Show that $\text{dim}(U) + \text{dim}(W) = \text{dim}(U + W) + \text{dim}(U \cap W)$

Question

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 $U\cup W$.

To roughly sketch the argument, I would say that when we combine U and W, some $r$ of the vectors in the union of the bases may be linear combinations of the others. So $\text{dim}(U) + \text{dim}(W) - \text{dim}(U\cap W) = \text{dim}(U\cup W)$. The result follows from adding $\text{dim}(U\cap W)$ to both sides. — Claycrusher, Sep 08 '16 at 03:37
$U+W$ is most certainly not equal to $U\cup W$. You must first understand the definitions to make any headway. — Alex Ortiz, Sep 08 '16 at 03:37
Yeah I didn't make the best choice of words. I know there is a difference, but am having trouble putting it into words. The idea of the union of the vectors as opposed to the union of the spans. — Claycrusher, Sep 08 '16 at 03:42
Possible duplicate of Given two subspaces $ U, W $ of vector space $ V $, how to show that $ \text{Dim}( U) + \text{Dim} (W) = \text{Dim}(U+W) + \text{Dim} ( U \cap W) $ — hardmath, Sep 08 '16 at 03:49

score 6 · Accepted Answer · edited Feb 05 '22 at 16:40

6

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.

edited Feb 05 '22 at 16:40

Martin Sleziak

53,687

answered Sep 08 '16 at 03:40

Nick R

1,513

So the $c$'s are essentially the rest of the basis of $U$ and the $d$'s are the rest of the basis of $V$? Also could you explain how $U + V$ is different from $U \cup V$ since I seem to be struggling to grasp the difference? – Claycrusher Sep 08 '16 at 03:46
$U \cup V$ is not a subspace. For example look at the subspaces $U = span { e_1 }$ and $V = span { e_2 }$, then $U \cup V$ is the union of two lines, which one can check is not a subspace. One can check that $U+V$ is infact the smallest subspace containing $U \cup V$. – Nick R Sep 08 '16 at 03:57

score 3 · Answer 2 · answered Sep 08 '16 at 03:43

Define a map $f$ from $U \times W \to U + W$, given by $f((u,w)) = u+w$. It is clear that $f$ is linear on $U \times W$.

The range of $f$ is the entire of $U+W$, since for every $v \in U+W$, $v$ can be written as $u+w$, hence $v = f((u,w))$.

Now,suppose that $f(u,w) = 0$. Then $u+w=0$ and $u=-w$, in which case $u \in U \cap W$ and $w \in U \cap W$. So $\ker f \times \ker f \subset (U \cap W) \times (U \cap W)$.

The other way, if $u \in U \cap W$, then $f(u,-u)=0$, so $(U \cap W) \times (U \cap W) \subset \ker f \times \ker f $

Hence, $\ker f = U \cap W$. Using the rank nullity theorem, $\dim(U \times W) = \dim(U+W) + \dim(U \cap W)$. But then, this just becomes $\dim(U) + \dim(W) = \dim(U+W) + \dim(U \cap W)$, since $\dim(U \times W) = \dim U + \dim V$.

Show that $\text{dim}(U) + \text{dim}(W) = \text{dim}(U + W) + \text{dim}(U \cap W)$

2 Answers2