# Show that $T_1=T_2S$ if $R(T_1)=R(T_2)$

Suppose $$W$$ is finite-dimensional and $$T_1,T_2 \in L(V,W)$$. Prove that if $$\operatorname{range}(T_1) = \operatorname{range}(T_2)$$, then there exists an invertible operator $$S \in L(V,V)$$ such that $$T_1 = T_2S$$.

Attempt: If $$V$$ is finite-dimensional the problem can be solved by taking basis for $$V$$, $$N(T_1)$$ and $$N(T_2)$$. But since the dimension of $$V$$ is not mentioned, how can the problem be solved when $$\dim(V)$$ is infinite? That's where I'm struggling.

Any help on the latter case would be appreciated!

• Can you please tell me, is this a book question? if yes then which book? Commented Nov 11, 2020 at 9:40
• @PNDas No, I guess it is not. At least I don't know if it is. But a similiar question with the difference that $V$ is finite-dimensional instead of $W$ is in Axel Sheldon's Linear Algebra Done Right. Commented Nov 11, 2020 at 9:42
• @Farhad Then where did you come across this problem? Commented Nov 11, 2020 at 14:34
• @Farhad Are you familiar with "quotient spaces"? Commented Nov 11, 2020 at 14:45

One approach is as follows: with the assumption of Zorn's lemma (or equivalently the axiom of choice), we can assume that $$\ker(T_j)$$ has a basis (for $$j=1,2$$), and that this basis may be extended to produce a basis of $$V$$. Let $$\mathcal K_j = (u^j_k)_{k \in \alpha_j}$$ denote a basis for the kernel of $$T_j$$ with $$\alpha_j$$ denoting an associated index set. Similarly, let $$\mathcal B_j = (v^j_k)_{k \in \beta_j}$$ denote a set of vectors that can be added to $$\mathcal K_j$$ to produce a basis $$\mathcal B_j \cup \mathcal K_j$$ of $$V$$.
• Argue that the vectors $$(T_j v^j_k)_{k \in \beta _j}$$ must form a basis of the range of $$T_j$$, and so $$\beta_j$$ must be finite. In particular, we can take $$\beta_j = \{1,\dots,r\}$$ without loss of generality.
• Define $$Q_j = \operatorname{span}\{v_1^j,\dots,v_d^j\}$$ and $$R = \operatorname{range}(T_1) = \operatorname{range}(T_2)$$. Argue that the maps $$T_j|_{Q_j} : Q_j \to R$$ are invertible (for $$j=1,2$$) ($$T|_Q$$ denotes the restriction of $$T$$ to $$Q$$).
• Define $$\bar S:Q_1 \to Q_2$$ by $$\bar S = [T_2|_{Q_2}]^{-1}T_1|_{Q_2}$$. Verify that $$T_2|_{Q_2}\bar S = T_1|_{Q_1}$$.
• To define $$S$$, extend the definition of $$\bar S$$ to a map on $$V$$ that is zero for all elements in $$\ker(T_1)$$. Equivalently, define $$S$$ on the basis elements with $$S(v^1_k) = \bar S(v^1_k) \quad k= 1,\dots,d, \qquad S(u^j_k) = 0 \quad k \in I_1.$$ Because we have $$T_2S(v) = T_1(v)$$ for all elements $$v$$ from our basis $$\mathcal B_1 \cup \mathcal K_1$$, we have $$T_2 S = T_1$$ as desired.