$T$ and $U$ have a common eigenvector, where $T$ and $U$ are linear operators on odd dimensional vector space $V$ and $T^2 = U^2 = I$. 
I am trying to prove that $T$ and $U$ have a common eigenvector, where $T$ and $U$ are linear operators on odd dimensional vector space $V$ and $T^2 = U^2 = I$.

I have been stuck on this problem for a long time. I have already proved that eigenvalues of $T$ and $U$ are either $-1$ or $1$. I have also found some invariant spaces with respect to $T$ and $U$ (e.g. $\big\{v+T(v)|v \in V\big\}$ and $\big\{v+U(v)|v \in V\big\}$) and tried to show the intersection between these sets is not ${0}$. But so far I haven't got any success in these areas either. This is probably because I am not using the assumption of odd dimensionality of $V$.  
Lastly, I have thought that each $T$ and $U$ must have at least $k+1$ linearly independent eigenvectors where $\dim(V) = 2k+1$. In this case, the intersection between these two sets must have a dimension greater than zero. I have tried to prove this thinking about the map between bases of each operator ($T$ and $U$ are injective and thus each base is mapped to another base) but the two bases are not necessarily the same. If this was the case the problem could be solved.  
I would appreciate any solution. Thank you in advance.

I just realized that your assertion requires that we assume the characteristic of the field is not 2, otherwise I'm pretty sure it's false.
Your last idea is the closest. Let $V_1=\ker (T-1)$, $V_{-1}=\ker (T+1)$, $W_1=\ker (U-1)$, $W_{-1}=\ker(U+1)$ be the eigenspaces. We'll show that $V=V_1\oplus V_{-1}=W_1\oplus W_{-1}$. Then one of $V_1$ or $V_{-1}$ will have dimension at least $k+1$, let the larger one be $V_\epsilon$, and similarly for $W_1$ and $W_{-1}$, and again let the larger one we $W_\eta$. Then $\dim V_\epsilon \cap W_\eta \ge 2k+2-(2k+1)=1$, so there is some common eigenvector of both $T$ and $U$.
Thus we just need to show that $V=V_1\oplus V_{-1}$ (by symmetry the same will follow for $W_1$ and $W_{-1}$).
For any vector $v\in V$, $v=\frac{v-Tv}{2} +\frac{v+Tv}{2}$, and 
$$T\frac{v-Tv}{2}=\frac{Tv-T^2v}{2}=-\frac{v-Tv}{2},$$
and $$T\frac{v+Tv}{2}=\frac{Tv+T^2v}{2} = \frac{v+Tv}{2}.$$
Thus $V=V_1+V_{-1}$, and we just need to check that $V_1\cap V_{-1}=0$, but this is essentially immediate. If $Tv=v$ and $Tv=-v$, then $v=-v$, so $2v=0$, and thus $v=0$.
Hence everything holds.
