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!

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.
