# Grassman formula for vector space dimensions

If $U$ and $W$ are subspaces of a finite dimensional vector space, $$\dim U + \dim W = \dim(U\cap W) + \dim(U + W)$$

Proof: let $B_{U\cap W} = \{v_1,\ldots,v_m\}$ be a base of $U\cap W$. If we extend the basis to $B_U = \{v_1,\ldots,v_m, u_{m+1}, \ldots, u_r\}$ and $B_W = \{v_1,\ldots,v_m, w_{m+1},\ldots,w_s\}$ then $S:=\{v_1,\ldots,v_m,u_{m+1},\ldots,u_r, w_{m+1},\ldots,w_s\}$ is a generating set of $U+W$. Now I have to prove that $S$ is linearly independent: $$0 = \sum_{i=1}^m a_i v_i + \sum_{j=m+1}^r b_j u_j + \sum_{k=r+1}^s c_k w_k \implies v = \sum_{i = 1}^m a_i v_i+\sum_{j=m+1}^r b_j u_j = -\sum_{k=r+1}^m c_k w_k$$

is a vector of $U\cap W$ and $b_j = 0$ since $B_U$ is independent. Therefore $0 = \sum_{i=1}^m a_i v_i + \sum_{k=m+1}^s c_k w_k$ and, since $B_W$ is independent we have that $a_i = c_k = 0$, and $\dim(U+W) = \dim U + \dim W - \dim(U\cap W)$.

Question: I can't understand why $v\in U\cap W$, since we have expressed it as linear combination of bases of $U$.

Your approach is fine, but I wonder if you might consider using an approach with Isomorphism Theorems as an exercise. If you convince yourself that that $(U\times W)/(U\cap W)\cong U+W$, then you should see that the dimension of the left side is $\dim(U)+\dim(W)-\dim(U\cap W)$ and the dimension of the right side is $\dim(U+W)$, and that by the isomorphism, the dimensions are equal. – rschwieb Dec 27 '12 at 14:44
Notice that we have two expressions for $v$. The first is $\displaystyle\sum_{i = 1}^m a_i v_i+\sum_{j=m+1}^r b_j u_j$ which is a linear combination of $\{v_i,u_j\}$ and so $v$ is an element of $U$. The other is $\displaystyle -\sum_{k=r+1}^m c_k w_k$ which is a linear combination of $\{w_k\}$ and so $v$ is an element of $W$. The second one looks like what you missed.