Show $A$ and $B$ have a common eigenvector if $A^2=B^2=I$ 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.
 A: 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.
A: For $M=A$ or $B$, define
$$V_M^\pm=\{v\in\Bbb R^n:v\pm Mv\}.$$
By definition, $V_M^\pm$ are linear subspaces of $\Bbb R^n$. From $M^2=I$ we know that 
$$w=v\pm Mv\in V_M^\pm \Rightarrow Mw=\pm w,$$
i.e. $V_M^+$(resp. $V_M^-$) is either $\{0\}$ or the eigenspace of $M$ associated to eigenvalue $+1$(resp. $-1$), and in particular $V_M^+\cap V_M^-=\{0\}$. Moreover, from
$$v=\frac{1}{2}(v+Mv)+\frac{1}{2}(v-Mv),\quad\forall v\in \Bbb R^n,$$
we know that 
$$V_M^+\oplus V_M^-=\Bbb R^n\Rightarrow \dim V_M^++\dim V_M^-=n.$$
Since $n$ is odd, it follows that one of $V_M^\pm$ has dimension strictly larger than  $\frac{n}{2}$, and denote it as $V_M$. Therefore,
$$\dim V_A +\dim V_B>\frac{n}{2}+\frac{n}{2}=n\Rightarrow V_A\cap V_B\ne \{0\},$$
which implies that  $V_A\cap V_B$ contains some common eigenvector of $A$ and $B$. 
