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I am working through Axler's Linear Algebra Done Right section 3D problem 5 which states "Suppose $V$ is finite-dimensional and $T_1, T_2 \in \mathcal{L}(V,W)$. Prove that range $T_1=$ range $T_2$ iff there exists an invertible operator $S\in \mathcal{L}(V)$ such that $T_1=T_2S$". There are other posts regarding this problem, such as the one here, however my attempted solution is different from that. I am not confident mine is correct, but I am more looking so to see why that is the case so I can understand the actual solution better. Here is my attempt:

First, suppose range $T_1=$ range $T_2$. Since $V$ is finite dimensional, we can apply the fundamental theorem of linear maps and conclude range $T_1$ and range $T_2$ are finite dimensional. Because that is the case, we can define the list $w_1,w_2,...,w_m$ to be a basis of range $T_1$ and range $T_2$. By definition of range, for each vector in our basis, we have: $T_1(v_1)=w_1, T_1(v_2)=w_2...T_1(v_m)=w_m$ for $v_1,v_2,...,v_m$ AND $T_2(v'_1)=w_1,T_2(v'_2)=w_2,...,T_2(v'_m)=w_m$ for $v'_1,v'_2,...,v'_m$. By problem 4 in section 3A (I will omit the full proof for brevity), we can conclude that both $v_1,v_2,...,v_m$ and $v'_1,v'_2,...,v'_m$ are both linearly independent lists in $V$.

Extend $v_1,v_2,...,v_m$ and $v'_1,v'_2,...,v'_m$ to bases of $V$: $v_1,v_2,...,v_m,v_{m+1},...v_n$ and $v'_1,v'_2,...,v'_m,v'_{m+1},...,v'_n$. Define the linear map $S\in \mathcal{L}(V)$ as: $S(v_i)=v'_i$ for $i=1,2,...,n,n+1,...,m$. The existence and linearity of $S$ stems from the existence theorem (3.5).

Now, to show $S$ is invertible, we will show it is injective. Assume $S(v)=0$ for some $v\in V$. $S(v)=S(c_1v_1+c_2v_2,...,c_nv_n)=S(c_1v_1)+S(c_2v_2)+...+S(c_nv_n)$ $=c_1S(v_1)+c_2S(v_2)+...+c_nS(v_n)=c_1v'_1+c_2v'_2+...+c_nv'_n$, where both $v_1,v_2,...,v_n$ and $v'_1,v'_2,...,v'_n$ are our bases from earlier. Thus, it follows $c_1=c_2=...=c_n=0$ by the linear independence of $v'_1,v'_2,...,v'_n$. This concludes the first part of the proof

For the second part, assume that there exists an invertible operator $S\in \mathcal{L}(V)$ such that $T_1=T_2S$. Now, let $w=T_1(v)\in$ range $T_1$. By our assumption, $T_1(v)=T_2S(v)$. As $S(v)$ maps to some $v'\in V$, $w=T_1(v) \in$ range $T_2$. Hence, range $T_1 \subseteq$ range $T_2$.

Now, let $w=T_2(v)\in$ range $T_2$. Since $S$ is an invertible operator, it is surjective, meaning that there exists $v_s \in V$ such that $S(v_s)=v$. By our assumption, $T_1(v_s)=T_2S(v_s) \rightarrow T_1(v_s)=T_2(v)$. Thus, $w=T_2(v)\in$ range $T_1$ which implies range $T_2 \subseteq$ range $T_1$. Altogether, range $T_1$ = range $T_2$.

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1 Answer 1

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Since no one has answered this, I'll give it a shot, though please bear in mind that I'm just working through this stuff for the first time and could be missing something. (:

As far as I can tell, the only thing missing from your proof is that indeed $T_1 = T_2 S$. Remember that we have only been given that the ranges agree; thus when we complete your $v_1 ... v_m$ and $v'_1...v'_m$ to bases of $V$, we need to ensure that nothing goes awry with the new basis elements, vis-a-vis the agreement between our two maps. Also, since I am doing this exercises out of Axler's book, I'm going to follow his notation and write our maps in $\text{Hom}(V,W)$ as $T$ and $S$ and look for $E \in \text{End}(V)$ with the desired properties, just to minimize the risk that I mix things up.

So, take $span\{v_1 ... v_m\}$, and observe that if $x \in V$ we have $Tx = \sum \lambda_j Tv_j$, since our $v_j$ span the range. So if you consider the vector $x - \sum \lambda_j v_j$ we have $T(x-\sum \lambda_j v_j) = Tx - \sum \lambda_j Tv_j = \sum \lambda_j Tv_j - \sum \lambda_j Tv_j = 0$, so $x - \sum \lambda_j v_j \in \text{null}T$, and clearly $x = x - \sum \lambda_j v_j + \sum \lambda_j v_j$, i.e. it is is the sum of a vector in the nullspace and a vector in $span\{v_1 ... v_m\}$. Furthermore one can see that $\text{null}T \cap span\{v_1 ... v_m\} = \{0\}$ by noting that if $y \in \text{null}T \cap span\{v_1 ... v_m\}$ then we have $Ty =0$ and $y = \sum \alpha_j v_j$, so $\sum \alpha_j Tv_j = 0$, and by linear independence of the $T_j$ (they were a basis of the range), that means all the $\alpha$ are zero, hence $y$ is the zero vector. Therefore $V = \text{null}T \oplus span\{v_1 ... v_m\}$ and by an identical argument, $V = \text{null}S \oplus span\{v'_1 ... v'_m\}$. So thus we may complete both of these to bases of $V$: $$\{v_1 ... v_m, l_1...l_m\}$$ $$\{v'_1 ... v'_m, l'_1...l'_m\}$$ where the $\{l_j\}$ and $\{l'_j\}$ are respectively bases of the null spaces of $T$ and $S$. Now we can define $E$ essentially as you did, and given $v \in V$ we can write it as: $$v = \alpha_1 v_1 + \cdots \alpha_m v_m + \beta_1 l_1 + \cdots \beta_n l_n$$ and we have $$Tv = \alpha_1 Tv_1 \cdots \alpha_m Tv_m = \alpha_1 w_1 \cdots \alpha_m w_m$$ where the other terms disappear because they are in the nullspace. And at the same time: $$SE(v) = S(\alpha_1 v'_1 + \cdots \alpha_m v'_m + \beta_1 l'_1 + \cdots \beta_n l'_n) = \alpha_1 w_1 + \cdots \alpha_m w_m$$ Where again the null terms vanish, we have used the definition of $E$ as you have defined it, and we have used the fact that, by construction, $S$ applied to the $v'$ basis vectors is exactly $T$ applied to the $v$ basis vectors.

I hope I'm not making mistakes, and hope this is helpful!

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