Mathematics - 2024-11-03

Derivative

00:17

for a map $f:S^n\to S^n$ is its degree the same as that of the suspension map $\Sigma f:\Sigma S^n\to \Sigma S^n$?

Ben Steffan

yes

this is immediate from the suspension isomorphism

Derivative

cool

copper.hat

01:41

@SineoftheTime nice gear

Simd

06:36

If I have a binary matrix over F_2, how much smaller can the number of ones in its inverse be?

leslie townes

06:59

if n is odd and you let I, 1, and U respectively denote the nxn identity matrix, the nx1 column vector with all 1s in it, and the nxn matrix with all 1s in it, then the matrix [[I,1],[1^T,0]], which has 3n ones in it, appears to be invertible with inverse [[U-I,1],[1^T,1]], which has (n+1)^2 - n ones in it. if that's right, it would show that for an (n+1)x(n+1) matrix with n odd, the ratio of ones in an invertible matrix to ones in the inverse matrix can at least be as small as 3n/((n+1)^2 - n)

no idea if that's close to optimal, but it's something

psie

11:02

Let $E$ be a set contained in a open ball of radius $K$, i.e. with finite measure. Consider now the open ball of radius $K+1$. By outer regularity of Lebesgue measure, we can find a decreasing sequence of sets $\{U_j\}$ belonging to the open ball with radius $K+1$ such that $E\subset\bigcap_1^\infty U_j$ and $m\left(\bigcap_1^\infty U_j\setminus E\right)=0$. My gut tells me $\chi_{U_j}\to\chi_E$ a.e., though I'm not sure how to show this.

Jakobian

@psie one can definitely take $U_j\subseteq W_K$ here, so I'm not sure either

psie

@Jakobian yeah, I guess the author did it to provide some wiggle room for the $U_j$, i.e. if $E=W_K$, then we can not really construct a sequence $U_j$.

Jakobian

We can

psie

sure, there is only one sequence then :)

Jakobian

@psie 2.40b is used, recall definition of $G_\delta$ set

Given such $G_\delta$ set $\bigcap_j V_j$ one can take $U_j = W_K\cap V_1\cap...\cap V_j$

Then $\bigcap_j U_j$ will be a $G_\delta$ set with the desired properties

psie

11:13

@Jakobian ok 👍 are you sure that $W_K$ is open?

Jakobian

@psie yes. All the things here are continuous

@psie consider $x$ outside that $G_\delta$ set and in $E$

psie

@Jakobian you don't mean $x\in \left(\bigcap_j U_j\right)^c\cap E$, do you?

Jakobian

No, the union of them

The complement of which is a null set

If $x$ is not in the $G_\delta$ set then from monotonicity assumption there is $N$ such that $x$ is not in $U_k$ for $k\geq N$

So the limit exists and is $0$

Same as $\chi_E(x)$

And if $x\in E$?

psie

ok, if $x\in E$, then yeah, $\chi_{U_j}(x)=\chi_E(x)$

@Jakobian I'm still doubting this though. $W_K$ is contained in an open ball and indeed $|\det D_xG|$ is continuous. The thing is, the set notation has two conditions which confuses me as to how to conclude openness of $W_K$.

$\Omega$ is of course open too.

Sine of the Time

11:41

@copper.hat :)

manual or automatic transmission?

psie

11:55

@Jakobian I think I see now why $W_K=\Omega\cap\{x:|x|<K\text{ and } |\det D_xG|<K\}$ is open; we can write $W_K=\Omega\cap B(K,0)\cap f^{-1}((-\infty,K))$, where $f(x)=|\det D_xG|$. These are all open sets.

Thanks!

Jakobian

12:31

@psie well sure. Its just that I see its open without even writing the equality $W_K = ...$ because I am so used to those type of arguments

pie

13:35

Hey everyone! I’ve noticed a lot of people I know IRL, especially high school students, struggle to find a starting point for self-learning math. I’m thinking about posting a question on MSE to ask for a community wiki-style roadmap that could guide beginners through different math subjects.

The idea would be for answers to cover prerequisites, recommended books, lectures, and problem books if available, as well as advanced levels for each topic. Do you think that’s a good idea? And would it be within the site’s guidelines?

nickbros123

that sounds like something that should already exist

Ben Steffan

that sounds like something that would get closed pretty quickly :)

nickbros123

lol yeah, i was speaking with respect to sites like art of problem solving, or chicago math bibliography or something broad like that

Ben Steffan

it's dubious whether this compatible with site guidelines, but my experience is that people don't take kindly to these questions

nickbros123

i can see how it can be closed as "opinion based"

Ben Steffan

13:37

it's not about mathematics per se and might be classified as personal advice or opinion based, with some justification

yeah

look at this search and see how many questions there are closed, say math.stackexchange.com/search?q=studying

Sine of the Time

math.stackexchange.com/search?q=self+learning

pie

14:19

I get where you’re coming from, but I think a community wiki post like this could really help a lot of users. Right now, we see a lot of repeat questions from people just starting out, all asking for guidance on where to begin. Having a structured, community-driven roadmap could serve as a go-to reference or FAQ for these questions, making it easier for new users to find answers and reducing the need for these types of posts.

Also, I’d like to get a moderator's opinion on this, since the question would ideally be a community wiki, and I believe only mods can convert a post to that format.

Ben Steffan

I don't disagree with the intention

I'm telling you that I doubt the post would live very long

sure, asking a mod is a good idea :)

pie

@XanderHenderson What do you think?

nickbros123

15:49

I really wish i had the time to solve dummit and foote's group theory properly. There seems to be a good serving of computational examples to build intuition, along with standard tricks sprinkled into the exercises. but unfortunately i gotta study for an exam so I am doing minimal exercises, just some proofs here and there, and the assignments. Exams really get in the way of ones learning

it sounds really weird to say that I will seriously study group theory after the semesters over lmao

Ben Steffan

16:05

does it? it sounds pretty normal to me

I will certainly use the winter break to study material form the term I deem important or interesting

The Empty String Photographer

Radical idea: $\infty=-\infty$

Ben Steffan

congratulations, you have discovered the 1-point compactification

The Empty String Photographer

Huh?

Ben Steffan

the 1-point compactification of $\mathbb{R}$ is the space $\mathbb{R}^+ = \mathbb{R} \cup \{\infty\}$ toplogized in a way so that a sequence in $\mathbb{R}$ which diverges to $\pm \infty$ converges to $\infty$ in $\mathbb{R}^+$

Alternatively it is the quotient of $\bar{\mathbb{R}}$ under the identification $\infty \sim -\infty$

The Empty String Photographer

Time for the 1-point compactification of $\mathbb{U}$ then!

Ben Steffan

16:14

what in god's name is $\mathbb{U}$

I don't think I have seen a bb U in my life

nickbros123

compactify you

Ben Steffan

I'm already compact

The Empty String Photographer

@BenSteffan I thought that was the universal set containing all numbers?

nickbros123

lol

Ben Steffan

The universal set containing all numbers is $\mathbb{C}$ :^)

maybe $\mathbb{H}$ if you're generous

nickbros123

16:20

@BenSteffan same. I also use it to sleep 12 hrs a day

Ben Steffan

haha me too

The Empty String Photographer

16:35

Time to measure the complexity of messages in this room using units of $f_\omega\left(\log_m(2^{N+g^l})\right)+1$.

Where $f_x(y)$ is the Wainer fast growing hierarchy, $m$ is metres, $N$ is newtons, $g$ is grams, and $l$ is litres.

Jakobian

16:57

one-point compactification is a unique compactification $Z$ such that $|Z\setminus X| = 1$

Marvelous thing about this definition is that you don't need to construct $Z$ in any way

Thorgott

it's only unique if compact includes Hausdorff and then you ought to require that $X$ is LCH

Jakobian

actually, its "compactification" that does here

in this definition I'm also not saying it has to exist

for non-Hausdorff spaces, a compactification is almost something that seems to be more useful to not require $X$ to be a subspace of $Z$ at all

at least, that's what I was thinking encountering all the constructions of the type

but then again I didn't encounter a setting where such compactifications matter

my definition is that a compactification of a space $X$ is a compact hausdorff space $Z$ and a map $f:X\to Z$ which is an embedding such that $f(X)$ is dense in $Z$

essentially... it doesn't really matter if we identify $X$ with $f(X)$ in this definition of course, but this is more categorical perhaps

it is pretty important how $X$ is embedded in $Z$ and not just the space $Z$, which people seem to neglect sometimes

Ben Steffan

17:13

is cheating at giving a talk by printing part of your material as a handout and asking people to just read it morally defensible

not that I have much of a choice here but

Jakobian

I've saw people giving handouts before as a way to better understand the presented material, not sure about this case though

leslie townes

17:41

ben: how are you using the material? if it's "trust me on this, this is as simple as i'm saying it is" or "this is complicated but believe me, there's a proof at the level of this talk," a link to a web page would be as good as a handout, although a handout might be nice.

Ben Steffan

it is a core definition of the talk :)

leslie townes

it only strikes me as cheating if you're somehow treating the audience as if they have knowledge of the thing you're handing out, like "there, i gave it to you, you know it, and using that information right now we see..."

Ben Steffan

maybe "core" is overstating it a little but it is very important

I'm going to give the audience a second to read the definition and then explain it, I just don't want to have to write it on the board

leslie townes

oh, so you are counting on them to somehow digest it in real time :) that feels a lot like cheating in my book, although a lot of people with access to typewritten slides will cheat on a projector, so it's a very common form of cheating

in any sense, probably preferable to having the audience watch you write for 10 minutes

Ben Steffan

really I'm asking a rhetorical question here haha

I know it's not great

leslie townes

17:46

this is just the standard problem with lecturing from anything other than a blank chalkboard, where the space and time limitations (in theory) push you toward crafting your talk so it is more understandable by human minds in real time

in practice, almost everybody is like "ooh, with as many slides as i want, space is free, they just go right up on the screen at the click of a button, time is free, DATA DUMP"

you have my blessing to dump data, just say 15 "i'm sorrys" to a picture of your favorite mathematician who didn't live to see cheap on-demand printing or computer based slide projection

and you aren't absolved unless you really mean it

Ben Steffan

sure, but this is very much a lecturing-from-a-blank-blackboard kinda situation

and also I will be graded on it so :)

leslie townes

18:01

@BenSteffan that is to say, it would be, if you weren't cheating

i'll stop, i'll stop :)

Ben Steffan

:-)

Xander Henderson

18:26

@leslietownes Ben is cheating?!

Ben Steffan

I am, in fact, cheating

Sine of the Time

time to get suspeded for rule violation

Ben Steffan

oh I'm a topologist I love getting suspended

being suspended only makes you nicer :)

and more stable

Xander Henderson

18:49

@pie About what?

Sine of the Time

5 hours ago, by pie

Hey everyone! I’ve noticed a lot of people I know IRL, especially high school students, struggle to find a starting point for self-learning math. I’m thinking about posting a question on MSE to ask for a community wiki-style roadmap that could guide beginners through different math subjects.

@Xander

Xander Henderson

Probably not on topic here.

Sine of the Time

5 hours ago, by pie

The idea would be for answers to cover prerequisites, recommended books, lectures, and problem books if available, as well as advanced levels for each topic. Do you think that’s a good idea? And would it be within the site’s guidelines?

psie

I've read the proof of the change of variables formula in Folland's book, but he only proves it for Borel measurable function and sets. The key identity he establishes is $$m(G(E))\leq\int_E|\det D_xG|\,dx$$and everything follows from this. I wonder, how do I show this inequality for Lebesgue sets $E\subset\Omega$, where $\Omega$ is the domain of $G$?

Here's my attempt; $$m(G(E))=m(G(A\cup M))\leq m(G(A))+m(G(M))\leq \int_A|\det D_xG|\,dx+m(G(M)).$$I don't know what to do with the term $m(G(M))$, where $M$ is a Lebesgue null set. Any ideas?

I don't know if I'm approaching this the right way. The change of variables formula is $$\int_{G(\Omega)}f(x)\,dx=\int_\Omega f\circ G(x)|\det D_xG|\,dx.$$Maybe there is a more direct way if this already holds for Borel functions $f$.

Xander Henderson

19:10

Is it not the case that every Lebesgue function differs from a Borel function on a set of measure zero? Isn't that, like, the whole idea?

(I haven't thought through the details of the theory in more than a decade, but, like... it's a null set. Who cares?)

psie

@XanderHenderson I get the same feeling, but I can only be certain once I've shown that no one cares :)

I think $G$ maps $M$ to another null set, but I don't know what's the justification (with the tools from the book, e.g. Lipschitz continuity has not been mentioned yet).

I guess your point makes more sense rather than my approach above in proving that inequality. $f$ equals a.e. to a Borel measurable function, and thus the same formula holds for $f$.

psie

19:40

I guess I need to use that the integral with respect to the complete measure equals the integral of the incomplete measure, as I was talking about some weeks ago.

imbAF

Can someone help me with simple rules that one should follow when making substitutions in an equation

Sine of the Time

@imbAF what do you mean?

imbAF

I am writing it

Shaun

Please may I have some feedback on the following?

-1

Q: What is the lattice of truth values in $\mathbf{Set}^2$?

Following on from this question of mine from a long time ago, I would like to know What is the lattice of truth values in $\mathbf{Set}^2$? Which lattice? I mean the Heyting algebra. I suspect that $(1,1)$ is at the top and $(0,0)$ is at the bottom. Since I think $(0,1)$ and $(1,0)$ are isomo...

imbAF

If I have this expression : $$H=\sum_{\mu,\nu=1}h_{\mu,\nu} c^\dagger_\mu c_\nu$$ and this $\gamma^\dagger_\alpha=\sum_{\nu=1}^n a_\nu c^\dagger_\nu$. And I substitute them here: $[H,\gamma^\dagger_\alpha]=E_\alpha \gamma^\dagger_\alpha$, Do I need to substitute the sums for $\gamma^\dagger_\alpha$ with the same index on both sides?

My attempt is to have a system of equations for the complex components $a_\nu$

But this is what happens to me:

$$[H,\gamma^\dagger_\alpha]=E_\alpha \gamma^\dagger_\alpha$$

$$[H,\sum_{\nu=1}^n a_\nu c^\dagger_\nu]=\sum_{\nu=1}^n E_\alpha a_\nu c^\dagger_\nu$$

$$\sum_{\nu=1}^n a_\nu[H,c^\dagger_\nu]=\sum_{\nu=1}^n E_\alpha a_\nu c^\dagger_\nu$$

$$\sum_{\nu=1}^n a_\nu[\sum_{\mu,\beta=1}h_{\mu,\beta} c^\dagger_\mu c_\beta,c^\dagger_\nu]=\sum_{\nu=1}^n E_\alpha a_\nu c^\dagger_\nu$$

Here I use the commutator relation for ladder operators: $[AB,C]=A\{B,C\}-\{A,C\}B$ and that $\{\gamma^\dagger_\alpha,\gamma_\beta\}=\delta_{\alpha\beta}$

Am I doing something wrong, math wise?

Ben Steffan

19:54

@imbAF wdym "same index on both sides"

index is only relevant for what's under the sum

imbAF

I mean that you have $\gamma^\dagger_\alpha$ in both sides of the equation

So I will substitute it with the suggested expression that contains a sum

in both sides

Ben Steffan

yes

imbAF

If I do that, in the end I come to an equation of sums, with different indices

so I cannot equate terms

since the component contain different notation

I wrote above, what the issue is here

Ben Steffan

well yes, you wrote that the issue is that you don't end up with what you should end up

imbAF

Yes, so I want to know

from a mathematical pov

Are my actions accurate?

Or am I mistaken somewhere ?

Ben Steffan

19:58

I don't know what a single of the terms in your equation means, how am I supposed to know :)

this very much smells like physics

imbAF

Basically what is c is an operator and the rest are numbers

and I wrote that if you have 3 operators in this way: $[AB,C]$ you can write it in another way

Which is written there

Ben Steffan

I don't know what the word "operator" means here, nor do I understand your summation conventions, nor anything else

imbAF

I see

Ben Steffan

this is better asked in the physics.SE chat, if such a thing exists I think

imbAF

I thought it was a math issue

Ben Steffan

20:01

it might be, but if it is it is buried under notation that's foreign to mathematicians

somebody familiar with the notation would be more help

imbAF

Just a math question

If you have an expression and you have a term, the same term on RHS and LHS

And the term can be expressed as a sum

would you write: $....\sum_n....=....\sum_n$ or $\sum_n....=....\sum_m$ ?

The latter right?

Ben Steffan

the latter is nicer, but the former is not wrong

Sine of the Time

it depends what you want to do

imbAF

Yeah but even if you write the 2nd one

Sine of the Time

it's not incorrect to write with the same index on both sides

imbAF

20:04

You should in the end come to an expression where the summations on both sides have the same index

if you want to compare coef. that multply the component of the sums, right?

Ben Steffan

no, not really. The indexing variable is just a helper.

imbAF

But if I am asked to compare the coef. multiplying the term e.g $c^\dagger_\alpha$ on both sides, then that implies that, no?

I have been spending 3 days with this particular problem and I can't seem to find a solution

But thanks

Ben Steffan

$\alpha$ is just an index here, no?

imbAF

yeah

Ben Steffan

it's a standin for 1, 2, 3, etc.

imbAF

20:08

yes

I mean I could explain what is going on from a math pov

and you are right this is physics

Ben Steffan

so you need to show that the coefficients of $c_1^\dagger$, $c_2^\dagger$, etc. have to be equal on both sides

but now the indexing variable is gone :)

imbAF

but I think it has more to do with math, I rather suspect that

Yeah

I end up with something like this:

$$\sum_{\mu,\nu,\beta=1}h_{\mu,\beta}a_\nu c^\dagger_\mu\delta_{\beta\nu}=\sum_{\nu=1}^n E_\alpha a_\nu c^\dagger_\nu$$

Because my aim is to have a system of equations for the complex number $a_\nu$

And the suggestion given is to compare the coef. of $c^\dagger_\nu$ on both sides of the equation

$c^\dagger_nu$ is something that mathematically can be written as a matrix

so not a number

Ben Steffan

sure, so you need to show that the coefficients are equal for $\mu = 1$ on the left and $\nu = 1$ on the right, then for $\mu = 2$ and $\nu = 2$, etc.

imbAF

exactly

I mean i could explain you there rules

that way you can

Ben Steffan

sorry, really not interested in that :)

imbAF

20:12

You can give me a definite answer as to whether the problem is not in the math

which would imply that I don't know something about the physics involved

Ok

Ben Steffan

I don't want to consider this further, sorry

imbAF

I understand

Ben Steffan

Again, I think the physics.se community would be of more help here. They also understand their maths quite well :)

imbAF

Ok, I will do that

Thanks

X4J

Suppose G is a group and H is a non-empty subset of G. By thinking of every g \in G as a permutation on G: is claiming that H is a subgroup of G would be precisely the same as claiming that each element in H satisfies that the restriction of it to H (restriction as for functions) is itself a permutation on H?

Ben Steffan

20:37

@X4J yes

you should try and prove that :)

Dannyu NDos

20:56

Do the math conventions actually conflate the outer product and the tensor product? Both are notated as $\mathbf{v} \otimes \mathbf{w}$.

Ben Steffan

which outer product?

Dannyu NDos

Outer product

In linear algebra, the outer product of two coordinate vectors is the matrix whose entries are all products of an element in the first vector with an element in the second vector. If the two coordinate vectors have dimensions n and m, then their outer product is an n × m matrix. More generally, given two tensors (multidimensional arrays of numbers), their outer product is a tensor. The outer product of tensors is also referred to as their tensor product, and can be used to define the tensor algebra. The outer product contrasts with: The dot product (a special case of "inner product"), which takes...

This one.

Ben Steffan

looks to be essentially the same thing as the tensor product

you just reinterpret $\mathbb{R}^n \otimes \mathbb{R}^m$ as $\mathbb{R}^{n \times m}$

so the notation seems reasonable, but I've never come across this

Dannyu NDos

The quirk is, the tensor product gives a vector, and the outer product gives a linear transformation.

Ben Steffan

the word vector is meaningless

the tensor product gives an element of a vector space, yes

a linear transformation is also an element of a vector space

i.e. a vector

also, to be somewhat pedantic, the outer product gives a matrix, not a linear transformation

but since we're working in $\mathbb{R}^n$ anyway I guess that doesn't really matter

Dannyu NDos

21:02

I'd prefer the physics convention for linear algebraic objects. $\vert v \rangle \langle w \vert$ for the outer product, and $\vert v \rangle \otimes \vert w \rangle$ for the tensor product.

And the quirk is, this extends to an arbitrary Hilbert space.

This is how quantum mechanics let those students' brain explode.

Ben Steffan

I somehow doubt this. The outer product in particular depends on a choice of basis.

Dannyu NDos

It doesn't. What vector is referred by $(\vert u \rangle \langle v \vert) \vert w \rangle = \vert u \rangle \langle v \vert w \rangle$ persists regardless of the basis.

Ben Steffan

That's precisely the point. Your expression doesn't, but the definition from the wikipedia article does.

So these are not the same

Dannyu NDos

The basis matters only for their calculation.

The Wikipedia page has a section for the abstract definition and that addresses the fact well.

Ben Steffan

fine

Jakobian

21:18

@psie how about writing $E = A\setminus N$ where $N$ is a null set and $A$ is a Borel set?

Then $$m(G(E))\leq m(G(A)) \leq \int_A |\det D_x G| dx = \int_E |\det D_x G| dx$$

That $G$ maps null sets to null sets is kind of what is being proven here as well, so maybe not the good route to take

psie

hmm, yeah, the way I've motivated it now is just that if $f$ is Lebesgue measurable it is $\overline{\mu}$-a.e. equal to $g$, where $g$ is a Borel measurable function and $\overline{\mu}$ is the complete Lebesgue measure. Then since $\int f\,d\overline{\mu}=\int g\,d\mu$, it follows that the change of variables formula also holds for $f$.

I will admit, this is probably not the way Folland intended for the theorem to be proved for Lebesgue measurable function, as he refers to an earlier theorem and says it follows like the arguments given there

Jakobian

Or like this: if $N$ is a null-set one can find a Borel null set $N\subseteq M$ so that $m(G(N))\leq m(G(M)) = 0$ from proved formula

So $G$ maps null sets to null sets

psie

yeah, that's nice

then I don't have to use that $C^1$ diffeo is Lipschitz on compact sets

I don't know how I'd use that $G$ would be Lipschitz anyway, not sure

copper.hat

21:42

i have dealt with analysis, etc, for more than 4 decades, not my main line of work. what is depressing is that if i back off mathematics for a few months (work, life, etc, interfering with my maths enjoyment) then all 4 decades seem to disappear and need an intense refresher. so depressing...

Sine of the Time

doing maths sucks

copper.hat

i wish, because that would be an additional source of enjoyment...

strange how "sucks" and "blows" can have similar meaning in some contexts.

Dannyu NDos

22:04

Tired of math? Study physics! (jk)

Ben Steffan

I know a person who's going to do just that haha

finishing their masters' in mathematics right now and already started on another undergrad in physics

as you do

Dannyu NDos

Sorry, that plan is under postponement.

Heck, when I was undergrad, my secondary major was electronic informatics.

Ben Steffan

That sounds useful

I personally am committed to uselessness :^)

Dannyu NDos

If I take another undergrad course into applied physics, my another secondary major would be creative writing.

That's 4 majors in total.

Money is all I need.

Ben Steffan

study finance :^)

Dannyu NDos

22:09

Finance? Nah, I don't mean I'm greedy.

Claudio

$f(\mathbf{x}):\mathbb{R}^n \to \mathbb{R},A \in \mathbb{R}^{m \times n}, \mathbf{b} \in \mathbb{R}^n$ and $f(x) = x^T A^T A x -2b^T A x +b^T b$. The matrix $A$ has full rank. I know that $f$ is convex. I proved that $A^T A$ is positive definite. Is the following line safe to write: $$f(x) \ge \lambda_{min}|x|^2-2b^TAx+|b|^2 \to \infty \text{ as } |x| \to \infty $$

Jakobian

@DannyuNDos Tired of math? Study math

7

Claudio

@Jakobian Insert understandable, have a great day meme

copper.hat

@Claudio If by $\lambda_\min$ you refer to $A^TA$, then yes.

Claudio

yeah the smallest eigenvalue

sorry forgot to mention that :P

Sine of the Time

22:14

@Claudio -1 and vote to close for lack of context

copper.hat

It just follows because for any symmetric (real) matrix you have $x^TBx \ge \lambda_\min(B) \|x\|^2$.

Dannyu NDos

I mean... I don't intend to take any undergrad course from the college that Political Finance belongs, and if I intended so, I'd rather take North Korean Studies.

copper.hat

It is true for any $x$.

Claudio

that much I know, I was worried about the middle term

Ben Steffan

what's an eigenvalue :(

copper.hat

22:15

who knows, they eat cats & dogs apparently

Dannyu NDos

Huh?

We don't.

Claudio

i was thinking that maybe I should take $|b| |Ax|$ instead of leaving it like that

copper.hat

@Claudio it really depends on what you are trying to do, hard to guess :-)

Dannyu NDos

@copper.hat I mean, I'm aware North Koreans do, but we South Koreans don't.

Claudio

I wanna show that $f$ is coercive

Ben Steffan

22:17

I have understood what a left module is
great success!

copper.hat

i don't have a philosophical objection to eating cats, dogs, horses, etc.

Claudio

namely $\lim_{|x|\to \infty} f(x) \to \infty$

Ben Steffan

I don't think I have time to understand what a module is unfortunately

copper.hat

@Claudio all you need is that $A^TA$ is invertible.

Claudio

wait I don't know

No wait, I need to know that $\lambda_{min} >0$

But I showed in a previous exercise that it is indeed the case

copper.hat

22:19

$A^TA$ is invertible iff $\lambda_\min >0$.

Think of $A^TA$ as diagonal.

Claudio

I'm not sure about the sign

copper.hat

yeah, i'm mixed about ace of base

Ben Steffan

...

Claudio

oh ok I showed that $A^TA$ is semidefinite positive so I used the fact that $A^TA$ is invertible

then I'm done

Fun story: I was talking after the lecture about this exercise and my professor realized that the exercise did have some problems since $A$ was a generic mxn matrix, therefore she added (today) the stronger assumption that $A$ has full rank lol

Ben Steffan

that happens

Sine of the Time

22:26

but it's not funny when happens during exams :(

Ben Steffan

good of you to notice and talk to her :)

@SineoftheTime true,

copper.hat

$x^T A^T A x -2b^T A x +b^T b \ge \lambda_\min \|x\|^2 - 2 \|b\|\|x\| -\|b\|^2 \ge \|x\|^2(\lambda_\min+{2 \|b\| \over \|x\|} -{ \|b\|^2 \over \|x\|^2})$.

Claudio

funnier fact: she discovered that the weaker assumption still works and to back this up she linked this

Ben Steffan

but in exams somebody always seems to notice

Claudio

@SineoftheTime dang

@copper.hat how did you get that mid term expression?

the matrix disappeared?

copper.hat

22:29

Oops, add in $\|A^Tb\|$, sorry

too late to edit

Claudio

is $\langle b, Ax\rangle \ge |A^Tb| |x|$ trivial to show?

copper.hat

i took an exam from Manuel Blum (complexity guy) and there was an ambiguity in the question. during the exam i asked him to elaborate but he refused so i answered both forms of the question. a few weeks later he made a fuss about it in class and wrote my name on the board.

i ask all range of questions from blindingly obvious to subtle.

strangely, people think i am stupid for asking the latter class

Ben Steffan

@copper.hat yikes

that's terrible

don't call people out in class

Claudio

@copper.hat I've found a useful result: $$ |Ax|\le |A||x|, |A| = \sqrt{\sum_{i}^{m}\sum_{j}^{n}a^2_{ij} }$$

copper.hat

it was neither positive nor negative

Ben Steffan

22:35

there's still absolutely no reason to put somebody's name on the board

copper.hat

@Claudio you need to be careful with that. what norm are you using on $x$

Claudio

euclidean

copper.hat

that follows because $\|A\|_2 \le \|A\|_F$, where the first is the induced Euclidean norm.

But, in general, the Frobenious norm is not submultiplicative.

Claudio

this might be too advanced for me hahaha

copper.hat

A cute result is $\|A\|_2 = \sigma_\max(A), \|A\|_F = \sqrt{ \sum_k \sigma_k(A) }$.

you should absolutely learn about the svd if you have not already.

the svd is not a disease

Claudio

22:43

@copper.hat I will drop the Frobenius norm result and just go with Cauchy-Schwarz, you kinda scared me there hahaha

I always get the two Schwar(t)z's wrong lol

copper.hat

sry, didn't mean to. sometimes more knowledge is not a good thing.

Claudio

no worries hhaha :p I greatly appreciate your help btw @copper.hat

copper.hat

@Claudio :-) just glanced at your profile. my 21 year old son is somewhere in Italy atm.

Dannyu NDos

I was playing with abelian groups and categorical limits.

Then I noticed that the direct limit of $\mathbf{0} \to \mathbb{Z}_2 \to \mathbb{Z}_4 \to \mathbb{Z}_8 \to \cdots$ is isomorphic to $\mathbb{Z}[1/2] / \mathbb{Z}$, where $\mathbb{Z}[1/2]$ is the adjoint ring with multiplication forgotten.

Claudio

@@copper.hat that's an awesome coincidence hahaha

copper.hat

22:51

he's working on a farm (WOOF?) he loves the food.

Claudio

@copper.hat wait how did you know Im 21 years old as well?

Dannyu NDos

And the inverse limit of $\mathbf{0} \leftarrow \mathbb{Z}_2 \leftarrow \mathbb{Z}_4 \leftarrow \mathbb{Z}_8 \leftarrow \cdots$? It's isomorphic to the 2-adic integers.

copper.hat

@Claudio I didn't :-)

Claudio

dang double coincidence :p

copper.hat

he was in Modena for a while, I insisted that he visit the Ferrari museum :-)

my first time in Italy was a long time ago, in Rimini.

Ben Steffan

22:54

a wild $\mathbb{Z}(2^\infty)$ appears

Claudio

@copper.hat that is awesome, farm work is awesome, it clears the mind and lets one appreciate what he eats truly

@copper.hat Modena is great for food, maybe great is a bit of an underestimation :p

copper.hat

i suppose, i grew up in Ireland in the 60's, farm work was pretty tiring & monotonous :-)

Jakobian

@copper.hat at least it only has a countable amount of discontinuities

copper.hat

i think i am missing some subtle thread here :-)

Dannyu NDos

It's funny that the direct limit has countable cardinality, while the inverse limit doesn't.

Claudio

22:57

@copper.hat We completed the olive harvesting on friday. My back still hurts, but I enjoyed it a lot. I don't know, after reading Vagabond's farming arc my whole view changed :p

but Ireland is a whole different beast hahaha

copper.hat

@Claudio I suspect the weather in IT is a little nicer to be outside than IRL :-)

Ben Steffan

@DannyuNDos is that surprising? :)

Claudio

where I live yeah, the north has been going through floodings and vexing hailstorms unfort

copper.hat

Sorry to hear that.

Claudio

@copper.hat but yeah Ireland is obv a bit rougher

copper.hat

22:59

Different, everywhere has its ups & downs...

Claudio

@copper.hat Italy, now Spain

copper.hat

Yeah, one of the lads in my group lives near Valencia.

Dannyu NDos

@BenSteffan I mean... I tried acquiring $\mathbb{Z}$ from finite abelian groups, and ended up with totally different objects.

Claudio

That is sad, hope he's doing fine. Btw, I realized it's 12 AM already, I need to go to bed, I have lessons at 8 tomorrow :|

copper.hat

Good night :-) I'm off to try cycling for a while, I managed to injure my hamstring a few days ago, I'm hoping I can at least cycle.

Claudio

23:01

have a good cycling session then :p

copper.hat

Buona notte (thx Google)

Ben Steffan

@DannyuNDos valiant endeavour, but ultimately doomed :P

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$Mathematics$ Mathematics