$\textbf{Exercise 9.}$ Suppose $V$ is finite-dimensional and $T: V \to W$ is a surjective linear map of $V$ onto $W$. Prove that there is a subspace $U$ of $V$ such that $T|_U$ is an isomorphism of $U$ onto $W$. Here $T|_U$ means the function $T$ restricted to $U$. Thus $T|_U$ is the function whose domain is $U$, with $T|_U$ defined by $T|_U(u) = Tu$ for every $u \in U$. $\textbf{Solution 9.}$ Let $w_1, \ldots, w_m$ be a basis of $W$. Because $T$ is surjective, for each $k = 1, \ldots, m$, there exists $v_k \in V$ such that $Tv_k = w_k$. Let $U = \operatorname{span}(v_1, \ldots, v_m)$. Thus $\dim U \leq m$. Now $\operatorname{range} T|_U = W$ (because $w_k \in \operatorname{range} T|_U$ for each $k = 1, \ldots, m$). The fundamental theorem of linear maps (3.21) thus tells us that $\begin{equation*} m \geq \dim U = \dim \operatorname{null} T|_U + m, \end{equation*}$ which implies that $\dim \operatorname{null} T|_U = 0$, which implies that $T|_U$ is injective. Thus $T|_U$ is an isomorphism of $U$ onto $W$. --- ![[sol-5.pdf#page=8]]