The following problem appears in Ch. 3 section 3.D of Axler's Linear Algebra Done Right
- Suppose $W$ is finite-dimensional and $T_1, T_2\in L(V,W)$. Prove that $\text{null }T_1=\text{null }T_2$ if and only if there exists an
invertible operator $S\in L(W)$ such that $T_1=ST_2$.
I'd like to prove the first conditional, ie from left to right. It seems some variant of this problem has been asked before here, here, and here but I am interested specifically in the attempt at a proof below.
Here is a picture I drew to try to understand this problem
Assume $\text{null }T_1=\text{null }T_2$.
Then $\text{dim range }T_1=\text{dim range }T_2$ so $\text{range }T_1$ and $\text{range }T_2$ are isomorphic.
That is, there is an invertible linear map between these subspaces of $W$. Let's call one such isomorphism $S$.
Now, we want to show that $T_1=ST_2$. At first glance, this makes a lot of sense: we can get from $T_1v$ to $T_2v$ using $S$.
However, the thing that I am struggling with is the possibility that two vectors from $V$ get mapped to the same vector in $\text{range }T_1$ but to two different vectors in $\text{range }T_2$.
Is this possible? Let's see.
Suppose $T_1v_1=T_1v_2=w$.
Then $T_1(v_1-v_2)=0$ and so $v_1-v_2$ is in the null spaces of $T_1$ and $T_2$ which are the same by assumption.
Hence, $T_2(v_1-v_2)=0$ and thus $T_2(v_1)=T_2(v_2)$.
If this is true, then for any $w_1\in\text{range }T_1$ we have $w_1=T_1(v)$ for possibly multiple vectors $v\in V$, but for each such vector we have $w_2=T_2(v)\in\text{range }T_2$.
Thus, we can define $S$ by $Sw_1=w_2$ with $w_1=T_1(v)$ and $w_2=T_2(v)$, $v\in V$.
This means that $ST_1v=T_2v$.
Ie, $ST_1=T_2$, and also $T_1=S^{-1}T_2$.
Finally, to make $S$ an operator on $W$ we need to define it on the vectors in $W$ that are not in the ranges of $T_1$ and $T_2$. We can make $S$ map such vectors to themselves.
I am very uncertain about the correctness of this attempt at a proof.
Best Answer
Your proof is almost correct. Yes, the idea is that you want $S$ to send $T_1v$ to $T_2v$ for all $v \in V$; in fact, the equation $ST_1 = T_2$ implies $S$ must be defined this way.
Unfortunately, we cannot simply define $S$ to fix all the vectors outside $\mathop{\mathrm{range}}T_1$. If $T_1$ and $T_2$ have different ranges, then $S$ will send vectors that are outside the range of $T_1$ to vectors that are inside the range of $T_2$. Can you see why this is a problem?
Even if $T_1$ and $T_2$ have the same range, your definition of $S$ is not necessarily linear. For example, suppose $T_1 : \mathbb{R} \to \mathbb{R}^2$ is the map $T_1x = (x, 0)$, and $T_2 : \mathbb{R} \to \mathbb{R}^2$ is the map $T_2x = (-x, 0)$. The range of both $T_1$ and $T_2$ is the $x$-axis in $\mathbb{R}^2$. As stated above, $S$ must send the points $(x, 0)$ to the points $(-x, 0)$, i.e., it must "flip" the $x$-axis. Can you see the problem in defining $S$ to fix all other points?
That being said, your definition does give us a way to define $S$ on the range of $T_1$, so we have a linear transformation $S : \mathop{\mathrm{range}} T_1 \to W$ such that $ST_1 = T_2$. To finish, can you find a way to extend $S$ to a linear operator from $W$ to itself?
Here's a tip. Often, when you want to define a linear transformation between two vector spaces, it is easiest to define where it sends basis vectors. This way, you don't run into issues where the map you have defined is not linear (as we did above). The above map $S : \mathop{\mathrm{range}} T_1 \to W$ currently specifies where some basis $\mathcal{B}$ of $\mathop{\mathrm{range}} T_1$ goes. So, to extend $S$ to a map on all of $W$, you can extend $\mathcal{B}$ to a basis of $W$, and then specify where the remaining basis vectors go.