Mathematics - 2019-01-13

1:38 AM

Hmmm, is it true that, given two ideals $(a,b)$ and $(a,c)$ in the ring $\mathbb{Z}$, that the ideal $(a,bc)=(a,b)\bigcap (a,c)$?

Leaky Nun

@Rithaniel think in terms of GCD

(and LCM)

you're asking whether gcd(a,bc) = lcm(gcd(a,b),gcd(a,c))

Rithaniel

Yeah, I'm actually trying to show that, if gcd(a,b)=gcd(a,c), then it also equals gcd(a,bc). Maybe I'm going about it backwards, though. I should maybe use a direct argument.

Ted Shifrin

I don't believe that.

Can you give me a counterexample?

Rithaniel

Hmmm, if $a=b^2$ and $b=c$?

Ted Shifrin

That'll do.

Rithaniel

1:46 AM

Okay, then more realistically: if gcd(a,b)=gcd(a,c)=1, then gcd(a,bc)=1.

Ted Shifrin

Aha. That sounds rather more reasonable.

CaptainAmerica16

I got some new math books today

Ted Shifrin

@Rithaniel: You should be able to do that very directly from the usual Bezout identity.

heya @CaptainAmerica

Like what?

CaptainAmerica16

A number theory book and "math popularization" book. It explains concepts from pretty much any subject you can think of. It's like 700 pages.

Rithaniel

Well, I've proven previously that gcd(m,n) is a linear combination of m and n, and I'd like to use ideals in the proof that if gcd(a,b)=gcd(a,c)=1, then gcd(a,bc)=1. Not seeing the immediately obvious proof, though.

CaptainAmerica16

1:50 AM

Well not "any subject" math is too vast for that.

Ted Shifrin

Write down those linear combinations and play, @Rithaniel. There are other approaches, but I like that one.

Who wrote the big book, CaptainAmerica?

CaptainAmerica16

@TedShifrin Jan Gullberg

Ted Shifrin

Never heard of it.

CaptainAmerica16

I found it by accident at the library and decided to buy it. amazon.com/Mathematics-Birth-Numbers-Jan-Gullberg/dp/039304002X

Ted Shifrin

I no longer have it, but there's a 3-volume Russian encyclopedia of math that I always found to be quite good.

CaptainAmerica16

1:56 AM

I'm trying to gather books purely for reference as well as textbooks.

Was it written in Russian?

Ted Shifrin

well, there are a lot of crappy books out there

I had it in English :P

CaptainAmerica16

oh, lol

I'm trying to go based on good Amazon reviews :P I'll probably get a dud eventually.

Ted Shifrin

LOL, I think most of my books have bad Amazon reviews :P Whining students who want answers to everything in the book.

Complaints about bindings falling apart are 1000% warranted.

CaptainAmerica16

XD

I haven't done as much math as I would have liked to lately. When I miss 3-5 days, I feel so rusty when I start working again.

Ted Shifrin

why'd you miss?

CaptainAmerica16

2:09 AM

@TedShifrin At first, a missed a few days because I had a lot of school stuff to catch up on. Now, I think I've just been a bit lazy.

Ted Shifrin

"a bit" :D

Being lazy is OK but don't bitch at us (me) later.

Mike Miller

A librarian once helped me reglue the binding on a book that totally fell apart. Told me it was a one shot deal though.

Ted Shifrin

Book rebinding is a good skill/art to learn.

CaptainAmerica16

@TedShifrin Lol, I'll just complain passive-aggressively

Or maybe that's the same thing...

Ted Shifrin

Yes, you are accomplished at that.

CaptainAmerica16

2:14 AM

Something I'm working on for the new year. I'm not calling it a "resolution" because then I'm likely to quit.

Ted Shifrin

Oh, and "working on" won't lead to quitting?

CaptainAmerica16

eeehhh

mathsssislife

hello!

CaptainAmerica16

I don't want to say "something I'm doing" that sounds like a commitment

mathsssislife

hi @TedShifrin

how are you?

Ted Shifrin

2:16 AM

you're commitment-phobic, CaptainAmerica

hi @maths

isn't it way past your bedtime, @maths?

mathsssislife

sleep isn't for math majors

Ted Shifrin

hmm

CaptainAmerica16

@TedShifrin Something else I'm working on for the new year :)

Ted Shifrin

another non-resolution? sleep?

oh, commitment phobia

CaptainAmerica16

lol

mathsssislife

2:20 AM

i've been able to prove that for a finite collection of open sets then their intersection is open, and for a finite collection of closed sets, their union is open

Ted Shifrin

Well, that sounds good.

CaptainAmerica16

I understand all of those terms except "open" and "closed"

mathsssislife

yeah, for some reason i'm attracted to topology

CaptainAmerica16

I had a feeling that's what this was :D

Ted Shifrin

they're generalizations of open and closed intervals in $\Bbb R$, @CaptainAmerica

if you make it to my book next year, you'll see 'em

CaptainAmerica16

2:22 AM

Ooh

I'll be back in like 15-20 minutes if you guys are still here.

Ted Shifrin

I won't be.

Leaky Nun

@TedShifrin hey

CaptainAmerica16

alright, talk to you later then

Ted Shifrin

have a good night, @CaptainAmerica

hey Leaky

Leaky Nun

I discovered a "totally original" proof of column rank = row rank

mathsssislife

2:25 AM

@TedShifrin, I have a proof for the closure of a set to be open, that was a proof where I asked you for help on before, and I did progress on it partially, but then I forgot about it. Anyways, I wrote a proof which I think is solid, is it okay to ask you for feed back on it, please?

Ted Shifrin

you mean closure is closed?

mathsssislife

yes

oops

i'm really tired

Ted Shifrin

Go to sleep.

Leaky Nun

@TedShifrin are you interested?

Ted Shifrin

How circular is it, @Leaky?

Leaky Nun

2:26 AM

not at all

mathsssislife

I will :)), I really need some sleep, but i've been working on it for quite a while and i'm excited about it

Ted Shifrin

I'm dubious that it's "totally original," but go on. I have to leave in 15 minutes.

I'd rather not talk math when you're asleep @maths

Leaky Nun

so it suffices to show that $\dim \operatorname{im} T^\ast \le \dim \operatorname{im} T$

and it suffices to have an injection $\operatorname{im} T^\ast \to \operatorname{im} T$

mathsssislife

@TedShifrin alright then, i'll send it tomorrow, is that okay with you?

Leaky Nun

well $T$ restricted to $\operatorname{im}T^\ast$ is the injection!

Ted Shifrin

2:28 AM

@maths: Not sure when I'll be around tomorrow, but sure.

Leaky Nun

i.e. $TT^\ast w = 0 \implies T^\ast w = 0$

Ted Shifrin

yeah, that proof is in my books, pretty much, Leaky.

Leaky Nun

oh great

Ted Shifrin

Want page numbers? :P

Leaky Nun

the proof we were given was 100 times worse

nah it's fine

Ted Shifrin

2:29 AM

In more casual language, restricting $T$ to the row space (of a matrix representation) is easily seen to give an isomorphism to the column space.

Leaky Nun

but my proof works for hilbert spaces :P

Ted Shifrin

but you don't get equality in Hilbert spaces

Leaky Nun

don't I?

if $T$ is continuous?

Ted Shifrin

only if the operator is reflexive?

Leaky Nun

googles reflexive operator

"All Hilbert spaces are reflexive"

is this irrelevant?

Ted Shifrin

2:32 AM

Hmm ... am I being stoopid? Maybe. You need $T^{**} = T$?

Maybe that is automatic for Hilbert.

My brain is dead.

Leaky Nun

well there is the Riesz representation theorem

$\langle T^{\ast\ast} v, w \rangle = \langle v, T^\ast w \rangle = \langle Tv, w\rangle$

Ted Shifrin

Huh?

Domains and ranges have to match up for starters.

No, no, this is crap.

How is that Riesz?

Leaky Nun

well Riesz says that $\langle v_1, - \rangle = \langle v_2, - \rangle \implies v_1 = v_2$

in this case we obtain $T^{\ast \ast} v = Tv$

Ted Shifrin

Nah.

That's not Riesz, seriously. That's just the fact that only the zero vector is orthogonal to every vector. That's just closure.

Oh, right.

So I see. Yeah, it is automatic for Hilbert. Just not automatic for Banach. That's where you need reflexive. Got it.

OK. cool.

Leaky Nun

cool

aha it's a trivial part of Riesz lol

Ted Shifrin

2:41 AM

That doesn't count as Riesz. That's just totally elementary. $\langle v,w\rangle = 0$ for all $w$ tells you $v=0$. Proof?

Mike Miller

Sorry, what is the question here?

Leaky Nun

yeah I know

Mike Miller

I need to decide if I'm interested. :D

Ted Shifrin

No.

You're NOT.

Mike Miller

Okay!

Ted Shifrin

2:42 AM

LOL

Leaky Nun

@MikeMiller well I found a "totally original" proof of column rank = row rank, turns out Ted's book has it

it's just that the proof I was given was 100 times worse

Mike Miller

Oh I see

zooms out

Ted Shifrin

The usual proof I have given in undergrad linear algebra is based on algorithms based on echelon form, but you don't need those.

Leaky Nun

yeah I've seen that also

it has certain beauty there

both ranks are the number of pivots

Ted Shifrin

Right, once you know that's well-defined.

Leaky Nun

2:44 AM

oh we don't know that RREF is unique?

Mike Miller

Rank is the same as largest size of minor with non-trivial determinant and this reduces to the standard fact $\det(A^\tee) = \det(A)$, no?

Ted Shifrin

That's a tricky proof, Leaky.

That's in some sense more sophisticated, Mike.

Leaky Nun

@MikeMiller so my proof also works for Hilbert spaces, and it is basically the fact that $T : \operatorname{im}(T^\ast) \to \operatorname{im}(T)$ is injective

i.e. $T T^\ast w = 0 \implies T^\ast w = 0$

Ted Shifrin

Night all.

Mike Miller

What is the rank of a map of Hilbert spaces?

Leaky Nun

2:50 AM

$\dim\operatorname{im}T$

Mike Miller

Dimension in what sense? Just literal vector space dimension?

Weird thing to consider but ok, I'll survive.

mathsssislife

@TedShifrin Goodnight Ted, i'm going to sleep as well, good night!

Leaky Nun

@MikeMiller well maybe $\dim \overline{\operatorname{im}(T)}$

idk I'm just exploring

oh well then I don't get the theorem do I

Mike Miller

I'm going to think about it quite slowly

For things of finite rank my proof is unchanged, working with a basis for your Hilbert space

Leaky Nun

sure

Rithaniel

3:08 AM

I feel like I'm running in circles trying to prove a basic property of gcd.

Leaky Nun

maybe run in squares?

Rithaniel

Nah, ellipses are the next option.

The current trail I've been running along: "As established in (1a), $\text{gcd}(k,m)$ can be written as the linear combination of $k$ and $m$. Supposing we have that $\text{gcd}(k,m)=1$ and $\text{gcd}(k,n)=1$, then we have there exist $a,b\in\mathbb{Z}$ s.t. $ak+bm=1$ or $akn+bmn=n$. Allow $x=an$, then we have $xk+bmn=n$."

I feel like this is a dead end, though. (And I see a chunk I can cut out)

Easily I can get to $xk+(bm-1)n=0$, but that's no help either.

I think I have an idea.

Well, it's much easier to show that if $k|mn$ and gcd(k,m)=1, then $k|n$.

Though, just to be sure, is this argument clear? "As established in (1a), $\text{gcd}(k,m)$ can be written as the linear combination of $k$ and $m$. Supposing we have that $\text{gcd}(k,m)=1$, then there exist $a,b\in\mathbb{Z}$ s.t. $ak+bm=1$ or $akn+bmn=n$. We have been given that $k|mn$, and we also know that $k|k$, so $k|(akn+bmn)$, which implies $k|n$."

mathsssislife

3:46 AM

since gcd(k,m)=1 , 1=ak+bm. Note since k divides mn, mn=ku, let n=nak+nbm, since mn=ku, n=nak+bku which implies that k divides n

Rithaniel

Much more efficient than mine.

Okay, but the same essential argument, so my logic is good.

Though, showing that gcd(k,m)=gcd(k,n)=1 implies gcd(k,mn)=1 is still proving difficult for me. Maybe I should sleep on it and come back tomorrow.

mathsssislife

i guess id start with saying gcd(k,mn)=e and then show that e=1

Math geek

4:19 AM

How would be the definition of a basis for a set of all closed subset of $X$(Topology of closed sets) look like? Will it be the complement of basis definition for a topology on $X$?

Definition of basis for a topology on $X$ given in the Foundation of Topology By Patty given by

Let $(X,\mathscr T)$ be a topological space. A basis for $\mathscr T$ is a subcollection $\mathscr B$ of $\mathscr T$ with the property that if $U\in \mathscr T$ then $U=\emptyset$ or there is a subcollection $\mathscr B'$ such that $U=\bigcup \{B:B\in \mathscr B'\}$

Let $(X,\mathscr T)$ be a topological space of closed sets. A basis for $\mathscr T$ is a subcollection $\mathscr D$ of $\mathscr T$ with the property that if $D\in \mathscr T$ then $D=X$ or there is a subcollection $\mathscr D'$ such that $D=\bigcap \{D:D\in \mathscr D'\}$

am I correct?

yiyi

5:37 AM

just double checking: If given a line L and a plane P, then L is not perpendicular to P if the dot product of the normal of P and L is not zero.

Akiva Weinberger

7:34 AM

@yiyi Wait, dot product of the normal of P and the what of L? The tangent vector of L?

@Rithaniel Say $ak+bm=1$ and $ck+dn=1$. Multiply the two together.

Math geek

11:42 AM

1

Q: Is my definition correct? How do I prove the finite union of elements of $\mathscr U$ lie in $\mathscr U$?

Definition of basis for a topology on $X$ given in the Foundation of Topology By Patty given by Let $(X,\mathscr T)$ be a topological space. A basis for $\mathscr T$ is a subcollection $\mathscr B$ of $\mathscr T$ with the property that if $U\in \mathscr T$ then $U=\emptyset$ or there is ...

general-topology

Akiva Weinberger

1:20 PM

1 nanocentury = pi seconds

Lucas Henrique

Hi chat!

ÍgjøgnumMeg

Hi @Lucas @Akiva

Lucas Henrique

@Daminark, giving continuity to that "advice" thing (about real linear algebra vs $K$ linear algebra): at a first sight (actually, its negation can also be true iff choice is not), the claim "every vector space has a basis" is not necessarily true; is it "interesting" to study vector spaces without stuff like $K^n$ on my mind? Personally, I'll always accept the axiom of choice.

(@everyone: feel free to give your opinions)

Shaun

in In the search of a question, 2 mins ago, by Shaun

I'm looking for the number of cyclic subgroups of the group of units modulo $n$. Sylow's Theorems might shed some light into the problem but I haven't thought about'm much in this context. Please let me know if you find anything :)

Akiva Weinberger

1:36 PM

I wonder how square roots work with Fourier series

Leaky Nun

@LucasHenrique every vector space has a basis

Akiva Weinberger

What's the Fourier series of $\sqrt{5-4\cos(x)}$, for example?

Leaky Nun

(under choice)

@AkivaWeinberger can't get wolframalpha to do it lol

ÍgjøgnumMeg

@Shaun how about decomposing $$(\Bbb Z/(n))^\times \cong \prod_{i=1}^g (\Bbb Z/(p_i^{e_i}))^\times$$

where $n = p_1^{e_1} \cdots p_g^{e_g}$

Lucas Henrique

@LeakyNun yup. I'm aware of this, as told in the big message above. The matter is that I'm studying real linear algebra, although I have basic knowledge on abstract algebra. So, even if I work with linear algebra on $K^n$, I'm not sure of how I'd generalize this to algebras over $K$, in general.

For example (I think), $V = (M_r(K), K^s)$ with usual sum (usual sum on each coordinate) and usual scalar product (usual scalar product on each coordinate) is a vector space but it's pretty distant from $K^n$.

user193319

1:52 PM

Prove Fatou's lemma: for a sequence of nonnegative functions $f_n : [0,1] \to \Bbb{R}$, $\int_{0}^{1} \limsup f_n(x)dx \le \limsup \int_{0}^{1} f_n(x)dx$...Question: Shouldn't the inequality point in the other direction?

Leaky Nun

@LucasHenrique I don't understand that notation

Shaun

@ÍgjøgnumMeg I think I see what you mean. Thank you. However, doing that would be somewhat of an anachronism because nothing of the sort has been done in the material I've covered so far. Nonetheless, I'll read a proof of the fact one can decompose such groups and build my understanding from it from the ground up :)

ÍgjøgnumMeg

@Shaun this is the Chinese Remainder Theorem and the fact that the group of units of a direct product is the direct product of the group of units in a ring

just for reference

@Shaun also there is a structure theorem for finitely generated abelian groups

Alessandro Codenotti

2:10 PM

@user193319 Yeah, either the inequality has to be reversed or the limsups should become liminfs

Shaun

2:20 PM

@ÍgjøgnumMeg Thank you. I recognise it now. It's been a couple of years since I've used it, s'all:)

user193319

2:34 PM

@AlessandroCodenotti Thanks!

Akiva Weinberger

2:59 PM

Never mind, I figured it out. So, all circles in the complex projective plane contain $[i,1,0]=[1,-i,0]$ and $[-i,1,0]=[1,i,0]$, and these are fixed by all rotations. Wild. — Akiva Weinberger Mar 24 '17 at 20:13

I completely forgot about this

Now I need to find out why it's true

So a circle in the complex plane is determined by $(\frac xz-j)^2+(\frac yz-k)^2=r^2$

or $(x-jz)^2+(y-kz)^2=r^2z^2$

If $z=0$ then that's just $x^2+y^2=0$, which means $x=\pm iy$

So you get $[y,iy,0]=[1,i,0]$ and $[y,-iy,0]=[1,-i,0]$

And that's independent of $j$, $k$, and $r$

Weird!

Similarly, a point in the complex projective plane is preserved by ninety degree rotations if $[x,y,z]=[-y,x,z]$.

$[1,i,0]=[-i,1,0]$ and $[1,-i,0]=[i,1,0]$, so they're preserved by those rotations.

Similarly, $[1,i,0]=[\cos\theta-i\sin\theta,\sin\theta+i\cos\theta,0]$ and similarly for $[1,-i,0]$, so they're in fact preserved by all rotations.

Solving $[x,y,z]=[-y,x,z]$ confirms that they're the only such points.

$\theta$ doesn't even need to be real

Mike Miller

3:30 PM

Since we were talking cyclic groups earlier: can you give a simpler description of $\Bbb Z/p^k$ as a $(\Bbb Z/p^k)^\times$ module than the extant description? For instance, is it number-theoretically clear what the fixed points are?

1010011010

3:47 PM

Hi all, very simple question about integration

So, suppose I compare different products under the integral sign with each other

For example

I am looking at

$$ \int \log x\log(\alpha -x) dx $$

versus

$$ \int \log x \log(\alpha -x) \log(\beta -x) dx $$

Given knowledge of the first integral I may use it in solving the second one during partial integration

Now there is an ambiguity in my solving the first integral with the constant inside it

And as such the indefinite integral in turn seems to have ambuigity in the usage of partial integration, or should I simply set the constant to zero in the first integral before using it in the second?

ÍgjøgnumMeg

@MikeMiller what do you mean by "extant description" ? The one where the $(\Bbb Z/(p^k))^\times$-action is just multiplication?

Mike Miller

I should have said "tautological description".

Yes.

I guess fixed points aren't too bad. A fixed point is something which has zero product with $(x-1)$ for all invertible $x$. For $p = 2$ the invertibles are odds, so this is $\Bbb Z/2$. For $p$ odd, both $p+1$ and $p+2$ are invertible mod $p^k$, so there are no fixed points.

user21820

4:43 PM

@Rithaniel When dealing with gcd, it's often more useful to first prove that it can be computed by the euclidean algorithm. Then the basic fact that you wanted (gcd(m,n) is a linear combination of m,n) falls out immediately; the algorithm calls itself on linear combinations of the original inputs. Also, when dealing with coprime numbers, it's always useful to use modulo relations. gcd(k,m) = 1 means m is invertible mod k (by the basic fact). Same for gcd(k,n) = 1. Then is m·n invertible mod k?

And here is an easy generalization: gcd(k,m·n) | gcd(k,m)·gcd(k,n) for every integers k,m,n such that k≠0.

Rithaniel

5:05 PM

Yeah, I'm going into an introductory course in number theory right now, so this stuff is all still fairly new to me.

Ah, they've signed out. I should probably ping them. @user21820

Thank you for the tips, by the way.

user21820

@Rithaniel Yes ping me if you want a response. Also, what I said above is what I think is the best way, which very unfortunately not many courses on discrete mathematics or elementary number theory adopt. Sadly, some never even teach students the euclidean algorithm.

Rithaniel

Yeah, I had to google it when you mentioned it, as I had not heard of it before.

It seems like it sets up a simple way to compute values.

vzn

@CaptainAmerica16 so whats the number-theory book? :)

user21820

5:25 PM

@Rithaniel And it's also in my opinion the way to really understand what gcd feels like, and do concrete computations. And many problems can be solved just by using the idea of the euclidean algorithm. For example, find gcd(x^m−1,x^n−1) where x is an integer and m,n are naturals.

Unrelated:

The most unexpected answer to a counting puzzle

►

Rithaniel

I just watched that video. It blew my mind a little bit.

user21820

@Rithaniel I haven't watched, because I want to figure out the answer for myself.

=)

Got to go. See you!

user193319

6:29 PM

Okay, evidently $\limsup \int f_n \le \int \limsup f_n$. But why isn't the following a counterexample to the inequality: $f_n : [0,1] \to \Bbb{R}$, $f_n(x) = n 1_{[0,\frac{1}{n}]}(x)$. Then $\int f_n = n \int 1_{[0,\frac{1}{n}]} = n \frac{1}{n} = 1$, so $\limsup \int f_n = 1$. It isn't hard to show that $\lim_{n \to \infty}f_n(x) = 0$ for $x \in (0,1]$, so $\int \limsup f_n = 0$. So $\limsup \int f_n = 0 < 1 =\int \limsup f_n$...What did I do wrong?

Hold on...I'm getting all the inequalities backwards I think...

@AlessandroCodenotti What am I misunderstanding?

Shaun

6:46 PM

The following is a terminology question for a certain subset of a group . . .

0

Q: What is $H_G^{(n)}:=\{h\in G: \operatorname{ord}(h)\mid n\}$ called for any fixed abelian group $G$ and $n\in\Bbb N$?

I'm reading "Contemporary Abstract Algebra," by Gallian. This is based on exercises 3.45 and 4.15 ibid. What is $H_G^{(n)}:=\{h\in G: \operatorname{ord}(h)\mid n\}$ called for any fixed abelian group $G$ and $n\in\Bbb N$? Here $\operatorname{ord}(g)$ is the order of $g$ in $G$. I suspect ...

group-theory terminology abelian-groups

Isa

Hello, does someone know if the scaled chi-squared distribution exists? Or is just the scaled inverse chi-squared the one that exists?

Shaun

I'd appreciate some help with it, please :)

@Isa I wish I could help. It's been over a decade since I did anything with the $\Chi$-squared distribution.

Isa

@Shaun ok don't worry. I've google it but I can't find anything, which makes me think that it does not exists.

Sir Cumference

7:05 PM

Logic is an underappreciated math

Alessandro Codenotti

7:31 PM

@user193319 I see nothing wrong here, I think you might still be confused about the direction of the inequalities

user193319

@AlessandroCodenotti So is $\limsup \int f_n \le \int \limsup f_n$ the right inequality?

Alessandro Codenotti

Yes

user193319

Okay. Thanks!

Alessandro Codenotti

(Note that while this is of course equivalent it's more common to write Fatou's lemma with liminfs and the reverse inequality)

Lozansky

How to prove $|\sum_{n=2}^{\infty} (-1)^{n+1} x^{2n}/n!| \leq x^4/2, x \in \mathbb{R}$?

Thorgott

8:43 PM

This looks like it wants you to use the Leibniz criterion, but the first couple terms can get ugly. There's probably a neat solution in this vein, but I don't see it right now. Either way, straightforward computation of the sum seems to suffice for demonstrating the inequality.

vesii

9:18 PM

Hey guys, consider the following theorem: Let $G=(V,E)$ a simple directed graph that every arrow is colored with black or pink. If $G$ has monochromatic cycle then $G$ hash simple monochromatic cycle. How to prove that theorem?

BTW monochromatic cycle means that all of its arrows has the same color.

Daminark

What's a simple monochromatic cycle?

(As in, what's a simple cycle?)

vesii

Its a path/cycle that there is no vertex which shown more then one time (only the first vertex shown twice). Like:
$V_1-V_2-V_3-V_1$

$V_1-V_2-V_3-V_2-V_1$ is not simple for example

Lucas Henrique

10:06 PM

@LeakyNun $M_r(K)$ is the ring of the $r\times r$ matrices over $K$ and $K^s$ is $\underbrace{K\times K\times\dots\times K}_{s \,\text{times}}$.

Leaky Nun

and what should $V = (M_r(K), K^s)$ mean?

Daminark

@vesii okay so, assuming we allow $V_1 - V_2 - V_1$ as a cycle, when only two vertices are involved, then just take your cycle and take two consecutive appearances of a vertex, what's between them is a simple cycle and it's monochromatic because the whole cycle is

I mean I guess if you don't allow that to be a cycle but you still have a non-simple cycle that means it can't just be of the form $V_i - V_{j_1} - V_i - V_{j_2} \ldots$

So in that case, there exists two consecutive instances of a single vertex such that there are at least two other distinct vertices between them, then that's a cycle. Anyway something to that effect is needed

Lucas Henrique

@LeakyNun the vector space defined this way, with sum distributing over coordinates and product too.

I think that this is a vector space, at least.

Daminark

$K^s$ isn't a field

Also wait why can $r$ and $s$ be different?

Lucas Henrique

Yes. What about it?

@Daminark Why wouldn't them be allowed to?

Daminark

10:21 PM

Okay so your notation still isn't clear, are you using $V = (S,K)$ where $S$ is a vector space over the field $K$?

Lucas Henrique

I picked this specific example because this vector space does not look like $K^n$

@Daminark no. $V$ is the set of the elements of the vector space

Daminark

Then what do $A$ and $B$ represent in $V = (A,B)$?

Lucas Henrique

The vector space is itself $(V, K, +, \times)$

Oh my, I've just realized why I'm retarded. Sorry guys, I meant $V = M_r(K)\times K^s$. Sorry.

Daminark

Oh

Well that's a vector space but the thing is, $M_r(K)$ is actually isomorphic to $K^{r^2}$

Lucas Henrique

I thought about this

Daminark

10:26 PM

I guess when you say "does not look like $K^n$" it depends on what you mean by "look like". Set-theoretically yeah it's different, but every vector space is isomorphic to $K^n$

Lucas Henrique

And I'm guessing everything will have some kind of isomorphism, as @Leaky cited (a long ago). I'm almost working around everything to get rid of the standard way we're taught because I don't want to be limited to $K^n$ when stating stuff like linear transformations, etc. For example, I don't know how to prove that every linear transformation is a matrix product outside of $K^n$. How could I do this?

Daminark

I mean, if you have an arbitrary vector space, choosing a basis is choosing an isomorphism with $K^n$. The nice thing in linear algebra is that this allows you to always reduce to $K^n$

So as for this particular problem

Lucas Henrique

@Daminark yeah IKR, I'm being not so rigorous but more practical with the generalizations, in the sense of literally proving everything without invoking $K^n$ unless I prove beforehand it's always possible to invoke it.

Daminark

Let $T:V\to W$ be a linear transformation. Let $\{v_1,\ldots,v_n\}$ be a basis for $V$ and $\{w_1,\ldots,w_m\}$ be a basis for $W$

We write $T(v_1)$ in terms of the $W$ basis, so $T(v_1) = a_{1,1}w_1 + a_{2,1}w_2 + \ldots + a_{m,1}w_m$

Lucas Henrique

@Daminark this works as desired, but I did not prove, yet, that every vector space has a basis - I've not seen the proof yet and, moreover, I know shit about choice :s

Daminark

10:32 PM

So, axiom of choice gives you every vector space, but in linear algebra you usually deal with finite-dimensional vector spaces

(In particular, between infinite-dim spaces you usually don't wanna try to write linear maps as matrices)

And in finite dimensions this is fairly straightforward. Let's take a vector space $V$. Obviously $V$ spans itself

And we're assuming $V$ is finite-dimensional (even if we haven't defined dimension, we can define finite-dimenisonal to mean spanned by a finite set)

So choose a minimal spanning set

(No need for axiom of choice when I say "choose", you take some finite spanning set which exists by assumption, and find a subset which still spans it and is minimal)

Now you want to show that this spanning set is linearly independent

Lucas Henrique

That's pretty straightforward.

mathsssislife

guys could anyone tell me if this proof is correct: Let a∈Ac∩(A′)c. Since a is not a limit point of A there exists a neighborhood of a such that for every q

in the neighborhood of a, q=a or q∉A. This implies that Nr(a) is contained within the complement of A. Now we would like to show that the neighborhood is contained within (A′)c, because then Nr(a) ⊂ Ac ∩ (A′)c which implies that the set is open, therefore the closure is closed. So assume that the neighborhood is not contained in (A′)c this implies that a ∈ A′ because Nr(a) ⊂ A′, which is a contradiction.

i'm trying to prove that the closure of a set is closed

i've had feedback, but i'm not sure if my proof is incorrect

Daminark

Let's call this spanning set $\{v_1,\ldots,v_n\}$. Assume it's linearly dependent, so we can find $a_i$ which aren't all $0$ such that $a_1 v_1 + \ldots + a_n v_n = 0$

mathsssislife

or correct

Daminark

There's some $i$ such that $a_i \ne 0$, let's relabel if necessary and assume $a_1 \ne 0$

vesii

10:37 PM

@Daminark Can you explain you proof a bit? I can't seem it yet.

Daminark

So now we write $-\frac{a_2}{a_1}v_2 + \ldots - \frac{a_n}{a_1} v_n = v_1$

So $v_1 \in \text{span}(v_2,\ldots,v_n)$

Which means $\text{span}(v_1,v_2,\ldots,v_n) = \text{span}(v_2,\ldots,v_n)$, which contradicts minimality

Lucas Henrique

I think I got it. My matter was mainly finding a basis, and infinite-dimensional linear algebra looks like "broken" in terms of working nicely like it does on finite dimensions. Thanks @Daminark.

Daminark

Oh infinite dimensional linear algebra makes every theorem die

That's why functional analysis is a subject

Have you seen rank-nullity?

Leaky Nun

is someone bashing infinite dimensional linear algebra?

hilbert spaces are so beautiful

mathsssislife

Let p$\in$ $(E \cup E')^c$ this implies that p$\notin$E and p$\notin$ E'. Since p$\notin$E', $\exists$ $N_r(p)$ : $\forall$ q$\in$ $N_r(p)$ , q$\neq$ p or q$\notin$ E. This implies thiat $N_r(p)$ $\cap$ E is the empty set which implies that $N_r(p)$ $\subset$ $E^c$.

Now we would like to show that $N_r(p)$ $\subset$ $E'^c$, so assume that is not the case this implies that $N_r(p)$ $\subset$ E' $\implies$ p$\in$ E' which is . a contradiction, therefore $N_r(p)$ $\subset$ $E^c$ $\cup$E' which means that it is open

Lucas Henrique

10:50 PM

@Daminark Nope. I have seen pretty much nothing about uni math with depth. Fun fact: I'm taking my entrance exams from today (01/13) to 01/15

mathsssislife

Thats my newer version of the proof, with the feedbacks included, may someone tell me if its correct, please?

Leaky Nun

you mean depth

Lucas Henrique

@LeakyNun non-native here. Thanks.

I'll probably have two main choices of uni, but I'm uncertain.. I'm sure that @ÉricoMeloSilva would have strong opinions about this. LOL

USP or Unicamp for a math undergraduate degree?

mathsssislife

@TedShifrin hi Ted!, how are you?

Lucas Henrique

They're both top-tier Brazilian universities, but my interest is pure. USP structure is shit and Unicamp focuses on computational mathematics.

Daminark

11:03 PM

@Leaky I mean more like, raw linear algebra in infinite dimensions. You lose so many theorems, and the point is when you have a Banach space and restrict to continuous or better yet, compact linear maps, it's nice again because it's nearly finite rank anyway

Leaky Nun

but it's still important

Daminark

I love functional analysis, just that it's literally there to restore order to chaos

Leaky Nun

we care about the category of vector spaces over a field

Daminark

Do we? Or do we care about the category of finite-dim ones, or separately that of Banach spaces and bounded linear maps?

Which includes the former

Leaky Nun

maybe

but we also care about that all the fields are morita equivalent or something like that

Daminark

11:06 PM

Not sure what that means so I won't contest lmao

Leaky Nun

probably isn't true

Morita equivalence

In abstract algebra, Morita equivalence is a relationship defined between rings that preserves many ring-theoretic properties. It is named after Japanese mathematician Kiiti Morita who defined equivalence and a similar notion of duality in 1958. == Motivation == Rings are commonly studied in terms of their modules, as modules can be viewed as representations of rings. Every ring R has a natural R-module structure on itself where the module action is defined as the multiplication in the ring, so the approach via modules is more general and gives useful information. Because of this, one often studies...

Alessandro Codenotti

@Daminark if you just have finite dimensional ones you're missing some (co)products which I guess categorical folks don't like

@LeakyNun it looks like such an equivalence cannot work on hom sets if you have vector spaces over a finite field on one side and over an infinite one on the other