Mathematics - 2022-12-21

copper.hat

01:27

i think of a functional as a scalar 'measurement'.

Ted Shifrin

That’s a valid intuition. :)

Akiva Weinberger

07:28

Just finished my final project for my smooth manifolds class

which was to write 8-10 pages about some topic that was (a) not covered in class (b) related to what we covered in class and (c) understandable to an audience of people who have taken the class

I wrote it in far less time than I would have liked. (There were just so many other class's finals I needed to deal with)

It's this if anyone's curious

one potato two potato

07:44

I'm planning to study Riemannian geometry through John Lee's book but... I don't know, it leaves too many theorems as an exercise.

@AkivaWeinberger This reminds me REU project at Chicago university.

Akiva Weinberger

Yeah the textbook for the class was Lee's Introduction to Smooth Manifolds

All the stuff I vomited onto that PDF I learned from do Carmo's Riemannian Geometry

one potato two potato

10:31

peep

arxiv.org/abs/2212.09835

A non-constructive proof of the Four Colour Theorem

Koro

11:04

Discrete metric space on non zero vector space can't be obtained from a norm.

Is this statement true for non zero vector spaces over any field?

It is certainly true if the field is R or C.

Koro

11:16

0

Q: Discrete metric on a non zero vector space is not obtainable from a norm.

This is an exercise problem from a textbook on Functional analysis. Show that the discrete metric on a vector space $X\ne \{0\}$ cannot be obtained from a norm. If the field over which $X$ is a vector space is $\mathbb R$ or $\mathbb C$, then for any $t\in X-\{0\}$, we have $d(t,0)=\|t\|=1$. Note...

Unknown x

11:43

could you suggest a good textbook for studying the basic properties of spectrum of closed operators?

Koro

14:22

2

Q: Find maximum value of $a$ such that the matrix has three linearly independent real eigenvectors .

The maximum value of $a$ such that the matrix $\begin{pmatrix} -3 & 0 & -2 \\ 1 & -1 & 0 \\0 & a & -2\end{pmatrix}$ has three linearly independent real eigenvectors . Please give me a hint Thanks

linear-algebra eigenvalues-eigenvectors

s there any alternative to the answer provided in the linked post?

I believe that there is a shorter answer which doesn't use WA.

Peter

Not enough that people over and over again claim to have proven (or disproven) a famous conjecture , after a famous conjecture WAS proven (like in the case of Fermat's last theorem or the Four Colour theorem) , they claim to have discovered a much easier proof. Not really better.

Koro

15:01

I think that the question is not correct.

No such maximum value of a exists- The characteristic poly. of the matrix A is $x^3+6x^2+11x+6+2a$, which can't have a 3 times repeated real root. So the poly. either has a real root repeated twice say p and an alike root q or it has all distinct roots (real).

we may not that if $a= -1/2( x^3+6x^2+11x+6)$, then the derivative of a vanishes at $\frac {-12\pm \sqrt{12}}6$ and the maximum value of a is at the positive root.

This max. value of a turns out to be $\frac 1{\sqrt {27}}$. At $\frac 1{\sqrt{27}}$, the char. polyn. has a real root repeated twice and an another real root. I checked that with this a, the matrix is not diagonalizable.

(hence does not have 3 linearly independent eigenvectors).

From some other source, it is a multiple choice question whose 'correct' answer is $a=\frac 1{\sqrt{27}}$.

leslie townes

15:46

@Unknownx i don't know that there is a lot that one can say in generality! reed and simon (all volumes) are pretty good for the basics of unbounded operators. kato's perturbation theory for linear operators has a lot of this material too.

Unknown x

@leslietownes Thank you. let me check

leslie townes

the core of reed and simon is vols 1 and 2

Koro

16:18

on rmp, some professors have very very bad rating and not to mention students' comments.

some of the comments are very impolite.

I have seen and interacted with some of them on mse. I think they are amazing and not at all how rmp projects them to the world.

leslie townes

16:46

sites like that generally have a big selection bias problem.

even if you control for that (e.g. schools that force all students to submit evaluations before they can get their grades) there is a raft of literature on how evaluations tend not to be very good at measuring what people would like them to measure. they can be strongly reflective of student biases, and also aspects of satisfaction with the experience that an instructor has no control over.

imagine yelp reviews of an italian restaurant that are basically "i hate italian food, why did someone make me eat here."

Goku カカロット

Are university departments allowed to use other departments to conduct some sort of research/survey using them? Voluntarily, of course.

leslie townes

i'm not sure i understand the question, but i don't generally know the methodology of papers that evaluate evaluations. presumably some of them involve giving different kinds of forms to different groups of people, and maybe outside of the ordinary course of what a school would otherwise be doing.

Ted Shifrin

17:02

Peer review … But that is extremely time-consuming. But I did more than my share with grad students and postdocs … and with faculty I wanted to nominate for teaching awards.

leslie townes

17:13

the schools i went to, whatever their process was, did seem to have it figured out for teaching awards. the people who got them were generally phenomenal. and they definitely didn't do it by computing a running average of student evaluations.

Ted Shifrin

I don’t remember hearing much about teaching awards in my 5 years at Berkeley.

leslie townes

there weren't a lot of them! L&S gave out one a year, i think. maybe one every two years

and it was usually enough influenced by student evaluations that math profs were not in the running :D

which is weird, because, OK, you have this award that is numerically more selective than an NSF grant, and yet, basically no money behind it (maybe a $1k honorarium), no changes in real life circumstances. so if you're junior faculty, what are you going to spend your time doing? maybe ignore teaching and buff up that NSF application.

Ted Shifrin

At UGA there was a permanent $6K raise (that went up after I’d won it) for the highest university teaching award. Up to 4 or 5 across the university each year.

leslie townes

that's cool and meaningful, i like it. at my wife's school it's a one-off $1000. one or two a year, university-wide.

they do allow it to be given to non-tenure track faculty, which is nice of them. i don't think iowa did that when i was there.

i totally get why they don't want to embarrass themselves by giving awards to people who immediately leave, but, it's funny when 'tenure-track' or 'tenured' basically excludes anyone deserving of the award who hasn't gotten it already.

Goku カカロット

17:40

@leslietownes well, for further context, I think the "research/survey" a joint project between education, psychology and some other similar majors. The information I was given says that their goal is to gather data from STEM departments in their study and teaching methodology, behavioral patterns in class, lab and on-campus generally. I thought it was weird, sounds like spying if you ask me

leslie townes

heh, yeah, but basically all research in psychology is something like spying. it has to be.

Goku カカロット

Only reason I said yes is because I'm stupid and curious

leslie townes

i opted in to a few psych studies as an undergraduate, and something involving pedagogy in chemistry.

Goku カカロット

So I'm guessing stuff like this is allowed for research? No one, specifically the faculty which are also part of it, seemed to think anything of it

leslie townes

well, i don't know how they get around the selection effects in those studies, and maybe they don't.

Goku カカロット

17:43

You mean selection bias?

Koro

Hi Leslie, do you have any views on the correctness of this question based on eigenvectors?

3 hours ago, by Koro

2

Q: Find maximum value of $a$ such that the matrix has three linearly independent real eigenvectors .

The maximum value of $a$ such that the matrix $\begin{pmatrix} -3 & 0 & -2 \\ 1 & -1 & 0 \\0 & a & -2\end{pmatrix}$ has three linearly independent real eigenvectors . Please give me a hint Thanks

linear-algebra eigenvalues-eigenvectors

leslie townes

well, yes. who opts into that stuff vs. who doesn't. and would undergrads at a university necessarily be representative of a population that includes all ages of people and people who do not go to any university.

one of my friends does research in social psychology that of necessity involves studies that fundamentally trick the subjects about what they are testing for. the field generally has a huge reproducibility problem and has been the site of a number of famous academic frauds.

koro if you have done the ground work on something like that i would trust your instincts and conclusions. i'm not about to solve a polynomial or fiddle with eigenvalues if there's a chance that someone or even several someones just happened to be wrong on the internet. it happens so often that i don't have the time.

it strikes me as a question that requires attention to detail regarding the solutions of polynomial equations. and people might be inclined to muddle up 'has a real eigenvalue' or 'has only real eigenvalues' with 'has a basis of real eigenvectors'. if you're paying attention to those kinds of details and the other answerers aren't, trust yourself.

Goku カカロット

@leslietownes I agree. Although, if this research or survey is simply limited to testing with in the institution itself then it likely accomplishes its goal, whatever that may be

Ted Shifrin

@Koro Surely the point is that if $a$ gets large, the cubic has only one real root. Find the value of $a$ where there’s a double root.

Koro

@leslietownes thanks Leslie. My observations match with the answer on the post and the answer there confirms them using wolfram alpha as well. :)

@TedShifrin Finding that point is easy in this case. It turns out that one of the solutions of p'(x), where p is the characteristic polynomial satisfies p(x). So p(x) must have a repeated real root.

But that's not the problem. The problem is- the question itself.

Ted Shifrin

17:59

But my brief argument gives all the justification you need.

What do you mean?

Koro

I believe that there is no such a that the question requires.

Ted Shifrin

You mean max versus sup.

It would be unlikely that the max gives diagonalizable, agreed. So the question is wrong.

Koro

Without going into too many details, I'll just give the reference of the answer there- at a= sqrt of (1/27), we lose diagonalizability (I checked this and found that G.M. for an eigenvalue is not same as its A.M.).

Ted Shifrin

Yup. So move on.

Koro

:)

What I find annoying however is that this question is at many places on the internet.

byjus.com/question-answer/…

leslie townes

18:03

wait, what? is someone wrong on the internet?

3

Koro

and here testbook.com/question-answer/…

all give the same wrong 'answer' and here gateoverflow.in/173468/mathematics-gate-ee.

leslie townes

so, you're putting forth a reality where it's almost as if a graduate student in mathematics is better at mathematics than google.

i'm not sure that i can accept that, but OK

Koro

:)

Ted Shifrin

Even more shocking, there are mistakes in books. I once got an email from someone who said that a statement in my algebra book contradicted an exercise in Fraleigh, and so I had to be wrong.

Sure, every number of degree 4 over $\Bbb Q$ is constructible. Have it your own way.

Xander Henderson

@TedShifrin Or Fraleigh was wrong. Or, most likely the emailer was wrong. :D

leslie townes

18:12

i think that differences like these are most commonly settled in the form of a fistfight in a parking lot somewhere.

my money's on ted.

Ted Shifrin

Fraleigh was wrong. He fixed it in a subsequent edition.

Xander Henderson

@TedShifrin *shock*

Thorgott

I love fistfighting people over mathematical inaccuracies

Goku カカロット

@leslietownes same. Ted is the main character on Math.SE. Rooting and betting on him

Ted Shifrin

I think it was likely a typo in the answers at the back of the book. But, sure, email me and tell me I’m wrong with that as your proof.

Betting is illegal.

leslie townes

18:15

we're betting in lesliecoin on an offshore exchange.

as i understand it, that makes anything we do entirely fine.

Ted Shifrin

Musk has ratted you out.

Goku カカロット

@leslietownes Leslie coin? Why not Leslie NFTs?

leslie townes

why not leslie NFTs? i'm glad you asked. we have a range of all of the best NFTs.

Ted Shifrin

Who else could possibly be the best?

Goku カカロット

Goku NFT

D.C. the III

18:18

@TedShifrin California being behind the curve once again.....

Goku カカロット

(They already exist)

robjohn

@Thorgott Did I miss a fight?

Ted Shifrin

@D.C.theIII Like not taking away the rights of women and LGBTQ ….

No, @robjohn. We were saying the internet and all math books are always right.

Goku カカロット

@robjohn You did

Death Battle, a YouTube channel that puts characters and even real people into fictional, animated battles, would love this kind of fight. May I propose this option to them?

Koro

18:48

I wish I knew Russian and Japanese.

and Spanish, German and some Chinese too.

and some Korean and Tagalog as well

Goku カカロット

@Koro Japanese? Is it because of the reason I think it is?

Koro

yes, that and because of my interest in languages.

Goku カカロット

@Koro did you read the Japanese transcript for the DBS chapter I shared?

Koro

no, I didn't understand it and am too lazy to paste that in translator :(.

Goku カカロット

@Koro Alright let me do it for you. You might be surprised

D.C. the III

18:54

@Koro I wish I knew all those languages too, except Spanish....... :P

leslie townes

do you already know spanish? it's a cool language.

D.C. the III

I do....speak it just as good as my english...although that might not be saying much...

Goku カカロット

I wish I knew Japanese well. Would help on all my visits there

D.C. the III

I grew up in Honduras. Father had a job as mining engineer, so I benfitted from a youthful plastic mind being exposed to it

Koro

@Gokuカカロット: I hear Hiragana is easier.

Why don't you start with that? I used to know it during my UG as I had taken a course on writing systems.

leslie townes

18:57

oh, cool! i don't know spanish too well, but i like what i know. and being able to understand mexican slang is very helpful in california even if you don't have 100% proficiency.

Koro

@D.C.theIII then French must be easy for you :).

(except the pronunciation part)

D.C. the III

Yes all the romance languages are relatively easy. I know basic French, took it in school. Portuguese has been the easiest for me to learn, but that's because I've had an interest in it

Italian has probably been the most difficult because I'm not exposed to it as often. Only if I watch some Italian soccer

Koro

that's nice. Please do let me know in case you come across a source where one can learn romanized Japanese.

D.C. the III

Romanian surprisingly sounds a lot like SPanish, but the words mean different things in most cases

@Koro Have you tried looking at the reddit Learn Japanese subreddit? Probably a good place to start for references

Koro

No, not yet. I'll check it out. Thanks. :)

D.C. the III

19:02

Knowing a second language provides one with such an enriched world view. It's something one under values when their younger.

Koro

you're absolutely correct. :)

Ted Shifrin

@D.C.theIII "I speak it just as good as ..."

Koro

I am bilingual so I understand :).

D.C. the III

@TedShifrin I said my English was debatable..... lol

Goku カカロット

@Koro Well I am learning it and yes Hiragana is easier. I had already gained some familiarity when I had to attend Japanese language classes back in high school over there

Koro

19:06

@Gokuカカロット I heard that one could hire families on rent in Japan using some app. Not sure how true that is. It's like suppose one wants to go watch some movie but their family has gone somewhere but they want to go with family so they can rent a family using that app.

Goku カカロット

@Koro you can hire all sorts of people there including girlfriends, one of my classmates, also a foreign student, did that. And as silly as that sounds, they actually make for great guides and someone that can accompany you if you're unfamiliar with the place

I didn't do it because.... I just couldn't get myself to stoop at that level

Yai0Phah

Second for Russian.

Especially in order to read some Russian mathematical texts.

Koro

yeah :). I want to be able to understand Zorich's book in Russian.

Jakobian

@AlessandroCodenotti something to think about: Let $X$ be a locally compact separable metric space. Does $X$ admit a compatible metric $d$ such that $A\subseteq X$ is compact iff $A$ is closed and $d$-bounded?

Alessandro Codenotti

21:25

@Jakobian surely $X$ needs to be complete

Jakobian

22:58

@AlessandroCodenotti it will be completely metrizable, since it's locally compact

Jakobian

23:20

@AlessandroCodenotti turns out that a metrizable space $X$ admits such a metric iff $X$ is locally compact and Polish

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