Exercise 1.6.22

The condition would be that $W_1\subset W_2$. Let $\alpha$ and $\beta$ be the basis of $W_1\cap W_2$ and $W_1$. Since $W_1$ and $W_2$ finite-dimensional, we have $\alpha$ and $\beta$ are bases with finite size. First if $W_1$ is not a subset of $W_2$, we have some vector $v\in W_1\setminus W_2$. But this means that $v\notin \mathrm{span}(\beta )$ and hence $\beta \cup \{v\}$ would be a independent set with size greater than that of $\beta$. So we can conclude that dim$(W_1\cap W_2)=$dim$(W_1)$. For the converse, if we have $W_1\subset W_2$, then we have $W_1\cap W_2=W_1$ and hence they have the same dimension.