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.
Proof looks good. Here is a shorter one (relies on some general facts about how linear maps may extend). Let $K$ be the kernel of $T_1$ and $T_2$, and let $P$ be any complement to $K$ in $W$, so that $V = K \oplus P$. Then $T_1$ and $T_2$ are completely determined by their effects on $P$, and they each restrict to isomorphisms of $P$ onto their images $T_i(P)$. There exists therefore an invertible linear map $S$ from $T_2(P)$ onto $T_1(P)$. Extend $S$ in any way to an isomorphism of $W$ onto itself. Then $S$ is an invertible linear operator on $W$ for which
$$ST_2v = T_1v$$
for all $v \in P$. But also $ST_2v = T_1v = 0$ for all $v \in K$, so in fact $ST_2 = T_1$ as operators on $V$.
Best Answer
HINT: $ \rm T(v-Tv) = Tv-T^2v = 0 $ so $ \rm v-Tv \in \ker T$, and $ \rm v= (v-Tv) + Tv$