# range $T_1=$ range $T_2$ $\iff$ there exists an invertible operator $\mathcal{L}(V)$ such that $T_1=T_2S$

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.

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.