Exercise 1.6.23

Let $\alpha$ and $\beta$ be the basis of $W_1$ and $W_2$. By the definition we have both $\alpha$ and $\beta$ are bases with finite size.

1.

The condition is that $v\in W_1$. If $v\notin W_1=\mathrm{span}(\alpha )$, thus $\alpha \cup \{v\}$ would be a independent set with size greater than $\alpha$. By Replacement Theorem we have dim$(W_1)<$dim$(W_2)$. For the converse, if $v\in W_1=\mathrm{span}(\{v_1,v_2,\dots ,v_k\})$, we actually have $W_2=\mathrm{span}(\{v_1,v_2,\dots ,v_k,v\})=\mathrm{span}(\{v_1,v_2,\dots ,v_k\})=W_1$ and hence they have the same dimension.

2.

Since we have $W_1\subset W_2$, we have in general we have dim$(W_1)<$dim$(W_2)$.