Mathematics - 2021-10-26 (page 1 of 3)

geocalc33

12:00 AM

Consider the map $f:A \to B$

leslie townes

i'm considering it. i have to say, it isn't very interesting.

geocalc33

What does this mean? $f_i : A \to B$ for $i=1,2,3,...$

leslie townes

@anakhro yes, over 10 years now.

unless subscripting has been given some kind of meaning, whereby subscripting an already-defined function does something goofy to the already-defined function, i'd assume that was just notation for a sequence of functions, each one of them from A to B.

geocalc33

I want to capture the set of linear maps between $A$ and $B$ and make a diagram of it

so if $V$ and $W$ are vector spaces then $L(V,W)$ denotes the set of linear maps between them

leslie townes

i dunno. i have never personally done that. it gets big even if V and W are of fairly small finite dimension so there is no hope of notationally capturing all of the individual maps. you have to choose what to abstract away.

"L(V,W)" is itself a pretty good abstraction of that set. i'm slightly kidding, but only slightly. it's the main one i've used.

when people have a small number of maps, each from the same V to the same W, they might draw a diagram with V and W as vertices, and then a family of oriented edges from V to W, perhaps labeled with letters or other symbols, each edge denoting an individual map within the family.

there are even fairly robust tex packages that help with stuff like that, although i have not used them.

geocalc33

12:13 AM

@leslietownes okay

Ted Shifrin

12:25 AM

@anakhro Did you ever respond to my response?

@leslietownes 6 1/2 for me! Time flees (or is that fleas?).

anakhro

@leslietownes How do you maintain a good breadth of knowledge on math? Did you find it never left, or do you have to keep reading, chatting on here, etc.?

@TedShifrin you mean your identification of the fact that I was trying to show a submanifold is totally geodesic?

Ted Shifrin

Yes … There are ways to do that having nothing to do with following geodesics. Context matters.

anakhro

Well I did provide my context. But it's okay, it was solved and I was being a BIG FAT LOSER.

Ted Shifrin

I didn’t scroll enough. Sorry for interrupting.

anakhro

It's okay, teddykins.

I love you regardless of how much you scroll.

Ted Shifrin

12:38 AM

What the hell is the $L^2$ metric? But we have an open set of a linear subspace, so that surely has zero second fundamental form.

anakhro

The metric tensor is given by $\langle\dot\varphi,\dot\varphi\rangle_\varphi = \int_{\mathbb R^n}\|\dot\varphi(x)\|^2\,\mu$ for a volume form $\mu$.

Ted Shifrin

Are these compactly supported diffeos?

In which case $GL(n)$ ain’t a submanifold.

Makes zero sense to me.

anakhro

No, I don't think they are compactly supported.

Ted Shifrin

Why is that integral finite?

anakhro

shrugs

All I know is Arnold does some weird stuff.

I am still but a stranger to this realm of infinite dimensional manifolds.

Ted Shifrin

12:54 AM

Arnol’d in my experience omits most details but is correct.

anakhro

I will download this 1966 paper of his and see if I can uncover the mysteries of this metric.

Did you ever meet Arnold, Ted?

AbstractSpacecraft

@geocalc33 fancy seeing you here! :)

My internet was off earlier today

Ted Shifrin

No. He was supposed to be at the conference I went to in honor of Elie Cartan’s 100th birthday, but the government wouldn’t let him go ….

Fabulous conference in Lyon,

AbstractSpacecraft

ghetto fabulous, son

^_^

Ted Shifrin

Huh?

anakhro

1:04 AM

@TedShifrin in the actual paper, Arnold seems to do only the L^2 metric for Diff D for a closed domain D.

leslie townes

@anakhro sorry for delay in response, i was picking up a demon. i have an OK memory for the basics and a lot of old books/notes around.

Ted Shifrin

Compact?

Munchkin is promoted to demon? Oh, we got the rains.

anakhro

At one point he does say bounded domain, but I am not sure if that was just a special statement.

Ted Shifrin

Well …. As I said, he’s sloppy with details. But my complaint still will hold.

anakhro

Demons are pretty cool, Leslie. Does yours have wings and a tail?

leslie townes

1:09 AM

ted, the california theater on kittredge in berkeley is closing. may have operated under a different name in your time.

following the UC theater into oblivion by a few decades.

Ted Shifrin

I remember it. The little arts theater on Telegraph (where I saw Bogart movie after Bogart movie, etc) closed ages ago.

Covid closed one of my favorite theaters in SD.

leslie townes

while i was there, a tiny arts theater on shattuck that had the projection "booth" in the aisle was demolished for expensive condominiums.

as a middle finger to history, the entryway was styled somewhat like that of a movie theater in the new building.

Ted Shifrin

I forget if the Albany theater on Solano near @copper is still there.

AbstractSpacecraft

SD is a nexus for MSE chatters

leslie townes

it was still there the last time i drove by it.

Joe Shmo

1:13 AM

does anybody know where algebra can come in useful in analysis

Ted Shifrin

All over?

Joe Shmo

so for example?

leslie townes

i like my theaters old and either huge and ornate or really small. not new and mid-size. no cup holders.

AbstractSpacecraft

Hilbert space is first some algebraic structure, I think a vector space

Joe Shmo

yeah not trivial examples

Ted Shifrin

1:14 AM

Depends on the algebra and on the analysis. Some fields of analysis are all algebra.

leslie townes

a lot of the elementary theory of operator algebras is pretty algebraic. you can prove fairly 'hard' early 20th century analysis with it very quickly.

Ted Shifrin

Lie groups is a good intermingling.

Lots of analysis in representation theory.

AbstractSpacecraft

ANT is heavy analysis once you get to L-functions

Ted Shifrin

SCV and residues overlaps AG and plenty of algebra.

leslie townes

there's the algebra of fourier series on the circle with absolutely convergent fourier coefficients. an element of that algebra without roots on the circle has a reciprocal in that algebra. norbert wiener, no slouch, proved it with a lot of tricky analysis. if you know the right algebra it's an exercise you could assign in week 3 of an operator algebras course.

Ted Shifrin

1:16 AM

Not much algebra there, Space.

leslie townes

in complex analysis, some of runge's theorems on uniform approximation by analytic functions or polynomials can be proved with general algebra techniques.

you do need topology and limits and stuff. it's maybe not "algebra" algebra.

Ted Shifrin

Lots of sheaf cohomology everywhere.

leslie townes

no True Analysis Statement can be proved with algebra.

Ted Shifrin

Ha ha.

Joe Shmo

fighting words, finally

anakhro

1:20 AM

@JoeShmo along the lines of what Leslie was saying, you should check out the Gelfand-Naimark theorem.

leslie townes

change my mind

Ted Shifrin

Define “true analysis.”

leslie townes

ted, if you can prove it with algebra, that ain't it.

Joe Shmo

well apparently all of early 20th century analysis?

Ted Shifrin

I like math that integrates different fields of math, not narrow shit.

Twink

1:24 AM

What's the toughest analysis? Complex analysis, real analysis, measure theory, functional analysis, fourier analysis, euclidean harmonic analysis, abstract harmonic analysis, etc....

Joe Shmo

lately I'm into concrete math, not abstract nonesense

harmonic analysis

leslie townes

analysis of the self.

anakhro

lol

Twink

lmao

Joe Shmo

oy vey

this escalated quickly

Ted Shifrin

1:26 AM

That’s super-narrow. Trompian.

Degenerated.

Joe Shmo

probabilistic proofs of analytic facts are black magic

Ted Shifrin

There’s plenty of those proofs in combinatorial number theory. Several masters at UGA.

Joe Shmo

probabilistic combinatorics is really cool

leslie townes

my daughter is demanding "raspberries and cheese sticks and milk" for dinner. this is pure free association based on what she could see in an open refrigerator.

Ted Shifrin

Yuck.

Joe Shmo

1:31 AM

you could game that to your advantage leslie

re: black magic. like try finding $\lim_{n \rightarrow \infty} e^{-n} \sum_{k=0}^n\frac{n^k}{k!}$ by hand. you could do it, after a long while, but the probabilistic proof is a couple of lines

leslie townes

she also wanted chocolate. that isn't visible, but she wants it. she had some on her birthday, over two weeks ago.

Joe Shmo

it feels like cheating

I always crave chocolate. always.

like you said chocolate. so now I want brownies. look what you did.

Ted Shifrin

tosses Shmo a dark-chocolate covered walnut

Joe Shmo

none of that healthy option shit

I want fudge

I actually do appreciate walnuts in my brownies

leslie townes

i like bernstein's probabilistic-ish proof of the weierstrass approximation theorem.

Twink

1:35 AM

talk about math please

leslie townes

although there is a functional analytic proof that is even shorter.

Ted Shifrin

Makes my teeth hurt. Nope.

Joe Shmo

and I don't indulge often either. hence I always crave.

yeah! bernstein's polynomials is another one

just popped up in the homework this week

leslie townes

a lot of baire category arguments have a probabilistic flavor, with genericity (or whatever it is called) in place of presence a.e. with respect to a measure.

AbstractSpacecraft

@leslie Given a smooth curve γ in C parametrized by z : [a, b] → C, and f a
continuous function on γ, we define the integral of f along γ by

Joe Shmo

1:37 AM

yeah its not really a probabilistic proof though is it. the only pseudo-probabilistic feature there is the binomial distribution

AbstractSpacecraft

Why do books say that?

Ted Shifrin

i don’t know all these proofs to which y’all refer.

Joe Shmo

the bernstein polys one you should definitely check out

its 3/4 of a page, if that

leslie townes

the existnece of a lebesgue integrable function whose fourier series diverges pointwise everywhere is something you can do with BCT although i think kolmogorov initially did it with a clever construction.

AbstractSpacecraft

Why not just a smooth curve $\gamma : [a,b] \to \Bbb{C}$?

Joe Shmo

1:38 AM

durrett has a proof

Ted Shifrin

Because to some of us a curve is a set, not a mapping. Done.

leslie townes

i think they may want to distinguish between the curve as a geometric object and the parametrization. i would have to see more context for more.

AbstractSpacecraft

I will get back to you on that :)

Reading from a $\Bbb{C}$ analysis book

Ted Shifrin

I agree with the book.

AbstractSpacecraft

I found the answer and it doesn't match Ted's

lol

Ted Shifrin

1:40 AM

”the answer”?

AbstractSpacecraft

The family of
all parametrizations that are equivalent to z(t) determines a smooth
curve γ ⊂ C, namely the image of [a, b] under z with the orientation
given by z as t travels from a to b.

Ted Shifrin

yes, for line integrals we need oriented curves. Read Ted’s book.

AbstractSpacecraft

So $\gamma$'s are equivalence classes, is the reason

Of a set of parameterizations

So $z$ is just one such rep

@TedShifrin where's your book?

Ted Shifrin

This formalism is crap.

Published.

AbstractSpacecraft

math.franklin.uga.edu/sites/default/files/users/user317/…

Is that it?

I thought you meant $\Bbb{C}$ analysis book though :)

Here's the book I'm reading:

fing.edu.uy/~cerminar/Complex_Analysis.pdf

Twink

1:45 AM

I prefer Palka's book

AbstractSpacecraft

Where's a link to that?

Twink

It's more for dummies like me

AbstractSpacecraft

Oh, cool, I got a copy of "Visual complex analysis". It's great

Twink

he explains with very much detail

AbstractSpacecraft

however, I like the formalness of the link

Twink

1:46 AM

he talks a lot

but I like that

it fits with my type of learning

AbstractSpacecraft

What type is that? Genius?

Twink

amazon.com/…

AbstractSpacecraft

Math genii, plural

Twink

no, it's like complex analysis for dummies I think

AbstractSpacecraft

@Twink oh you mean this free book here:

gbv.de/dms/ilmenau/toc/216098203.PDF

Twink

1:48 AM

he explains everything with a lot of details

yes, that one

that's my favorite book

AbstractSpacecraft

Actually I couldn't find a free version. That link was just the TOC

Twink

I bought it and it didn't arrive to my address, so I made a complain, and they sent another copy. After I received it, I got the other one, so I have two books

AbstractSpacecraft

Nice, amazon is good for that

Twink

I want to sell the other book

No, I bought it on springer's site

AbstractSpacecraft

I mean for selling book

Twink

1:51 AM

it was cheaper than in amazon

should I return the other copy of the book or should I keep it?

and sell it?

AbstractSpacecraft

I was thinking on that

Twink

xD

AbstractSpacecraft

I think maybe have a nice copy and a copy you can treat badly

Twink

I love Palka, I can't treat it badly

AbstractSpacecraft

Cuz I have old mathbooks that were just treated like shit (by me)

and even missing pages

Twink

1:53 AM

I never treat books badly

I treat them with love and care

AbstractSpacecraft

You're a sait of bookdom

*saint

I even wrote in the covers of some books

Twink

I'm just very carefully with my things

that's a sacrilege

AbstractSpacecraft

Ikr

Twink

to write in the covers with pen

AbstractSpacecraft

They are my Fermat notes

When I die, nothing will be conjectured though :|

Twink

1:55 AM

use post-it

AbstractSpacecraft

The greats stole all the good maths from us

Derivative

are take home tests the norm in the second half of a math undergrad?

AbstractSpacecraft

Define norm

on the space of undergrads

Twink

actually it's not a norm

it's a seminorm

Avra

Hello! To locate item in the list, then possibility the item is there on average is $\Sigma_i (i)/n = 1/n + 2/n+ \cdots+n/n = (n+1)/2$. Possibility of element is not in the list is simply $n$ meaning search through the entire list. No on average, we need half the times the item is not there and half the time the item is there, which equals to $n/2 + (n+1)/4$. What do you think about last statement please?

i.e., n/2+(n+1)/4

AbstractSpacecraft

1:57 AM

Did Bateman or Horn invent that?

Ted Shifrin

@Derivative Assuredly this is COVID most places.

@Twink smack

Derivative

and in the beforetimes? How was it like?

AbstractSpacecraft

You spread germs like a petri dish and everyone got sick

Ted Shifrin

I gave take-homes almost never. I learned that even with “honors students” cheating took place.

Twink

@TedShifrin Ouch!

AbstractSpacecraft

2:01 AM

Cheating can be honorable in some circles, if you don't get caught :|

Ted Shifrin

Great.

Sounds like one of our political parties.

AbstractSpacecraft

Yes, probably them

Twink

@AbstractSpacecraft which circles?

AbstractSpacecraft

Circles of cheaters

Twink

the unit circle?

hmm, I had never heard about that

AbstractSpacecraft

2:02 AM

It's actually the whole world

the whole world cheats somewhere

in there lives

Little cheating never hurt no body less they got caught

lol

Twink

I don't think Ramanujan cheated

AbstractSpacecraft

He cheated us of his great maths in older age

And he also was known to not give a proof

Twink

@TedShifrin did you ever cheat on an exam?

AbstractSpacecraft

at all :)

anakhro

Note to self: Ted does not like poop jokes.

Twink

2:05 AM

I have never cheated, I'm too anxious for that

user10478

For me it depends on what you consider cheating.

Twink

copying

user10478

I don't think so

AbstractSpacecraft

I never cheated either. I think everyone in this room is cool

They would never cheat.

user10478

I have terrible handwriting, and at least once I did a problem the incorrect way and wrote random gibberish as my work because I didn't know how to do it the way we were supposed to; I just knew it involved summations.

I got full credit for the problem.

AbstractSpacecraft

2:08 AM

Goin to hell, cheater!

^_^

j/k

user10478

I didn't consider it cheating but some might.

AbstractSpacecraft

Yeah, so cheating is a relative phenomena

user10478

I got paid back slightly either way.

It was something to do with power series, and later when I went back to reteach myself calc cause I didn't remember much from school, I went over my exams, and I spent some time trying to reason through my work on the problem, before the incident all came back to me.

AbstractSpacecraft

Nice, I know how it is now...

I can't even comprehend my notes later, but somehow other people in the world can comprehend theirs

So I usually just throw away my accumulated notes

user10478

Now that I'm self studying I take great notes.

When I was in school I was awful at it.

Twink

2:13 AM

Hmmm, too bad, I always understand and keep my notes

you should be more organized

I always take good notes, unless the teacher is disorganized

and I use different colors of pens and highlighters

Ted Shifrin

2:50 AM

@Twink Yes, once. A 7th grade history class where we had to memorize pages of dates and names. I never cheated again. But I did hate classes like that

Joe Shmo

I don't take notes

I didn't take notes all grad school

or undergrad for that matter

can't do it. I like the book better anyway

Ted Shifrin

Some of us don’t copy books on the board. For advanced courses there often is no book.

So you may have to grow up.

leslie townes

i'm not even a person, i'm a chatterbot trained on my designer's excellent notes.

Ted Shifrin

Channeling munchkins.

AbstractSpacecraft

3:09 AM

@geocalc33

geocalc33

3:56 AM

@AbstractSpacecraft

copper.hat

5:58 AM

@TedShifrin Hi Ted, the landmark twin beside kains on Solano is still going strong!

leslie townes

copper did you have any memories of the california theater? just east of campus, shutting down now.

copper.hat

7:00 AM

@leslietownes on kittridge?

yes

there used to be some veggie restaurant on the oxford corner. good something or other i think?

leslie townes

oh, yeah.

i forget what it was. i got food poisoning there, or near there, once.

porridgemathematics

8:42 AM

hi, im trying to understand the conformal map $f$ in this image imgur.com/a/O1N4JIy , it maps the upper half plane to some polygon. I understand the authors argument saying that $\arg(f')$ jumps at the $z_k$ , but he later then says that since $f_k = (z-z_k)^{- \beta_k}$ jumps the same amount at $z_k$, $\frac{f'}{f_k}$ is holomorphic in a neighbourhood of $z_k$, this would make sense to me if we knew $|f'|$ did not jump at $z_k$, but im not sure how to see this

Semiclassical

12:50 PM

Schwarz–Christoffel mapping, yum

shintuku

Let $U \subset \Bbb R^n$ be open and let $\vec a \in U$, and let $f:U \to \Bbb R^m$ be differentiable at $\vec a$. Prove that the Jacobian at $\vec a$ is identical to the derivative at $\vec a$.

We say $f$ is differentiable at $\vec a$ if there exists a linear map $f'(\vec a): \Bbb R^n \to \Bbb R^m$ such that $\lim \limits_{\vec h \to \vec 0} \frac{f(\vec a + \vec h) - f(\vec a) - f'(\vec a)\vec h}{\|\vec h\|} = 0$. Notice this can be rewritten as the equality $f(\vec a + \vec h) - f(\vec a)= f'(\vec a)\vec h + \epsilon(\vec h)$ if we suppose $\epsilon(\vec h)$ satisfies $\lim \limits_{\ve…

Let $\vec e_j$ be a standard vector of $\Bbb R^n$ and let $\mu(t\vec e_j)$ satisfy both $\lim_{t \to 0} \frac{\|\mu(t\vec e_j)\|}{|t|}=0$ and $f'(\vec a)t\vec e_j + \mu(\vec e_j) = f(\vec a+t\vec e_j) - f(\vec a)$. With some algebra, we obtain $\frac{f(\vec a+t\vec e_j) - f(\vec a) - f'(\vec a)t\vec e_j}{|t|} - \frac{\| \mu(t\vec e_j) \|}{|t|} = 0$, and taking the limit as $t$ goes to $0$, we obtain: $\lim \limits_{t \to 0} \frac{f(\vec a+t\vec e_j) - f(\vec a) - f'(\vec a)t\vec e_j}{|t|} = 0$.

Using the linear map properties of $f'(\vec a)$, we obtain $\lim \limits_{t \to 0} \frac{f(\vec a+t\vec e_j) - f(\vec a)}{|t|} - f'(\vec a)\vec e_j =0$, i.e., $\lim \limits_{t \to 0} \frac{f(\vec a+t\vec e_j) - f(\vec a)}{|t|} = f'(\vec a)\vec e_j$, and therefore the jth column of $f'(\vec a)$ is the jth partial derivative of $f$ at $\vec a$, which is the definition of the Jacobian.

mein gott that is a bit longer than i expected

well, if anyone has the grace and charity to read it, is it believable?

argh i need to prove the limits exist

ignore the above, my bad for spamming chat

Matthias Herrmann

2:04 PM

Hey,
$exp(x_{1},x_{2})$ - what does the author mean by that expression? I thought exp only has one argument

exp(x1, x2)

pdf page 159 https://mml-book.github.io/book/mml-book.pdf

ah nvm its not a comma but a multiplication

wklm

Hi folks, I'm working on my assignment for decision theory. I got iid random variables and need to show that the given estimator is inadmissible under the quadratic loss. Does it suffice to compare its risk (expected loss) to an unbiased estimator (sample mean) whose risk would be 0? I wrote my attempt in latex here: cdn.discordapp.com/attachments/554278933554659328/…

Could someone please let me know if that makes sense?

Koro

2:42 PM

Suppose that I want to prove that L=R, where L and R are some mathematical expressions that evaluate to something. I consider: L is finite and equals some value say m, then using some methods I show that R is also equal to m. Then can I say that L=R? Or should I do the same from R to L also?

Koro

3:12 PM

Is my question clear?

shintuku

not sure what you mean: $L=m$, $R=m$ together imply $R=L$ and $L=R$

Balarka Sen

@wklm I'm no expert, but seems correct. You have shown the sample mean dominates your estimator, right? You need to know $(a - 1) \mu \neq b$

You have written $a \mu \neq 1/b$. Typo?

shintuku

@Koro equality is transitive and commutes

wklm

@BalarkaSen I assumed that without the loss of generality we can set $\mu = 0$ and then it suffices to show that there are a and b for which $ (b + (a - 1)\mu)^2 > 0$ while for an unbiased estimator it is 0

Koro

@shintuku L=R is to be proven. Doesn't that mean that I have to prove that "$L=m\implies R=m$ and $R=m\implies L=m$"?

The confusion arose while trying to prove equality of limit of two sequences...

What am I missing?

nevermind, I think my confusion is clear now.

If L has say three possible values u,v,w then accordingly R will also have u,v,w if the equality is to hold.

Then, I think that my proof here is correct:

shintuku

3:28 PM

it is faster to use instead $R = m \land L = m \implies R = L$ citing transitivity of equality. It is transitivity that is doing the work here

Koro

Balarka Sen

@wklm Why there exists $a, b$? Don't you want to show for all $a > 1$ and for all $b$?

Koro

shintuku

@Koro also $R = m \implies L = m$ if and only if $R=L$, so you don't need $L=m \implies R = m$

wklm

@BalarkaSen imgur.com/QVm9B0R

this is the definition I am using

Koro

3:31 PM

@shintuku in my proof shared above, I used $\iff$ wherever possible.

shintuku

$L = m \implies R = m$ is equivalent to $R = L$, and so is $R = m \implies L = m$

furthermore, $L = m \implies R = m$ if and only if $R = m \implies L = m$

Koro

hmm, right.

I got confused earlier.

wklm

it says that for a rule A(P) to dominate rule B(P) for every P, So if there's a P for which Risk(B(P)) < Risk(A(P)) then A doesn't dominate B

Koro

Thanks @shin :)

Peter John

math.stackexchange.com/q/4288020/922120 I have a question on free $\Bbb Z_2$-space

Balarka Sen

3:36 PM

@wklm OK, but that means the quantifier is on $\mu$, not $a, b$.

I agree it suffices to take any $\mu$ s.t. $(a - 1)\mu \neq b$.

$\mu$ is the parameter. $a, b$ are some arbitrary constants someone fixed for you.

wklm

ahh!! correct! Many thanks for pointing this out!

thank you @BalarkaSen

Balarka Sen

Welcome.

Or "wklm"

wklm

:D

leslie townes

koro, as an exercise, you might try finding an argument that fits on one page and doesn't involve cases :)

Koro

I'm glad you saw my proof. Thank you! I think the cases may be reduced if I use Bernoulli's inequality instead of using log and exponent function.

leslie townes

3:46 PM

i mean, maybe some case analysis is necessary, but my instinct is that even if you use cases, you can wrap some of them up in a few lines, instead of repeating variations on the same thing

liminfs always exist, for example. in this way they are better than limits, and often one can avoid separating out the case where the limit does exist

a lot of the problem seems to be in relating limiting behavior of a_n log(1+1/n) to a_n/n (i'm not sure what you wrote fully explains the relationship, although i think everything you wrote is true). that might be the focus for the simplification, even if you still use cases

just thinking out loud

Koro

liminf always exists -yes so I had to consider cases on liminf only (liminf finite non zero, infinite and zero).

leslie townes

you really love your cases

Koro

I used $\lim \frac{\log (1+x)} x=1$ as $x\to 0$ to relate the two limiting behavior.

leslie townes

i guess maybe it's necessary depending on the setup. rudin (for example) spends what some would regard as too much time with the extended real number system to avoid this

Koro

I'll think about reducing no. of cases and let you know if there's a progress in that.

shintuku

3:54 PM

The projection matrix $A(A^TA)^{-1}A^T$, where the matrix $A$ has as column vectors a basis for the $V$ onto which we want to project, is equal to the identity matrix if $A$ is invertible. does this have any geometrical significance? I'm trying to figure out why it collapses into the identity matrix

leslie townes

i mean, do whatever you want :) i just note a lot of analysis arguments can take up a whole lot of space if they use cases and it can be helpful to minimize them. for example the triangle inequality |x-z| <= |x-y| + |y-z| encodes a lot of true information regardless of the relative positioning of x, y, z on the line, which would come up if you think of |t| as a 'piecewise-defined function' and break down in terms of whether |t| is t or -t in each of those component | |s

shin, that's orthogonal projection onto CS(A)? orthogonal projection onto the full space is I? if you want to find the nearest vector in the full space to a vector itself, there's nothing to do? i dunno if any of this counts as 'geometrical'

shintuku

right, it's orthogonal projection onto the column space of $A$

leslie townes

algebraically, if A is invertible just expand (A^T A)^(-1) as a product of the inverses of A and A^T

not what you asked for but worth keeping in mind at all times. the hypothesis is what lets you bring ^{-1} inside the product

shintuku

right, if $A$ is invertible we get $A(A^TA)^{-1}A^T = AA^{-1}(A^T)^{-1}A^T = I$

Koro

@shin I revised it about two days ago or so. First find projection of vector a onto vector b. Then, project a onto space U (Leslie mentioned what is meant by it -nearest vector ...) and note that the formulae come almost similar.

leslie townes

4:01 PM

in terms of the arguments that realize that formula as a geometric projection, i think it's:if you want to find the closest vector in [a subspace of R^n] to v in R^n, that's sometimes some work, but if the subspace of R^n happens to be all of R^n, you just return v

shintuku

i'm trying to interpret why the matrix would collapse, when we usually think of projection as getting the closest vector in the column space to a vector that is not in the column space "plane"

oh, right leslie

oh: if a matrix is invertible, the row space and column space are identical, no? i.e. have the same span

leslie townes

if the matrix is regarded as an operator on R^n, yes

sometimes people use matrices to represent operators on different spaces of the same dimension :) where that doesn't quite work out, but you get "whole space" for the row space and "whole space" for the column space, whatever those are, in any event

shintuku

that means the column space spans the domain! so that explains that

yeah i'm thinking R^n

hm so this isn't always the case then

leslie townes

just think R^n. people hate on it but for finite dimensional linear algebra it (and C^n) can't be beat

shintuku

right, thanks for the tips

shintuku

4:25 PM

ah, we can also notice that if $A$ is invertible (which makes it square), then $P = A(A^TA)^{-1}A^T$ will also be invertible and square, meaning both $P$ and $A$ have the same column space

which is obvious when we set $P=I$

hm, but that's not exactly the fact that makes a projected vector not move

leslie townes

projected vectors move?

shintuku

hm, its shadow instead

leslie townes

they can move, i'm not particular about this.

all sorts of interpretive narratives help or not. whether it's the thing changing or the label of the thing changing. that comes up a lot.

moving vs. casting shadows is different but also kind of the same.

shintuku

ok we need to work on these interpretative narratives

this can become an allegory for the cold war kind of thing

$P$ is the soviet union in all of this

this will get people learning

$\vec x$ is trying to save democracy

leslie townes

there was a trend in the mid 20th century where some math publications under particularly authoritarian regimes would include front matter explaining how some vague quote from lenin or mao led to the work in question.

maybe in a few years we'll have that in the USA.

shintuku

4:41 PM

$\vec y$ is trying to save $\vec b$, and the only way he can do it is by finding a projection onto $\vec A$

$\vec c$ is not quite right in the mind, in all of this. he goes on a long run and gets a big following, from vectors in $\vec A$

$\vec y$ has a fear of heights after having his work partner fall while pursuing a criminal

that complicates things

leslie townes

gosh it's weird to see low-rep accounts asking questions about stuff that most people don't see until halfway through graduate school if ever. particularly when the questions are, for lack of a better word, [the type of question that people say there's no such thing as].

shintuku

the trap as a newbie is that you can understand the words since its in english

so there's an illusion of proximity and semblance

leslie townes

that's true. it's also fairly standard and even reasonable in many fields to simply skip over stuff that seems like it might be too involved on a first, second, or third pass. 'i can come back to it later if i need to.'

shintuku

they need to work on their interpretative narratives

Slate

guess it depends on where and how you were raised, too - for at least a subset of environments I would guess that math education is shifting quite a lot younger, and so the topics are growing more compact (and content slough is high)

leslie townes

4:53 PM

that's an interesting point. i've seen that first hand a little, even a long time ago.

TheReal__Mike

Seems chat has changed a lot

leslie townes

this was a criticism of the 'moore method' of teaching. it can produce knowledge along some narrow curated dimension, and maybe even novel work along that dimension, while also manufacturing ignorance of anything else.

TheReal__Mike

did old members like Ted, Mike, Balarka Sen etc left ?

Leaky Nun is there

leslie townes

ted and balarka have both been around very recently although they are not present presently.

TheReal__Mike

Okay thanks

leslie townes

4:55 PM

i don't know which mike you're referring to. you've been around fairly recently.

TheReal__Mike

Mike Miller

Koro

@shintuku haha

TheReal__Mike

We had conflicting names, thus I changed my name

leslie townes

i haven't seen him in a while.

Slate

@leslietownes Not familiar with the Moore method - not a math educator, as fun as I think it'd be. Skimming it, the conclusion makes sense (but eh, a naive perspective is worth the price).

leslie townes

4:58 PM

the moore method is problematic for other reasons. excluding black people from the classroom was part of the moore method.

Slate

...seriously?? good lord

TheReal__Mike

Akiva Weinberger was another old member. I am talking about this Mike mathoverflow.net/users/40804/mme

He changed his username

Koro

Mr. Akiva has also been around lately

leslie townes

i don't think most math educators would be familiar with the moore method. it is too extreme for K-12 deployment and probably has never made much headway into ed schools.

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$Mathematics$ Mathematics