The problem states:
Show that for $U,W$ subspaces of a finite dimensional vector space $V$
$$\text{dim}(U) + \text{dim}(W) = \text{dim}(U + W) + \text{dim}(U \cap W)$$
Hint: Let $\mathscr{B}_U = \{u_1, …, u_s\}$ be basis for U and $\mathscr{B}_W = \{w_1, …, w_t\}$ be a basis for W.
Since intersection of two spaces might not be empty, let $\mathscr{B}_{U \cap W}= \{u_1, …, u_r\}$ where $r \leq \text{min}\{s,t\}$. Argue how many linearly independent vectors should be in the basis
for $U\cap W$.
Obviously there are grammar errors but that is the question verbatim. I understand the argument that is being suggested, but I'm not sure how to argue it formally. I suspect that it is related to the union rule $$n(U\cup W) = n(U) + n(W) – n(U\cap W)$$ but I am not sure how to bridge the gap between cardinality of the sets and the dimension of the subspace. I know that the dimension of the subspace is the number of linearly independent columns. I'm also assuming that $U + W$ could be written $U\cup W$.
Best Answer
Let $\{b_1, \ldots, b_k\}$ be a basis for $U \cap W$. Then first you need to argue you can extend this basis to a basis $b_1, \ldots, b_k, c_1, \ldots, c_l$ of $U$ and $b_1, \ldots, b_k, d_1, \ldots, d_n$ of $W$.
Now argue that $b_1, \ldots, b_k, c_1, \ldots, c_l, d_1, \ldots, d_n$ is a basis of $U+W$, so $\dim(U) + \dim(W) = (k + l) + (k + n) = (k + l + n) + k = \dim(U+W) + \dim(U \cap W)$.
I'll gladly fill in any details you are unclear about.