Let $W_1$ and $W_2$ be subspaces of a finite-dimensional vector space V.
Let $\dim(W_1) =n_1$ and $\dim(W_2) = n_2$.
Claim: Show that $W_1 \cap W_2 \leq $ the smaller of $n_1$ and $n_2$.
My attempt at the proof:
We will consider all the cases such claim holds.
Case 1: $W_1 = W_2$
Since $W_1 =W_2$ then $W_1 \cap W_2 = W_1=W_2$.
This further implies that the dimension of both subspaces are equal. i.e. $n_1=n_2$ which further implies that $\dim(W_1 \cap W_2)=n_1=n_2$.
Case 2: (a) $W_1 \cap W_2 = W_1$ (b)$W_1 \cap W_2 = W_2$
(a) By the nature of the case we are considering it's given that $W_1 \subset W_2$. Because $W_1 \cap W_2$ is just all of $W_1$ means that the $\dim(W_1 \cap W_2)=\dim(W_1)=n_1$.
(b) On the other hand, by the same logic above $\dim(W_1 \cap W_2)=\dim(W_2)=n_2$.
Case 3: $W_1 \cap W_2 = B \cup A$ where B and A are some arbitrary subsets of $W_1$ and $W_2$ respectively. It also splits into two separate cases as follows:
(a) $W_1 \subset W_2 \implies$ $\dim(W_1 \cap W_2)<n_1$ since $W_1 \subset W_2$ and it's a "<" sign because it contains both some elements from $W_1$ and some from $W_2$.
(b) $W_2 \subset W_1 \implies$ $\dim(W_1 \cap W_2)<n_2$ by the same reasoning above.
Question: I would very much appreciate if you can point out any of the small details that I might be missing and check over my proof. Also, if there's another approach to this proof a suggestion would be appreciated also. I would appreciate a stronger acknowledgement towards my proof but willing to take suggestions for different paths. Thank you in advance!
Edit #1: The question is solved, however can anyone suggest me the cases that I would have to consider? Do I have all the cases and are they enough to prove the claim? Only reason why I am asking for this is because it will help me understand the intuition around the claim. Thanks!
You may do this problem without separating the problem into three cases.
Hint: Prove that if $U \subset V$ is a vector subspace of a vector space $V$, then dim$U\leq$ dim$V$ (What if the contrary happens? Consider the basis....). Then the theorem follows (how?).
Modification of your proof: The third case should be phrased as the symmetric difference of $W_1$ and $W_1$ is non-empty. Then you can indeed to use case 2 to finish the proof, since you can write $W_1 \cap W_2=W_1 \cap (W_1 \cap W_2)=W_2 \cap (W_1 \cap W_2)$. The first equality gives you dim$(W_1\cap W_2) \leq$dim$W_1$, and so on.