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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.

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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.

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    $\begingroup$ 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

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