TITLE: Invariant Subspace of Two Linear Involutions
QUESTION [8 upvotes]: I'd love some help with this practice qualifier problem:
If $A$ and $B$ are two linear operators on a finite dimensional complex vector space $V$ such that $A^2=B^2=I$ then show that $V$ has a one or two dimensional subspace invariant under $A$ and $B$.  
Thanks!

REPLY [6 votes]: Consider the linear transformation $AB:V\to V$.  Since $V$ is a complex vector space, $AB$ has at least an eigenvector, call it $X$, i.e. $ABx=\lambda x$. Note, $AABx=Bx=A\lambda x=\lambda Ax$ and  then $x=BBx=BA\lambda x=\lambda BAx$  So consider the space $\langle x, Ax\rangle$. Note, this space is invariant under $A$, clearly, $Bx=\lambda Ax$ and $BAx=x$ so it's invariant under $B$. This space is at least one dimensional.