Been working on the section about inner product spaces in Axler. This is where it starts getting a bit newer for me. Obviously, I know what the dot product is between real vectors, and I've even worked with inner products when I learned about orthogonal functions in my Quantum Mechanics class, but my professor was so obtuse about his explanations that while I made some connection to the dot product, it was basically just "Ok, so this is like the dot product, but for functions" without really understanding the underlying structure nor the importance of the "conjugate symmetry" property of inner products, among other things. Definitely haven't worked with the theory for them, anyway.
Axler's Linear Algebra Done Right contains this problem in chapter 6A:
Suppose $\mathbb{F} = \mathbb{R}$ (That is, that we're working with a real vector space and potential inner products which map into the reals and not the complex numbers) and $V \neq \lbrace 0 \rbrace$. Replace the positivity condition (which states that $\langle v, v \rangle \geq 0 \quad \forall \quad v \in V$) in the definition of an inner product with the condition that $\langle v, v \rangle > 0 $ for some $ v \in V$. Show that this change in the definition does not change the set of functions from $V\times V$ to $\mathbb{R}$ that are inner products on $V$.
It is clear that the positivity and the definiteness condition in conjunction with the fact that $V \neq \lbrace 0 \rbrace$ would imply this weaker condition, so basically all there is to do here is to prove that the apparently-weaker condition implies the positivity condition in a real vector space.
Of course, I use all the properties of an inner product (page 166) except for the positivity property, in conjunction with the property that $\langle v, v \rangle > 0$ for some $v \in V$. I also take the liberty of using any properties of the inner product which do not depend on positivity.
I thought originally that the solution to this problem would be due to just the conjugate symmetry and additivity, fixed an arbitrary $w\in V$ and played around with expressions of the form $\langle v + w, v - w \rangle$ which got me expressions that included $\langle v, v \rangle$ and $\langle w, w \rangle$ but they didn't seem to work. Then I decided to mess around with $\langle v+\lambda w, v + \lambda w \rangle$ for $\lambda \in \mathbb{R}$ and it kind of fell out.
I checked the best-known solution website to check my answer, and what do you know, the solution isn't in the solution manual nor is it on the website, so I figured I'd write it up!
So, below is my proof.
Disclaimer: This blog is not for you to copy off my solutions, but rather for me to improve my solution-writing, so if your goal is to copy off me, keep in mind it is your own education you're harming! Not only am I merely a learner as well, but you're more likely to bomb your next test if you copy. So follow the hint from above first and then go talk to your professor during office hours before you even think about copying this.
Let $V$ be a real, nontrivial vector space. Suppose $\langle , \rangle : V \times V \rightarrow \mathbb{R}$ satisfies all properties of an inner product except possibly the positivity condition, and suppose that there exists some $v \in V$ such that $\langle v, v\rangle > 0$.
Let $w \in V$ as well. I wish to show that $\langle w, w \rangle \geq 0$
Case 1: If $w = \lambda v$ for some $\lambda \in \mathbb{R}$ then ...
$$ \begin{align} \langle w, w \rangle & = \langle \lambda v , \lambda v \rangle \\
&= \lambda \langle v, \lambda v \rangle \qquad \text{first-slot homogeneity} \\
&= \lambda \overline{\langle \lambda v, v \rangle} \qquad \text{conjugate symmetry}\\
&= \lambda \langle \lambda v, v \rangle \qquad \text{The image is in } \mathbb{R} \\
&= \lambda^2 \langle v, v \rangle \qquad \text{first-slot homogeneity} \\
&\geq 0 \qquad \text{Since } \lambda \geq 0 \in \mathbb{R} \text{ and } \langle v,v \rangle > 0
\end{align}$$
Case 2: If $w \neq \lambda v$ for any $\lambda \in \mathbb{R}$ then...
This implies that $ v + \lambda w \neq 0$ for all $\lambda \in \mathbb{R}$. This means that, by the definite property of the inner product, we must have $\langle v + \lambda w, v + \lambda w \rangle \neq 0$ for all $\lambda \in \mathbb{R}$.
I will show that if $\langle w, w \rangle < 0$ it would imply that there is some $\lambda$ for which $\langle v + \lambda w, v + \lambda w \rangle = 0$, and thus by the converse we must conclude that $\langle w, w \rangle \geq 0$.
First note that we can rewrite $\langle v + \lambda w, v + \lambda w \rangle$. I occasionally use additivity in the second slot, which follows from additivity in the first slot and conjugate symmetry and does not require positivity. Additionally, since $\lambda$ is real according to our initial conditions, we must have homogeneity in the second slot. The argument parallels the steps in case 1 and again only relies on homogeneity in the first slot, conjugate symmetry, and the fact that we are working in a real vector space. Lastly since we are in a real vector space, $\langle u, v \rangle = \langle v, u \rangle$, which I will call "symmetry" to replace "conjugate symmetry."
$$\begin{align} \langle v + \lambda w, v + \lambda w \rangle &= \langle v, v + \lambda w \rangle + \langle \lambda w, v + \lambda w \rangle \\
&= \langle v, v + \lambda w \rangle + \langle \lambda w, v + \lambda w \rangle \qquad \text{ left additivity} \\
&= \langle v, v \rangle + \langle v, \lambda w \rangle + \langle \lambda w, v\rangle + \langle \lambda w, \lambda w \rangle \qquad \text{right additivity} \\
&= \langle v, v \rangle + \lambda \langle v, w \rangle + \lambda \langle w, v \rangle + \lambda ^2 \langle w, w \rangle \qquad \text{homogeneity} \\
&= \langle v,v \rangle + 2 \lambda \langle v, w \rangle + \lambda ^2 \langle w, w \rangle \qquad \text{symmetry} \\
\end{align}$$
Note that, with $v$ and $w$ fixed, this is a second-degree polynomial in $\lambda$ with coefficients $a = \langle w, w \rangle, \quad b = 2 \langle v, w \rangle, \quad c = \langle v, v \rangle$. We can easily determine that if $ a = \langle w, w \rangle < 0$ then, using the fact that $c = \langle v, v \rangle > 0$ the quadratic discriminant $b^2 - 4 a c$ would be positive, and there would be a real value of $\lambda$ satisfying $ \langle v,v \rangle + 2 \lambda \langle v, w \rangle + \lambda ^2 \langle w, w \rangle = 0$.
Of course, since we have that $\langle u,u \rangle = 0$ iff $u = 0$ and we know that $v + \lambda w \neq 0$ for all $\lambda$, which means, by the converse of the paragraph above, that we must conclude $\langle w, w \rangle \geq 0$. This is the positivity condition, and we are done!
Thursday, August 17, 2017
Thursday, August 10, 2017
Creating Unity: The Partial Ring of Quotients
So, if you have taken introductory abstract algebra, you have probably learned about something called a field of quotients, which is basically a generalization of the idea of rational numbers (aka "fractions") on certain rings called integral domains. But not every ring is an integral domain, so what if you want to create something like "fractions" using any ol' commutative ring--even a rng, that doesn't have a multiplicative iiidentity? Well, you can -- kind of -- with some restrictions, as long as it has at least one element that is not a zero divisor.
In this post, I'll remind you how the field of quotients is constructed and how it works (but I won't prove that it works, since that's done in every textbook ever), and then we will use similar ideas to create a structure called the Partial Ring of Quotients, and find the key to creating unity!
(Added later: Some have told me this is also called the localization of a ring. My approach differs from Wikipedia's in notation, since I made this post based on an exercise in Fraleigh's algebra. I also aim to explain a bit more thoroughly than Wikipedia does!)
I figure it would probably help to talk about the field of quotients a bit first, since I know that Wikipedia can be a bit terse. As a refresher, if you have an integral domain $D$, which is a commutative ring with unity (1) and no zero divisors (such as $\mathbb{Z}$), you can construct its field of quotients $F$ fairly intuitively!
The field of quotients' operations are defined on a set of equivalence classes of all ordered pairs of D. This sounds complicated, but it's not really! In $\mathbb{Q}$, for instance, we know that $\frac{1}{2} = \frac{3}{6}$. Those could be represented as ordered pairs, under some equivalence relation $\sim$ -- we'd like to generalize the idea that $(1, 2) \sim (3,6)$ to an arbitrary integral domain D. So, if D is an integral domain, we choose to say that $(n_1, d_1) \sim (n_2, d_2) \Leftrightarrow n_1 * d_2 = n_2 * d_1$, where $n_1, n_2, d_1, d_2 \in D$ (with $d_1, d_2 \neq 0$). Is this equivalence relation arbitrary--no! This is just like "cross-multiplying" to check if two fractions are equal, and disallowing fractions with a denominator of 0. This equivalence relation relies on $d_1$ and $d_2$ also not being zero divisors, and on the commutativity of $D$. Proof that it is an equivalence relation can be found in any introductory algebra text (or might even be relegated to the exercises).
The equivalence classes $[(n,d)]$ (read: numerator, denominator) form the elements of $F$, our field of quotients. The elements are equivalence classes because we want to make sure we make it clear that, for example, even though $(1, 2)$ and $(3, 6)$ may look different, they are actually representatives of the same exact element of $\mathbb{Q}$. You can add two equivalence classes in $F$ via $[(a,b)] + [(c,d)] = [(ad+bc, bd)]$ and multiply the equivalence classes by $[(a,b)]*[(c,d)]=[(ac, bd)]$. That these operations are well-defined regardless of representative ordered pairs, commutative, satisfy all the field axioms, etc., is, again, a topic textbooks go over immensely. But, you should be able to verify that if you represent two fractions in $\mathbb{Q}$ as ordered pairs, these operations work the same way as addition and multiplication in $\mathbb{Q}$.
In the field of quotients (which, to be fair, I haven't justified is actually a field), we now have a multiplicative inverse for every element not equivalent to 0, for if $[(a, b)]$ is an element of $F$ such that $a \neq 0$, the element $[(b,a)]$ is clearly also in $F$ and forms its multiplicative inverse -- just like taking the reciprocal. So even though we don't get multiplicative inverses in the integers, every element of $\mathbb{Q}$ does have a multiplicative inverse.
Can we think of $F$ as containing $D$? Of course. Simply define a map from $d \in D$ to $[(d, 1)]\in F$. It is a simple task to check that this gives an isomorphism from $D$ to a subset of $F$.
So above is basically the process you can use to turn any integral domain into a field that contains that integral domain -- extending it to permit division by any nonzero element.
So, say we've got a commutative rng $R$. There's a couple problems with the construction of $F$ above and commutative rngs, the most obvious being that $1$ might not be a member of our rng, so there goes the easy isomorphism we had and an easy way to define unity with $[(1,1)]$. Also, remember how I mentioned, briefly, that the equivalence relation between ordered pairs actually providing an equivalence relation relies on the denominators $d_1$ and $d_2$ not being 0 divisors? Arbitrary rings might have a plethora of 0 divisors. So ... we can't quite create a field containing any old rng (after all, fields must have no zero divisors), and we lose out on some possible denominators.
But hope is not lost! As long as R has at least one element which is not a zero divisor, we can create a ring called $Q(R,T)$, the partial ring of quotients of $R$ with respect to a fixed set $T$. With it, we will create unity, (which may be the only type of unity the mathematician can hope to create, but one can dream) and we can create multiplicative inverses for at least some elements of our rng. But what is this set $T$, and why do we need it? It must be multiplicatively closed, and made up of non-zero divisors (you'll see why later).
If R has any element which is not a 0 divisor, then there is a multiplicatively-closed set of non-zero-divisors in R.
We can now define the Partial Ring of Quotients $Q(R, T)$ on the set of equivalence classes of $R\times T$ the same way as we did for the field of quotients, with the same exact addition and multiplication operations. (Note that, as is convention, $R\times T$ denotes all ordered pairs $(r, t)$ with the first element in $R$ and the second element in $T$).
At this point, I will bother justifying that we still have an equivalence relation when describing the elements of $Q(R,T)$, since the potential existence of 0 divisors should make you uneasy about any claim regarding multiplication!
The relation $\sim$ defined as $(r_1, t_1) \sim (r_2, t_2) \Leftrightarrow r_1 t_2 = t_1 r_2$ is an equivalence relation (and thus, the elements of Q(R,T) are well defined).
Though this argument is not materially different than the argument that the field of quotients' elements are well-defined, the last line illustrates the primary difference between the partial ring of quotients and the field of quotients: It is absolutely essential that the denominators are not zero divisors, or the last step of the argument would not work. This is why we can only allow the denominators to be members of a set which contains no zero divisors (but does not yet explain why that set must be closed under multiplication).
The proof that addition and multiplication are well-defined (at least in the sense that regardless of representatives, you get the same equivalence class) is not materially different from the proof for the field of quotients, except again it does hinge on the fact that T has no zero divisors. I will not repeat it here, since the crux of the differences is illustrated above.
On the way to ensuring $Q(R,T)$ is a ring, we need that $<Q(R,T), +>$ is a commutative group:
Woo hoo, we just created a ring out of elements of some other ring that has some elements that aren't zero divisors. Can we even consider $Q(R,T)$ to contain the elements of $R$ in some way, when there might even be an element $1$ to stick in the denominator for the easy isomorphism? What does this construction even add? So from here on, I'll be talking about cool things that the construction of the Partial Ring of Quotients adds to our original ring.
First cool thing: even if R does not have unity, $Q(R, T)$ does: If $t^*$ is any element of T, the element $[(t^*, t^*)]$ acts as unity in $Q(R, T)$. It's our multiplicative identity!
So, with that, that is pretty much all of the advanced exercises from the section on the partial ring of quotients, from an old version of Fraleigh's Algebra. This took forever to write up. It's particularly infuriating to write something up nicely when it's already neatly in your head and needs no new organization.
At some point, I might have the energy to write up an example of a partial ring of quotients, say, $Q(\mathbb{Z}_9, \lbrace 1, 2, 4, 5, 7, 8\rbrace)$. But for right now, let's leave it at that.
In this post, I'll remind you how the field of quotients is constructed and how it works (but I won't prove that it works, since that's done in every textbook ever), and then we will use similar ideas to create a structure called the Partial Ring of Quotients, and find the key to creating unity!
(Added later: Some have told me this is also called the localization of a ring. My approach differs from Wikipedia's in notation, since I made this post based on an exercise in Fraleigh's algebra. I also aim to explain a bit more thoroughly than Wikipedia does!)
The Field of Quotients
The field of quotients' operations are defined on a set of equivalence classes of all ordered pairs of D. This sounds complicated, but it's not really! In $\mathbb{Q}$, for instance, we know that $\frac{1}{2} = \frac{3}{6}$. Those could be represented as ordered pairs, under some equivalence relation $\sim$ -- we'd like to generalize the idea that $(1, 2) \sim (3,6)$ to an arbitrary integral domain D. So, if D is an integral domain, we choose to say that $(n_1, d_1) \sim (n_2, d_2) \Leftrightarrow n_1 * d_2 = n_2 * d_1$, where $n_1, n_2, d_1, d_2 \in D$ (with $d_1, d_2 \neq 0$). Is this equivalence relation arbitrary--no! This is just like "cross-multiplying" to check if two fractions are equal, and disallowing fractions with a denominator of 0. This equivalence relation relies on $d_1$ and $d_2$ also not being zero divisors, and on the commutativity of $D$. Proof that it is an equivalence relation can be found in any introductory algebra text (or might even be relegated to the exercises).
The equivalence classes $[(n,d)]$ (read: numerator, denominator) form the elements of $F$, our field of quotients. The elements are equivalence classes because we want to make sure we make it clear that, for example, even though $(1, 2)$ and $(3, 6)$ may look different, they are actually representatives of the same exact element of $\mathbb{Q}$. You can add two equivalence classes in $F$ via $[(a,b)] + [(c,d)] = [(ad+bc, bd)]$ and multiply the equivalence classes by $[(a,b)]*[(c,d)]=[(ac, bd)]$. That these operations are well-defined regardless of representative ordered pairs, commutative, satisfy all the field axioms, etc., is, again, a topic textbooks go over immensely. But, you should be able to verify that if you represent two fractions in $\mathbb{Q}$ as ordered pairs, these operations work the same way as addition and multiplication in $\mathbb{Q}$.
In the field of quotients (which, to be fair, I haven't justified is actually a field), we now have a multiplicative inverse for every element not equivalent to 0, for if $[(a, b)]$ is an element of $F$ such that $a \neq 0$, the element $[(b,a)]$ is clearly also in $F$ and forms its multiplicative inverse -- just like taking the reciprocal. So even though we don't get multiplicative inverses in the integers, every element of $\mathbb{Q}$ does have a multiplicative inverse.
Can we think of $F$ as containing $D$? Of course. Simply define a map from $d \in D$ to $[(d, 1)]\in F$. It is a simple task to check that this gives an isomorphism from $D$ to a subset of $F$.
So above is basically the process you can use to turn any integral domain into a field that contains that integral domain -- extending it to permit division by any nonzero element.
So what about those rngs, and what's the partial ring of quotients?
But hope is not lost! As long as R has at least one element which is not a zero divisor, we can create a ring called $Q(R,T)$, the partial ring of quotients of $R$ with respect to a fixed set $T$. With it, we will create unity, (which may be the only type of unity the mathematician can hope to create, but one can dream) and we can create multiplicative inverses for at least some elements of our rng. But what is this set $T$, and why do we need it? It must be multiplicatively closed, and made up of non-zero divisors (you'll see why later).
If R has any element which is not a 0 divisor, then there is a multiplicatively-closed set of non-zero-divisors in R.
See, multiplying any non-zero-divisor $a$ by another non-zero-divisor $b$ (which may just be $a$ again) in our rng must yield a third non-zero divisor. Otherwise if $(ab)c = 0$ for nonzero c, then, using associativity, $a(bc)=0$, which since $a$ is not a 0 divisor, implies that $bc=0$, which is a contradiction -- $b$ was not a zero divisor.Thus the set of all non-zero-divisors in any ring or rng is multiplicatively closed; let's call this set T. (In fact, any multiplicatively closed set that contains no zero divisors will suffice throughout the argument, but this simply proves that if there are any non-zero-divisors in a rng, such a set must exist!)
We can now define the Partial Ring of Quotients $Q(R, T)$ on the set of equivalence classes of $R\times T$ the same way as we did for the field of quotients, with the same exact addition and multiplication operations. (Note that, as is convention, $R\times T$ denotes all ordered pairs $(r, t)$ with the first element in $R$ and the second element in $T$).
At this point, I will bother justifying that we still have an equivalence relation when describing the elements of $Q(R,T)$, since the potential existence of 0 divisors should make you uneasy about any claim regarding multiplication!
The relation $\sim$ defined as $(r_1, t_1) \sim (r_2, t_2) \Leftrightarrow r_1 t_2 = t_1 r_2$ is an equivalence relation (and thus, the elements of Q(R,T) are well defined).
Clearly, $\sim$ is reflexive, since R is commutative, so for any element $(r, t)\in R\times T$ we have $r*t = t*r$ and $(r,t)\sim(r,t)$.
That it is symmetric is also simple: If $(r_1,t_1) \sim (r_2, t_2)$ then $r_1 t_2 = t_1 r_2$. A quick rearrangement using R's commutativity yields that $r_2 t_1 = t_2 r_1$ so $(r_2, t_2) \sim (r_1, t_1)$
The transitivity is where we need that the elements of T are not zero divisors. (This argument is parallel, by the way, to the original argument that the elements of F the field of quotients are well defined) If $(a, b) \sim (c, d)$ and $(c, d) \sim (e, f)$, then we can write the equations:
$$ ad = bc \text{ and } cf = de $$
Multiplying both sides of the first equation by f, and both sides of the second equation by b, we get:
$$ adf = bcf \text{ and } bcf = bde $$
Rearranging and subtracting both sides of the equation from each other, we now get:
$$ \begin{align}
adf - bde &= bcf - bcf \\
daf - dbe &= 0 \quad \text{using commutativity here} \\
d(af - be) &= 0 \quad \text{left distribution works in all rings}
\end{align}$$
Since $d$ is an element of T, and not a 0 divisor, we must conclude that $(af-be) = 0$. Thus $af = be$ and $(a, b) \sim (e, f)$.
Though this argument is not materially different than the argument that the field of quotients' elements are well-defined, the last line illustrates the primary difference between the partial ring of quotients and the field of quotients: It is absolutely essential that the denominators are not zero divisors, or the last step of the argument would not work. This is why we can only allow the denominators to be members of a set which contains no zero divisors (but does not yet explain why that set must be closed under multiplication).
The proof that addition and multiplication are well-defined (at least in the sense that regardless of representatives, you get the same equivalence class) is not materially different from the proof for the field of quotients, except again it does hinge on the fact that T has no zero divisors. I will not repeat it here, since the crux of the differences is illustrated above.
On the way to ensuring $Q(R,T)$ is a ring, we need that $<Q(R,T), +>$ is a commutative group:
One thing that is not immediately obvious is that we need to make sure that $[(r_1, t_1)] + [(r_2, t_2)] = [(r_1 t_2 + t_1 r_2, t_1 t_2)]$ is actually an element of Q(R, T) -- this is one place where it is crucial that T is multiplicatively closed. Because of this, $t_1 t_2$ is an element of $T$, so the operation $+$ is closed in Q(R,T). If we did not require that T were multiplicatively closed, "+" might yield an element outside of T!
The argument for the associativity and commutativity of $+$ is identical to the argument for F, so I omit it here.
Using the additive identity $0 \in R$, the element $[(0, t)] \in Q(R,T)$ (for any representative $t$) acts as the additive identity. For $$[(r_1, t_1)] + [(0, t)] = [(r_1 t + 0 t_1, t_1 t)] = [(r_1 t, t_1 t)] = [(r_1, t_1)]$$ because $r_1 t t_1 = r_1 t_1 t$ by commutativity. Note that regardless of representative from $T$, $[(0, t_1)] = [(0, t_2)]$ since $0 t_2 = 0 = 0 t_1$.
Lastly, the existence of additive inverses in $R$ guarantees it in $Q(R,T)$ as well. For $[(r, t)] \in Q(R,T)$, there exists $[(-r, t)] \in Q(R, T)$ which satisfies the properties of the additive inverse. $$[(r, t)] + [(-r, t)] = [(rt + (-r)t, t^2)] = [(rt + -(rt), t^2)] = [(0, t^2)]$$ which is the additive identity in $Q(R,T)$.It doesn't take much more work to conclude that $Q(R,T)$ is a ring:
Multiplicative closure is, again, our biggest problem here, and it again hinges on T being multiplicatively closed. $[(r_1, t_1)] * [(r_2, t_2)] = [(r_1 r_2, t_1 t_2)]$ must give an element of $Q(R,T)$ because both R and T are closed under multiplication.
Multiplicative associativity is trivial (read: simply symbol-pushing) and hinges on its associativity in R. The argument is again the same as for the field of quotients.
$$[(a,b)]*([(c,d)]*[(e,f)]) = [(a,b)]*[(ce, df)] = [(a(ce), b(df)] \\
\qquad = [((ac)e, (bd)f)]=[(ac, bd)]*[(e,f)] = ([(a, b)] * [(c,d)]) * [(e,f)]$$
Thus, $Q(R,T)$ is a ring.
By this point you're probably wondering what the point is.
First cool thing: even if R does not have unity, $Q(R, T)$ does: If $t^*$ is any element of T, the element $[(t^*, t^*)]$ acts as unity in $Q(R, T)$. It's our multiplicative identity!
First, it is trivial to verify that we get the same element in Q(R,T) regardless of representative from T: $(t_1, t_1) \sim (t_2, t_2)$ because $t_1 t_2 = t_2 t_1$ in $R$, which is commutative.
If $[(r, t)]$ is an element of $Q(R, T)$, then $[(t^*, t^*)] * [(r, t)] = [(t^* r, (t^*)^2]$. In checking the equivalence relation we will see that this is the same element as $[(r, t)]$. For $(r, t) \sim (t^* r, =(t^*)^2$ because $ r (t^*)^2 = t^*r t^*$ via commutativity in $R$.Second cool thing: We really can consider $Q(R,T)$ to "contain" all of $R$. Fix an element of $T$, called $t^*$. Then the map $\phi : R\rightarrow T$ defined by $\phi(r) = (r t^*, t^*)$ is an isomorphism of $R$ into $Q(R,T)$.
Think about this as just multiplying any element of $R$ by our version of 1 in $Q(R,T)$!
First of all $\phi$ is one-to-one. If $\phi(r_1) = \phi(r_2)$ then $[(r_1 t^*, t^*)] = [(r_2 t^*, t^*)]$. Following the defined equivalence relation, $r_1 (t^*)^2 = r_2 (t^*)^2$, and since $t^*$ is not a 0 divisor, we can cancel and determine that $r_1 = r_2$.
Additionally, $\phi$ preserves $*$:Third cool thing: We can now think of $Q(R,T)$ as containing a multiplicative inverse for every element of $T$!
$$\begin{align} \phi(r_1 r_2) &= [(r_1 r_2 t^*, t^*)] \\
&= [(r_1 r_2 (t^*) ^2 ), (t^*) ^2] \quad \text{multiplication by 1} \\
&= [(r_1 t^*, t^*)] * [(r_2 t^*, t^*)] \quad \text{rearranging, using commuatativity.} \\
&= \phi(r_1) * \phi(r_2)
\end{align}$$
And $\phi$ preserves $+$:
$$\begin{align} \phi(r_1 + r_2) &= [( (r_1 + r_2 )t^*, t^*)] \\
&= [(r_1 + r_2)(t^*) ^2 ), (t^*) ^2] \quad \text{multiplication by 1} \\
&= [(r_1(t^*)^2 + r_2(t^*) ^2 ), (t^*) ^2] \\
&= [(r_1 t^*, t^*)] + [(r_2 t^*, t^*)] \\
&= \phi(r_1) + \phi(r_2)
\end{align}$$
Thus $\phi$ is an isomorphism (note that it might not be onto, just into).
So, for $t \in T$ we know its image under $\phi$ is $[(t t^*, t^*)] \in Q(R,T)$. But since T is multiplicatively closed, $t t^* \in T$ so we can rightfully say that $[(t^*, t t^*)] \in Q(R,T)$. If we multiply, $$ [(t t^*, t^*)] * [(t^*, t t^*)] = [t (t^*)^2, t (t^*)^2]$$ which is, as mentioned before, unity in $Q(R,T)$.
Notice that this might not be the image of an element in $R$, so the elements of our multiplicatively-closed subset T have inverses in $Q(R,T)$ even if they didn't in $R$.
So, with that, that is pretty much all of the advanced exercises from the section on the partial ring of quotients, from an old version of Fraleigh's Algebra. This took forever to write up. It's particularly infuriating to write something up nicely when it's already neatly in your head and needs no new organization.
At some point, I might have the energy to write up an example of a partial ring of quotients, say, $Q(\mathbb{Z}_9, \lbrace 1, 2, 4, 5, 7, 8\rbrace)$. But for right now, let's leave it at that.
Thursday, July 27, 2017
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:
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):
Now, we check scalar multiplication:
Thus T is a well-defined and linear transformation.
Next, we show that U is the kernel of T:
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 $p$.
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 $p$.
Friday, July 21, 2017
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.
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.
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.
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 $\phi(\mathscr{U})$.
$$\begin{align}
\phi(\mathscr{U}) &= P( (P|A)^{-1} (\mathscr{U}))\\
&= P(V\cap A)
\begin{split} &\quad \text{for some V} \subset X \text{ that is open, since}\\
&\quad \mathscr{U}\text{ being open implies its preimage in A is open in A by definition}\\
\end{split}\\[1em]
&= P(V) \cap P(A)
\begin{split}&\quad P(V\cap A) \subset P(V \cap A) \text{is well known; and, in this case,}\\
&\quad F\in P(V)\cap P(A) \Rightarrow F\subset A \text{ (by inclusion lemma) and }\\
&\quad \exists x \in F \text{such that} x\in V \text{and} x \in A \text{ so }P(F) \in P(V\cap A)\\
\end{split}
\end{align}$$
($\Leftarrow$) 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.
Thursday, July 20, 2017
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?
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 ;)
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