Prove that there exist invertible operators R and S such that $T_1=ST_2R$ iff dim null $T_1$ = dim null $T_2$

Question

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 $T_1=ST_2R$.

Answer 1

First, linear independence of $T_1(r_i)$ is obvious, because if $\sum c_iT_1(r_i) = 0$, then $T_1(\sum c_ir_i) = 0$, but then $\sum c_ir_i$ belongs to the kernel of $T_1$. Hence, it is equal to a linear combination of the $u_i$, which contradicts the fact that these are linearly independent.

If a linear transformation $S$ on $W$ maps a basis to itself, then it is both injective and surjective. To see this, if it maps a basis $e_i$ to a basis $f_i$, that is if $T(e_i) = f_i$, then if $T(\sum c_ie_i) = 0$, then $\sum c_i T(e_i) = 0$, but then $\sum c_if_i = 0$, so this can only happen if the $c_i$ are zero, but this implies that $\sum c_ie_i = 0$. Hence, $T$ is injective, and by rank nullity theorem(if you do not know this, then you can show it directly as well) $S$ will be surjective as well.

The fact that any linear transformation must take $0$ to $0$ follows from the fact that $S(0) = S(0+0) = S(0) + S(0) = 2S(0)$, now subtract $S(0)$ from both sides.

Answer 2

Suppose $a_1T_1(r_1)+\dots+a_nT_1(r_n)=0$. Then $T_1(a_1r_1+\dots+a_nr_n)=0$ and therefore $a_1r_1+\dots+a_nr_n\in\operatorname{null} T_1$; by assumption, $$ a_1r_1+\dots+a_nr_n=b_1u_1+\dots+b_mu_m $$ and therefore $$ -b_1u_1-\dots-b_mu_m+a_1r_1+\dots+a_nr_n=0 $$ forcing $a_1=a_2=\dots=a_n=0$.

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 $V$.

These should be bases of $W$, not $V$ (most likely a typo, but better being precise).

Now, by definition, $R$ maps a basis of $V$ to a basis of $V$; thus its image is $V$, because it contains a spanning set (so $R$ is surjective) and the rank-nullity theorem tells you that $R$ is also injective. Similarly for $S$.

A linear map always maps $0$ to $0$, so $S(0)=0$: the null space is a subspace, isn't it? When you proved it, you surely used this fact. Anyway, set $w=S(0)$; then $w=S(0+0)=S(0)+S(0)=w+w$. Hence $0=w+(-w)=w+w+(-w)=w+0=w$.

Finally, two linear maps that act the same on a basis (or, more generally, on a spanning set) are equal; if you don't recall this fact, write $v=a_1u_1+\dots+a_mu_m+b_1r_1+\dots+b_nr_n$ and compute $T_1(v)$ and $ST_2R(v)$, showing they're equal.