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Problem: Suppose $W$ is finite dimensional and $T_1,T_2 \in \mathcal{L}(V,W)$.Prove that null$T_1$=null$T_2$ if and only if there exists an invertible linear operator $S \in \mathcal{L}(W)$ such that $T_1=ST_2$

I looked at an answer to this proof. They started it by assuming range$T_2$ is finite dimensional. They let $w_1,...,w_m$ be a basis for this range. Thus there exists $v_1,...,v_m$ such that $T_2(v_i)=w_i$ for $i \in \{1,...,m\}$ Thus the linear map is defined by $T_2(a_1v_1+\ldots +a_mv_m)=a_1w_1+\dots +a_mw_m$

My question here is how can we define this linear map? How do we know that $v_1,...v_m$ form a basis for $V$?

What if there are basis vectors in $V$ in the null space of $T$ that get mapped to zero such that the dimension of V is greater than m?

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  • $\begingroup$ It can be shown that the $\{v_1, \dots, v_m\}$ are linearly independent. Also, are you assuming $V$ is finite-dimensional? and what is $T$ in this question? $\endgroup$
    – peek-a-boo
    Commented Jan 4, 2020 at 17:16
  • $\begingroup$ No the problem does not say $V$ is finite dimensional Even if $v_1,...,v_m$ are linearly independent isn't it true we can only define this map if $v_1,...v_m$ is a basis of $V$? $\endgroup$
    – user736276
    Commented Jan 4, 2020 at 17:17
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    $\begingroup$ yes, that's right. what we can do is extend this basis to be a basis of $V$ (this is possible even in infinite dimensions I believe if you assume the axiom of choice) but I don't see how this $T$ is helpful. Is the $T$ supposed to somehow be $S$? (I still haven't figured out this problem yet lol so I can't see the idea yet) $\endgroup$
    – peek-a-boo
    Commented Jan 4, 2020 at 17:27
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    $\begingroup$ We don't know that $v_1,\dots,v_m$ forms a basis for $V$, but we do know that it can be extended to a basis $v_1,\dots,v_m,v_{m+1},\dots$. Alternatively we know that there exists a complementary subspace to the span of $v_1,\dots,v_m$ within $V$. $\endgroup$ Commented Jan 4, 2020 at 17:27
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    $\begingroup$ I don't see a difference here between $T$ and $T_2$ unless we specify something like "$T$ is only defined on its domain which is the span of $v_1,\dots,v_m$" $\endgroup$ Commented Jan 4, 2020 at 17:31

1 Answer 1

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Below I've posted a solution to the original problem in your post. I've broken down the proof into a series of exercises. I've done this in order to highlight the main ideas, and so that I wouldn't bore you with details that you are already comfortable with. If it would be helpful to have solutions to any of these exercises, then let me know. I'll be happy to help.

The original problem was:

Problem: Suppose $W$ is finite dimensional and $T_1,T_2\in\mathcal{L}(V,W)$. Prove that $\text{null }T_1=\text{null }T_2$ if and only if there exists an invertible linear operator $S\in\mathcal{L}(W)$ such that $T_1=ST_2$.

Solution, Part 1: First suppose we have an invertible $S\in\mathcal{L}(W)$ with $T_1=ST_2$. Our goal is to show that $\text{null }T_1=\text{null }T_2$.

Exercise 1: Prove that $\text{null }T_1=\text{null }T_2$, given that $T_1=ST_2$ for some invertible $S\in\mathcal{L}(W)$. $\square$

Part 2: Suppose that $\text{null }T_1=\text{null }T_2$. Our goal is to show that there is an invertible $S\in\mathcal{L}(W)$, with $T_1=ST_2$.

Note that the image of $T_2$ is finite-dimensional, since it is a subspace of $W$, which is finite-dimensional. Let $\{w_1,\ldots,w_m\}$ be a basis for the image of $T_2$. For each $w_i$, pick a $v_i\in V$ with $T_2(v_i)=w_i$.

Exercise 2: Show that $\{v_1,\ldots,v_m\}$ is linearly independent.

Let $V_1=\text{null }T_1=\text{null }T_2$, and let $V_2=\text{span }\{v_1,\ldots,v_m\}$.

Exercise 3: Show that $V_1\cap V_2=\{0\}$.

For $1\le i\le m$, let $u_i=T_1(v_i)$.

Exercise 4: Show that $\{u_1,\ldots,u_m\}$ is linearly independent. You may find it helpful to use the fact that $V_1\cap V_2=\{0\}$.

Since $\{w_1,\ldots,w_m\}$ is linearly independent, it can be extended to a basis $\{w_1,\ldots,w_m,\ldots,w_n\}$ for $W$. And since $\{u_1,\ldots,u_m\}$ is linearly independent, it can be extended to a basis $\{u_1,\ldots,u_m,\ldots,u_n\}$ for $W$. Let $S:W\to W$ be the unique linear transformation that sends $w_i\mapsto u_i$ for $1\le i\le n$. Then $S\in\mathcal{L}(W)$, and $S$ is invertible since it sends a basis to a basis. All that remains is to show that $T_1=ST_2$.

Exercise 5: Show that $V_1+V_2=V$. In other words, show that for every $x\in V$ there exist $y\in V_1$, $z\in V_2$ such that $x=y+z$.

Exercise 6: Show that $T_1=ST_2$. You may find it helpful to use the fact that $V_1+V_2=V$. $\square$

Comment: The fact that $W$ is finite-dimensional simplified notation in Part 2, but this assumption is not necessary in either part of the proof.

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