$\operatorname{dim}(W_1\cap W_2)\le n$ so $\operatorname{dim}(W_1\cap W_2)≤\operatorname{dim}(W_2)$

Question

Let $W_1$ and $W_2$ be subspaces of a vector space $V$ having dimensions $m$ and $n$, respectively, where $m\ge n$.

$(a)$ Prove that $\operatorname{dim}(W_1 \cap W_2)≤n$

$(b)$ Prove that $\operatorname{dim}(W_1+W_2) \leq m+n$

For $(a)$ I have:
$1. (W_1\cap W_2)$ is a subspace of $W_1$ and $W_2$
so $\operatorname{dim}(W_1\cap W_2)\le \operatorname{dim}(W_2)=n$

or

$2.$ The conclusion is $W_1 \cap W_2 \subseteq W_2$

You have to state your question clearly, do you want us to check your proof? Are you stuck somewhere? Also, learn mathjax to type maths. — Siong Thye Goh, Nov 04 '17 at 08:49
And for (a) 1.- I would like to know how the proof came to dim(W1∩W2)≤dim(W2)=n @SiongThyeGoh Because if 1.- is correct we can forget about 2.- — Ayd, Nov 04 '17 at 09:28

score 0 · Answer 1 · answered Nov 04 '17 at 08:45

Hint for part b:

Starting from the basis of $W_1 \cap W_2$, extend it to a basis of $W_i$, call it $B_i$. Try to construct a set that can span $W_1+W_2$ using $B_i$.

Remark: about your second approach, the conclusion is $W_1 \cap W_2 \subseteq W_2$? That is the reason for your first approach.

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