# Prove that $\text{null}(T_1)= \text{null}(T_2)$ if f $\exists$ $S \in L(W,W)$ such that $T_2 = ST_1$

If $$W$$ 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 $$S \in L(W,W)$$ invertible such that $$T_1 = ST_2$$

I don't understand the thought process behind this proof and was hoping for someone to illuminate for me the reason behind certain actions from a solution that I found for the proof.

Solution: If we assume $$\text{null}(T_1)= \text{null}(T_2)$$. Since $$W$$ is finite dimensional, so is range$$(T_2)$$. Let $$w_1,\cdots,w_n$$ be a basis of range$$(T_2)$$, then there exists $$v_1, \cdots, v_n \in V$$ such that: $$T_2v_i = w_i, \; i = 1, \cdots, n$$

The first thing the author wants to establish is $$V = \text{null}(T_2) \oplus \text{span}(v_1,v_2,...v_n)$$

So we work through this (which I omitted) which I understand the mechanics. Then we arrive at the declaration:

Similarly, $$T_{1}v_{1},....T_{1}v_{n}$$ is linearly independent. For if $$a_1T_1v_1+\cdots+a_nT_1v_n=0$$ we have $$T_1(a_1v_1+\cdots+a_nv_n)=0.$$

Note that $$\text{null}(T_1)= \text{null}(T_2)$$ it follows that: $$0=T_2(a_1v_1+\cdots+a_nv_n)=a_1w_1+\cdots+a_nw_n.$$

Thus $$a_1=\cdots=a_n=0$$. Now extend $$w_1,\cdots, w_n$$ to a basis of $$W$$ as $$w_1,\cdots, w_n,e_1,\cdots,e_m$$ and $$T_1v_1, \cdots,T_{1}v_{n}$$ to a basis of $$W$$ as $$T_1v_1, \cdots,T_{1}v_{n},f_1, \cdots, f_m$$.

Define $$S\in L(W,W)$$ by: $$Sw_i=T_1v_i, \; Se_j=f_j, \; i=1,\cdots,n; \; j=1,\cdots,m.$$

Note that $$V=\text{null}(T_2) \oplus \text{span}(v_1,\cdots,v_n)$$ any $$v\in V$$ can be expressed as: $$v=v_{\text{null}}+a_1v_1+\cdots+a_nv_n,$$

where $$v_{\text{null}} \in \text{null}(T_1)= \text{null}(T_2)$$ and $$a_1=\cdots=a_n \in \mathbb{F}$$. Hence we have:

\begin{align*} ST_2(v)=&ST_2(v_{\text{null}}+a_1v_1+\cdots+a_nv_n)\\ =&ST_2(a_1v_1+\cdots+a_nv_n)\\ =&S(a_1w_1+\cdots+a_nw_n)\\ =&a_1T_1v_1+\cdots+a_nT_1v_n\\ =&T_1(a_1v_1+\cdots+a_nv_n)\\ =&T_1(v_{\text{null}}+a_1v_1+\cdots+a_nv_n)=T_1(v) \end{align*}

So my questions:

1) What is the purpose of showing that the set $$V$$ is a direct sum?

2) How did we conclude that $$T_{1}v_{1},....T_{1}v_{n}$$ is linearly independent?

I understand the mechanics at the very end, but not the reason behind setting everything up in order to arrive at our conclusion.

Source of the solution if you wanted to refer to the original document (It is question 4):

http://linearalgebras.com/3d.html

1. $$V= \text{null}(T_2) \oplus \text{span}(v_1, \ldots, v_n)$$ ensures $$S$$ is invertible. Since $$\text{null}(T_2) \cap \text{span}(v_1, \ldots, v_n)= \emptyset$$, no linear combination of $$v_1, \ldots, v_n$$ can be in $$\text{null}(T_2)= \text{null}(T_1)$$. So since $$\{ w_1, \ldots w_n \}$$ are linearly independent, this ensures $$\{ Sw_1, \ldots, Sw_n \}= \{ T_1v_1, \ldots, T_1v_n \}$$ remains linearly independent. An invertible map must send a linearly independent set to an independent set of vectors.
2. Actually, you showed this as part of your proof. You assumed $$a_1T_1v_1+\cdots+a_nT_1v_n=0$$ and showed this implied $$a_1= \ldots =a_n=0$$, the definition of linear independence.