\left(A-\frac{a+b}2I\right)$ and $\widetilde{B}=\frac2{a-b}\left(B-\frac{a+b}2I\right)$, we have $\widetilde{A}^2=\widetilde{B}^2=I$. Therefore, by our previous argument (with extension to the algebraic closure of $F$ if necessary), $\widetilde{A}$ and $\widetilde{B}$ have a common eigenvector. Since $A,B$ are affine transforms of respectively $\widetilde{A},\widetilde{B}$, we conclude that $A$ and $B$ also share a common eigenvector.