Intersection of two three-dimensional subspaces Let $S$ and $T$ be three-dimensional subspaces of $\Bbb R^4$. 
Does it follow that $\dim(S \cap T ) \geq  2$?
Intuitively I would say yes, but its unclear how that could be proven? 
What would be a good way to approach this proof?
 A: Without loss of generality we can represent the intersection of the two subspace with a system of the form:
$$
\begin{cases}
s_1x+s_2y+s_3z+u=0\\
t_1x+t_2y+t_3z+u=0
\end{cases}
$$
that gives the solution:
$$
(s_1-t_1)x+(s_2-t_2)y+(s_3-t_3)z=0
$$
and this is the equation of a subspace of dimension at least $2$
A: Let $\{s_1,s_2,s_3\}$, $\{t_1,t_2,t_3\}$, $\{s_4\}$ and $\{t_4\}$ be bases for $S,T,S^\perp$ and $T^\perp$ respectively.  $S^\perp + T^\perp$ is spanned by $\{s_4,t_4\}$, so $\dim(S^\perp+T^\perp) \le 2$ and $\dim((S^\perp+T^\perp)^\perp) \ge 2$.
It remains to show $(S^\perp+T^\perp)^\perp \subseteq S \cap T$.  Let $v \in (S^\perp+T^\perp)^\perp$.  Then for all $a,b \in \Bbb{R}$, $\langle as_4+bt_4, v \rangle = 0$.  In particular, let $b = 0$ (resp. $a = 0$) so that $v \in (S^\perp)^\perp$ (resp. $v \in (T^\perp)^\perp$).  Since $S$ and $T$ are finite dimensional, $S = (S^\perp)^\perp$ and $T = (T^\perp)^\perp$.  Therefore, $v \in S$ and $v \in T$, and $\dim(S \cap T) \ge \dim((S^\perp+T^\perp)^\perp) \ge 2$.
