Exercise 1.6.14

For $W_1$ we can observe that by setting $a_2=p$, $a_3=q$, $a_4=s$, and $a_5=t$ we can solve $a_1=q+s$. So $W_1=\{(q+s,p,q,s,t)=p(0,1,0,0,0)+q(1,0,1,0,0)+s(1,0,0,1,0)+t(0,0,0,0,1):p,q,s,t\in {𝔽}^5\}$. And

is the basis. The dimension is four. And similarly for $W_2$ we may set $a_4=s$, $a_5=t$. And then we have $a_1=-t$, $a_2=a_3=a_4=s$ and

. And hence

is the basis of $W_2$. The dimension is two.