(Context) To enhance my capacity to write good mathematical proofs and simultaneously familiarize myself with the fundamental ideas of elementary linear algebra, I have attempted to solve the third problem in this document [http://math.northwestern.edu/putnam/filom/Linear_and_Abstract_Algebra.pdf] and justify my solution in a proof-based manner. The question and my attempted solution to it are provided below.
(REQUEST) Can you tell me if my final answer and my proof thereof are correct? Additionally, if possible, can you provide me with some feedback on the style of my proof, as well as with some suggestions as to how I might be able to make my proof more clear, concise, and correct? In advance, thank you very much.
(QUESTION) Let $V$ be a 10-dimensional real vector space. Let $U_1$ and $U_2$ be real subspaces of $V$ such that $\dim U_1 = 3$ and $\dim U_2 = 9\space.$ In order to cause the problem to admit only one solution, we shall assume that $U_1$ is a subspace of $U_2\space.$ And let $\epsilon$ be the real vector space that consists of every linear transformation $T\in\mathcal{L}(V)$ such that $U_1$ and $U_2$ are both stable under $T$ ; in that case, compute the value of $\dim\epsilon$ .
(MY SOLUTION) Since $U_1$ is a 3-dimensional real vector space, we may let $\{v_1\space,v_2\space,v_3\}$ be a basis of $U_1\space.$ Since $U_1$ is a subspace of $U_2$ and $\{v_1\space,v_2\space,v_3\}$ is a linearly independent set in $U_1$ ( because $\{v_1\space,v_2\space,v_3\}$ is a basis of $U_1$ ), it follows that $\{v_1\space,v_2\space,v_3\}$ is a linearly independent set in $U_2\space.$ As $\{v_1\space,v_2\space,v_3\}$ is a linearly independent set in $U_2\space,$ and as $3<9=\dim U_2\space,$ it follows that there exist some $v_4 \dots v_9$ such that $\{v_1 \dots v_9\}$ is a basis of $U_2\space.$ And likewise, as $\{v_1 \dots v_9\}$ is a linearly independent set in $U_2$ ( because $\{v_1 \dots v_9\}$ is a basis of $U_2$ ), and as $U_2$ is a subspace of $V$, it follows that $\{v_1 \dots v_9\}$ is a linearly independent set in $V$, and therefore, since $9<10=\dim V$, we see that we can find some $v_{10}\in V$ such that $\{v_1 \dots v_{10}\}$ is a basis of $V$. Let $\zeta:U_1\times U_1 \times U_1 \times U_2 \times U_2 \times U_2 \times U_2 \times U_2 \times U_2 \times V \rightarrow \mathcal{L}(V)$ such that $\zeta(u_1\space,u_2\space,\dots,u_{10})$ is a linear operator on $V$ such that we have $\zeta(u_1\space,u_2\space,\dots,u_{10})(v_i)=u_i$ for all $i\in\{1\dots10\}$ ; because $\{v_1 \dots v_{10}\}$ is a basis of $V$, by the construction principle of linear maps, we see that such $\zeta$ is well-defined. Let $u$ be an arbitrary element of $U_1\times U_1 \times U_1 \times U_2 \times U_2 \times U_2 \times U_2 \times U_2 \times U_2 \times V$. Let $\eta \in U_1$ and $\phi \in U_2$ ; then as $\eta \in U_1$ and $\{v_1\space,v_2\space,v_3\}$ is a basis of $U_1\space,$ we see that there exist some $\eta_1\space,\space\eta_2\space,\space\eta_3\in\mathbb{R}$ such that we have $\eta=\eta_1v_1+\eta_2v_2+\eta_3v_3\space;$ and likewise, as $\phi \in U_2$ and $\{v_1 \dots v_9\}$ is a basis of $U_2\space,$ it follows that there exist some $\phi_1 \dots \phi_9 \in \mathbb{R}$ such that we have $\phi = \phi_1v_1+\dots+\phi_9v_9$ ; in that case, we have $$\begin{aligned} \zeta(u)(\eta)&=\zeta(u)(\eta_1v_1+\eta_2v_2+\eta_3v_3) \\ &=\eta_1\zeta(u)(v_1)+\eta_2\zeta(u)(v_2)+\eta_3\zeta(u)(v_3) \\ &=\eta_1u_1+\eta_2u_2+\eta_3u_3\space, \end{aligned}$$ and as $u_1\space, u_2\space, u_3 \in U_1$ ( because $u\in U_1\times U_1 \times U_1 \times U_2 \times U_2 \times U_2 \times U_2 \times U_2 \times U_2 \times V$ ), and as $U_1$ is a linear subspace that is closed under linear combination, it follows that we have $\zeta(u)(\eta)\in U_1\space,$ and therefore, $U_1$ is stable under $\zeta(u)$ . And likewise, we have $$\begin{aligned} \zeta(u)(\phi)&=\zeta(u)(\phi_1v_1+\dots+\phi_9v_9) \\ &=\phi_1\zeta(u)(v_1)+\dots+\phi_9\zeta(u)(v_9) \\ &= \phi_1u_1+\dots+\phi_9u_9\space, \end{aligned}$$ and as $u_1 \dots u_9 \in U_2$ ( because $u\in U_1\times U_1 \times U_1 \times U_2 \times U_2 \times U_2 \times U_2 \times U_2 \times U_2 \times V$ and $U_1\subseteq U_2$ ), and as $U_2$ is a subspace that is closed under linear combination, it follows that we have $\zeta(u)(\phi)\in U_2$ , and therefore, $U_2$ is stable under $\zeta(u)$ . We see that $\zeta(u)$ is a linear operator on $V$, and that $U_1$ and $U_2$ are stable under $\zeta(u)$ , and thus, we see that $\zeta(u)$ is an element of $\epsilon$ , and hence, as $\zeta(u)\in\epsilon$ for all $u\in U_1\times U_1 \times U_1 \times U_2 \times U_2 \times U_2 \times U_2 \times U_2 \times U_2 \times V=\mathrm{domain}(\zeta)$, it follows that $\epsilon$ is a valid codomain for $\zeta$ , and that we may thus view $\zeta$ as a map from $U_1\times U_1 \times U_1 \times U_2 \times U_2 \times U_2 \times U_2 \times U_2 \times U_2$ into $\epsilon$ . Again, let $u$ and $w$ be elements of $U_1\times U_1 \times U_1 \times U_2 \times U_2 \times U_2 \times U_2 \times U_2 \times U_2 \times V$. Let $\lambda$ and $\mu$ be real scalars. And let $v\in V$. As $v\in V$ and $\{v_1 \dots v_{10}\}$ is a basis of $V$, it follows that there exist some $x_1 \dots x_{10} \in \mathbb{R}$ such that we have $v=x_1v_1 + \dots + x_{10}v_{10}\space;$ and thus, $$\begin{aligned} \zeta(\lambda u + \mu w)(v) &= \zeta(\lambda u_1+\mu w_1\space,\lambda u_2 + \mu w_2\space,\dots,\lambda u_{10} + \mu w_{10})(x_1v_1 + \dots + x_{10}v_{10}) \\ &= x_1\zeta(\lambda u_1+\mu w_1\space,\lambda u_2 + \mu w_2\space,\dots,\lambda u_{10} + \mu w_{10})(v_1) + \dots +x_{10}\zeta(\lambda u_1+\mu w_1\space,\lambda u_2 + \mu w_2\space,\dots,\lambda u_{10} + \mu w_{10})(v_{10}) \\ &= x_1(\lambda u_1 + \mu w_1) + \dots + x_{10}(\lambda u_{10} + \mu w_{10}) \\ &= (x_1\lambda u_1 + \dots + x_{10}\lambda u_{10}) + (x_1\mu w_1+\dots+x_{10}\mu w_{10}) \\ &= \lambda(x_1u_1 + \dots + x_{10}u_{10}) + \mu(x_1w_1+\dots+x_{10}w_{10}) \\ &= \lambda(x_1\zeta(u)(v_1) + \dots + x_{10}\zeta(u)(v_{10})) + \mu(x_1\zeta(w)(v_1)+\dots+x_{10}\zeta(w)(v_{10})) \\ &= \lambda\zeta(u)(x_1v_1 + \dots x_{10}v_{10}) + \mu\zeta(w)(x_1v_1 + \dots + x_{10}v_{10}) \\ &= (\lambda\zeta(u)+\mu\zeta(w))(v) \\ \zeta(\lambda u + \mu w) &= \lambda\zeta(u) + \mu\zeta(w)\space, \end{aligned}$$ and therefore, $\zeta$ is a linear map. Suppose that $u\in\ker\zeta\space;$ then as $i$ ranges from $1$ to $10$ , $$\begin{aligned} \zeta(u)(v_i)&= O(v_i) \\ u_i &= 0\space, \end{aligned}$$ and therefore, we have $u=0$ and thus $\ker\zeta=\{0\}$ , and as $\ker\zeta=\{0\}$ and $\zeta$ is linear, it follows that $\zeta$ is injective. Let $A$ be an arbitrary element of $\epsilon$ . Then as $v_1\space,v_2\space,v_3\in U_1$ and $U_1$ is stable under $A$ ( because $A$ is in $\epsilon$ ), it follows that we have $Av_1\space,\space Av_2\space,\space Av_3\in U_1\space;$ and likewise, as $v_4 \dots v_9 \in U_2$ and $U_2$ is stable under $A$ ( because $A$ is an element of $\epsilon$ ), it follows that we have $Av_4 \dots Av_9 \in U_2\space;$ and as $v_{10} \in V$ and $A$ is an operator on $V$, we have $Av _{10}\in V$ ; and therefore, we have $(Av_1\space,\space Av_2\space,\dots,\space Av_{10}) \in U_1\times U_1 \times U_1 \times U_2 \times U_2 \times U_2 \times U_2 \times U_2 \times U_2 \times V$, and hence, $\zeta(Av_1\space,\space Av_2\space,\dots,\space Av_{10})$ is defined. Now let $v\in V$. Then as $v\in V$ and $\{v_1 \dots v_{10}\}$ is a basis of $V$, there exist some $x_1 \dots x_{10} \in V$ such that $v=x_1v_1 + \dots + x_{10}v_{10}\space,$ and therefore, we have $$\begin{aligned} \zeta(Av_1\space,\space Av_2\space,\dots,\space Av_{10})(v) &= \zeta(Av_1\space,\space Av_2\space,\dots,\space Av_{10})(x_1v_1+\dots+x_{10}v_{10}) \\ &=x_1\zeta(Av_1\space,\space Av_2\space,\dots,\space Av_{10})(v_1)+\dots+x_{10}\zeta(Av_1\space,\space Av_2\space,\dots,\space Av_{10})(v_{10}) \\ &=x_1Av_1+\dots+x_{10}Av_{10} \\ &=A(x_1v_1+\dots+x_{10}v_{10}) \\ &=Av \\ \zeta(Av_1\space,\space Av_2\space,\dots,\space Av_{10})&=A\space, \end{aligned}$$ and therefore, for all $A$ in $\epsilon$ , we have $A\in\mathrm{Im}\space\zeta$ , and therefore, $\zeta$ is a surjective map from $U_1\times U_1 \times U_1 \times U_2 \times U_2 \times U_2 \times U_2 \times U_2 \times U_2 \times V$ into $\epsilon$ . And as $\zeta$ is a linear bijection from $U_1\times U_1 \times U_1 \times U_2 \times U_2 \times U_2 \times U_2 \times U_2 \times U_2 \times V$ into $\epsilon$ , it follows that $\zeta$ is an isomorphism between $U_1\times U_1 \times U_1 \times U_2 \times U_2 \times U_2 \times U_2 \times U_2 \times U_2 \times V$ and $\epsilon$ , and therefore, we have $$\begin{aligned} \dim\epsilon&=\dim(U_1\times U_1 \times U_1 \times U_2 \times U_2 \times U_2 \times U_2 \times U_2 \times U_2 \times V) \\ &=3\dim U_1 + 6\dim U_2 + \dim V \\ &= 3\cdot3 + 6\cdot9 + 10 \\ &=73\space. \end{aligned}$$ Q.E.D.
This proof appears valid :)
Stylistically, the proof is extremely weak. That said, I think it would be missing the point to write a harsh critique, because your proof reads exactly like a proof written by you should read. By this I mean: it shines through in every line of this proof that you have put a tremendous amount of effort into learning linear algebra and logical reasoning. As such your mental energy is focused on being correct rather than on being insightful. Please understand this as a good thing.
At some point you will be more confident about your ability to write correct things if you are absolutely forced to, and you will feel more comfortable leaving bigger gaps for the sake of clarity. But you do not have that confidence yet, which is for now not a problem but a good assessment of reality: you are learning. Such confidence right now would be unwarranted at best.
That said, there are three things worth keeping in mind even at this stage. In decreasing order of importance:
Finally, a couple much more fiddly concerns that probably shouldn't bother you, but I can't resist writing here:
Weigh the benefit of introducing notation. There is simply no reason to write $U_1\times U_1\times U_1\times U_2\times U_2\times U_2\times U_2\times U_2\times U_2\times V$ more than once; just name it $X$ the first time it comes up. Leaving it spelled out does nothing but leave you open to making typos (indeed, this did happen: at the end of #8)
Pay attention to the stock phrases your references use. The results of this take longer to show, of course, and will vary depending on what you notice. For instance, I learned early on the power of the words "by definition". Those words can be overused, but the two parenthetical comments that you make in #14 are textbook use cases: "... and $U_1$ is stable under $A$ by definition."