Suppose $T_1, T_2\in L(V,W)$. Prove that if $\text{null }T_1=\text{null }T_2$ then there exists invertible operator $S\in L(W)$ such that $T_1=ST_2$

Question

The following problem appears in Ch. 3 section 3.D of Axler's Linear Algebra Done Right

4. 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.

Answer 1

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.

Answer 2

This answer is based on the answer by Frank, and is my attempt at understanding that answer by writing it out in more steps.

First let me address why the solution in the OP was incorrect.

In the OP, the proposed definition of a map $S$ was that

$$Sw_1=w_2$$

for $w_1=T_1v \in \text{range }T_1$ and $w_2=T_2v \in \text{ range }T_2$.

and $$Sw=w$$ for $w \notin \text{range }T_1$.

The problem with this definition is that for vectors $w$ in $\text{range }T_2$ that are not in $\text{range }T_1$ we have that $T_2v=w$ for some $v$ in $V$ and

$$ST_2v=Sw=w\neq T_1v$$

Now, as Frank pointed out, even if the ranges were the same we could still have the problem illustrated in the following example.

$$T_1(x)=(x,0)\tag{1}$$

$$T_2(x)=(-x,0)\tag{2}$$

We might depict these maps as

These maps have the same range (the x-axis in $\mathbb{R}^2$.

In the original question, something akin to the following map was proposed

$$S(x,y)=\begin{cases}(-x,0), \text{ if } y=0 \\ (x,y), \text{ if } y \neq 0 \end{cases}\tag{3}$$

Turns out $S$ is nonlinear.

Here is why

Take a vector $a=(x_1,0)$ and $b=(x_2,y)$ with $y\neq 0$.

Then

$$S(a+b)=S(x_1+x_2,y)$$ $$=(x_1+x_2,y)$$ $$=(x_1,0)+(x_2,y)$$ $$=a+b$$ $$=S(-a)+S(b)$$ $$=-S(a)+S(b)$$

We can see this with the following picture

Now let's look at what might be a correct solution.

We have a vector space $V$ of dimension $m+n$, a subspace null ${T_1}=$ null $T_2$ of dimension $m$, and another subspace, which I'll call $R$, which is the complement of null $T_1$ and has dimension $n$.

Thus $V=R\oplus (\text{null }T_1)$.

Let $(v_1, v_2,...,v_n)$ be a basis for $R$. Then, $(r_1,...,r_n)=(T_1v_1,...,T_1v_n)$ is a basis for range $T_1$ and $(t_1,...,t_n)=(T_2v_1,...,T_2v_n)$ is a basis for range $T_2$.

We need to define a linear operator $S\in L(W)$ such that the specific relationship

$$T_1=ST_2\tag{4}$$

is satisfied. This means that $S$ is defined on all of $W$, even though (4) is true for vectors $v \in V$, and represents an isomorphism between range $T_1$ and range $T_2$.

Let the subspace range $T_1$ $\cap$ range $T_2$ be called $Y$.

I will define $S$ for three subspaces that form a direct sum for $W$. These three subspaces are

1. range $T_1$
2. $C_2$ = complement of $Y$ in range $T_2$
3. $U$ = complement of $(\text{range }T_1 + \text{range }T_2)$ in $W$

Let's also define the subspace $C_1$ = complement of $Y$ in range $T_1$.

Thus, we have

$$\text{range }T_1 = C_1 \oplus (\text{range } T_1 \cap \text{range } T_2)$$

$$\text{range }T_2 = C_2 \oplus (\text{range } T_1 \cap \text{range } T_2)$$

$$W= U \oplus (\text{range }T_1 + \text{range }T_2)$$

$$=U\oplus C_1 \oplus C_2\oplus (\text{range } T_1 \cap \text{range } T_2)$$

$$=U\oplus \text{range }T_1\oplus C_2$$

Note that since range $T_1$ and range $T_2$ have the same dimension, then $C_1$ and $C_2$ have the same dimension, let's call it $d$.

First let's define $S$ on range $T_1$ by

$$Sr_i=t_i$$

for $i=1,...,n$.

Second, let's define $S$ on $C_2$.

Let $(x_1,...,x_d)$ be any basis for $C_2$ and $(z_1,...,z_d)$ be any basis for $C_1$.

Let

$$Sx_i=z_i$$

for $i=1,...,d$.

Finally, let's define $S$ on $U$.

Let $(u_1,...,u_l)$ be a basis for $U$, where $\dim U=l$ and define

$$Su_i=u_i$$

for $i=1,...,l$.

Then our $S$ is formed by an isomorphism of range $T_1$ onto range $T_2$, an isomorphism of $C_2$ onto $C_1$, and an isomorphism of $U$ onto $U$.

For any vector $w\in W$ we have

$$w=\sum_{i=1}^n a_i r_i+\sum_{i=1}^d b_i x_i+\sum_{i=1}^l c_iu_i$$

$$Sw=\sum_{i=1}^n a_i t_i+\sum_{i=1}^d b_i z_i+\sum_{i=1}^l c_iu_i$$

which is an invertible linear map (ie an isomorphism) since the null space of $S$ is the zero vector.

As a bonus, let's prove the converse of the statement proved above. That is, let's prove that if there exists an invertible linear operator $S\in L(W)$ such that $T_2=ST_1$ then null $T_1=$ null $T_2$.

Assume, for proof by contradiction that null $T_1\neq$ null $T_2$. Let $v\in V$ such that $v\in $ null $T_1$ and $v \notin$ null $T_2$.

Then,

$T_1v=0$

$$ST_1v=0\neq T_2v$$

which contradicts our original assumption that $ST_1v=T_2v$ for every $v\in V$.

Thus, null $T_1$ = null $T_2$.