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Suppose $V$ is finite dimensional and $T_1, T_2 \in \mathcal{L}(V,W)$. Show that range $T_1=$ range $T_2$ $\iff$ there exists an invertible operator $S \in\mathcal{L}(V)$ such that $T_1=T_2S$. I would like to know if my proof holds, please. Thank you in advance for you feedback!

($\impliedby$) Suppose first that there exists an invertible operator $\mathcal{L}(V)$ such that $T_1=T_2S$. We would like to show now that range $T_1=$ range $T_2$. We will proceed by double inclusion:

($\subset$) Let $w \in$ range $T_1$. Then, there exists $v \in V$ such that $T_1(v)=w$. Thus, by our supposition, $T_1(v)=(T_2 \circ S)(v)=w$. As $S$ is invertible, then $S$ is surjective. So, there exists $v' \in V$ and $w' \in$ range $S$ such that $S(v')=w'$. So, $T_1(v)=T_2(w')=w$ and we conclude that $w \in $ range $T_2$. Thus, range $T_1\subset$ range $T_2$

($\supset$) Let $w \in$ range $T_2$. Then, there exists $v\in V$ such that $T_2(v)=w$. As $S$ is surjective, then there clearly exists $v' \in V$ such that $S(v')=v$. Then, by our hypothesis, $T_1(v')=T_2(S(v')) \iff T_1(v')=T_2(v)=w$. So, $w \in$ range $T_1$ and we conclude that $T_1\supset$ range $T_2$

So, range $T_1$=range $T_2$

($\implies$) Let's suppose now that range $T_1=$ range $T_2$. We would like to show that there exists an invertible operator $S\in\mathcal{L}(V)$ such that $T_1=T_2S$.

By fundamental theorem, $\dim V=\dim$ null $T_1+\dim$ range $T_1$ and $\dim V=\dim$ null $T_2+\dim$ range $T_2$. As range $T_1=$ range $T_2$, then $\dim$ range $T_1=\dim$ range $T_2$. Thus, $\dim$ null $T_1=\dim$ null $T_2$. As $V$ is finite dimensional, we can consider a basis of null $T_2$ $(u_1,...,u_n)$ and a basis of null $T_1$ $(u'_1,...,u'_n)$. Both basis are linearly independent in $V$, so we can extend $(u_1,...,u_n)$ to a basis of $V$: $(u_1,...,u_n,v_1,...,v_m)$. We extend as well $(u'_1,...,u'_n)$ to a basis of $V$: $(u'_1,...,u'_n,v'_1,...,v'_m)$.

Now we can define $S$ as the following:

$S(u'_1)=u_2, S(u'_2)=u_1, S(u'_i)=u_i, i=3,...,n$ and $S(v'_j)=v_j, j=1,...,m$.

Clearly $S$ is injective and surjective by construction. So $S$ is invertible.

Then, $T_1(u'_1)=T_2(S(u'_1))=0, T_1(u'_2)=T_2(S(u'_2))=0, T_1(u'_i)=T_2(S(u'_i))=0, i=3,...,n$ and $T_1(v'_j)=T_2(S(v'_j)), j=1,...,m$ which is true as we supposed that range $T_1=$range $T_2$. We conclude that there exists an invertible operator $S \in\mathcal{L}(V)$ such that $T_1=T_2S$

I feel like I complicated too much the second implication, if there is a simpler approach I would like to know what it is, please.

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Basically I think you have the right idea. However, in your second part you cannot show that $T_1(v'_j)=T_2(S(v'_j))$ since $S(v'_j)=v_j$ and $v'_j$ were chosen independently, so there is no reason why $T_1(v'_j)$ should be equal to $T_2(v_j)$; the fact that both are vectors of the same (range) subspace does not suffice. You must therefore constrain the choice either of the $v_j$ or of the $v'_j$ so as to be sure that $T_1(v'_j)=T_2(v_j)$ for all $j$; this is possible since (assuming you chose the $v_j$ first) $T_2(v_j)$ lies in the common range subspace, and therefore has a pre-image by $T_1$. A subtle point then is to show that such a choice can be made while ensuring that together with the $u'_i$ you get a basis of $V$; this is a bit tricky, but you can show that any possible choice (with $T_1(v'_j)=T_2(v_j)$) will indeed give a basis.

There are also a few unnecessary quirks in your argument: why use the primed variables with $T_1$ and the unprimed ones with $T_2$, but most of all, why swap indices $1$ and $2$ when matching $u_1,u_2$ with $u'_1,u'_2$. This is a real problem, since nothing ensure there will even be two indices available, i.e., that $n\geq2$.

If you get a bit experienced, you can do some stuff easier by recalling general facts; for instance, right-composition with an invertible map (here $S$) never changes the image of a map, which immediately takes case of your first part. But the second part does require some real work.

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