Determine necessary and sufficient conditions on $W_{1}$ and $W_{2}$ so that $\dim(W_{1}\cap W_{2}) = \dim W_{1}$. Let $W_{1}$ and $W_{2}$ be subspaces of a finite-dimensional vector space $V$. Determine necessary and sufficient conditions on $W_{1}$ and $W_{2}$ so that $\dim(W_{1}\cap W_{2}) = \dim W_{1}$.
MY ATTEMPT
The necessary and sufficient condition is given by $W_{1}\subseteq W_{2}$.
We can consider a basis $\mathcal{B}_{12} = \{w_{1},w_{2},\ldots,w_{k}\}$ for $W_{1}\cap W_{2}$. Then it is possible to extend it to a basis $\mathcal{B}_{1} = \{w_{1},\ldots,w_{k},\ldots,w_{m}\}$ for $W_{1}$ and a basis $\mathcal{B}_{2} = \{w_{1},\ldots,w_{k},\ldots,w_{n}\}$ for $W_{2}$
But, according to the given hypothesis, $k = m$. Thus $\mathcal{B}_{1} = \mathcal{B}_{12}$. In other words, one has that
\begin{align*}
W_{1} = \text{span}(\mathcal{B}_{1}) = \text{span}(\mathcal{B}_{12}) \subseteq \text{span}(\mathcal{B}_{2}) = W_{2}
\end{align*}
just as desired.
The converse implication is clear. Indeed, if $W_{1}\subseteq W_{2}$, then $W_{1}\cap W_{2} = W_{1}$ and the condition on the dimensions holds.
Can someone double-check my arguments?
 A: Your argument is correct. 
Perhaps a more straightforward way (in my opinion) is using the fact that
$$\dim(W_1\cup W_2) = \dim(W_1)+\dim(W_2)-\dim(W_1\cap W_2).$$
If  $\dim(W_1)=\dim(W_1\cap W_2)$, then this becomes
$$\dim(W_1\cup W_2) = \dim(W_2),$$
and since $W_1\cup W_2\supseteq W_2$, then $W_1\cup W_2=W_2$, i.e., $W_2\supseteq W_1$ (because they are linear spaces).
A: This is basically correct. A few suggestions: 
You have written $\mathcal{B}_1=\{w_1,\ldots, w_k,w_{k+1},\ldots, w_m\}$ and $\mathcal{B}_2=\{w_1,\ldots, w_k,w_{k+1},\ldots, w_n\}$ as extensions of $\mathcal{B}_{12}$ to bases of $W_1$ and $W_2,$ respectively if I understood correctly (you did not specify what $\mathcal{B}_2$ was a basis for). In particular, $\mathcal{B}_1\cap \mathcal{B}_2$ contains $w_{k+1}$, which is not true a priori$^*$. A (perhaps) better way to write this is: 

Extend $\mathcal{B}_{12}$ to $\mathcal{B}_1=\{w_1,\ldots, w_k,w_{k+1},\ldots, w_m\}$, then because $\dim W_1\cap W_2=\dim W_1$, we know that $m=k$. It then follows that $W_1=W_1\cap W_2\subseteq W_2$. 

Otherwise, it's very well done!
$^*$ For instance take $W_1$ to be the $xy-$plane in $\mathbb{R}^3$ and $W_2$ the $yz-$plane. Then $\mathcal{B}_1=\{e_1,e_2\}$ and $\mathcal{B}_2=\{e_2,e_3\}$. Then $\mathcal{B}_{12}=\{e_2\}$, but as we can see there should be no common elements besides $e_2$ in the extension of the basis $\mathcal{B}_{12}$ to a basis for $W_1$ and $W_2$.
