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!
 A: 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.
A: 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.
A: This is not true in general, I guess you omitted part of the question (something like "$V$ is a complex vector space").
The dimension of subspaces stable under linear operators is something that depends much on the base field $K$.
Let us assume we are in a situation where your argument (more precisely the variation thereof pointed out by Qiaochu and Arturo) does not work, i.e. $E^A_{i} \cap E^B_{j} = \{ 0 \}$ for all $i,j$.
As Arturo explained, there is no one dimensional subspace invariant under $A$ and $B$ in this case.
Moreover, any subspace invariant under $A$ and $B$ is not contained in any of the $E^A_i$, $E^B_j$.
The linear maps $p_{i,j} : E^A_i \rightarrow E^B_j$ which are the restriction to $E^A_i$ of the projection along $E^B_{-j}$ onto $E^B_{-j}$, are isomorphisms.
A two-dimensional subspace of $V$ invariant under $A$ but not contained in $E^A_{\pm 1}$ is of the form $Kx \oplus Ky$ where $x \in E^A_1$, $y \in E^A_{-1}$ are not zero.
The same is true for $B$ instead of $A$, so $p_{1,1}(x)$ and $p_{-1,1}(y)$ are colinear.
We can scale $y$ and assume that $y = p_{-1,1}^{-1} \circ p_{1,1}(x)$.
Similarly, $p_{1,-1}(x)$ and $p_{-1,-1}(y)$ have to be colinear, so $p_{1,-1}^{-1} \circ p_{-1,-1} \circ p_{-1,1}^{-1} \circ p_{1,1} (x) = \lambda x$, and so such an $x$ exists iff $p_{1,-1}^{-1} \circ p_{-1,-1} \circ p_{-1,1}^{-1} \circ p_{1,1}$ has an eigenvector.
Over $\mathbb{R}$, it can happen that this invertible operator has no eigenvector, for example if we take $A=\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{pmatrix}$ and $B=\begin{pmatrix} 0 & -1 & -1 & 1 \\ 1 & 0 & -1 & -1 \\ -1 & -1 & 0 & 1 \\ 1 & -1 & -1 & 0 \end{pmatrix}$, the aforementioned operator has matrix $\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$ which has no real eigenvalue.
