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?
Best Answer
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).