Exercise 2.1.20

To prove $A=T(V_1)$ is a subspace we can check first $T(0)=0\in A$. For $y_1,y_2\in A$, we have for some $x_1,x_2\in V_1$ such that $T(x_1)=y_1$ and $T(x_2)=y_2$. Hence we have $T(x_1+x_2)=y_1+y_2$ and $T(c{x}_1)=x{y}_1$. This means both $y_1+y_2$ and $c{y}_1$ are elements of $A$.

To prove that $B=\{x\in V:T(x)\in W_1\}$ is a subspace we can check $T(0)=0\in W_1$ and hence $0\in B$. For $x_1,x_2\in B$, we have $T(x_1),T(x_2)\in W_1$. Hence we have $T(x_1+x_2)=T(x_1),T(x_2)\in W_1$ and $T(c{x}_1)=\mathit{cT}(x_1)\in W_1$. This means both $x_1+x_2$ and $c{x}_1$ are elements of $B$.