Suppose V, W are finite dimensional and $T\_1,T\_2\in \mathscr L(V,W)$. Prove that if dim null $T\_1$ = dim null $T\_2$, then there exist invertible operators $R\in \mathscr L (V)$ and $S \in \mathscr L(W)$ such that $T\_1=ST\_2R$. 

Here's what I have. I just need to fill in some missing gaps. 

Let dim null $T\_1$ = dim null $T\_2$. Let $u\_1,\ldots,u\_m$ and $v\_1,\ldots,v\_m$ be bases for null $T\_1$ and null $T\_2$, respectively, which we will extend to the following bases of $V$: $u\_1,\ldots,u\_m,r\_1,\ldots,r\_n$ and $v\_1,\dots,v\_m,s\_1,\ldots,s\_n$. **Consider $T\_1(r\_1),\ldots,T\_1(r\_n)$ which is linearly independent in $W$.**

This is one of the pieces that I need some clarification on. Why is this linearly independent? 

Then I can extend this to a basis of $V$: $T\_1(r\_1),\ldots,T\_1(r\_n),p\_1,\ldots,p\_k$. Similarly, we do this for $T\_2(s\_i)$. Namely, $T\_2(s\_1),\ldots,T\_2(s\_n),q\_1,\ldots,q\_k$ is a basis of $V$. 

I define $R\in\mathscr L(V)$ in the following way: $R(u\_i)=v\_i$ and $R(r\_j)=s\_j$. 

I define $S\in \mathscr L(W)$ in the following way: $S(T\_2(s\_j))=T\_1(r\_j)$ and $S(q\_l)=p\_l$.

**Since S and R map one basis to another basis this makes S and R invertible.** 

This is another part that I'm not sure about. 

Since the $u\_i$'s form a basis of null $T\_1$ then $T\_1(u\_i)=0$. I would like to show that $ST\_2R(u\_i)=ST\_2(v\_i)=S(0)=0$, but **I'm not sure how to conclude that $S(0)=0$**. Furthermore, $T\_1(r\_j)=ST\_2(s\_j)=ST\_2(R(r\_j))$. Thus, we have $T\_1=ST\_2R$.