# Given two linear transformations T1 and T2, show that range T1 = range T2 if and only if there is an invertible operator S such that T1=T2S.

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 $$T_2$$.

$$\square$$

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.