Exercise 3.D.3

Proof. Suppose $T\in L(V)$ is an invertible operator such that $\mathit{Tu}=\mathit{Su}$ for every $u\in U$. Let $u_1,u_2\in U$ such that $S{u}_1=S{u}_2$. Then

Because $T$ is injective, it follows that $u_1=u_2$. Therefore $S$ is also injecitve.

Conversely, suppose $S$ is injective. Let $u_1,\dots ,u_n$ be a basis of $U$ and extend it to a basis $u_1,\dots ,u_n,v_1,\dots ,v_m$ of $V$. Because $\dim <!--\; FUNCTION\; APPLICATION\; -->S=0$, it follows that $S{u}_1,\dots ,S{u}_n$ is linearly independent and we can extend it to a basis $S{u}_1,\dots ,S{u}_n,{\nu }_1,\dots ,{\nu }_m$ of $V$. Define $T\in L(V)$ by

Because $S{u}_1,\dots ,S{u}_n,{\nu }_1,\dots ,{\nu }_m$ is linearly independent, it follows that $\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->T=0$. Hence $T$ is injective and, therefore, invertible. □