Different proof that two subspaces' intersection cannot be $\{(0,0,0)\}$ Larson Edwards Falvo - Elementary Linear Algebra





For $b$, how to argue in terms of bases rather than in terms of geometry?
Also, the book has not yet given $$-\dim(S_1 \cap S_2) = \dim(S_1+S_2) - \dim(S_1) - \dim(S_2) \le 3 - 2 - 2 = 1$$
I was thinking that $$S_1 \cap S_2 = \{v | v=c_{11}v_{11} + c_{12}v_{12}=c_{21}v_{21}+c_{22}v_{22}\}$$
where $S_1$ has basis $\{v_{11}, v_{12}\}$
where $S_2$ has basis $\{v_{21}, v_{22}\}$
so if $$S_1 \cap S_2 = \{0\}$$
then 
$$0 = c_{11}v_{11} + c_{12}v_{12} = c_{21}v_{21}+c_{22}v_{22}$$
$$\to c_{11} = c_{12} = c_{21} = c_{22} = 0$$
I think? I'm not sure what to do with that.
And then $$v_{11}, v_{12}, v_{21}, v_{22}$$ can't be linearly independent. I'm not sure what to do with that.
I'm not sure how to conclude here. How can I approach this?
 A: Since $R^3$ has dimension $< 4$, the set $\{v_{11}, v_{12}, v_{21}, v_{22}\}$ must be linearly dependent, which means $$c_{11}v_{11} + c_{12}v_{12} + c_{21}v_{21}+c_{22}v_{22} = 0$$ for some $c_{ij}$ not all zero. So $$c_{11}v_{11} + c_{12}v_{12} = -(c_{21}v_{21}+c_{22}v_{22})$$ is a nonzero element of both $S_1$ and $S_2$.

A more abstract way to put it: The (external) direct sum of two vector spaces $V$ and $W$ is the set of pairs $V \oplus W = \{(v,w) : v \in V, w \in W\}$; it is a vector space by componentwise addition and scalar multiplication. Given a basis $\{e_i\}$ of V and $\{f_j\}$ of W, the set of all $(e_i,0)$ and $(0,f_j)$ is a basis of $V \oplus W$, so $\dim(V \oplus W) = \dim V + \dim W$.
Now define a linear map $T \colon S_1 \oplus S_2 \to R^3$ by $T(s_1,s_2) = s_1 - s_2$. Since $\dim(S_1 \oplus S_2) = 4 > 3 = \dim R^3$, there is some nonzero element $(s_1,s_2)$ of the null space of $T$. But that means $s_1 - s_2 = 0$, so $s_1 = s_2$ is a nonzero element of both $S_1$ and $S_2$.
(In fact, the rank-nullity theorem can be applied to $T$ show that for two subspaces $S_1$ and $S_2$ of $V$, $$\dim(S_1 \cap S_2) = \dim S_1 + \dim S_2 - \dim(S_1 + S_2) \ge \dim S_1 + \dim S_2 - \dim R^3$$
and the result follows.)
