Invariant Subspace of Two Linear Involutions 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!

Arturo's answer can be condensed to the following:
Let $U_1$, $\ldots$, $U_4$ be the eigenspaces of $A$ and $B$. Letting the simple cases aside we may assume that $U_i\oplus U_j=V$ for all $i\ne j$. We have to produce four nonzero vectors $x_i\in U_i$ that lie in a two-dimensional plane.
For $i\ne j$ denote by $P_{ij}:V\to V$ the projection along $U_i$ onto $U_j$; whence $P_{ij}+P_{ji}=I$. The map $T:=P_{41}\circ P_{32}$ maps $V$ to $U_1$, so it leaves $U_1$ invariant. It follows that $T$ has an eigenvector $x_1\in U_1$ with an eigenvalue $\lambda\in{\mathbb C}$. Now put $$\eqalign{x_2&:=P_{32}x_1\in U_2\ ,\cr x_3&:=x_1-x_2=P_{23} x_1\in U_3\ ,\cr x_4&:=x_2-\lambda x_1=x_2- P_{41}P_{32}x_1=x_2-P_{41}x_2=P_{14}x_2\in U_4\ .\cr}$$
It is easily checked that all four $x_i$ are nonzero.
