Mathematics - 2017-04-24 (page 3 of 3)

Daminark

22:00

@Zach I have to work on manifolds for now, but I may play eventually

Ted Shifrin

Zach has things to think about, but he's just being procrastinative, as usual.

Meow Mix

$(x + y + z)^2 = 2(x^2 + y^2 + z^2)$

Huh?

Balarka Sen

me too

Meow Mix

I have tacos to eat.

Daminark

Isn't everyone procrastinative?

Ted Shifrin

22:01

You're supposed to be asleep, Balarka.

PVAL-inactive

I sure am

Topologicalife

@TedShifrin with ur subtitution or @MeowMix's substitution?

Balarka Sen

am i now

Daminark

"Work on manifolds" for me means "initiate the process of procrastinating on manifolds"

Ted Shifrin

No, Zach's substitution was not good. Mine.

Meow Mix

22:01

Huh?! I tried my hardest!

Ted Shifrin

You have manifold ways of procrastinating on manifolds, Demonark, we know.

Topologicalife

But I still have to expand that ugly terms.

Daminark

For sure

Topologicalife

Well, the original are more ugly than the eq I got with ur substitution.

PVAL-inactive

With classwork the easiest thing for me was to spend some time immediately after being given the assignment, going through them and seeing which ones I can think through immediately and memorizing the ones I didn't know how to do.

Meow Mix

22:02

Multiply each side by 0. Therefore, it is true. braces for smack

Balarka Sen

I am forgetting Zorn's lemma. halp

PVAL-inactive

I'd usually come up with the solution while doing whatever.

at least with sort of challenging rigorous mathematics type exercises.

That's probably not a good strategy for the busywork my calc students get.

Alessandro Codenotti

@BalarkaSen it's just the axiom of choice

:P

Meow Mix

Oh yeah Ted's is kinda better lol

Balarka Sen

not helpful man

Daminark

22:04

If you have a partially ordered set, whether or not every chain has an upper bound, there is a maximal element

Topologicalife

Another problem: how to prove $x^2+1=0$ has no solutions in $\mathbb{R}$.

(I already proved it, no need help)

Alessandro Codenotti

It might be more helpful (understandable) than "if in a poset every chain has an upper bound then there is a maximal element"

Meow Mix

prove that $x^2 \geq 0$ for $x$ in R will suffice

Topologicalife

\geq

Balarka Sen

@Daminark "whether or not"? are you trying to troll

PVAL-inactive

22:06

The way Zorn's lemma is stated is super useful and a lot easier for me to understand then AoC or WOP.

Topologicalife

Yeah, that's the easy way :P

Balarka Sen

Ok, got it. Thanks everyone.

PVAL-inactive

There's like one or two problems in Atiyah and MacDonald where its easier to use the straight AoC

Daminark

Yeah @Balarka, it was if every chain has an upper bound

Topologicalife

I did some manipulations.

PVAL-inactive

22:07

and I always felt those were really challenging.

Daminark

I have only seen AoC used in ways that were very much... off to the side/intermediary

It's easy to miss AoC

But if Zorn's lemma comes up it looks you in the face

Balarka Sen

eg in the proof that Noetherian module iff every submodule is finitely generated

Alessandro Codenotti

there's plenty of proofs that use Zorn very explicitely

PVAL-inactive

I think generally in specific applications its difficult to find products where existence of elements is interesting, but posets on the other hand are plentiful and many of your techniques not involving AC (likes bases of vector spaces) already use posets pretty explicitly.

@Balarka That one's probably easier to do with Zorn.

Balarka Sen

Ya it's obvious with Zorn

Daminark

22:26

That makes sense @PVAL

The main scenario I've seen thus far that used the axiom of choice was in Banach-Tarski, in designating certain points as the starters for the orbits

Little Rookie

23:06

I've written a proof on an open cover for $\mathbb{Q}\cap [0,1]$ with no finite subcover
https://math.stackexchange.com/questions/2250567/an-open-cover-for-mathbbq-cap-0-1-that-does-not-contain-a-finite-subcover

Does the line : Then, the finite subcover has the form $\{(\frac{\sqrt 2}{2},2), O_{n_{1}}, O_{n_{2}}, ... ,O_{n_{k}}$ for some $k\in \mathbb{N} \big \}$.

require more justification or it's clear enough?

Ted Shifrin

23:31

@LittleRookie: It seems fine to me.

Little Rookie

Okay :)

Ted Shifrin

Of course, @LittleRookie, what's going on is that $O_j\subset O_k$ when $j\le k$, so finitely many $O_j$ might just as well be replaced by $O_{\max j}$.

Evan

Hey folks, if there exist some upper triangular matrices A, B, and C which solve AA^\dagger + BB^\dagger = CC^\dagger, is there a cheap way to calculate C given A and B?

Meow Mix

Hi I'm here

Ted Shifrin

@Evan: What's \dagger?

Evan

23:34

adjoint

conjugate transpose, rather

Ted Shifrin

Got it.

Upper triangular makes it easy, I guess.

Evan

I'm not even sure C is unique or whatever, but does this look like anything familiar to anyone?

Ted Shifrin

Can't we just take $C=A+B$?

Mike Miller

h

Evan

you'll get cross products, right?

Ted Shifrin

23:36

gn, @MikeM

Mike Miller

sleepy

Ted Shifrin

Well, let's write $A=D+U$ and $B=D'+U'$, @Evan.

Topologicalife

Hi.

I have a question about the system: $\begin{cases} x'=-y \\y'=x \end{cases}$

I got as eigenvalues $\lambda = \pm i$ and I got as eigenvectors $\mbox{col}(is,s)$ where $0\neq s \in \mathbb{R}$

Ted Shifrin

Yeah, I take it back, @Evan.

The solution curves are circles centered at the origin, @Topologicalife.

Topologicalife

so if I take $s=1$ I got as eigenvectors $(0,1)$ and $(1,0)$, is that correct?

Ted Shifrin

23:40

No, where did your $i$ go?

Topologicalife

Yeah, I know, but I got a problem I can't solve in my resolution.

Ted Shifrin

$(\pm i,1)$.

Topologicalife

Yeah but I'm writing $(\pm i,1)$ as $(0,1) + i(1,0)$

Ted Shifrin

You can't diagonalize working over $\Bbb R$, @Topologicalife, only working over $\Bbb C$.

Topologicalife

so I can build my real basis with $(0,1)$ and $(1,0)$

Ted Shifrin

23:42

With respect to the real basis you wrote, you'll get a matrix of the form $\begin{bmatrix} \cos t & -\sin t \\ \sin t & \cos t\end{bmatrix}$, not a diagonal.

Meow Mix

Fun fact: getting cut with fine glass hurts.

Ted Shifrin

In other words, you'll get back to your original skew-symmetric matrix ($t=\pi/2$).

Sergey Dylda

Hello, everyone. Got some small question: If $G$ is a Lie group. Does the set of all left invariant vector fields on $G$ consitute a $C^{\infty}(G)$-module or $K$-vector space?

Topologicalife

Okay, let me rewrite it...

Ted Shifrin

Very dangerous game, Zach.

Meow Mix

23:43

It wasn't on purpose...

I had to pick up the large shards of glass that fell.

I just finished vacuuming.

Ted Shifrin

I take it you broked something glass, Zach?

Meow Mix

Mhm. A bowl of fruit.

Luckily, I already ate all the fruit.

So it was just a bowl.

Ted Shifrin

@Sergey: The latter. Putting in a general function on $G$ will ruin invariance.

You need constants.

Sergey Dylda

@TedShifrin Thanks

Topologicalife

To get the complex eigenvectors I solve $(A- \lambda I)w_1 = 0$ where $w_1 = \mathbf{u} + i\mathbf{v}$. Now I obtain ther real basis $\mathbf{u},\mathbf{v} \in \mathbb{R}^2$. If $w_1 = (a+bi, c+di)$ then $\mathbf{u} = (a,b), \quad \mathbf{v} = (c,d)$ so that ...

Ted Shifrin

23:47

@Topologicalife: I know the game.

You're going to get back to your original matrix.

Meow Mix

@Ted Sorry I haven't been doing a lot of math as of late. Sometimes I like to focus on programming and general computer stuff more than math.

Ted Shifrin

No, you mean to use $(a,c)$ and $(b,d)$.

Mike Miller

@SergeyDylda Note that a left-invariant vector field is determined by its value at a point.

Ted Shifrin

The key point being that $G$ acts transitively on $G$ (i.e., one orbit).

Meow Mix

And for that reason, I'm conflicted about CS and pure math...

Ted Shifrin

23:50

Zach, wait 5 more years to worry about it.

Topologicalife

now I can construct a matrix $C=\begin{bmatrix}{a}&{b}\\{c}&{d}\end{bmatrix}$ and the solution will be $Ce^{\mbox{Re}(\lambda)} \begin{bmatrix}{\cos (wt)}&{-\sin(wt)}\\{\sin(wt)}&{\cos(wt)}\end{bmatrix} \begin{bmatrix}{c_1}\\{c_2}\end{bmatrix}$

Sergey Dylda

@TedShifrin Can we say that push-forward is not $C^\infty$ linear thus left-invariance is ruined?

Mike Miller

agreed

Topologicalife

Oh @TedShifrin, that is what my book says. So I suppose is an error.

Mike Miller

@SergeyDylda No, fX is just no longer an invariant vector field.

Ted Shifrin

23:52

Well, @Topologicalife: You can check the details of what I said directly. You want real and imaginary parts of the complex eigenvector.

Topologicalife

You are right, yeah.

Ted Shifrin

@SergeyDylda: Pushforward by whom?

Topologicalife

But I didn't notice it until you said it.

Thank you so much, I'm going to give it a look.

Daminark

@Ted With the way things are heading it seems like in 50 years you'll have to know what you want to do from when you're 5 :P

Ted Shifrin

Demonark: At the rate things are going right now, the world may not be here in 5 years.

Meow Mix

23:53

I'm probably going to be going to the college all my siblings went to.

Daminark

Eh, I'm a bit more optimistic than that

Sergey Dylda

@TedShifrin By derivative induced by left translation map

Ted Shifrin

For it to be linear over $C^\infty$ functions, the function has to be constant on orbits.

Meow Mix

I hate being sad and gloomy. But we're all doomed to die some day.

Ted Shifrin

Yeah, my fault for being less than optimistic ...

Meow Mix

23:55

And the news doesn't help either.

Daminark

I suppose. Though if the world ends in 5 years there will be no need to spend my 30s panicking about tenure so I dunno

Meow Mix

If the world ends in 5 years I won't be able to do any algebraic geometry stuffs.

Topologicalife

@TedShifrin I checked what you said and I still have the same problem

Ted Shifrin

@Topologicalife: Can you succinctly and clearly tell me what the problem is?

Topologicalife

$Ce^{\mbox{Re}(\lambda)} \begin{bmatrix}{\cos (wt)}&{-\sin(wt)}\\{\sin(wt)}&{\cos(wt)}\end{bmatrix} \begin{bmatrix}{c_1}\\{c_2}\end{bmatrix}$ isn't a solution for my problem.

Yeah.

Daminark

23:56

@Meow Are you referring to the news in general or is there some meteor coming that I haven't heard about? I basically don't know what's happening in the world right now

Psets leave no time for such things

Meow Mix

Sensationalism is what I meant.

Ted Shifrin

@Topologicalife: The eigenvalues are $\pm i$, whose real part is $0$.

Topologicalife

The solution of a system of linear ode's is $Ce^{\mbox{Re}(\lambda)} \begin{bmatrix}{\cos (wt)}&{-\sin(wt)}\\{\sin(wt)}&{\cos(wt)}\end{bmatrix} \begin{bmatrix}{c_1}\\{c_2}\end{bmatrix}$ where $C$ is the matrix I said a few lines before. Are you agree with this?

My problem is that solution doesn't hold for my system.

Daminark

Also ha ha @Zach if the world ends in 5 years I'll still be able to do algebraic geometry in 2 so :P

Topologicalife

$C=\begin{bmatrix}{0}&{1}\\{1}&{0}\end{bmatrix}$

Daminark

23:57

Lol jk I'm sure you'll do great

Meow Mix

@Ted Is there eigenv[alue/ector] stuff in your LA / MV book?

Ted Shifrin

@Topologicalife: It does too. No, you left out a minus!

Yes, Zach, all of chapter 9.

Topologicalife

Where?

Ted Shifrin

Isn't $C$ supposed to be your original matrix giving the ODE?

Meow Mix

Sometimes I wonder if I'm going to be a complete failure at math. Like not know how to prove anything. Basically, I'm scared of doing my own research without a Ted to consult for a hint.

Topologicalife

23:59

No, $C$ is the matrix formed by $\mathbf{u}$ and $\mathbf{v}$

Ted Shifrin

Zach: You have a long way to go.

Daminark

"a Ted"

Meow Mix

I do.

10 years+

Actually, maybe not that much.

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