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!

  • $\begingroup$ Can you please tell me, is this a book question? if yes then which book? $\endgroup$
    – PNDas
    Commented Nov 11, 2020 at 9:40
  • $\begingroup$ @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. $\endgroup$ Commented Nov 11, 2020 at 9:42
  • $\begingroup$ @Farhad Then where did you come across this problem? $\endgroup$ Commented Nov 11, 2020 at 14:34
  • $\begingroup$ @Farhad Are you familiar with "quotient spaces"? $\endgroup$ Commented Nov 11, 2020 at 14:45

1 Answer 1


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.

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