Show that $\dim(U + V) = \dim(U) + \dim(V) - \dim(U \cap V)$

Question

Let $W$ be a vector space and let $U$ and $V$ be finite dimensional subspaces. Not sure how to go about solving this.

Answer 1

Take a basis $(v_1,...,v_m)$ of $U \cap V $. This basis is linearly independen in both $U$ and $V$ so extend this basis to a basis of $U$: $ (v_1,...,v_m,w_1,....,w_j)$ and to a basis of $V$: $(v_1,...,v_m,u_1,...,u_k)$. Notice $\dim (U \cap V) =m$, $\dim U = m + j $ and $\dim V = m + k $. We want $U + V $ to have dimension $m + j + k $. So our goal is to check that $(v_1,...,v_m,w_1,....,w_j,u_1,....,u_k)$ is a basis for $U + V $ We show this list is independent. Suppose

$$ \sum \alpha_iv_i + \sum \beta_i w_i + \sum \gamma_i u_i = \Theta $$

where $\alpha_i, \beta_i, \; \; and \; \; \gamma_i \in \mathbb{F} $

Rewrite equation above as follows:

$$ \sum\gamma_i u_i = - \sum \alpha_iv_i - \sum \beta_i w_i $$

$$ \therefore \sum\gamma_i u_i \in U $$ since in the right hand side we have a combination of the basis elements of $U$. But we know that $u's$ are in $V$. Therefore $\sum\gamma_i u_i$ must belong to $U \cap V $. But, we know $(v_r)_{r=1}^{m}$ is basis for the intersection, there take $\lambda_i \in \mathbb{F}$ such that

$$ \sum^k\gamma_i u_i = \sum^m \lambda_i v_i \iff \sum^k\gamma_i u_i - \sum^m \lambda_i v_i = \Theta$$

But we know list of the $u's$ and the $v's$ is linearly independent. Therefore, $\gamma_i = \lambda_i = 0 $. Therefore our original equation becomes

$$ \sum \alpha_iv_i + \sum \beta_i w_i = \Theta $$

But, again list $v's$ and $w's$ are linearly independent, and hence $\alpha_i = \beta_i = 0$

Therefore, by definition the list $(v_1,...,v_m,w_1,....,w_j,u_1,....,u_k)$ is linearly independent.

All we need to check now is that $(v_1,...,v_m,w_1,....,w_j,u_1,....,u_k)$ spans $U + V $. I will leave to you this part.

The problem is now solved.

Answer 2

Hint Pick a basis $B$ for $U \cap V$. Extend it to bases $B_U, B_V$ for $U$ respectively $V$. Prove that $B_U \cup B_V$ is a basis for $U+V$.