# Equal null space of two linear operators implies existence of an invertible operator

The full question goes as follows: "Suppose W is finite-dimensional and T1, T2 $$\in$$ L(V,W). Prove that null(T1) = null(T2) if and only if there exists an invertible operator S $$\in$$ L(W) s.t. T1 = ST2.

I know that there already have been questions and answers for this particular question, but I am wondering if my proof is correct if I assume V and W are both finite-dimensional. Thanks in advance for reading and answering my question!

(->) Suppose null(T1) = null(T2). Then, dim(null(T1)) = dim(null(T2)).

According to the fundamental thm,

dim(range(T1)) = dim(V) - dim(null(T1)) = dim(V) - dim(null(T2)) = dim(range(T2)).

dim(range(T1)) = dim(range(T2)) implies that range(T1) and range(T2) are isomorphic, which warrants the existence of a isomorphism S that is invertible.

(<-) Suppose that such invertible operator S exists. Then, it implies that range(T1) and range(T2) are isomorphic. If so, dim(range(T1)) = dim(range(T2)).

Again, apply the fundamental thm to obtain that dim(null(T1)) = dim(null(T2)).

** I see that the above proof is very simple, but is this correct assuming that both V and W are finite-dimensional?

• $S$ being an isomorphism between $\text{range}(T_1)$ and $\text{range}(T_2)$ does not necessarily imply $T_1 = ST_2$. Commented Apr 4, 2022 at 2:20

You need slightly more, for each direction of the proof. For the forward direction, you need more than just the existence of an isomorphism between $$R(T_1)$$ and $$R(T_2)$$. Your choice of $$S$$ needs to be more explicit so that you can actually get $$T_1 = ST_2$$. Whereas right now you've only shown that $$R(T_1) = R(ST_2)$$.
A hint: a missing ingredient here is that you might want to describe the linear maps $$T_1, T_2$$ in terms of some set of linearly independent vectors. A useful result might be the more explicit form for rank-nullity theorem:
Let $$T: V \rightarrow W$$ be linear. If $$\dim(V) = n$$ and $$\dim(N(T)) = k$$, then there's a basis for $$V$$ written as $$\{v_1, \ldots, v_n\}$$ such that $$\{v_1, \ldots, v_k\}$$ is a basis for $$N(T)$$ and $$\{T(v_{k+1}), \ldots, T(v_n)\}$$ is a basis for $$R(T)$$.
• This is the right approach. I think you should write out exactly how $S$ behaves on the "other" basis vectors (the ones that aren't in $R(T_2)$. But overall you've got the right idea: the best way to describe any linear transformation is to specify how it acts on each element of a basis. Commented Apr 5, 2022 at 5:06