@Timbuc sod off and go lay into the OP, the bastard gave two Eigenvectors! Go tell him how you don't understand what they are! – Alec Teal just now edit

Too mean?

Eh, I'm inclined to agree with Timbuc personally. Instead of wording the original post as "If the remaining Eigenvector is one you already have (that is one is repeated) the matrix cannot be diagonalisable." I would word it instead as "If you cannot find a third linearly independent eigenvector then..."

Is that literally all he's on about @JMoravitz

Either way, it seems to me that the going back and forth has gone on a bit too long.

That's my biggest gripe, and appears to be his too, yes.

I want to know who's encouraging him with upping his comments...

The whole bit about vacuous truths, if you accept a definition where you cannot find "a remaining eigenvector" in the first place

Well the OP gives two eigenvectors... I figured he'd understand?

Anyway @JMoravitz I don't want to spread the time-wasting, but can you just confirm something for me: "I'm not mental, and what is easy to interpret?"

Fucker is a skilled troll, because now he's like "Whoa, keep it civil"

Again, I wouldn't think of either party as being labeled as a troll, and I think both sides could have perhaps handled the situation better. I think it is a good idea that you stop responding in the comments either way.

His concerns (assuming they are the same as mine) I see as being warranted, but I also see how your attempts to appease his concerns were good and should have been adequate.

Thank you! I was questioning myself.

Does anyone have an ideas about this question: math.stackexchange.com/questions/1363578/…

@JMoravitz okay now I've lost rep for it - I wouldn't say that was needed @JMoravitz

@AlecTeal he seems to have picked up on the fact that you said "any point" rather than "any point on that line" and is making a big deal about it. However, not having picked up on that and getting a bit angry, you might have helped fan his flame.

@robjohn I was attracting downvotes (EVEN THOUGH DOING EXACTLY WHAT THE OTHER ANSWERER DID) so I've reposted. I shouldn't be rolling in rep for that but I shouldn't be downvoted.

Also @robjohn is that really it? I'm just glad it's been confirmed that I'm not missing some huge chunk of information that I've filled in with craziness without realising. I just don't get who (and there are 4 of them) was encouraging his comments!

Thanks

Now the new one is getting downvoted, - but it's EXACTLY what the answerer did!? WTF

I hate this site, no effort gets rewarded sometimes, but my effort of responding to that tit just gets ploughed into the ground

@AlecTeal I don't see that what you said is exactly what the other answerer is doing. He is showing that there are only two eigenvectors. You are saying that if there are only two eigenvectors, then it is not diagonalizable, which is true, but does not answer the question.

@robjohn don't you dare say "hints are frowned upon here"

You get rep back when you delete stuff right?

sigh I hate when a question leaves you feeling very dumb. I bet it has a simple answer, too

@PhilipHoskins It's always only simple when you know where to look.

@DanielFischer Nice answer! Thanks. What about the problem made you think to use convolution. I was trying to get a handle on the intersection of {(t,theta) in S^1 x S^1: |t-theta|<0.1} and M x S^1, so your answer makes me think that was on the right track

I've just given a downvote to an math.stackexchange.com/questions/1363998/… here but an upvote to a correct answer, will I ever get the 1 rep point back?

@PhilipHoskins The $m(M\cap (\theta - 0.1, \theta + 0.1))$. That's just the value of the convolution of the two characteristic functions at $\theta$. In problems about the measure of sets, convolutions often occur, so with those two points, it just remains to check whether it leads to the desired conclusion.

@DanielFischer if I get my brother onto this site will that be a problem? (We'll have the same IP)

@AlecTeal If you don't interact, no problem. If you vote for each other, it will be a problem. The occasional vote won't be a problem, but if there is a trend, that's bad.

@DanielFischer Ahh, I see. If I try to write the measure as an integral, the convolution appears. Thanks again

@PhilipHoskins You're welcome.

@DanielFischer I lost an account of mine once (I'd really like that back!) can you note "Moss is my brother" on the account or something?

@AlecTeal Not sure what you mean?

@DanielFischer I had an account called "I'mGettingThere" which was... destroyed. I'd like that undone, but also don't want my brother getting destroyed for association with me!

@AlecTeal Technical note: deleted, not destroyed. I don't think that would be undone (restoring accounts is, as far as I gathered, a major pain in the rear, will be done only in special situations, not for deletions that appear correct). To keep your brother safe, take care that it doesn't look like you're doing something illicit. Cross-voting: bad.

@DanielFischer the circumstances for that was talking in a chat. If anything does happen can I say "look, I disclosed it" or something?

Depends on what happens.

I'm not saying "immunity" I'm saying making a note that if (especially over the holidays where we're both home) that "IP $\not{\implies}$ same person" @DanielFischer

@AlecTeal We need more than just the same IP to conclude two accounts are one person anyway. And as long as there is no interaction or account-abuse, having multiple accounts is tolerated (though not officially supported by the software, it can happen that accounts are automatically merged).

Then why was I'mGettingThere purged?

@AlecTeal I guess you used it to do something forbidden. But since I was not involved, I don't know what.

@DanielFischer they concluded it was a "sock account" but the only evidence could have been my IP address (it was in a TOTALLY different circle to maths)

Anyway, I'm going to bed now. Further enquiries to somebody else.

...comforting

@AlecTeal The deletion came from totally different circles too.

@AlecTeal I think hints are fine. However, you said that you were doing exactly what the other answerer did, and that is not really so.

Ah, so I blundered into drama ... imagine that.

Dear moderators, could you please delete my accepted answer here: math.stackexchange.com/questions/1058658/…

It unfortunately answers a different question than the one asked, and it has already misled one person. I feel bad about this.

Poll for you guys

I do some education work on the side. No conflict of interest whatsoever with my current employer.

Needless to say, I get paid for it

When I took this job, I signed an agreement saying that I don't have any employment agreement that would conflict with my current employment, which is true, even time-wise (side job is completely on my own schedule)

I found out that "actual" company policy is that I cannot be paid for anything outside of my current employment, and that if my employer finds out, they will send a Cease and Desist letter

What does one do in this situation? ugh

is this "actual" company policy on the thing you signed?

@anon No. However, everything's at-will, so you would think I basically have to abide by how they run things anyway

Or at least I don't recall it being on the agreement

(I am still waiting for them to send me my copy...)

It's weird how I found this out. The company has an internal online forum

Someone asked if they could write fiction on the side and get it published, get royalties from that. They found out the above

even though they also indicated that the agreement was unclear

That person has since left; I don't know her at all

@Potato I don't think a moderator can do that. I think you would need to contact Stack Exchange support.

I doubt internet forum posts are valid supplements to signed agreements. wait for your copy of it back. if it's not explicit and nobody explicitly mentioned it to you (i.e. you just found someone saying this on some forum post) then I wouldn't worry about it.

I agree^

Perhaps @Clarinetist take your copy of the agreement to a lawyer, just in case.

how's things @anon

good

just proctored last final today

just started on netflix

get a month off

you?

just finished my mini-vacation, Netflix included

finished unpacking and moving in to my new apartment today, minus getting the Internet to actually work

so now I have to get reading again

Anything good on Netflix?

<---not a big movie fan

:-)

after just reading through it, I feel compelled to say that urysohn's lemma is pretty awesome

which one is thag

"Suppose $X$ is a normal topological space. If $A, B \subseteq X$ are disjoint closed subsets, then there is a continuous function $f:X \to [0,1]$ such that $f(A) = \{0\}$ and $f(B) = \{1\}$."

the proof was pretty cool to me because it was done by induction

induction on?

you basically construct a growing sequence of neighbourhoods $\{U_i\}_{i \in \mathbb{Q} \cap [0, 1]}$ around $A$ such that $U_0 \supseteq A$, $U_1 = X\setminus B$, and $p < q \implies \overline{U_p} \subseteq U_q$, by letting $(r_n)_{n \in \mathbb{N}}$ be a sequence bijecting $\mathbb{N}$ with $\mathbb{Q}$ and inducting on $n$

creating the set $\{U_{r_1}, \dots, U_{r_n}\}$ for each $n \in \mathbb{N}$ which satisfies the properties

and then define the right function based on this thing

it's pretty convoluted but I like how weird the approach was

apparently I can't edit chat messages twice but the set used in the proof was actually $\{U_0, U_{r_1}, \dots, U_{r_n}, U_{1}\}$

also "bijecting $\mathbb{N}$ with $\mathbb{Q}$" should be "...with $\mathbb{Q} \cap (0, 1)$"

cute

I think it's pretty hype

Haha that comment makes me feel like I wandered into Twitch chat XD

Urysohn HYPE Kappa

@Alex: Somehow it turned out that I want to know a bit about modules so uh :)

:P I guess I don't want to distract you (or me) too much with modules

But it's tempting

(I don't know anything besides what I can guess from algebra I)

:P

Ah

I don't know what they're for except representation theory.

Which is to say, I don't know what represenation theory is for :)

So my dilemma is that I want to produce a Binet formula mod $9$, but $9$ isn't prime so the standard theory doesn't work. If you go back to the derivation you consider the shift operator on the vector space $\Bbb F^\infty$. Well, that could be the shift operator on $(\Bbb Z_9)^\infty$ but can you do anything from there?

@SamuelYusim cute

welp

O.O

That sucks

:( :( :(

@robjohn Fixed my post .. see if it's okay now

@AlexClark 36 hours? Why?

Had that before, never want it again.

That's nasty food poisoning @AlexClark maybe you should go to the hospital and get your stomach pumped

Omg thats HORRIBLE

I agree

Presumably you can drink water by now @AlexClark

I hope so :/

How much wieght have you lost @AlexClark

aw

Go see a doctor

You never know

11 times is EXTREME

Please

You're killing skill patrol here

So stubborn ;-)

Tough guy

Check out the extensions maths.kisogo.com/index.php?title=Special:Version - finally made it mine

Will strip not-mathjax stuff and start an extension for Mediawiki later :)

Templates inside mathtags FTW!

Nice work pal :-)

latex is native on hack.chat/?math

Very cool room colours^

math.stackexchange.com/questions/1364170/… .Can someone solve this

someone dere to solve?

@BenDover Many thanks!! Seems the OP finally decided to comment and accepted my answer here !!!!! :D

gosh, @AlexClark, what the heck did you eat for meal?

What is this object called in English? It's just what one is left with after cutting a spherical shell.

Spherical cap

hemisphere

(assuming you cut it in half)

@Szabolcs It looks like the northern hemisphere.

@anon i have a silly algebraic topology question. can you help me?

iunno

was going through a revision, and stumbled upon Hatcher's 16. (c).

if X retracts onto A, then \pi_1(A) --> \pi_1(X) is an injection. however, clearly any loop in A can be homotoped to a point by homotoping along the disk A bounds.

so \pi_1(A) --> \pi_1(X) is zero, a contradiction.

but, what if X is the solid torus and A is the unknot in X that winds twice around? that bounds no disk, but it seems that X doesn't retract onto A. how to prove it?

you've already done it. it represents $2 \in \mathbb Z = \pi_1(D^2 \times S^1)$, so the map $\pi_1(A) \to \pi_1(D^2 \times S^1)$ is multiplication by two

that's similar to how the mobius band doesn't retract onto its edge it sounds like

oh, alright, @Mike. told ya it's silly.

one way of showing it, yeah

i guess really the only way. w/e

i'm somehow more comfortable with proving these retract thingys using the homology split exact sequence, which is of little use here. or rather, the splitting information is of no use.

wait a second. if $X$ retracts to $A$, then $\pi_1(A) \to \pi_1(X)$ is an injection. multiplication by two is surely an injection!

so how does that prove anything?

i am being dumb.

@Vortico Thanks! That's what I needed.

oh, I see. $\pi_1(A) \to \pi_1(X) \stackrel{\pi_1(r)}{\to} \pi_1(A)$ cannot be identity.

@anon it's essentially the moebius strip problem. $A$ bounds a moebius strip inside the solid torus. the equatorial circle of the moebius strip represents the generator of the torus.

mmhmm

i'm trying to find a knot inside $X = D^2 \times S^1$ which bounds no disk inside $X$, and yet there is no easy description of $\pi_1(A) \to \pi_1(X)$

take a piece of string, make a trefoil knot but don't join the endpts. wind the endpts around the hole of $X$ and then join it.

how to prove that $X$ doesn't retract onto this ?

hey supersonic

:p

comment vas tu

quiet room, is everyone busy doing maths ?

Shush

hmm ?

this is interesting

plz no more topology

that has nothing to do with topology, @Agawa001

what's that lec about, B?

have a look yourself. it's combinatorics.

"Given n points in the plane, an ordinary line is a line that contains exactly two of these points, and a 3-rich line is a line that contains exactly three of these points. An old problem of Dirac and Motzkin seeks to determine the minimum number of ordinary lines spanned by n noncollinear points, and an even older problem of Sylvester (the "orchard planting problem") seeks to determine the maximum number of 3-rich lines.

In recent work with Ben Green, both these problems were solved for sufficiently large n, by combining tools from topology (Euler's formula), algebraic geometry (the Cayley-Bacharach theorem, and the classification of cubic curves), additive combinatorics (via the group structure of said cubic curves), and even some Galois theory (through the theorem of Poonen and Rubinstein that a non-central interior point in the unit disk can pass through at most seven chords connecting roots of unity).

We will discuss how these ingredients enter into the solution to these problems in this talk."

well, wow.

it's a pretty cool problem.

how did you not "meh" at this?! :P

why would I meh at very beautiful mathematics?

you just gave me the impression that you don't really like combinatorics, is all. I was wrong. :)

whatcha doing?

I'm free!

After so, so long.

@SohamChowdhury yes, you were. I just don't know enough combinatorics to appreciate it.

appreciate what?

this?

combinatorics.

I mean, I am perfectly willing to appreciate any mathematics once I know it and understand it well enough.

Since I don't know combinatorics, you've naturally never seen me appreciating it.

nice to hear.

@SohamChowdhury I? Goofing. I have to study, but I don't want to.

Cool statement on /r/math: "Pomegranate seeds are the elements of a 3D Voronoi diagram"

hahahaha

theguardian.com/science/alexs-adventures-in-numberland/2015/jul/…

look at that.

did you figure out the index 2 subgroup problem?

nope. I'll think about it today.

ok.

oh, wow, the proof of $N_2 \geq 3$ is pretty slick.

Get some rest, @AlexClark

Let's hope you'll not throw up again in these 40 minutes.

nah, I'm good with tea.

which link, @AlexClark?

oh. sure.

Elliptic curves give a lot of examples of sets with a lot of 3-rich lines, apparently, due to the group structure on ell curves. Sylvestor posed that this was the best possible bound.

I'm glad you went to the doctor @AlexClark no need to torture yourself :-)

Get some sleep.

is ellyptic curve stronger than triple aes ?

Hey @SohamChowdhury I am free now.. I figured out your question

@BalarkaSen though I dont understand much of that talk .. It feels very interesting.. Since it talks about incidence combinatorics ...

@Rememberme show me.

yup

@AlexClark eh, that makes two of us. :P

this is so cool.

($t \overset{\lambda_r}{\longrightarrow} rt$)

so we know that (from the description of the group)$\forall ,a=a^{-1}$ left multiply by $b$ that gives you $ab=a^{-1}b \implies ab=a^{-1}b^{-1} \implies ab=(ba)^{-1} \implies ab=ba$ . Hence the group is abelian

right, not left.

good.

Ya right... My dyslexic problem.. :p

don't misuse that word. it's not funny.

I know that ...

I literally mean my dyslexic problem

you're dyslexic?

I was dyslexic... Went for some very hard counseling and was revived

Why did I always said to Jasper"Fight you can win", I had gone through that state

Well this is going off math.. So i don't want to discuss it anymore

oh. apologies.

No worries :)

I hate people who go "oh im sooooo ocd rite????" all the time, that's why. anyhoo.

So was your proof also the same as mine?

i prefer this: $(gh)^2 = e = g^2 = ghhg = (gh)(hg)$. cancel, profit.

doesn't really matter.

it's ultra-trivial.

@Soham can you refer me some books for olympiadish problems?

you hate them toh.

I have to gear up for ISI... Though I can do most of their questions but who knows what might come

ah.

Engel is awesome.

I will surely look at it

I do a few problems on the bus everyday.

This should be shown to all those people who hate maths

hi all

hey.

shower and then study rings, hopefully get to modules today. bye for now.

Hello :)

My name is Alice x

<3

I love mathematics :)

hi @Alice

Being online scares the **** out of me :/

you know your name appears automatically?

:P

hi chat

I'm terrified of discrimination for being transgender.

This isn't the right forum for such a conversation though.

I'm sorry.

I can't hide it any more.

It's really depressing :(

So yeah . . .

But hey: at least I'll be happier in the long run, right? :)

I just hope my ideas are seen in the same light as anyone elses . . .

*else's./

I <3 Maths :)

I'm sorry.

@Alice I understand your feelings but please stop this... Others might not happen to like this behavior of yours

Okay.

@Alice this is the wrong place to discuss anything but math

I'm sorry.

however, if would like to discuss math.. go ahead!

What do you know about Algebraic Groups?

(I don't know what's with those capital letters . . . )

it's best just to ask the question directly and then someone can answer if they want to

I know.

I'm sorry.

This is stupid of me.

Let's move on.