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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?

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

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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 $R(T_1) = R(ST_2)$.

For the reverse direction, again you need a little more, as you've only shown that the dimensions are equal, rather than the subspaces themselves.

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 $R(T)$.

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  • $\begingroup$ Thanks for the reply. Following your suggested thm, can you comment if this is a way to go?: There exists a basis {v1, ... vn} of V s.t. {v1, ..., vk} is a basis for N(T) and {T1(vk+1), ... T1(vn)} that is a basis for R(T). Similarly, there exists a basis {u1, ... un} s.t. {u1, ..., uk} is a basis for N(T) and {T2(uk+1), ... T2(un)} that is a basis for R(T). Then, because the two bases for R(T) is each linearly independent, I extend each of it as bases for W. Then, I define linear operator S between the two basis for W (mapping T2(uk) to T1(uk) and the other extended basis according to index) $\endgroup$
    – Chapel
    Commented Apr 4, 2022 at 4:23
  • $\begingroup$ 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. $\endgroup$ Commented Apr 5, 2022 at 5:06

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