Exercise 3.D.5

Proof. Suppose $\mathrm{range}<!--\; FUNCTION\; APPLICATION\; -->T_1=\mathrm{range}<!--\; FUNCTION\; APPLICATION\; -->T_2$. Let $u_1,\dots ,u_n$ be a basis of $\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->T_1$ and extend it to a basis $u_1,\dots ,u_n,v_1,\dots ,v_m$ of $V$. By the same reasoning used in the proof of 3.22, it follows that $T_1{v}_1,\dots ,T_1{v}_m$ is a basis of $\mathrm{range}<!--\; FUNCTION\; APPLICATION\; -->T_1$. Let ${\nu }_1,\dots ,{\nu }_m\in V$ be an inverse image of this last basis when $T_2$ is applied and ${\mu }_1,\dots ,{\mu }_n$ a basis of $\mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->T_2$. Since $T_2{\nu }_1,\dots ,T_2{\nu }_m$ is linearly independent, we have that $\mathrm{span}<!--\; FUNCTION\; APPLICATION\; -->({\nu }_1,\dots <!--\; FUNCTION\; APPLICATION\; -->,{\nu }_m)\cap \mathrm{null}<!--\; FUNCTION\; APPLICATION\; -->T_2=0$. Therefore ${\nu }_1,\dots ,{\nu }_m,{\mu }_1,\dots ,{\mu }_n$ is a basis of $V$ and, because of this, the following linear map is invertible.

Define $S\in L(V)$ by

Let $v\in V$. There are $a_1,\dots ,a_n,c_1,\dots ,c_m\in 𝔽$ such that

Then

Hence $T_1=T_{2}S$.

For the implication in the other direction, suppose $S\in L(V)$ is an invertible operator such that $T_1=T_{2}S$.

Clearly $\mathrm{range}<!--\; FUNCTION\; APPLICATION\; -->T_1\subset \mathrm{range}<!--\; FUNCTION\; APPLICATION\; -->T_2$.

Let $w\in \mathrm{range}<!--\; FUNCTION\; APPLICATION\; -->T_2$ and $v\in V$ such that $T_{2}v=w$. Since $S$ is invertible, there is $\nu \in V$ such that $\mathit{S\nu }=v$. Thus

Hence $w\in \mathrm{range}<!--\; FUNCTION\; APPLICATION\; -->T_1$ and therefore $\mathrm{range}<!--\; FUNCTION\; APPLICATION\; -->T_2\subset \mathrm{range}<!--\; FUNCTION\; APPLICATION\; -->T_1$, implying $\mathrm{range}<!--\; FUNCTION\; APPLICATION\; -->T_1=\mathrm{range}<!--\; FUNCTION\; APPLICATION\; -->T_2$, as desired. □