Is it true that for $T$ a symmetric operator on Hilbert space, the map $(T-aI)^2 + b^2I$ is surjective?

I'm trying to prove the spectral mapping theorem for polynomials in the real case because in class we kinda cheated and used FTA

Oh also in the above, $a$ can be whatever and $b\ne 0$

If it's at all helpful I have proven it's injective, and thus has dense image, so showing it's a closed map would suffice

I'm confused, Demonark. In the application I would expect, your $\lambda = a+bi$ and you're doing $(T-\lambda)^*(T-\lambda)$, so it's not surjective when $\lambda$ is an eigenvalue.

It's only operating right now on a real Hilbert space so I'm not sure that will hold

This is actually the proof I give in my linear algebra book (in the finite-dimensional case, of course). This operator is singular.

hello chat

Heya @EricSilva.

so i just got stuck in an elevator AMA

what kind of elevator?

I used to have fears of getting stuck in the elevator in Evans Hall at Berkeley (going up 11 floors), so I tried to walk whenever possible.

idk how to answer that question

@Ted the stairs were locked :(

Isn't locked stairs against all sorts of codes/laws?

i assume it probably is

the elevators help button also didnt work

Well, we live in unconstitutional times altogether.

and the open door button didnt work

Yikes. Glad you survived!

i just had to wait in there until someone called the elevator (which was 45 minutes luckily

it was bad

do a john mcclane

@TedShifrin Depends. A lot of the buildings on my campus have stairways that are basically outside. Access to the stairs from the ground floor is typically impossible, but you can get into the stairwell from higher floors and go down to the ground.

Heya, mr anon!!

heya

I don't like it, @Xander.

Is your eye OK, anon?

works fine

Great. When you didn't respond, I was afraid you'd gone blind.

maybe even better than before, as my left eye was slightly blurry compared to my right one before the injury

volleyball = eye surgery

Oh, wait, you did have surgery?

no

OK.

I got it.

I wouldn't try that theory again ...

lol

You still teaching this year?

yep. trig and calc, plus an extra thing with promising HS students (a bit of honors calc / fun probs of my design). and a seminar algebraic geo thing

Holy cow. This sounds like a full-time job. Did you apply to grad schools for the fall?

nah. holding off for now.

Wow.

it's not too much of a job really, since the lectures are planned out and the students work is checked by computer.

the HS students thing and the seminar thing I design the probs/lectures though

My teaching for AoPS is all planned out and I hate that. Especially when the people in charge make less than intelligent choices for what's reasonable for the students.

yeah

I've written out a bunch of supplementary problems that are more geometric.

You must be enjoying the HS thing, though. How did they round up the students for you?

I pointed out we could explain multiplying out (a+b)(c+d+e) by writing a 2x3 array and then combining terms at the end, but my boss vetoed that cuz it's not in the ebook

by reaching out to high schools for bright students. many of them are taking uni classes early.

they are brighter than all of my actual uni students.

I occasionally had good high school kids in my Spivak (Calculus with Theory) or Multivariable Math honors courses.

Yes, sure, of course they are!

Just like most of the undergrads taking grad courses are better than most of the grad students.

I'm amazed they gave that group to you rather than one of the faculty members who might have enjoyed it.

also, our department just heard research / student engagement talks from five prospective new faculty members over that last two weeks.

Oh, is the high school thing not a whole year-long actual course?

no, it's not. although for a certain segment of students it's an ask-questions period for the uni number theory & crypto course, which I wrote the problem sets for

(actually saw my problems posted on MSE at one point, made me lol)

@TedShifrin my honors analysis class had a high school junior in it

she was unreal

There was a 13 year old in some grad course I took at berkeley. Lol

Heya @Antonios

hi @TedShifrin

still haven't had time to think about that lin alg problem :/ sorry

this rep theory homework is quite challenging.

I have had a number of high school kids (or younger). Probably the most amazing was the son of one of my colleagues, who took the Spivak course from me when he was 14-15, then algebra the next year, then diff geo while he was in graduate stuff. He went off to MIT, ultimately a PhD from Princeton.

That's ok, @Antonios. I worked on it a while, then emailed the kid, who's not even replied. I'm pissed off at that.

@anon: Yes, a lot of my diff geo problems have shown up on MSE ... even when I was teaching. There's a guy in Brazil now posting some and needing copious hints.

@TedShifrin It seems like math ability is more a function of years of experience than age. A 13 year old that started young can easily be better than the average highschooler/undergrad, it seems.

Do you agree with that?

and that's annoying that they didn't respond.

There's a 10th grader in my AoPS class who's incredibly impressive. There's a lot to be said for native talent and passion.

native talent probably has a nontrivial amount to do with early childhood socialization and circumstance

to be sure. Seems like you can get quite far at a young age if you just have proper training, too, though.

[not that that's necessarily what's best for a child]

@EricSilva: Certainly in the case of my student at UGA, he had two very gifted mathematician parents. I don't know about my current student. My parents didn't have anything particularly mathematical in their backgrounds, but there's no denying my sister and I were raised in an intellectual household.

I've always believed that decent teachers and hard work can get anyone through a good portion of mathematics curriculum. But that's different from just sheer gift.

seems like one can get through undergrad just fine on hard work

i mean what you end up good at probably has a lot to do with the preferences you form very early on cause those play a role in how you form habits. I think moooost talent (aside from utter freaks of nature maybe) comes from that pivotal time, which is why being the child of academics makes you way more likely to be an academic

@TedShifrin yeah i agree with this. most of the rhetoric around mathematics and giftedness is kind of bleh, we have too many bad teachers

@TedShifrin just the person that I need

We also have way too many people pushed into AP courses and unqualified teachers teaching many of those.

Why's that, Karim?

@Antonios-AlexandrosRobotis I disagree. Back in home country, people didn't care about education and I did 2 years of bullshit art things. I didn't know any knowledge in Science or math. I studied them by myself learned programming and reverse engineering and did SAT and AP by self-studying

i.e. I think being a nerdy kid probably made me better at nerdy things and it snowballed and now im like p decent as a student

and aced them and graduated with high-honours in math, but definitely I didn't do math in highschool

@TedShifrin how are you

@Adeek i dont think this contradicts anything @Antonios said, in fact i think it supports what he's saying

Better than the last 36 hours. I had to have a root canal done (through a crown) earlier ... but my tooth has stopped hurting like crazy :P

Karim: You have incredible amounts of self-motivation, as do most of the people who spend hours in chat here. But I would argue that's atypical, based on my century of college teaching.

@TedShifrin I need to do new eye glasses test. I think the glasses that I have is making my eye sight worse

yeah self-motivation I would say is key to doing Science

I hope one day people can teach their kids to be self-motivated

as everyone has potential from my opinion

I go to the eye doctor on Wednesday. My eyes are definitely deteriorating. ... But I'm ancient.

Anyhow, was there something mathematical?

Hi @PVAL.

hi @ted

can a noncompact manifold admit a nonexact symplectic form? my guess is no

of course

yeah just want to check with you a proof that I am reading

T^2 \times C^n

i was trying to find one on the cylinder but couldnt

or S^2 \times C^n etc

hmm

@Adeek is that Stein

H^2_dr better be \ne 0

@TedShifrin so the issue I wanted to ask I agree with the whole story that this is analytic

Programmer finds himself in need of understanding math and I was depressed because I was too dumb to understand math behind required math for my program

or of course you'll fail

but I am back and opened my book to study some math!

That's why the cyllinder wont work\

but I am not sure what about the point $z_0$ itself

I think that the integral at point z_0 itself still defined and analytic right @TedShifrin ?

ahhh @PVAL there's the key i was being dumb thank you

or cylinder

Karim: That's the point of the proof, yes.

It's one of those.

I see can we say the picture is like as follows

@ted I bet you'll outlive your root canaled tooth.

so we start with the point $z_0$ and draw lines from it surrounding our open set

and as we fill everything out it becomes an analytic function right @TedShifrin ?

The way I'm deteriorating, I can't argue with you, @PVAL. I have already had 2 implants done.

I think that is the correct picture ?

I have no idea what you're talking about, Karim.

I was wondering if we can visualize this extension ?

I actually prefer to prove it thinking about Laurent series. ... Where in the proof did he use that $f$ is bounded on that punctured disk? What happens if you try to do this with $f(z)=1/z$ and $z_0=0$?

oh yeah it is over here

You just need that the integral of $f$ over a small circle around $z_0$ is small

There's an exercise where you prove it under the assumption that $f$ blows up slower than $1/(z-z_0)$

oh

I see your point @TedShifrin

@TedShifrin You end up with a $2\pi i$ that doesn't go to zero as $\epsilon$ does.

0celo, I'm asking Karim for his understanding, not you!

I have an exam Wednesday!

yeah I see your point

yeah I like laurent proof than this one

I just checked my lecture notes from when I taught out of Stein/Stakarchi. I did this proof using Laurent series ...

ohh

do you have your notes online ?

I mean on pdf ?

No, just handwritten scribbles.

So did you try the question I just gave you? What happens with $f(z)=1/z$?

You still get an analytic function when you have a continuous function on $C$.

Is there an easy way to compute the metric in the round $S^n$ around some point $x$ in geodesic coordinates that go up to the antipodal point?

normal coordinates w/e

That's what the usual spherical coordinates do for you, more or less. The spherical coordinate $\phi$ is geodesic distance from the north pole.

is $\phi$ the one that goes from $-\pi$ to $\pi$?

No, $0$ to $\pi$.

I'm not European or a physicist. :P

if f is bounded on the puncturned disk then I think we approach zero

I'm both according to some people.

So in spherical coordinates the metric is $d\phi^2 + \sin^2\phi \,d\theta^2$ ... and that's geodesic polar coordinates at the north pole.

Oh.

The circumference of a circle is not $4\pi$

whoops

@TedShifrin mhm, thanks

The circumference of a circle is $C r$, where $C$ is some constant that noöne can ever remember.

I'm not following you, Karim ...

I'll be back in a moment. It's martini time.

man... I wish I had gin. :(

@XanderHenderson What's the best way to say "here's a calculation that I don't want to do, but can be done"

I have moderately good vermouth at the moment, and some killer olives, but no GIN!

without saying "as the reader can check..."

ohh wait you will not get that is undefined you will get that it is $2\pi * i$ we could extend things using cauchy formula

@Eric Gromov proved that as long as an open manifold has an almost complex structure it has a symplectic form for each class in H^2.

@0celo7 The circumference of a circle is $Cr$, where $C$ is a constant that can be determined by standard computational tricks.

No, for something else

and I'm serious

"It can be shown that..."

?

yeah but that doesn't give you a sense for how difficult the proof is

"After a long and tedious---but fairly straightforward---computatation, it can be shown that..."

nice

thx

heh

if I say tedious no one will call me out on it

I would never put that in a peper

it's not a paper

oh, well then

go for it

I have a long, tedious computation that didn't fit on the page the right way

I mean, I know how to do it. But I really don't want to do induction and all the other crap I'd need

so I have turned it into a figure, and rotated it 90 degrees

lol

found it!

@Xander: I'm exceedingly picky about vermouths. My favorite I can only get up near where you are (in Costa Mesa).

I'm picky about gin, but have not yet found a vermouth that I am really happy with

and I am super, super picky about my whiskey

@PVAL: That's an interesting result.

It aint an easy one.

I am picky about gins, but I usually have about 10 in stock :) ... Vermouth — the best is Boissière.

It's in the book that at some point @Balarka was claiming to read.

I have way too much scotch in my cabinet right now

I only drink scotch in the winter, and it's never winter in CA.

lol

heh

heya @Faust! :)

@XanderHenderson AH

(Eliashberg-Mishachev)

I found a ref for the calculation

yay!

yeah, horrible integrals.

i cant stand scotch i have a bar at my place with around 400 bottles of alcohol none scotch. thought many of everything else

scotch should be old enough to drink itself!

I'll have to see if I can find the result/proof, @PVAL.

@PVAL-inactive I read reading that book

@Faust: A serious student can't be an alcoholic :P

Pretty hard

I enjoy lots of young scotches.

@TedShifrin i drink once every 3-6 months

@PVAL-inactive o dam

Wow ...

@PVAL-inactive reported.

There aren't many 21yo scotches around here under $200, and the ones that are aren't usually the greatest.

@PVAL-inactive Yeah, I blow about $3-400 on a bottle once or twice a year

then drink very slowly

gimme some examples so i can be envious.

I definitely don't pretend to be a scotch aficionado. (I always have trouble spelling that word.)

I have trouble spelling all words.

@XanderHenderson take a look at this awful font arxiv.org/pdf/1701.01460.pdf

I've got about half of an Auchentoshan 24 year old left

Spelling is eazy

sounds delicious.

and a rather nice Glenkinchie

and a couple of random bottles of cask strength scotches

oh, and a somewhat young Glenfarclas which is quite good

Demonark: Did you sort out your spectral issue?

(it isn't allowed to drink itself)

Is that like the barber who shaves herself?

yes, very much like that

I think if I bought scotch that expensive I'd probably just drink it.

actually, it could probably drink itself in Europe, but not in the US

@XanderHenderson youtube.com/watch?v=Rr8ljRgcJNM

I'm trying to convince the wife that we need this in the house: thewhiskyexchange.com/p/9335/…

The scotch I'm most happy about owning is a 11 yo

its a cask strenght first fill sherry ledaig from signatory.

nice

okay, dinnertime

my favourite alcohol of all time is actually a bottle of 30 year rum thats not particularly expencive

Probably the most expensive scotch I ever had was like some 20 year old IB laphroaig.

Heya @MikeM

@Faust: What's new with you and math?

honestly i have only tried scotch 4 or 5 times but everytime its been terribru

terribru = terrible + brew ... I like that :P

A lot of the lower end blends have a significant amount of neutral grain alcohol.

@TedShifrin it's a meme memegenerator.net/instance/38297185/impossibru-guy-terribru

mathb.in/22556 this was an exam question I had today that I totally screwed up bc I didn't read it closely but now I'm trying to figure it out...any suggestions? I think the answer is yes but I'm not sure why

@TedShifrin gongshow midterms all over the place assignment everywhere recently went through almost 100 research papers trying to find something on number theory that i wanna write about

I didn't think Faust was a meme regurgitator, but perhaps he is.

being around Balarka has that effect

@kanderson8: Good question. Why do you think it's yes?

what a meme?

that guys face is retarded

@TedShifrin sadly everything intresting is too dam long

@Faust: I only know uninteresting things :)

I mainly think it's yes because I can't come up with any counterexample...any interval will sum to larger than one. The set of (1/2^n) won't sum to larger than one but it's countably infinite so it doesn't work as a counterexample

Right. If we have a convergent series with sum $\le 1$, we can certainly have a countable example where this holds.

This is sort of tricky, @kanderson8. If you have an uncountable set $S$, for any $n\in\Bbb N$, there must be uncountably many elements of $S$ in some interval $[k/n,(k+1)/n]$, for $k=0,\dots,n-1$.

I think I thought of this question and answered it while TAing some calculus class.

I've never thought about it before, @PVAL. It's a cool question.

(I'm assuming $S\subset [0,1]$.)

Obviously, otherwise, one element of $S$ will mess us up.

@kanderson8: Do you agree?

yeah just chop up R into countable pieces and use the fact that a countable union of finite pieces is countable.

so one of the pieces is infinite and you win.

I'm not quite following.

We may as well assume $S\subset [0,1]$ to start with.

Yeah I agree, I just don't know think I would've came up with this on my own

This is a sneaky problem, @kanderson8. You see how to finish?

If I were teaching real analysis (which I assume this came from), this would be my problem to challenge the best students ... unless I'd discussed this already in class.

should i stop talking probably

?

@PVAL: So you're considering the partition into $\{0,\dots,1/n,1/(n-1),\dots, 1/3,1/2,1\}$, rather than what I did?

I guess it's sorta equivalent.

well

the point is mine are bounded below

I deleted to not give away the answer.

not because I thought it was wrong btw

I think both ways work.

But, @kanderson8, I've been around for a long time and this is the first time I've seen this question. So you shouldn't think it's just "totally trivial" and you're "stooopid" for not seeing it on a test.

I don't see how you excluded the (0/n,1/n) interval

which of course doesn't tell you anything if thats infinite.

Oh, right, more than countably many in there isn't yet a contradiction. Although it is.

Right. You win.

Yup, this problem is very sneaky.

@kanderson8 do you want the answer?

I'm not sure i have a good hint.

Well, I already gave the hint, I guess, for yours.

Very cool question, @kanderson8, and very clever, @PVAL. I bow. :)

note the sequence a_0=sup(S), a_{i+1}=Sup(S-{a_1,...,a_i}) actually gives a construction of such a series.

but the proof (that I know) that that series diverges feels equivalent to the same proof.

You mean a series contradicting the bound?

yeah

Wow, I wish I had known this question to have Spivak put it into his book. :P

@TedShifrin: I'm not sure I'm clear on where to go next...I can pick an interval [0,1/n] which is uncountable but how do I show it won't sum to something larger than 1?

@kanderson8: It will, of course. But @PVAL is right that my hint was slightly insufficient. So listen to him, please.

write $[0,1]= \{0\} \cup_n [1/n,1/{n+1})$

You mean $n-1$.

And then $\{1\}$ at the other end, if we care.

as a countable union of finite sets is again countable $S \cap [1/{n+1},1/n)$ is infinite for some n

Ugh, you left out parentheses. Very confusing.

I don't see the code.

huh

You mean $[1/(n+1),1/n)$ for $n\ge 1$.

what parentheses

yeah

And then $\{1\}$, I suppose, not that it matters.

err

Anyhow, I'm gone.

then $(1/{n+1},1/n]$

so anyway take a series with all the terms in $(1/{n+1},1/n]$.

@0celo7 here ?

can we discuss some small analytic thing

I don't understand that thing about IVT

can someone discuss this with me little more ?

looking

If n_t is not constant, you can find two different values

Say it achieves both 6 and 7

I don't understand why would imply the existence of some $t_0$ with $n_{t_0}$ not integral

is it from the book "winding number" or something like that?

Well, it's a continuous function of t

So it has to hit 6.5

yeah sure

I agree with you

but what about this integral situation

But that's not an integer

I don't understand that

oh

yeah

"integral" in that usage is integer

I thought integral as integration

yeah that makes sense

Not like, $\int$

yeah yeah

that is silly of me

hahaha

I propose calling integers (the numbers) zahls like the old German masters did

ich kan nichttt

lol