This comes from Sheldon Axler's Linear Algebra Done Right 3e, chapter 3.B #25.

As a novice math student I struggled with this proof for a long time and would be both grateful and interested for an evaluation of this proof and perhaps a few hints toward a better way to prove this result.

Suppose that $V$ is finite-dimensional and $T_1, T_2 \in L(V,W).$ Prove that range $T_1 \subset$ range $T_2$ if and only if there exists $S \in L(V,V)$ such that $T_1 = T_2S.$

My proof:

First, suppose that range $T_1 \subset$ range $T_2$. Since $V$ is finite-dimensional, the fundamental theorem of linear maps implies that range $T_1$ and range $T_2$ are both finite-dimensional as well.

Let dim $range T_1 = n$ and dim $range T_2 = m$, so that $n \le m$. Furthermore, let $T_1v_1, ..., T_1v_n$ be a basis for $range T_1$. Now, since $range T_1 \subset range T_2$, we can extend our basis of $range T_1$ to a basis of $range T_2$, so that $T_1v_1,...,T_1v_n,T_2u_{n+1},...,T_2u_{m-n} = range T_2 = w_1,...,w_m$ (where $w_i = T_1v_i$ for $1 \le i \le n).$

Then by definition of range, there exist $v_1,..., v_n \in V$ such that for any arbitrary $T_1v \in range T_1$, we have that $v \in span\{v_1,...,v_n\} = V'.$ Since $Tv_1,...,Tv_n$ is a linearly independent list in $W$, it is easy to show that $v_1,...,v_n$ is a linearly independent list of vectors in $V$. Since $V$ is finite-dimensional, we can extend this list of vectors to a basis for $V$ in the form: $B_1 = v_1,..., v_n,u_{n+1},...,u_{m-n}, y_{m+1},...,y_{p}$, where $dim V = p$.

Now for arbitrary $T_2$ such that $range T_1 \subset range T_2$, we have, for arbitrary $v \in V$, a basis for $T_2$ such that: $T_2v = T_2(c_1v_1 ,...,c_1v_n, c_{n+1}u_{n+1},...,c_{m-n}u_{m-n},c_{m+1}y_{m+1},...,c_py_{p}) = c_1T_1v_1 + ... + c_nT_1v_n + w_{n+1} + ... + w_{m-n}$ , where $Ty_j = 0$ for $m+1 \le j \le p$. (and where $w_j \in W$ for $ n+1 \le j \le m-n).

Next, referring to the basis $B_1$ defined above, define $S$ such that $Sv_i = v_i$ for $1 \le i \le n$ and $Su = 0$ for $u \in B_1$ such that $u \notin span\{v_1,..., v_n\}$.

Hence, we have $T_2S(v) = T_2(S(B_1)) = T_2(c_1v_n + ... + c_nv_n) = c_1T_1v_1 + ... + c_nT_nv_n = T_1(v).$ This proves one direction of implication.

To prove the other direction of implication, suppose there exists $S \in L(V,V)$ such that $T_1 = T_2S.$

Then for $T_1v \in range T_1$ we have $T_1v = T_2S(v) = T_2(Sv)$, which implies $range T_1 \subset range T_2$, as desired.


1 Answer 1


It's a little unclear what you mean when you say things like "we have $T_1 v$ such that ...". Do you mean "there's a $v$ such that $T_1 v$ ..."? Keep in mind that there may be a $u \neq v$ such that $T_1 v = T_1 u$, so $v$ is not unique.

In particular, this section seems problematic:

by definition of range, there exist $v_1,..., v_n \in V$ such that for any arbitrary $T_1v \in range T_1$, we have that $v \in span\{v_1,...,v_n\} = V'.$

What is $v$ here? Is it an arbitrary vector in $V$? Then this is certainly false if $T_1$ is uninvertible.

  • 1
    $\begingroup$ You're right; I thought my idea took into account the fact that different vectors in V could map to the same vector in W but that's not the case. Ultimately, I caved and studied the solution in the manual; thank you for your response. $\endgroup$
    – Meta
    Jul 3, 2017 at 12:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.