$S$ ,$T$ are two-dimensional subspaces of $\mathbb{R}^3$. Prove $\dim(S\cap T) \ge 1 $. 
Let $S$ and $T$ be two-dimensional subspaces of $\mathbb{R}^3$. Prove that $\dim(S\cap T) \ge 1 $. 

My attempt: $\dim(S+T)= \dim S+\dim T-\dim(S\cap T)$. We are given $\dim S =\dim T = 2$. Also $\dim(S+T) \le 3$, since $S+T$ is a subspace of $\mathbb{R}^3$. It follows that $\dim(S \cap T ) \ge 1$. Which finishes the proof. 
But I want a different approach that doesn't use the above formula. Now since $S$ and $T$ are two-dimensional subspaces of $\mathbb{R}^3$, they are planes passing through origin. How do I show from here, that their intersection have only two possibilities, either they will be a plane or a line in $\mathbb{R}^3$? 
 A: Since $S$ is a two-dimensional subspace of $\mathbb R^3$, it is a plane going through the origin, so it can be represented by an equation $ax+by+cz=0$. Also, $T$ can be represented by $ex+fy+gz=0$. If these two planes are identical, i.e., there is $\lambda,\mu\in\mathbb R$ such that $\lambda a=\mu e$, $\lambda b=\mu f$, $\lambda c=\mu g$, then $S\cap T=S=T$ is a plane going through the origin so it has dimension $2$. Otherwise, $S\cap T$ can be represented by the equations
\begin{cases}
ax+by+cz=0,\\
ex+fy+gz=0,
\end{cases}
which is a line going through the origin so it has dimension $1$.
A: $S$ and $T$ are two dimensional in $R^3 \implies$ two planes passing through origin. Now when we take the intersection of these two planes - 
Case_1 :-  If $S=T \implies S \cap T = S$ or $T$ which is the same plane.
Case_2 :-  If $S != T \implies$ it will be a line passing through the origin.
Now you cannot have the $dim(S \cap T) < 1$ because it means their dimension is $0 \implies$ only zero vector in the subspace. This is only possible when $S$ and $T$ are orthogonal. For the subspace $S$ to be orthogonal to $T$ their dimension should be $r$ and $n-r$ respectively which is not the case here. The orthogonal subspace for a plane in $R^3$ should be a line through origin ($1$-dimensional) but $T$ is $2$ dimensional so it can never be completely orthogonal subspace to $S$.
A: Here is an abstract approach from general vector space theory:
Recall the following formula: Let $V$ be a finitedimensional vector space with subspaces $V_1$ and $V_2$. Then
$$\operatorname{dim} V_1 + \operatorname{dim} V_2 = \operatorname{dim}(V_1 \cap V_2) + \operatorname{dim}(V_1 + V_2)$$
Applying this with $V_1 = S, V_2 = T$, we get, since $\operatorname{dim}(S+T) \leq \operatorname{dim}(\mathbb{R}^3) = 3.$
$$4= \operatorname{dim}(S \cap T) + \operatorname{dim}(S+T) \le \operatorname{dim}(S \cap T) + 3$$ forcing $\operatorname{dim}(S \cap T) \geq 1$
But this statement is geometrically clear also. You have two planes through the origin in $\mathbb{R}^3$. Either the planes fall together, and the intersection is two dimensional, or the planes don't fall together. In this case, since the planes intersect in $(0,0,0)$, they must share a line and the intersection is one-dimensional.
A: Let $S$ and $T$ be generated by the orthonormal sets $\{u_1,u_2\}$ and $\{v_1,v_2\}$, respectively.
The statement $\dim(S\cap T)\ge1$ is equivalent to there being a nonzero element in $S\cap T$.
Let $u_3$ a normal vector orthogonal to $\{u_1,u_2\}$, and decompose $v_1,v_2$ in the associated basis:
\begin{cases}
v_1 = a_1 u_1 + a_2 u_2 + a_3 u_3, \\
v_2 =b_1 u_1 + b_2 u_2 + b_3 u_3.
\end{cases}
Now, if $a_3=0$ then $v_1\in S$ and therefore $v_1\in S\cap T$.
Let us therefore assume $a_3\neq0$. For the same reason we can assume $b_3\neq0$. But then consider the linear combination
$$w\equiv b_3 v_1 - a_3 v_2.$$
Such $w$ clearly doesn't contain any $u_3$ component, and thus $w\in\mathrm{span}(\{u_1,u_2\})=S$. We conclude that $w\in S\cap T$, and therefore $\dim(S\cap T)\ge1$.
A: Consider the linear map from $S \oplus T \to \mathbb R^3$ given by $(s, t) \mapsto s-t$.  Since the dimension of $S \oplus T$ is 4, this map has a non-zero kernel.  What does that tell you?
