A favorite problem from Linear Algebra Done Right

Solutions to Linear Algebra Done Right are all over the Internet, and I'm not really saying that's a good thing, but it means that I, in my independent study, am well able to verify my approaches.

That is, until I approach a problem differently than others have, and then I have to post my solution on math.stackexchange.com and hope that someone bothers to read it and confirm my approach. These are some of those problems, and incidentally, they were some of the more interesting ones I've enjoyed.

For some common notation: 

$\mathscr{P}(\textbf{F})$ : The vector space of (finite) polynomials on the field $\textbf{F}$
$\mathscr{P}_{n}(\textbf{F})$: The vector space of polynomials of degree $\leq n$
$\mathscr{L}(V, W)$: The vector space of linear maps from the vector space V to the vector space W (These maps can all be represented by matrices).

So, with that out of the way... 

Problem 4.11: 

Fix some $p \in \mathscr{P}(\textbf{F})$ with $p \neq 0$, where $p$ has degree $m$. Let $U = \lbrace pq | q \in \mathscr{P}(\textbf{F})\rbrace$.

Notice that U is a linear subspace of $\mathscr{P}(\textbf{F})$, composed of all polynomials with p as a factor. Since our vector space is commutative, we can talk about the quotient space $\mathscr{P}/U$, the space of cosets (or "affine subsets") of the form $r + U$ for some $r \in \mathscr{P}(\textbf{F})$ under typical coset addition.

The most important question we might want to ask is, what is the structure of this quotient space, and what is its dimension? It turns out that this quotient space is isomorphic to $\mathscr{P}_{m-1}(\textbf{F})$, and we will show this this by showing that U is the null space of a linear transformation, which we will set up in the next line.

Let $T \in \mathscr{L}(\mathscr{P}(\textbf{F}), \mathscr{P}_{m-1}(\textbf{F}))$ be the transformation defined as $Ts=r$  for $s \in \mathscr{P}(\textbf{F})$, where $r$ is the unique polynomial remainder of degree $<m$ defined via the division algorithm; the unique $r$ where $s = p q + r$ for some $q \in (\mathscr{P}(\textbf{F})$. (This ensures that T is well-defined and maps into $\mathscr{P}_{m-1}$)

Now, a natural question to ask would be, are you sure that T is actually a linear transformation? After all, all of the structural arguments that follow about kernels of maps rely on having a linear transformation.

First, we check additivity (homogeneity):

Let $s$ and $t$ be polynomials on $\textbf{F}$. 

By the division algorithm, $s=p q_1 + r_1$ and $t = p q_2 + r_2$ for $q_1, q_2$ polynomials, $r_1, r_2$ of degree $< m$. Evaluating T, we see that $Ts = r_1$ and $Tt = r_2$, and $Ts + Tt = r_1 + r_2$. 

Now we need to ensure that $Ts + Tt = T(s+t) $

Evaluating $s + t$, we see that: 

$s + t = p q_1 + p q_2 + r_1 + r_2 = p(q_1 + q_2) + (r_1 + r_2)$. 

Since $(r_1+r_2)$ also has degree $< m$, this is thus the unique representation of $s + t$ in this form, and $T(s+t) = r_1 + r_2$.

Now, we check scalar multiplication:

Let $\alpha \in \textbf{F}$, and consider $\alpha s$ for $s \in  \mathscr{P}(\textbf{F}$.
$\alpha s = \alpha(p q_1 + r_1) = p(\alpha q_1) + \alpha r_1$ 

Since $\alpha r_1$ is of degree $< m$ and of course $\alpha q_1$ is a polynomial over F, this again is the unique representation guaranteed by the division algorithm and $T(\alpha s)= \alpha T(s)$

Thus T is a well-defined and linear transformation.

Next, we show that U is the kernel of T: 

To show $U \subset null T$: Suppose $u \in U$. Then by the definition of U, $u = pq = pq + 0$ for some $q$, so $Tu = 0$ and $u \in null T $. 

To show $null T \subset U$: Suppose $u \in null T$. Then $u = pq + 0$ for some q, so $u = pq$ and $u \in U$.

And T is onto ... 

If $r$ is any polynomial in $\mathscr{P}_{m-1}(\textbf{F})$, then the inclusion $i(r) = r$ is the identical polynomial in $\mathscr{P}(\textbf{F})$ and $Tr = r$, since $ r = 0p + r$ and deg r $< m$. Thus all polynomials in $\mathscr{P}_{m-1}(\textbf{F})$ are in range T, and T is onto.

Now we can put it all together!

Using Theorem 3.91d, we see that $\mathscr{P}(\textbf{F})/null T \cong range T$.

Thus $\mathscr{P}(\textbf{F})/U \cong \mathscr{P}_{m-1}(\textbf{F})$, using the substitutions that U=null T and range T = $\mathscr{P}_{m-1}(\textbf{F})$.

This confirms the initial statement, and indicates that the dimension of the quotient space between these two infinite-dimensional vector spaces is $(m-1)$, where, again, $m$ is the degree of the polynomial $p$.

Quotients of Subspaces and Subspaces of Quotients

I'm currently at Section 9 in Willard, on quotient spaces and the strong topology.

I've still got an open question on stackexchange for 9A.2, which seems to require using compactness to prove the second part. (The text hasn't covered compactness yet! So that's a pain.)

Exercise 9D took me a while, and it feels messy because of notation. So I'll write it up here, which should make it neater and force me to explain it better.

Requisite background: 

Now, in this text quotient spaces are understood in the context of decompositions. I greatly dislike this word, and would much rather use the word partition. So if you're like me and have only ever heard the word "partition" before, and not "decomposition," then just read the word "partition" in your head every time they word "decomposition" shows up.

A decomposition $\mathscr{D}$ of a topological space $X$ is a collection of disjoint subsets of $X$ whose union is $X$.

In other words. It's a partition. A subset of the decomposition is a set of equivalence classes, effectively.

We assign a topology to the decomposition by determining that a set $\mathscr{E}$ of equivalence classes (a subset of $\mathscr{D}$) is open in $\mathscr{D}$ iff the union of those equivalence classes is an open set in $X$. That is, $\bigcup \lbrace E | E\in \mathscr{E} \rbrace$ is open in $X$.

We also note that this endows $\mathscr{D}$ with the quotient topology induced by the map P(x), which maps x to the equivalence class containing x. (I call this the Partition map in my head, though it appears nowhere in the text.)

If we have endowed the decomposition with such a topology, we naturally call it the decomposition space.

Okay, on to the problem...

When we're talking quotient spaces, there's two ways you could naturally think of to "restrict" the decomposition to some subset A of the original space X.

The first way:
Let's denote as $\mathscr{D}_A$ the decomposition of A defined by $F\in \mathscr{D} \Rightarrow F\cap A \in \mathscr{D}_A$. The topology on $\mathscr{D}_A$ thus follows from the subspace topology on A.
That is, $\mathscr{U}\subset \mathscr{D}_A$ is open iff $\bigcup \lbrace F | F\in \mathscr{U}\rbrace$ is open in A, as a subspace of X.

The second way:Let's denote our second alternative, $\mathscr{D}|A = \lbrace y \in \mathscr{D} : A\cap y \neq \emptyset \in X \rbrace$ . This is a subset of our original space $\mathscr{D}$ endowed with the subspace topology from $\mathscr{D}$, composed of all equivalence classes which "touch" our set A. 

Evidently, sometimes these two topologies are not identical. The text gives an example which makes this obvious: Consider the set $\lbrace (x, 0) \in \mathbb{R}^2| x < 0\rbrace \cup \lbrace (0,1)\rbrace $; for which $\mathscr{D}_A$ has an isolated point, and $\mathscr{D}|A$ does not.

However, if A in X is a union of elements of our original $\mathscr{D}$, then $\mathscr{D}_A$ is homeomorphic to $\mathscr{D}|A$, which I will prove here. 

For this homeomorphism, I'm going to explicitly define a map. It may help to remind yourself what $P(x): X \rightarrow \mathscr{D}$ means. Also remember common restriction notation; I will abuse it a little and use $(P|A)(x): A\rightarrow \mathscr{D}_A$. That is, $(P|A)(x)$ maps x onto the equivalence class considered as a subset of A, not necessarily the full equivalence class in $\mathscr{D}$.

Now for our homeomorphism! Define $\phi: \mathscr{D}_A \rightarrow \mathscr{D}|A$ by:
$$ \phi (x) = P( (P|A)^{-1}(G)) \text{ for } G\in \mathscr{D}_A$$
Let's take a moment to make sure the map is well-defined, one-to-one, and onto.

Well-defined: If $G\in \mathscr{D}_A$ then $(P|A)^{-1}(G)$ is actually a big ol' subset of X (well, the subset G!), and I'd want to make sure that every element in this subset gets sent to the same element of $\mathscr{D}$, and more specifically, an element of $\mathscr{D}_A$. Since G in $\mathscr{D}_A$ is defined to be equal to $F\cap A$ for some set $F \in \mathscr{D}$ then every element in G is also in F. Every element of F gets sent to $F\in \mathscr{D}$ by P() so regardless of the element we consider in G, it will get sent to this equivalence class F. Additionally, since $G=F\cap A$ is nonempty, F must be in $\mathscr{D}|A$ by definition.

Onto: If $F\in \mathscr{D}|A$, this means $F\in \mathscr{D}$ and $F\cap A$ is nonempty. But then $F\cap A$ is an element of $\mathscr{D}_A$ which maps to F, so F is in the image of $\phi$

One-to-One: Suppose $\phi(G_1) = \phi(G_2) = F \in \mathscr{D}|A$ for $G_1, G_2 \in \mathscr{D}_A $ Then $(P|A)^-1(G_1)$ and $(P|A)^-1(G_2) \subset F$ in X.

We didn't need that A is a union of elements of D in order to prove that the map is well-defined, one-to-one, and onto! We only need it to prove that it is a homeomorphism.

Quick lemma I am going to call the lemma of inclusion: With the assumed information in this problem, $F\in \mathscr{D}$ and $F \cap A$ is nonempty, then $F \subset A$. Since A is a union of elements of $\mathscr{D}$ and the elements of $\mathscr{D}$ are disjoint, any element $F\in \mathscr{D}$ is either entirely contained in A, or has no intersection with A. This follows immediately.

So, now that we have that squared away, to ensure it's a homeomorphism, I will prove that a set is open in $\mathscr{D}_A$ iff its image is open in $\mathscr{D}|A$

($\Rightarrow$) Let $\mathscr{U}\in \mathscr{D}_A$ be an open set, and consider $\phi(\mathscr{U})$. 

$$\begin{align}
\phi(\mathscr{U}) &= P( (P|A)^{-1} (\mathscr{U}))\\
&= P(V\cap A)
\begin{split} &\quad \text{for some V} \subset X \text{ that is open, since}\\
&\quad \mathscr{U}\text{ being open implies its preimage in A is open in A by definition}\\
\end{split}\\[1em]
&= P(V) \cap P(A)

\begin{split}&\quad P(V\cap A) \subset P(V \cap A) \text{is well known; and, in this case,}\\
&\quad F\in P(V)\cap P(A) \Rightarrow F\subset A \text{ (by inclusion lemma) and }\\
&\quad  \exists x \in F \text{such that} x\in V \text{and} x \in A \text{ so }P(F) \in P(V\cap A)\\
\end{split}
\end{align}$$

Since $P(V)$ is open in $\mathscr{D}$, and $P(A) = \mathscr{D}|A$, we see that $P(V) \cap \mathscr{D}|A$ is therefore open in the subspace topology inherited from $\mathscr{D}$.


($\Leftarrow$) Let $\mathscr{V} \subset \mathscr{D}|A$ be open in $\mathscr{D}|A.$ Then there exists some $\mathscr{W} \in \mathscr{D}$. Consider $\phi^{-1} (\mathscr{V})$

$$\begin{align}
\phi^{-1}(\mathscr{V}) &= \phi^{-1}(\mathscr{W}\cap \mathscr{D}|A) \\
&= P|A( P^{-1} (\mathscr{W}\cap \mathscr{D}|A)\\
&=P|A(P^{-1}(\mathscr{W})\cap P^{-1}(D|A)) \quad\\
&= P|A(P^{-1}(\mathscr{W}) \cap A) \quad  \text{using the inclusion lemma}\\
\end{align}$$

Now since $\mathscr{W}$ is open in $\mathscr{D}$ its preimage in X is open, which implies  $P^{-1}(\mathscr{W}) \cap A$ is open as a subset of A under the subspace topology!

Therefore $\phi^{-1}(\mathscr{V})$ is open in $\mathscr{D}_A$

We conclude that $\phi$ is a homeomorphism.

PHEW! Now that we are done, I bet you can see why the notation really messed with me in this proof. This was hell to type up, but it definitely helped me make sure I really knew why what I was doing worked.

Introduction

Right now, I'm doing a lot of studying. 

This summer, my schedule is: 

Mondays: Topology (Willard)
Tuesdays: Linear Algebra (Axler) and "Baby Rudin"
Wednesdays: Topology (Willard) and Algebra (Fraleigh)
Thursdays: Topology (Willard) and Linear Algebra (Axler)
Fridays: "Baby Rudin" and Algebra (Fraleigh)

My goal? 

• Become entirely proficient with, at the very least, undergrad linear algebra and algebra. I took these courses almost in their entirety as an undergrad and mostly need to become more familiar with the theory (the courses were not as proof-heavy as I would have liked).

• Work through as much of Rudin and Willard as possible. These texts are heavier and take a lot longer to work through. My goal at the start of the summer was to complete them, but I'd probably have to devote my entire summer to working only on these texts, and that's not worth it. I'm not even doing all of the exercises--just around half! 

This blog is planned, mostly, to be a record of my progress and a way to motivate myself to typeset some proper proofs. I'll be posting my favorite proof of the day, or perhaps two! 

If you stumble across this in your studies feel free to use this to help you, but don't copy! You don't learn by copying. You don't even get better grades.

Image source: Abstruse Goose

Don't just read it; fight it! Ask your own questions,
look for your own examples, discover your own proofs.
Is the hypothesis necessary? Is the converse true?
What happens in the classical special case? What
about the degenerate cases? Where does the proof
use the hypothesis?

--- Paul R. Halmos 

Also, keep in mind I'm learning so my proofs may be incorrect anyway ;)