@AsafKaragila A year? Congrats.

Thanks. It seems that just yesterday we were a day short from a full year and I was on this very chat...

Very good.

Lewis Carroll (Oxford mathematician)

Mind helping me with som elementary linear algebra?

@N3buchadnezzar i'll give it a shot ( but i'm sure so will everyone else in here)

...wll, too)

And away we go. See you guys later, or tomorrow... whatever.

@AsafKaragila bye

"Prove: If V and W are finite-dimensional vector spaces such that dim(W)<dim(V), then there is no one to one linear transformation $T:V \to W$"

Does anyone have that mathjax link?

@asaf, look at the bases elements

Hint: Injective linear transformation converse linear independence.

sorry, that was @N3buchadnezzar

@EricGregor I will make sure the waiter doesn't bring me any of those. Thanks!

:P

@AsafKaragila, would you possibly be able to help me with a manifolds problem?

No way. Not even if I would have stayed (I am really about to leave)... I know nothing about manifolds.

HAH!

i understand. just thought i'd ask

Well. Now I'm with my coat and whatnot. So this is for real, bye!

tchau

I thought that for a linear transformation to be one to one, we have to have dim(V)=dim(W)

of course not. consider the inclusion of $\mathbb{R}^2$ into $\mathbb{R}^3$

that is certainly injective

@N3buchadnezzar Well, I can imagine the answer: It probably has to do with that the $dim(W)$ is the number of LI columns, say $n$. Therefore, it can only affect $n$ components in $V$'s space (where $dim(V)>n$). As a result, a vector in $V$ could "point to" more than one vector in $W$.

@AsafKaragila ...which I think is similar to @Asaf's comment

That's the most I can help without breaking out my old textbooks.

Thanks I guess =)

@N3buchadnezzar heh. did it at least make sense? even if it's not too rigourous?

it did make sense, buuut I had the same idea. but problems with the rigour :p

oh. hehe. same problem i always have in class

I usually have problems in lectures with the man throwing theorems at me and the pretty girls distracting me. The combination of those two seems to be lethal.

@N3buchadnezzar ditto! plus, the theorems come so fast and they build on each other during the class...

i mean, it's one thing when they intro a theorem this class, then next class they build on that. it's a whole new ballgame when the theorem from this class builds on an earlier one from this class!

:D

@N3buchadnezzar right... except no pretty girls ever waived at me (we didn't have too many pretty girls in my math school)

Neither have we ;)

and the one i liked the best just got engaged 2 months ago :(

what does the circle between vector spaces mean ?

if there were a linear transformation such that only $T(0)=0$, then the images of the basis vectors of $V$ shouldn't have dependency relations. but they do

isn't that sufficient?

@N3buchadnezzar i can't remember the word, but isn't that when one function's input is the output of the other function? you know g(f(x))?

it is thanks

It was stated in theorem 83.1 ...

theorem 83.1?

@EricGregor In my book

I think my book was infected with theorems, the book is filled with them, and they seem to multiply.

what the heck is the name of that... darn

so you are reading chapter 83 huh? wow! :D

Now I only need to figure out what the dilation means, than I can perhaps understand what they are asking for.

jeff, composition?

@jeff

@EricGregor ty

@KannappanSampath I'm here and will be till late. If you want to talk, just ping me.

hi hi hi!!!

Hi!

@ymar I can haz halp plox?

@N3buchadnezzar Can haz. If I'm able to help.

@ymar It has to do with vector spaces

What is wrong with them?

Everything! they hit me and laugh of me!

@N3buchadnezzar :)

Tell them their momma is in the field.

So what's the problem besides that?

YHAGGH

I deleted what I just typed

I tried to write my problem here second try

Let $T_1: V \to V$ be the dilation $T_1(v)=4v.$

I dare someone to tell OP "it's obvious."

Find a linear operrator $T_2:V \to V$ such that $T_1 \circ T_2 = I$ and $T_2 \circ T_1 = I$

What is your field?

huh?

@anon No problem!

R_n

OK.

You mean R. Then just v->v/4, no?

@anon LOL

I do not know, it is not given in the problem

He actually did it. We've got a badass here.

Perhaps they mean generall vectorspaces. :/

What's not given? The field? If it's not given, then the solution may not exist.

well, you can't do it over char 2, but otherwise division by 4 is possible.

@anon why don't you do it in here? you could say [ATSIGN]ItsNotObvious: It's OBVIOUS!!!!

@ymar I wrote the problem as stated in my book

He isn't a regular here, I'll leave him alone.

@jonas wtg

So anon gave you the answer.

"well, you can't do it over char 2, but otherwise division by 4 is possible. " ?

I have 10 starred messages somewhere in here????

oh. I see

Where? How do I see them?

@Jeff they are in my basement, you can come around to look if you want

I think you have to manually Ctrl+F through the pages of 1836 starred messages.

It includes free candy and chloroform as a bonus.

@N3buchadnezzar could you just scan 'em and send 'em to me?

@anon Then they become ItsNotObvious...

@N3buchadnezzar Do you know how to solve the problem now?

@t.b. That's exactly the thread I was thinking of :)

@anon where are the starred message pages?

Read closely the links on the right sidebar; one says "show all 1836"

Indeed

thanks

@anon @tb actually, i like that guys username. i mean, haven't you all, at one time or another, read a math text or attended a math lecture where the author/speaker said "it is apparent" (or something like that), and you had no idea why? I sure have!

@Jeff Aaaall the time, is it not obvious?

phew

Tomorrow I'll buy an ergonomic keyboard. 8-).

@N3buchadnezzar hehe

Incidentally, my frustrations with authors usually occur where they don't even bother with an "it's obvious," they merely assert something.

@anon i tried the first three pages and didn't find much.

why do people star a message in chat?

@N3buchadnezzar hehe

it's like special upvoting, I guess

@Jeff that is stilla mistery, just look at the starred messages.

@JonasTeuwen i tried one of those once. didn't like it

Asaf obtained many votes for drinking and JM for leaving.

I'm trying to remember why it is obvious that you should come up with that series if you think hard enough.

stars mean someone found a comment especially funny or helpful or insightful. or they were smashed and had a completely irrational thought process.

@Jeff If I don't, my hands will become completely useless.

@anon or Asaf forced them to star messages.

and sometimes I think people star messages arbitrarily just to spite the star gods.

@anon ha

Stares at the starred "hi" comment I guess you are right

Wow, I didn't notice that.

@anon: it's 3k you need for voting to close

Let $T:R^3 \to R^3$ be the orthogonal projection of $R^3$ onto the $xz$-plane. Show that $T \circ T = T$.

Any tips?

That's obvious right?

What does projection mean?

What is the projection of something already on the $xz$-plane?

ops

do you mean R^3 to (an embedded copy of) R^2?

@N3buchadnezzar No problem. It took me a while before I noticed that when you think mathematics goes much better.

Is not projection a linear transformation P from a vector space to itself such that P2 = P. ? =)

@N3buchadnezzar That's one definition.

Sorry for asking such silly questions, I find the booki hard to read and I am a beginner...

@anon Its OBVIOUS! :D

Yes, that is the best definition.

But, you can obviously see that it is the same.

I tried procastinating by proving the weierstrass product formula for sine, that did not work out great.

@JonasTeuwen Better -- yes. Much better? Depends on who's thinking.

I don't understand my own answer from two years ago anymore.

lol

Imagine this

@N3buchadnezzar That's obvious.

I've looked at the LaTeX in some of my earliest answers. What an idiot that guy was!

When you die, you meet a former version of yourself every ten years. So if you lived to be 30, your 10 old self and 20 old self would meet you.

@N3buchadnezzar You should use $\sin(\pi x) = \frac{\pi}{-x \Gamma(x) \Gamma(-x)}$.

@JonasTeuwen I am trying to prove that formula!

:p

Oh, right. Define $f(x) = \Gamma(x) \Gamma(x - 1) \sin \pi x$.

ah! why did you take a -x outside of the $\Gamma(1-x)$?

@anon To mess with his head.

@JonasTeuwen Could not fool me, it is not that obvious.

@N3buchadnezzar And then it is obvious that $f(x + 1) = f(x)$.

So, that's cool. What does that tell you?

That f is constant?

No it does not.

Periodic =/= constant

Or does it?

what the holy dicks...

I hope his teacher hits him... :/

@JonasTeuwen So f is periodic.

Why does it surprise me that $\Gamma(x)/Gamma(x + 1) = \frac1x$?

Oh.

No, it does not surprise me.

@JonasTeuwen its obvious.

Yes, it is.

I'll blaim my wrists.

that's just xG(x)=G(x+1), which is just by-parts or whatev

@JonasTeuwen wrists make you forget functional equations? :)

@anon No it is done by obviousness not parts =)

No, the pain does 8-).

@anon Aaaah. I do know that! I was biking and trying to divide those integral definitions, then I thought: "I'll plot it when I'm home to see what it is!", then I just did this and then I was like: Huh!

@JonasTeuwen So we could plug in x=\pi/2 for an example ?

And only when I wrote it down I thought about it.

Hmm.

I do not know if that will do much good.

Don't overload the pies. Try x=1/2 for something nicer.

so

$f(1/2) = \Gamma(1/2)^2$

What, just fill in $x + 1$.

whew

Hi, Matt.

@JonasTeuwen Hey, what sort of pain is it? Since last week the usual wrist pain I had from typing seems to have spread to my hands. At least some of the time. Do you have the same?

@JonasTeuwen Yes, that will show that $f(x)=f(x+1)$, but plugging in $x=1/2$ gives what I wrote above.

Hi ymar!

Hi, Benjamin.

@MattN Yes. And my lower arm.

Crap : (

And in the morning my neck.

Too.

Shoulder, neck.

And a numb feeling.

: (

Then the numb feeling goes away and pain comes into place 8-)!

@JonasTeuwen comfortably numb or?

What do you do about it?

hi

Well, now I cannot ignore it anymore!

I'll after I gave a lecture buy an ergonomic keyboard. If I have to wait until the university does it...

@BenjaminLim Did you find that extension degree?

@ymar In the link you gave me yesterday Patrick tacitly assumed that $\Bbb{Q}(a) \supset \Bbb{Q}(\sqrt{2})$. How does one know that?

@MattN If it is the same, I can guarantee you it will only become worse! 8-).

I need to do something now before it gets worse.

I really need my hands.

But without pain.

There are things you can do about the pain, but they stop working too! 8-)!

The crappy thing is that since last week I also get it when I write not just when I type.

Yes.

@BenjaminLim Wait a minute, I have to read what he wrote.

I can write for 2 minutes maybe.

This sounds as if we had the same.

Cool. That gives some kind of bond.

Lucky me, I'm friends with a hand surgeon.

Going to have this checked out.

Would a friend operate on you? 8-).

@MattN He could give you a helping hand or two ;)

Until now I only had it from time to time and only in my wrists so I just ignored it.

@JonasTeuwen You crazy? I'd never have it operated. But they can have a look and tell me what it is and how to prevent it from getting worse.

Okay, the order is like: wrists, then hands, then arms or fingers.

Then it does not go away anymore.

You guys talkin' about repetitive. stress injuries?

Hmm, carpal tunnel surgery is not that bad of what I've heard and it can completely get rid of all the symptoms.

@MattN Do you still have the same power in your hands?

@J years ago there were mixed reports about that surgery. maybe it's improved

Hello everybody, if you can upvote my certainly rate as good question recently, i would be certain that you will be friend of mine, America and Isreal, since this is much more like a disaster as much as the isreal lose the palestine

Nah. I don't trust people. They're usually incompetent. So it's bound to be screwed up. I'd rather have pain than screwed up hands.

@MattN The risks are quite low.

@JonasTeuwen I think so. But I'm not sure how to test it.

The right is much worse than the left hand for some reason.

I think you will get back from that once it gets worse enough that you can't sleep of the pain.

Are you right-handed?

Yes.

@BenjaminLim I think you're right. That should be checked. I accepted his answer without reading it very carefully, because I already had my problem solved and I didn't want to leave the question without an accepted answer.

Does that surprise you?

@Victor: Please leave the politics out of math. I commented on your question trying to explain it a bit, but it's not a good idea to go around soliciting upvotes from strangers. When life hands you lemons you make lemonade or move on.

@BenjaminLim But I check that in my question.

@JonasTeuwen Yes since I use them both the same when I type and I thought typing was the killer.

Also, that flag made no sense.

@MattN Mouse!

@BenjaminLim Do you understand that part?

@JonasTeuwen Trackpad... on laptop. I don't use a mouse. But yes.

Laptops are horrible.

Your wrist should be flat.

I know. But that's hard to implement.

For the love of monkey's take a separate keyboard.

I never use a laptop anymore without a separate keyboard and screen :-).

So why do I have one, you could ask.

Because then I can take all my stuff!

have you seen this

Deserves an IG Nobelprize for mathematics.

@leo a few more of those are mentioned after this comment

classic games are really hard!

Hi Teddy.

hi teddy

see you all!

bye

Bye leo.

Hi all

hi teddy

bye teddy

@MattN Whatever you do, don't take painkillers too often. They give you some kind of permanent kopfschmerz!

@JonasTeuwen Which did you take?

Oh my.

You have 8 hours to live.

Cyanid?

Drugs are bad, mkay?

I'm not taking anything at the moment. The pain is not permanent.

No, I was more specific. Cyanid is not a painkiller.

Good.

Well, in a manner of speaking..

@JonasTeuwen Cyanid kills the pain, eventually.

@tb How's the head?

Yes, but I do believe that killing someone does not really qualify as a "painkiller".

@N3buchadnezzar But not without increasing it for a while.

mary owes me an accept.

Maybe I'm just a whiner. It is only a little bit of pain.

Let's see!

@MattN a bit better. Still not good, thanks for asking. :)

@MattN Which question?

@tb What's that 3 days in a row? Or more?

@JonasTeuwen Euler method.

@JonasTeuwen i had some pain in my elbow. went to a Dr who said the description of it didn't sound like RSI, but like regular old tendon inflammation....

So you're not really hip are you.

i made it go away by taking a break and then making sure to use better form when mousing (not good form, but just doing better form was enough).

You should work harder.

@MattN third day.

I didn't get to sleep that much...

I saw.

3.5 = 0?

@tb that's not good! how can you do difficult math without a clear mind?

Q/(7/2)Q

@Jeff Practice!

The problem isn't doing difficult math inebriated. It's figuring out what the hell you were doing when you get sober again. :)

@JonasTeuwen well, if you can do it that tired, imagine how good you'll be when you get some good sleep!

@anon i thought that's what happens when you do cerebral stuff while stoned?

@Jeff That's why I stick to the easy stuff :)

Good answer : D

Yes, that is also why I do analysis.

@JonasTeuwen What would you do otherwise?

@tb 33k reputation of easy questions? i dunno... i've been looking for easy questions for a month now.

Me? Make coffee. Sell wine. Be a whisky critic.

@tb Thanks for earlier today btw.

@JonasTeuwen That's what you'd do if you had enough sleep and a clear mind?

No, if I wouldn't do analysis!

hey

Have anyone here taken galois theory ?

@Asaf 's mother probably did.

I had it by Hendrik Lenstra. Is seems like he still teaches the course: math.leidenuniv.nl/~mkosters/algebra3

@JonasTeuwen She has probably taken everything, but on a more serious note?

I was thinking about taking it

@JonasTeuwen Is it strange I can read that just fine ?

Hey David! I didn't understand what kind of function you were looking for earlier.

@N3buchadnezzar No.

@JonasTeuwen And our courses are in english... Later on.

@David: You can't have a continuous bijective function $f: \mathbb{R} \to \mathbb{R}$ such that $\lim_{x \to \pm \infty} f(x) = 0$ by the mean value theorem. But then you mentioned the Bell curve, so I was puzzled. Would $f(x) = \frac{1}{1+x^2}$ do?

This is a bachelor course, so yes.

wow^^

The MSc courses are in English.

Well yeah, it is technically the same here. I guess

@tb Yes, the bell curve didn't make much sense, and my question was not as clear as it could have been. Let me get my thoughts together and rephrase that.

third year

What's up with Victor again? He's back at his NARQ habit but this time worse than it was when he first showed up here.

@JonasTeuwen No she didn't.

Is Victor Israeli?

We have all these crazy Israeli's here.

I don't know about his questions but this doesn't make sense to me.

@JonasTeuwen No, that's porton.

@tb What I'm really trying to do is define a homeomorphism $f:\mathbb{R}\to(a,b)$ where $a$ and $b$ are in $\mathbb{R}$ and $a<b$. My original idea was to try and find a function $f(x)$ that goes to zero as $x$ tends to $\pm\infty$. That way I could just have $f(x)+a$. Of course $f$ would also have to somehow approach $b$ as an absolute max for all $\mathbb{R}$.

@AsafKaragila He said something about Israel this evening... I think.

For some reason I can't render MathJax right now...

@MattN Sadly this actually makes more sense to me than most of his questions... That's not to say it makes any sense.

: )

That sounds like a slight exaggeration.

@AsafKaragila I guess she took "Group theory", if you know what I mean?

@tb But I think I've found a (somewhat complicated) function that works. Would $f(x)=(b-a)x e^{-\frac{1}{2}x^{2}+\frac{1}{2}}+a$ work?

How about adjusting $\arctan$?

Yes, squeeze, that is multiply by $(b - a)/\pi$ and then move it up.

@DavidK First map $R$ to $(0,1)$ via a modified version of $x \mapsto \frac{x}{1 + |x|}$. Then adapt to $(a,b)$. Is that not a homeomorphism?

Is anyone else having trouble rendering latex right now?

There are uncountably many of them. Fortunately we've named a few!

it's getting crowed in here

I love $\arctan$.

@DavidK Works for me.

@DavidK did you click the bookmark? it works fine here, too.

@MattN K. I'll be right back then. Gonna restart Chrome.

@DavidK You can try shift reload.

I'm hungry.

eat

I had a 400g steak, and some of my better half's steak... but I could eat another one.

400g : D Now that's a steak.

@DavidK no

She had a 300g sirloin, it was pretty great.

@tb Did you count any of the two questions as obnoxious, btw?

I regretted ordering the 400g entrecote.

@ymar You assumed that those 6 elements you gave were linearly independent over $\Bbb{Q}$.

@MattN no.

@tb Good. : )

@BenjaminLim I don't think I did.

When solving the linear system you said it yourself @ymar

I was planning to finish my presentation today. But I only managed to write one page. I'm just way too slow : (

Hrmm... Still not working. Oh well.

@BenjaminLim No, I just want to express $\sqrt 2$ as a linear combination of those. I don't care whether they're independent or not.

@ymar I don't understand, you said suppose $\sqrt{2} = a_0 + a_1\alpha + \ldots a_5\alpha^5$.

So then did you not compare coefficients?

@MattN What is it about?

@ymar Which is already using the fact that they are linearly independent over $\Bbb{Q}$?

Fourier transform on finite Abelian groups.

@BenjaminLim How is that using this fact?

@Dav but it's not rendering automatically, like it usually does. i have to click my bookmarklet (or whatever that's called).

@BenjaminLim Take $(1,0)$ and $(2,0)$ in $\mathbb R^2$

@Jeff Mine actually just started working again.

They're not independent, but $(3,0)$ is a linear combination of them.

@DavidK it's like magic

@tb How would I adjust arctan so that it would approach $b$ as an 'upper bound' and approach $a$ as a 'lower bound'?

@ymar Because you are using the fact that the coefficient of each $\sqrt{2}$ or whatever is zero. How do you compare coefficients then?

@DavidK Squeeze it such that it's range only has measure $1$, then multiply by $b - a$, then shift it up/down.

@BenjaminLim Yes, that's what I said. I don't understand where I'm using that some coefficients are zero...

@BenjaminLim Shall we move to the CA room?

ok

Just dropping in for a moment. I am proctoring a final at UCLA.

Bye guys

Wow! I want to go here: conferences.science.unsw.edu.au/IWOTA2012 !

@robjohn Hi!

@robjohn hi, how are you?

@JonasTeuwen Heh, my measure theory professor is in the scientific committee.