How do I prove that $\dim(V+W) = \dim(V) + \dim(W)?$

Question

How do I prove that $\dim(V+W) = \dim(V) + \dim(W)$, where $V$ and $W$ are subspaces? The base of $V$ is given as $\{v_1,v_2,...,v_r\}$ and the base of $W$ is given as $\{w_1,w_2,...,w_m\}$. $V\cap W = \{0\}$ is also given.

I've proved in a previous part that $S$, the set that contains the vectors in the bases of $V$ and $W$ is linearly independent. I am not sure how to prove $\dim(V+W) = \dim(V) + \dim(W)$ since the only way I've previously proved this sort of problem is by showing the number of non pivot columns.

Answer 1

The set $S=\{v_1,\dots, v_r,w_1,\dots w_m\}$ is linearly independent (as you proved already) and also generating. This means it is a basis. The dimension of a vector space equals the number of basis vectors needed to span it. This is $r+m$ in your case. Consequently, $\dim(V+W)=r+m$. Now, since $\{v_1,\dots, v_r\}$ is a basis for $V$, $\dim(V)=r$ and likewise for $W$. This proves your assertion.

Answer 2

Use the fact that $$ Dim(W+V)=Dim(W)+Dim(V)-Dim(W\cap V). $$

This fact should be demonstrated in your textbook.