Mathematics - 2016-07-25 (page 3 of 4)

0celo7

9:00 PM

@BalarkaSen I have a proof that uses Riemannian geometry if you're curious

Balarka Sen

Get a nice enough intersection for them to be convex subsets of the charts.

0celo7

It's sketched in Bott & Tu

@TedShifrin probably knows what I'm talking about

Balarka Sen

Then convex contractible subsets of R^n are diffeom to R^n.

Danu

By the way @TedShifrin, I asked this before but didn't really get a clear answer: Is there a good reason why we talk about the "cocycle condition" for transition functions on bundles (other than that the equation looks the same as the cocycle condition I know from sheaves on Riemann surfaces)? Balarka mentioned somethingsomething sheaves too, but nothing cleancut. Do you know something?

0celo7

@BalarkaSen Yes, that's actually the hardest part to prove, imo.

Balarka Sen

9:01 PM

It's an exercise in Hirsch. No idea how to prove it for n > 2

0celo7

I found two proofs

Ted Shifrin

Yeah, @0celo. I've never actually worked that out myself.

0celo7

@TedShifrin would you mind reading this then? math.stackexchange.com/questions/1869739/…

Balarka Sen

@0celo7 Not right now.

0celo7

One proof is in an old Gromov paper

the other is in some French calculus book

Ted Shifrin

9:02 PM

yes, @Danu, it's still a Čech-cocycle condition for $GL(k)$.

Balarka Sen

@0celo7 Link me to it.

Ted Shifrin

I'll look later, @0celo, but thanks!

Balarka Sen

Gromov is a cool guy.

0celo7

@BalarkaSen The easier proof is in the post I just linked Ted.

Ted Shifrin

I actually need to get out of here and work on a recommendation.

0celo7

9:03 PM

Lemma 4.

Balarka Sen

Sure, but I am interested in seeing Gromov's method.

0celo7

Ok, let me find it.

I found the reference in Jeff Lee's geometry book

the other Lee that I recommended

Ted Shifrin

Officially, @Danu, just as line bundles are $H^1(X,\mathscr S)$ for the sheaf $\mathscr S$ of nowhere-zero continuous, smooth, or holomorphic functions, there's an analogous description of vector bundles where you have the sheaf associated to $GL(k)$.

Balarka Sen

@0celo7 Wow, your proof of manifolds admitting good cover seemed to go in the same direction I sketched.

0celo7

@BalarkaSen Wow?

Yeah you had the right intuition

Maybe it works with triangulations

But with Riemannian geometry it's pretty immediate

M. Gromov, Convex sets and Kahler manifolds, in Advances in Differential
Geometry and Topology, World Sci. Publ. (1990), pp. 1~38.

Ted Shifrin

9:05 PM

I'm shocked that Gromov would bother to do something that's so much folklore ...

0celo7

I should note that I do not know for sure if the paper contains the proof

Balarka Sen

@0celo7 I don't think they are distinct, geodesically convex neighborhoods is how you get triangulations, if I overheard right.

0celo7

Jeff Lee says it does

Ted Shifrin

Gotta love the internet age. I now have a pdf of Gromov's article. :)

0celo7

@BalarkaSen Oh?

Balarka Sen

9:06 PM

Note that I dunno anything about that. I just overheard it.

0celo7

My prof said the best reference for triangulations is Munkres "Elementary Diff Top"

Oh, I lied. There's a third proof in some German analysis lecture notes

Danu

@TedShifrin I never did learn that line bundles are classified by $H^1$ :\

Ted Shifrin

I no longer have that book, either. It sucks giving away almost my whole library.

0celo7

@TedShifrin A prof at my school is retiring, my advisor got like 20 books from him

A whole set of Spivak

Ted Shifrin

It's a fancy way of saying bundles are determined by transition functions satisfying — guess what? — the 1-cocycle condition, @Danu. Nothing more.

0celo7

9:08 PM

I'm trying to cop a Vol. 1, although it's Vol. 2 that's impossible to get any more

Ted Shifrin

I had people walking out of my office with stacks of books taller than they were, @0celo. So, yeah.

I kept my nicely LaTeXed Spivak, but gave away two sets of the old ones.

Danu

You* had several sets? lol

0celo7

I don't know why Vol 2 is so hard to find

Balarka Sen

@Danu By $H^1$ do you mean sheaf cohomology? For singular cohomology, I learnt a neat proof.

Well, came up with a proof, more like, with sufficiently many hints from Mike.

Ted Shifrin

Yes, @Danu. And Spivak didn't give them to me, although he gave me other books.

It does appear that Gromov proves the result for a $C^2$ convex diffeomorphism.

0celo7

9:10 PM

@TedShifrin Have you heard of Fernando Schwarz?

@TedShifrin Which page?

Danu

@BalarkaSen Sheaf

Ted Shifrin

The conclusion is on p. 8, @0celo, but he uses a bunch of stuff before. I haven't sat down to understand it.

0celo7

Ah, ok.

Ted Shifrin

It's based on the Legendre transform. This will make @Semiclassic ecstatic!

Danu

lol

Gromov is such a legend

Balarka Sen

9:13 PM

Actually.

Danu

I wish math. was a bit more accessible, like physics, so I could understand why people are so highly regarded

Krijn

Physics is accesible?

Balarka Sen

@TedShifrin @0celo7 Take a manifold $M$, possibly noncompact. Cover it with a bunch of charts. Take a sequence of finite subcollection $C_i$ of charts such that union of charts in $C_i$ contains union of charts in $C_{i+1}$ (so you keep adding chart-after-chart).

0celo7

@TedShifrin do you know if you can write in to Spivak's publisher and ask for the books?

Vol 2 isn't on Amazon

I'd like to have it...it seems easier than Kobayashi

Balarka Sen

Union of charts in $C_i$ is a submanifold with boundary of $M$, call it $W_i$. So $M$ is direct limit of $W_i$'s.

Ted Shifrin

9:16 PM

Spivak is his own publisher, @0celo (Publish or Perish). I believe Volume 2 disappeared from print years ago. You can google Publish or Perish and see.

0celo7

@TedShifrin Why did it disappear?

Ted Shifrin

Books go out of print. I think he quit publishing.

Balarka Sen

Homology preserves direct limit. And direct limit of countably generated stuff is countably generated, no?

0celo7

@TedShifrin I mean why did that one go out of print but the others did not

@TedShifrin Oh, interesting thing. I was in the library the other day and needed some Riemannian geometry. So I went and grabbed Petersen (we've talked about this before). Turns out he flat out deleted a whole chapter on hypersurfaces in the latest edition

Ted Shifrin

Here is the link. He's only got his relatively new physics book active on there and says to go to Amazon for everything else.

Danu

9:18 PM

@TedShifrin, you asked earlier what I was drinking or smoking...

0celo7

It seems like he cuts more and more stuff with each release of the book

Danu

So you get to choose which whisky I drink tonight

Ted Shifrin

LOL, @Danu, now that you've eaten?

Danu

1. Ardbeg 10
2. Tobermory 10
3. Lagavulin 12
4. Springbank 10

Take a pick

Krijn

Easy, 1.

Ted Shifrin

9:19 PM

I don't know any of those, sorry.

heya @Krijn.

Danu

@Krijn So mainstream ;D

Krijn

Hello!

Danu

@TedShifrin Pick one!

Ted Shifrin

I'm more of a connoisseur of gins ...

Danu

@Krijn the laga 12 is really something---it's cask strength, too

Ted Shifrin

9:19 PM

What do the 10 and 12 signify?

Danu

@TedShifrin Years

Ted Shifrin

so the 12 you should save for celebrating something special ... :)

ok, I pick #2.

Danu

Cool!

Tobermory it is

Balarka Sen

One morning in a fit of pique / she drowned her father on the creek / the water tasted bad for a week / and we had to make do with gin

re "gin"

Krijn

Wait long enough and the 12 will be an 18

Ted Shifrin

9:20 PM

Um, gee, thanks, @Balarka.

0celo7

@Krijn age has got to be on the clock :)

Danu

So what's a good gin, @Ted?

I'm very interested in "understanding" high quality liquor

Krijn

Ketel 1, obviously

Danu

lol

Ketel 1 is jenever, no?

Krijn

It is.

Danu

9:21 PM

That's not the same as gin

Krijn

The worst one possible, really

Kopstootjes! :D

Ooooooooooh noooo.

Hellllll yes

That was the last thing I remember from this weird night in november.

Danu

9:22 PM

Excellent drink

Ted Shifrin

Lots of options ... I buy more American ones these days. But I love Citadelle from France. Hendrick's from Scotland is amazing.

Danu

@TedShifrin Okay. What kind of flavors should I be looking for in a good gin?

Ted Shifrin

Depends on your taste. I don't like floral; I like more juniper berry and herbal flavors.

Hendricks actually has a slight cucumber tinge, so it's good muddled with fresh cucumber (rather than vermouth and olives) :)

Danu

Right, I think I probably agree

Hmm, I like the idea of cucumber

Ted Shifrin

I guess I need to come visit so we can test them all out? :D

Danu

9:24 PM

Come on over to Munich any time ;)

Krijn

Also try a kopstootje then

Danu

^

Ted Shifrin

OK, I really must get some work done. Bye, guys.

Danu

Classic Dutch drink

Bye Ted

0celo7

Dutch sounds so weird

Danu

9:24 PM

You sound so weird

0celo7

Terrible comeback, you don't know what I sound like.

Danu

(when trying to pronounce Dutch)

^but I do know that for certain

0celo7

@Danu Oh, I don't disagree :D

If I try to do it as a German, it sounds weird

Alessandro

since I know nothing about Dutch it kinda looks like German with vocals to me...

0celo7

If I try to do it as an American, it sounds stupid

Krijn

9:26 PM

Try it as a Dutch person

0celo7

I am not a Dutch person

Krijn

It's pronounced a bit like "cop stoat G"

Danu

hahaha

More like

0celo7

Yeah, I got the "cop stoat"

Danu

"cop stoat ch" where "ch" is as in "chips"

and a bit of an "uh" sound after the "ch"

0celo7

9:27 PM

Ah, I was trying to pronounce the j as in German

cop-stoat-ye

Danu

oh that's fine

But with a t it sounds a bit different

"(t)je" is the Dutch analogue of "chen" in German

0celo7

Maybe if I did it in Pfälzisch it would sound better

Then I get a ch sound

@Danu Aww, Danutje

Krijn

Almost

You'd have to add an extra u

0celo7

Whut

Krijn

Danuutje

0celo7

9:30 PM

I should probably figure out the homology of CW complexes

isaac9A

Do all uncountably infinite sets have the same cardinality? If I have an uncountably infinite set S, then surely even though it is uncountable the power set of S should be strictly greater than S itself right?

*cardinality of the power set of S should be stricktly greater than the cardinality of S

Danu

@0celo7 :)

@Krijn That's not true :P

This is not how it works for proper names, AFAIK

Krijn

What?!

Danu

@0celo7 This is a really delightful topic

Balarka Sen

I am not enjoying it.

Danu

9:32 PM

Mapping degrees, so nice

Everything by inspection (in dim $\leq 3$)

Balarka Sen

Oh, homology of CW complexes

Danu

hahaha you grump

Balarka Sen

I thought by topic you meant Dutch pronunciation.

Alessandro

@isaac9A no, $|\mathcal{P}(S)|>|S|$ for every set, actually there are so many cardinalities that the collection of all the cardinals ends up being a proper class in ZFC, but that's a topic I don't know too much about

Danu

@BalarkaSen Feel left out? ;)

Balarka Sen

9:33 PM

Sort of kind of

Danu

So tell us something about your language(s)

Balarka Sen

It's weird.

Danu

lol

Alessandro

Is there a difference in pronunciation in Dutch between a repeated vocal/consonant and a single one?

Krijn

Sometimes.

isaac9A

9:35 PM

Mandarin Chinese does not really have one constant equivalent of the verb 'to be'. 我很傻（I very stupid) is a complete sentence in Mandarin. There is a character 是 which at some times is roughly equivalent to 'to be' but it is not completely.

Random Variable

@Semiclassical Using the functional equation $\sum_{n=-\infty}^{\infty} e^{-\pi n^{2}x} = \frac{1}{\sqrt{x}} \sum_{n=-\infty}^{\infty} e^{-\pi n^{2}/x}$, it appears that $\vartheta_{3}(0,q) \sim \sqrt{\frac{-\pi}{\log q}}$ as $q \to 1^{-}$.

Alessandro

that's a verbless sentence though @isaac9A (which is interesting in itself since most languages need a verb)

Agawa001

its really relieving yo step onto one place among thousands where all chat subjects turns automatically into interreligious fight

Balarka Sen

We don't get inter-religious fights here a lot but we do get irreligious fights.

Krijn

Russian leaves out words like "am", "is" and "are" all the times, I believe

Alessandro

9:38 PM

@Krijn are there words that are spelt in the same way except that one has a repeated letter and the other one doesn't? (There is a proper term from linguistics for when repeated letters influence pronunciation like in Italian, but I can't remember it right now and it's something I find curious)

isaac9A

@Alessandro ok but it still a gramatically and semantically correct sentence in mandarin chinese

Danu

@Agawa001 what?

Alessandro

@isaac9A indeed, I wasn't arguing with that (also because I remember very little of Mandarin so I'm not really qualified to discuss it)

Danu

@Alessandro Usually, yes

@Alessandro Yup, many

"veel, vel"

"raap, rap"

etc

Alessandro

Since we're sharing language fun facts, in Italian the subject is omitted if it is a personal pronoun (because you can work it out from the conjugated verb)

David Zhang

9:42 PM

Hey guys, I've got a quick algebra question. Suppose somebody hands you the multiplication table of a (small) finite group. Is there a reasonable algorithm to compute (the multiplication table of) its automorphism group?

isaac9A

In Mandarin Chinese, there are two different words for aunt/uncle, depending on if the aunt/uncle is on your moms side or dads

David Zhang

The problem is really just listing the group automorphisms, since once you know those it's easy to compose them. But I don't know a better way to do this than some sort of backtracking search

Krijn

@Alessandro I once read that Italian has certain rules to how to pronounce words (i.e. no trouble in English like with steak and beak) and that they therefore don't need a word for "to spell"

Alessandro

@krijn that's true, we use the English word "spelling" when we really need to

Balarka Sen

@0celo7 Not sure if you saw my proof of manifolds having countable homology.

I am quite certain it works.

Alessandro

9:47 PM

also that is something I find very annoying about English, apart from the "i" whose pronunciation you can never guess, but how is one supposed to know that the verb "bow" is read differently from the "bow" with arrows? And sometimes read and lead rhyme with each other, but not always...

isaac9A

@Alessandro context :)

0celo7

@BalarkaSen uhh

I didn't, sorry

Alessandro

ok, last language fact since this was a mathematics chat once, pen and pencil have completely unrelated etymologies

Krijn

Yes, and pencil comes from the same word as "penseel" in Dutch @Danu!

isaac9A

This is a mathematics chat?

Krijn

9:51 PM

I prefer to call it a mathematicians chat

isaac9A

Then can someone explain this sentence to me please?

Let R be the set of all sets that are not members of themselves. If R is not a member of itself, then its definition dictates that it must contain itself, and if it contains itself, then it contradicts its own definition as the set of all sets that are not members of themselves.

more specifically how does its definition dictate that it must contain itself

i thought its definition dictates that it cannot contain itself

Alessandro

@isaac9A this is known as Russel's paradox

Balarka Sen

@0celo7 Cover your manifold $M$ by countably many open sets (consequence of 2d countability). Index them by $\Bbb N$, and consider the subspace $A_i$ which is union of first $i$ of the open sets. $A_i \subset A_{i+1}$, $M$ is union of $A_i$ aka direct limit of $A_i$'s.

Alessandro

the point is that whether it contains itself or not you always reach a contradiction

it shows that the naive approach to set theory doesn't really work

isaac9A

@Alessandro yeah Im trying to understand the part "If R is not a member of itself, then its definition dictates that it must contain itself..."

Balarka Sen

9:53 PM

You can replace $A_i$ by $B_i = \text{cl} A_i$ easily. $B_i$ is a compact manifold with boundary.

So $B_i$ has finitely generated homology.

0celo7

Why is $B_i$ compact?

Krijn

If $R$ is not a member of itself, then it is not a set that is not a member of itself, so it is a set that is a member of itself.

Balarka Sen

oops lol

oh, but that's easy to fix

@0celo7 Instead of $A_i$ being the whole chart in your open cover, choose it to be the unit open ball inside.

Danu

@isaac9A Just reduce it to strict logical operations: Language is tripping you up.

Assume $R$ does not contain itself.

Then, because $R$ must contain all sets that do not contain themselves, $R$ must in particular contain itself.

0celo7

Oh, balls

Danu

9:56 PM

Conclusion: $R\notin R\implies R\in R$

This is a contradiction

Alessandro

$R$ has as member every set which is not a member of itself, so if $R$ isn't a member of itself it must be a member of $R$

Balarka Sen

$B_i$ then is just homeomorphic to a closed ball :)

0celo7

Yes, so it's compact.

And we have a countable number of them, right

What next

isaac9A

@Danu and @Alessandro Thanks yeah it was the language :)

Danu

In these simple questions it is often less confusing to "talk in symbols" :P

Balarka Sen

9:58 PM

@0celo7 Sorry, I meant, choose $A_i$ to be union of balls inside the charts. Note that $A_i$ was union of first $i$ of the charts, not the charts themselves.

0celo7

I understood that

Balarka Sen

But anyway, $B_i$ is compact.

$M$ is direct limit of $B_i$.

0celo7

Yes, the directed set is ordered by inclusion

Balarka Sen

Homology of $M$ is direct limit of homology of $B_i$, because homology commutes with direct limit.

0celo7

Yes

Balarka Sen

9:59 PM

Homology of $B_i$ is finitely generated.

0celo7

I believe you on that one

Balarka Sen

Direct limit of countably generated stuff in a countable directed set is countable.

Because it's just a quotient of direct sum of the stuff in the directed set.

0celo7

uhhh

Balarka Sen

Direct sum of countably many countably generated stuff is countable, hence so is any quotient.

isaac9A

Ambitious question of the day: Can someone explain eigenvalues and eigenvectors to me?

Everything I read online I'm like wtf

0celo7

10:00 PM

@isaac9A $Mv=\lambda v$

isaac9A

I have a basic understanding of linear algebra and undestand concepts like basis, linear independence and how they all relate to eachother, but I am by no means an linear algebra expert

0celo7

@BalarkaSen I don't know how to prove that

Krijn

@isaac9A Do you know enough about linear maps/matrices and vector spaces?

isaac9A

How much is enough?

Krijn

Do you understand what I mean by $T: V \to V$

isaac9A

10:02 PM

I understand the basics about what a linear transformation is and what a vector space is

Krijn

Or $T(v) = \lambda v$ for some vector $v \in V$ and some scalar $\lambda$

isaac9A

How do I read those symbols?

do I need to install something to read TeX?

Krijn

$T$ is the linear map, $V$ is the vector space. We apply $T$ to some vector $v$.

Ah yes, see the thingy on the right

$\LaTeX$ in chat: tinyurl.com/cfqcvpc

isaac9A

got it thanks

much easier to read now!!!

<3

@Krijn you are saying T is a linear transformation from V to V

correct?

Krijn

Yes

Also $\heartsuit$ is \heartsuit

Balarka Sen

10:06 PM

@0celo7 It is sufficient to prove that countable direct sum of finitely generated groups is countable for us. $\{G_i\}$ be countably many abelian groups. Choose generators $g_1^i, g_2^i, \cdots, g_r^i$ of $G_i$. $\oplus G_i$ is generated by countably infinitely long tuples with $0$'s on each slot except the $n$-th one, which has any of $g_k^n$'s in it.

There are clearly countably many of such tuples.

Typo, apologies.

isaac9A

So eigenvalue/eigenvector is a corresponding scalar and matrix that when either is applied to the vector space creates the same resulting vector space?

@Krijn

Krijn

No, no, not the vector space

0celo7

@BalarkaSen Yeah, ok.

Balarka Sen

This proof pushes through for $G_i$ countably generated too.

But I dare not say it lest you get "confused" and ask me to be "rigorous"

0celo7

Pushes through?

Krijn

10:08 PM

If we have some vector $v \in V$ such that when we apply $T$ to that $v$, we get $v$ multiplied by some scalar $\lambda$ (i.e. $T(v) = \lambda v$) then we call $v$ an eigenvector for the transformation $T$. @isaac9A

0celo7

@BalarkaSen What

Balarka Sen

@0celo7 Synonymous to "works".

isaac9A

I see, and the scalar would be the eigenvalue?

Krijn

Indeed!

Balarka Sen

@0celo7 See, this is exactly what I was referring to.

isaac9A

10:09 PM

so eigenvalues are just scalars that would produce the same result on a vector that a corresponding matrix (eigenvector) multiplication would do?

0celo7

@BalarkaSen Now I am confused

isaac9A

sorry my english sucks

0celo7

I was not before

Balarka Sen

I am sleepy.

isaac9A

I speakaaaaa daaaaa chineeeeeeese

Balarka Sen

10:10 PM

Anyhow, you're convinced about the proof?

Krijn

@isaac9A Eigenvalues are always associated to their eigenvectors.

0celo7

I could not reproduce it, but I'm about as convinced as I care to be, yes.

Krijn

No, sorry, not always

Balarka Sen

OK, great.

Alessandro

@isaac9A geometrically you can think about eigenvectors as directions that are only stretched or compressed but not rotated by a linear transformation

Balarka Sen

10:10 PM

The point was, indeed, second countability.

0celo7

Yeah.

Balarka Sen

(thanks @TedShifrin)

I have to go to school tomorrow but I know I can never sleep, even if I try to. Suggestions? (no troll suggestions please, this is serious)

Alessandro

@balarka stay awake the whole night if it doesn't damage your productivity too much the following day and reset your sleep schedule tomorrow night

that's what I usually do at the end of holidays or such to get used to the school timetable at least

Krijn

Drink warm milk, go to bed, read (something not too in depth) with just a small light for an hour, then turn off the small light and count sheep

Balarka Sen

@Alessandro That's impossible though, because I have to go to school.

0celo7

10:14 PM

I try to dream, and that puts me to sleep

I try to imagine some cool scenario or something

That keeps my mind occupied and then I wake up

Works most of the time

Balarka Sen

@Krijn, @0celo7 I'll try those out.

Thanks.

0celo7

@BalarkaSen I replay movies or games in my mind

Doing things my way

I've been having crazy dreams lately

I had one strange one where my gf's brother was a super-genius spider

and we were best friends

but I got jealous of him and stepped on him

Alessandro

@isaac9A now if we denote with $T$ the square matrix associated to the linear transformation and suppose that $v$ is an eigenvector we have $Tv=\lambda v$ so $Tv-\lambda v=0$ and $(T-\lambda I)v=0$ where $I$ is the identity matrix with the same dimension as $T$

(by the way eigenvectors are nonzero vectors by definition) so this product is 0 iff the determinant of $T-\lambda I$ is 0, and that's how you calculate the eigenvalues, you literally compute this determinant and end up with a polynomial in $\lambda$, the so called characteristic polynomial

Krijn

@0celo7 ...

0celo7

@Krijn there was also a sexy nurse involved, don't ask how

I just remember the highlights

Alessandro

10:20 PM

I feel I'm missing out now since I never remember my dreams

Krijn

I'm off to see if I remember my dreams tonight.

G'bye

0celo7

cheerio

Alessandro

good night

isaac9A

@Alessandro thanks man Im still dissecting your second response but the first one thinking about it geometrically is very helpful

you guys rock <3

@Alessandro so T is the matrix which is the linear transformation, v is the eigenvector and λ is the eigenvalue? So the linear transformation multiplied by the eigenvector is the same as multiplying the eigenvalue by the eigenvector?

Alessandro

that works in $\mathbb{R}^n$, but generally speaking the vectors in your vector space could be functions, matrices or other weird things where the geometric intuition doesn't really work

If $L$ is a linear transformation and $T$ is the matrix associated to $L$ in some fixed basis ,$v\in V$ then $L(v)=Tv$, they are different ways to write the same thing

to saying that $Tv=\lambda v$ is restating the definition of eigenvector using another notation

isaac9A

10:34 PM

I was just copying and pasting what you wrote earlier about Tv=λv

Alessandro

yes, that's just the definition of eigenvector (since we assumed $v$ to be one)

isaac9A

I thought the eigenvector itself was T, the matrix associated to some linear transformation L, not the vector iteslf but I guess duhh the eigenvector is a vector not a linear transformation

Thanks for the clarification

@Alessandro so the eigenvalues are the roots to the polynomial T−λI ??

Alessandro

that is a matrix, the polynomial is the determinant of that matrix

and yes, the eigenvalues are the roots of this polynomial

isaac9A

the determinant of the matrix T−λI is just a number right? Where does the polynomial come from?

or do you get the polynomial because you dont know λ beforehand

?

Thanks for your patience

Alessandro

you treat $\lambda$ as an unknown

so you end up with a polynomial with $\lambda$ as variable

isaac9A

10:42 PM

gotcha

Alessandro

wait a moment while i figure out how to write matrices in LaTeX

isaac9A

so in the equation to calculate the determinant you will have to solve for λ

Alessandro

ok, let's use this matrix as an example \begin{pmatrix}
2 & 0 \\
4 & 3 \\
\end{pmatrix}

it is the matrix associated to the transformation $\mathbb{R}^2\to\mathbb{R}^2$ that sends the generic vector $(x,y)$ in $(2x,4x+3y)$

if we subtract $\lambda I$ from it we get \begin{pmatrix}
2-\lambda & 0 \\
4 & 3-\lambda \\
\end{pmatrix}

and the determinant of this matrix is $(2-\lambda)(3-\lambda)$ so the eigenvalues are $2$ and $3$

isaac9A

Now im confused. How do the 2 eigenvalues work

?

Alessandro

hm, what do you mean?

isaac9A

10:49 PM

how can we reconstruct Tv=λv

arctic tern

they have to submit a w2 form first

isaac9A

If T is the 2x2 matrix you outlined above, how would multiplying a vector by it be equivilant to multiplying a vector by either 2 or 3?

Alessandro

well, so far we only know that there exist some $v$ such that $Tv=2v$ and some other $w$ such that $Tw=3w$, but we don't know anything yet about $v$ and $w$

isaac9A

(3, 2) multiplied by 2 is (6, 4) where as by 3 is (9, 6)

???

arctic tern

@isaac9A nobody said multiplying by 2 and multiplying by 3 would be the same as each other

isaac9A

10:51 PM

so the result means that some vectors multiplied by 2 will be equivalent to multiplying by the linear transformation where as others have to be multiplied by 3?

Alessandro

yes, and some other vectors (those which are not eigenvectors) will be completely changed by the transformation

isaac9A

so we solved for the eigenvalues, but we still don't know which vectors are eigenvectors? what was the point of that?

Alessandro

we are going to find the eigenvectors from the eigenvalues

it might seem to be backward but that's the easiest way

isaac9A

gotcha

0celo7

is there a harder way

isaac9A

10:55 PM

why would you want to find eigenvalues and their corresponding eigenvectors in the first place?

When you can just multiply the vector by the linear transformation?

seems like a lot of work to me just to find scalar multiples that when multiplied by a vector produce the same result as multiplying said vector by a given matrix of linear transformation

Alessandro

uhm, do you know what a diagonal matrix is and agree that computations with diagonal matrices are much easier than with normal matrices?

isaac9A

yes

I dewwwwww

XP

XD

still seems like a lot of work to me just to find scalar multiples that when multiplied by a vector produce the same result as multiplying said vector by a given matrix of linear transformation

Alessandro

@0celo7 not that I know of, but life and maths are full of surprises

arctic tern

@isaac9A yeah, but can you apply a matrix 1000 times by paper? you can with eigenstuff.

0celo7

@Alessandro It's not so much the "easiest way" as it is the "only way"

isaac9A

10:59 PM

seems like a lot more computationally expensive than just multiplying the matrix of linear transformation with the vector

Alessandro

it turns out that if a linear transformation has "enough" eigenvectors than it can be rewritten in another basis as a diagonal matrix

isaac9A

having to solve for eigenvalues, then solve which vectors can be used with these eigenvalues

0celo7

@isaac9A try calculating $M^{1000}v$ without using eigenvalues

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$Mathematics$ Mathematics