Exercise 3.D.6

Proof. Suppose there are $R\in L(V)$ and $S\in L(W)$ such that $T_1=S{T}_{2}R$. Then

Where the first equation follows from Exercise 4, the second from Exercise 22 in 3B and the third because $R$ is invertible. Hence $\dim <!--\; FUNCTION\; APPLICATION\; -->\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->T_1\le \mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->T_2$. We also have

Where the first equation follows from Exercise 5 and the second from Exercise 23 in 3B and because $\dim <!--\; FUNCTION\; APPLICATION\; -->\mathrm{range}<!--\; FUNCTION\; APPLICATION\; -->S=\dim <!--\; FUNCTION\; APPLICATION\; -->W\ge \dim <!--\; FUNCTION\; APPLICATION\; -->\mathrm{range}<!--\; FUNCTION\; APPLICATION\; -->T_2$. Thus

Hence $\dim <!--\; FUNCTION\; APPLICATION\; -->\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->T_1\ge \dim <!--\; FUNCTION\; APPLICATION\; -->\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->T_2$. Therefore $\dim <!--\; FUNCTION\; APPLICATION\; -->\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->T_1=\dim <!--\; FUNCTION\; APPLICATION\; -->\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->T_2$.

Conversely, suppose $\dim <!--\; FUNCTION\; APPLICATION\; -->\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->T_1=\dim <!--\; FUNCTION\; APPLICATION\; -->\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->T_2$. Let $u_1,\dots ,u_n$ and ${\mu }_1,\dots ,{\mu }_n$ be bases of $\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->T_1$ and $\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->T_2$, respectively. Extend them to the bases $u_1,\dots ,u_n,v_1,\dots ,v_m$ and ${\mu }_1,\dots ,{\mu }_n,{\nu }_1,\dots ,{\nu }_m$ of $V$. By the same reasoning used in the proof of 3.22, we have that $T_1{v}_1,\dots ,T_1{v}_m$ and $T_2{\nu }_1,\dots ,T_2{\nu }_m$ are bases of $\mathrm{range}<!--\; FUNCTION\; APPLICATION\; -->T_1$ and $\mathrm{range}<!--\; FUNCTION\; APPLICATION\; -->T_2$, respectively.

Define $R\in L(V)$ by

And $S\in L(\mathrm{range}<!--\; FUNCTION\; APPLICATION\; -->T_{2}R,W)$ by

It is easy to show that $T_1=S{T}_{2}R$ (it is almost the same as in the proof of Exercise 4 and Exercise 5) and that $S$ injective. By Exercise 3, there is an invertible operator on $W$ that satisfies the same property as $S$, which completes the proof. □