Proving the "Complement Subspace" Theorem using $V = (\text{Range} R) \oplus (\text{Null} S)$ where $R$ and $S$ are linear maps This problem is intended to prove the "complement subspace" theorem in a particular way, with the theorem being that for every subspace $W \subseteq V$, there exists a subspace $W'$ such that $W \oplus W' = V$. We can assume that $V$ is finite dimensional. 
In a previous part of the problem, we were 
 given two linear maps $R: U \to V$ and $S: V \to U$ with the composite $$ S \circ R: U \xrightarrow{R} V \xrightarrow{S} U $$ being the identity map on $U$. We then proved that $V = \text{Range} R \oplus \text{Null}S$. We are now asked to use this result to prove the "complement subspace" theorem above. 
We were also given a hint to think of a space $U$ and a function $R: U \to V$ for which $W = \text{Range}R$, or to think of $U$ and $S: V \to U$ for which $W = \text{Null}S$.
My attempt at a solution:
I'll be somewhat brief, as some elements of my proof rely on results from previous homework, which we are encouraged to use. First, I want to construct a vector space $U$ that is defined as the range of $S$ when acted on vectors in $V$. I apply $S$ to basis vectors $v_1, ..., v_k \in V$ to get $S(v_1), ..., S(v_k)$. $S$ is surjective by definition, so $S(v_1), ..., S(v_k)$ span $U$ (we proved in a previous problem that surjective linear maps send spanning lists to spanning lists).
Now we cut the vectors $S(v_1), ..., S(v_k)$ to a basis of $U$, call it$S(v_1), ..., S(v_n)$. By a previous homework problem we proved the existence of a linear map $R: U \to V$ such that $$ S \circ R: U \xrightarrow{R} V \xrightarrow{S} U $$ is the identity map on $U$. 
So, a linear map exists that takes $S(v_1), ..., S(v_n)$ to $RS(v_1), ..., RS(v_n)$ and the list $S(v_1), ..., S(v_n)$ spans $U$ so we can also show that $RS(v_1), ..., RS(v_n)$ spans the range of $R$, by definition. So, we now have vectors in $\text{Range}R$ to make a subspace $\text{Range}R$. We can now apply $V = \text{Range} R \oplus \text{Null}S$ and show that letting $W = \text{Range}R$, there must be a complementary subspace $W'$ in which $W \oplus W' = V$, and that $W'$ is exactly $\text{Null}S$. 
Is this a valid proof? I'm worried it might be convoluted or could be simplified, but it may be what was desired given the hint, and another hint elsewhere saying previous homework problems may also be invoked.
 A: It's a little difficult to follow, due to the reliance on multiple previous proofs, but I think there's a serious structural issue with the proof. You should be assuming that $W$ is some subspace of $V$, and then constructing $R$, $S$, and $U$ in terms of $W$. At this point, you seem to be defining $W$ in terms of $R$ (which is, in turn, defined in terms of $S$, which appears to be arbitrary). You really should be starting with $W$, and using it to define $R$ and $S$. What if there were subspaces $W$ that were not the range of any transformation $R$? It's not true, but this possibility should be (implicitly) eliminated by your proof.
Even if we could fix up the proof, I would definitely call this circuitous. I don't like the idea of using the given result to prove the existence of complemented subspaces. In my opinion, there are three results that are utterly fundamental to finite-dimensional linear algebra:


*

*Every linearly independent list can be extended to a basis,

*Every spanning list can be reduced to a basis, and

*Every basis is of the same length.


By the time you finish a finite-dimensional linear algebra course, you should know and be able to apply these three results.
In this case, the first result can be used. Start with a basis for $W$. This is a linearly independent list in $V$. Then, extend to a basis of $V$. The list of vectors added to the original list will span a complementary subspace.
Since this is an assignment question (not that you haven't done plenty towards solving it yourself), I'll leave you to work out the details of the above argument. Feel free to let me know by comment if you'd like further guidance.
