Mathematics - 2023-11-01

A. P.

00:24

Hi, I am trying to understand this part of the following proof: $m(A_n)^{1/p}\leq C m(A_n)^{1/q}$. But I cannot follow it. Can you explain me that part?

71

A: $L^p$ and $L^q$ space inclusion

In Rudin's book Real an complex analysis, we can find the following result, shown by Alfonso Villani: Let $(X,\mathcal B,m)$ be a $\sigma$-finite measure space, where $m$ is a non-negative measure. Then the following conditions are equivalent: We have $L^p(X,\mathcal B,m)\supset L^q(X...

If this it is true for $f=1_{A_n}$ then is it also true for $f$?

user 726941

03:07

Answering machine message at a mathematics department : "The number you have dialed is imaginary, please rotate the keypad 90 degrees and try again."

Xander Henderson

03:28

@ioch nope. Only snakes.

Ted Shifrin

@Xander I saw a snake walking down the street earlier this afternoon. And a toucan — I think — driving by in a car.

user 726941

Watch out he's a diamond back supporter

Ted Shifrin

Well, these days neither state is high on my personal list, but Texas is down at the bottom with FL.

copper.hat

i programmed my outside light's brightness to ramp up & down to add some Halloween flavour

user 726941

🎃🎃🎃

Ted Shifrin

03:39

Does it moan, too, copper?

user 726941

🎇🎆🎇🎆

copper.hat

unfortunately no, just myself

Ted Shifrin

Well, we knew that.

copper.hat

at some point i wanted to have a gas power flame sort of thing outside but the regulatory issues were too much

think ring of fire

user 726941

👻☠️👻☠️

copper.hat

03:41

for reasons unknown, our porch is currently infested with mosquitoes, never seen this density before

Ted Shifrin

It’s the imaginary climate change.

copper.hat

just rotate right and it will all be real

user 726941

Johnny Cash did a good ring of fire

copper.hat

for some reason i thought he served time in san quentin, but he didn't, he just played there

user 726941

Yup

copper.hat

03:46

i'm off looking for psqs...

and trying to figure out how to work ansatz into a sentence

user 726941

They're great for creating chatrooms if the OP is a willing leaner

Ted Shifrin

Make sure you capitalize. You know … noun in German.

user 726941

For those unable to afford online tutoring

copper.hat

i will capitalise appropriately...

user 726941

would that make you an opportunistic capitalist 🤔

copper.hat

04:00

a catalytic capitaliser

Thomas Finley

Suppose $G$ is a finite abelian group and $p|o(G),$ where $p$ is a prime number. Then there is an element $a\neq e\in G$ such that $a^p=e.$------- This is what the Cauchy's Theorem for Abelian Groups say. My question is, does it also gurantee somehow, that $a^m\neq e(,m\in\Bbb Z)$ when $m\lt p$ ?

Ted Shifrin

You know enough to answer that.

robjohn

@ThomasFinley Have you heard of Bezout's Identity?

Thomas Finley

@robjohn yes.

@TedShifrin ha ha, the point is I last worked with group theory one year and 6 months ago, and while doing ring theory, I needed this theorem, so I am in need of clarifications here.

Ted Shifrin

No excuses. Figure it out.

If you think you belong doing research, stop asking us for help and work on it.

This is completely elementary.

robjohn

04:13

@ThomasFinley If $m\lt p$, what can you say about $m$ and $p$?

Thomas Finley

@robjohn We can say that $gcd(m,p)=1.$

Thomas Finley

04:25

@TedShifrin Maybe I have figured it: No, $a^m=e$ when $m\lt p$ and $a\neq e$ can never be true if $a^p=e.$ This is because, say $a^p=e$ and $\exists m\lt p$ such that $a^m=p$ then, $o(a)\lt p$. We call $o(a)=m'\lt p$ and note that, $a^b=e$ for some integer $$ iff $m'|b$ which means that, $m'|p$ as, $a^p=e$ which further implies that, $m'=1$ and so, $a^{m'}=a^1=e=a,$ a contradiction because $a\neq e$ by assumption. ------ This ends the argument. @robjohn Is it correct?

Oh, I made a typo. That should be $a^m=e$

copper.hat

04:41

@ThomasFinley you can edit your comments for a few mins after entering them.

robjohn

@ThomasFinley If $a^m=e$ and $a^p=e$, where $p$ is prime and $m\lt p$, what is also true?

Thomas Finley

@copper.hat I checked it, but the time was over :/

@robjohn I didnot get what you are saying. Are you saying that my argument has a flaw?

@robjohn Here is a repost of my argument with the typo fixed: $a^m=e$ when $m\lt p$ and $a\neq e$ can never be true if $a^p=e.$ This is because, say $a^p=e$ and $\exists m\lt p$ such that $a^m=p$ then, $o(a)\lt p$. We call $o(a)=m'\lt p$ and note that, $a^b=e$ for some integer $b$ iff $m'|b$ which means that, $m'|p$ as, $a^p=e$ which further implies that, $m'=1$ and so, $a^{m'}=a^1=e=a,$ a contradiction because $a\neq e$ by assumption.

robjohn

05:06

You say that $a^b=e$ iff $m'\vert b$. Why?

Thomas Finley

@robjohn This is because of the theorem: "Let G be a group, and let g ∈ G have order m. Then $g^n = e$ if and only if m divides n."

user 726941

1 hour ago, by Ted Shifrin

If you think you belong doing research, stop asking us for help and work on it.

Thomas Finley

@user726941 If you think, you understand what I am saying then stop posting things in here that is irrelevant at the moment and work on understanding the conversation.

TedShifrin said it to me, when I was asking for the clarifications. Presently, as things went, I solved the problem and now, robjohn inquires something about my approach, to which I am replying now.

So the message you posted here becomes totally irrelevant, out of context and weird.

robjohn

05:22

The basic thing I was trying to get to was to say that if $a^m=e$ and $a^p=e$ was that since $(m,p)=1$, we have $x,y\in\mathbb{Z}$ so that $xm+yp=1$. This means that $e=(a^m)^x(a^p)^y=a^1$

Thomas Finley

@robjohn Ah, that's a simpler approach.

user 726941

1 hour ago, by robjohn

@ThomasFinley Have you heard of Bezout's Identity?

Very important^

and basic

:-)

Thomas Finley

@user726941 yeah, but as I did, this can be verified using Bezout's identity as well ;)

robjohn

I've used Bernoulli and Bezout to prove many things.

Thomas Finley

@robjohn Bernoulli's inequality?

robjohn

05:29

$(1+x)^m\ge1+mx$ where $m\ge1$ or $m\le0$

it is reversed if $0\le m\le1$

Thomas Finley

If it's the inequality, then I can agree with you as well. They both are really handy tools.

@robjohn oh, yeah I agree.

TheGhostOfGaussPossessesMe

0

Q: A tight positivity conjecture about sums over divisors of square-free integers.

Let $p_n$ be the $n$th prime number and all variables, unless otherwise specified, are natural numbers. Conjecture: For all fixed $k \geq 0$ and square-free $n \gt k$, the following function evaluates to a positive integer: $$F(n) = \sum_{d \mid n} \mu(d) \sum_{a^2 = 1\mod d, \ \\ 0 \leq a \lt d...

Hi, @Koro @shintuku

This is a conjecture of mine I just discovered using some code verification

You can change the inner group and also the numerator to different things for different problems

But the sum's always got to be over square-free ints

@LukasHeger I started the chapter on complex integration, will ping you soon. I should pay you for last time at least since you did take some time to create problems

I don't like those problems though. Let's start out next on the integral theorems etc.

The meat of $\Bbb{C}$ analysis

Koro

06:03

I find Munkres manifolds text to be very chatty.

So much verbose

leslie townes

this text, is it talking to you right now?

Koro

Lot of statements in the book seem to be bloated with verbose to the point that the presentation becomes sleep inducing.

TheGhostOfGaussPossessesMe

06:27

@Koro pretty soon you'll get into cohomology, that's exciting!

TheGhostOfGaussPossessesMe

06:43

K made some edits, the conjecture is starting to make more sense. No tight constant present

$$F(n,m) = \sum_{d \mid n} \mu(d) \sum_{a^2 = 1\mod d, \ \\ 0 \leq a \lt d}\left\lfloor\frac{m-a}{d} \right\rfloor$$ always positive!

user 726941

It's the texts that talk to me after I close the book that worry me @leslietownes

or that creep into my dreams...

TheGhostOfGaussPossessesMe

If you could only remember them lol

user 726941

psychologists say that memory is the residue of thought

TheGhostOfGaussPossessesMe

Hopefully they're residues modulo a large prime number so that you have inverses

user 726941

They justify memorization by saying that if you don't remember it you haven't thought about it enough

Koro

07:01

$क^2+ख^2+2.क.ख=(क+ख)^2$ :-)

user 726941

Looks like a perfect square trinomial to me

Koro

I thought latex won't render it but it does.

user 726941

$$क^2 - ख^2 = (क +ख)(क - ख)$$

Koro

:)

Let $फ :म \to \mathbb र$ be a map. (goes well with \mathbb too)

冥王 Hades

@ioch it means those equations are circular will tell you nothing that you don’t already know about the diagram, because that’s not how they’re meant to be solved

user 726941

07:15

Or you could've made an algebraic mistake setting them up

冥王 Hades

Or that, depends on the problem. We can’t give a meaningful answer without seeing it

user 726941

Exactly

If you want to hand-wave go to the physics chatroom please

冥王 Hades

Or be a cheerleader

user 726941

Or that^

user 726941

07:32

> The opposite of hand-waving in mathematics (and related fields) is sometimes called nose-following, which refers to the unimaginative development of a narrow line of reasoning that—while correct—can also end up making the subject dry and uninteresting.

Soumik Mukherjee

07:48

@Koro 2.क.ख is the extra energy

Lukas Heger

08:33

@TheGhostOfGaussPossessesMe thanks for the feedback. How do you want us to proceed? Should I do exercises on complex integration? If yes, do you want more computational exercises or more proofs? I'm also fine with just answering questions, if you prefer that

TheGhostOfGaussPossessesMe

Let's work with questions at first, then we'll see what exercises @LukasHeger

and proofs too

Lukas Heger

ok!

TheGhostOfGaussPossessesMe

So no need to create anything yet

Just wait! I want to read a little into the integration parts

Lukas Heger

sure

TheGhostOfGaussPossessesMe

@Shaun thanks for the alignment of my equations :)

RÜFÜS DU SOL - Always [Official Audio]

►

@Peter did you see my comment in some post about the $X^2 + 1 \in \Bbb{P}$ i.o. problem?

The comment is in contradiction with the answer given

So is that problem solved or ...?

I think most think that it's unsolved, I do too

Peter

08:48

I wanted to look up the latest status in wikipedia , but aborted because of an annoying donation appeal. The last time I looked up Bunyakovsky conjecture , the claim was that no poylnomial with degree 2 or more has been solved. No idea why the paper claims otherwise.

The iceland mathematician made an astonishing discovery but it does not cover the claimed proof , as I mentioned.

TheGhostOfGaussPossessesMe

😎

@Peter have you seen this yet (mine): math.stackexchange.com/questions/4798144/…

It relates to twin primes, but I failed to mention that because of bad rap

It actually relates to many problems, many problems have counting functions of this form

Peter

Not yet carefully read. The expression is complicated.

TheGhostOfGaussPossessesMe

Yes it is. But also it's quite simple! :> Simple but very hard to deal with

You can use other groups of units modulo the $d$

those would be for other problems. The $a^2 = 1$ group is for twin primes

cosets of the group is for $2k$-separated primes for $k \geq 1$

They touch on such functions in sieve theory but never attack them directly / with arithmetic functionology

As we know sieve theory suffers from "parity problem" and even Zhang noted his approach can't quite reach twin primes, i.e. has the theoretical limit of $k = 3$

I want to use multiplicativity, but it seems hopeless heere

Peter

Maybe , it is worth mentioning that the twin prime conjecture is open , but the not much easier claim that infinite many prime gaps do not exceed 246 , is proven ! For me, absolutely baffling.

TheGhostOfGaussPossessesMe

Yes, it is!

Very strange

Peter

09:04

Similar baffling that the weak Goldbach conjecture is proven , but the strong still open. What makes the strong conjecture so more difficult than the weak conjecture ?

TheGhostOfGaussPossessesMe

Ikr. And why can't they find some reasoning on top of those results that finishes it u

up

Certainly a proof could gain a lot of footing if that were possible

Peter

Ikr ? I know some chat-abbreviations , but not this.

TheGhostOfGaussPossessesMe

I know, right

It's the primes. They're maddening 😆

Peter

And that we do not know whether there are infinite many NON-Wieferich primes is the weirdest I ever heard.

TheGhostOfGaussPossessesMe

You have unique factorization multiplicatively, but when you add those generators, essentially change the operation, we can't figure out the problems

Terry Tao will prove these problems one day :)

Peter

09:10

The one with the famous arithmetic-progression-proof ? In fact , he is a candidate to solve this , if it is even feasible.

TheGhostOfGaussPossessesMe

Yes he's worked on several of those arithmetic progression problems

He met Erdos as a kid

His book "Additive Combinatorics" is pretty awesome

Peter

Erdoes was a truely fascinating mathematician. Many beatiful conjectures , some solved.

TheGhostOfGaussPossessesMe

Erdos invented the probablistic method

It's like probablistic vs deterministic algorithms. Sometimes probablistic algorithms are just easier to code for certain problems or out perform their deterministic counterparts

Peter

It is also a very useful tool to estimate the expected number of primes one finds in some project or to predict how many prime factors will be in a particular range if one tries to factor numbers.

robjohn

09:28

@Koro $\unicode{x0915}^2+\unicode{x0916}^2+2\cdot\unicode{x0915}\cdot\unicode{x0916}=(\unicode{x0915}+\unicode{x0916})^2$

Koro

10:01

@robjohn cool :-), how did you create/find the codes?

found it :)

Nasser

10:24

math gurus: if an integrand have singularity (or not defined) at a point, say x0. Does this mean the antiderivative will also have singularity or not be defined at same point? I can't find counter example. For example integral of 1/x is ln(x) and both of these "blow" up at x=0. same for integral of 1/x^2. Same for integral 1/sin(x) and so on. Is this true in general?

Astyx

$1/\sqrt{x}$

Nasser

@Astyx thanks. Good counter example. So just because the integrand has singularity at one point, this does not mean the antiderivative will also. got it.

Astyx

Glad to help

Jakobian

11:10

@Nasser this is the reason why Riemann integration sucks. You can only integrate bounded things

Nasser

@Jakobian at school that is all what I learned :) I am not math major.

robjohn

11:44

@Nasser $\log(x)$ blows up at $x=0$, but the integral $x\log(x)-x$ does not.

$x^{-a}$ for $0\lt a\lt1$ blows up at $x=0$, but $\frac1{1-a}x^{1-a}$ does not.

Nasser

@robjohn but x*ln(x) is not defined at x=0 ?

robjohn

the integral is $0$ as is the limit

blowing up at $0$ has nothing to do with the value at $0$.

Shaun

@TheGhostOfGaussPossessesMe No worries :)

Paramanand Singh

12:07

@Koro that's a nice identity using variables from Hindi language. Long back my primary education was in Hindi, but I never saw any variable names in Hindi. In particular polynomials were always in variable $x$.

Mad

12:21

"A topological manifold of dimension n is a nonempty Hausdorff space M satisfying the second axiom of countability axiom, such that any
point p ∈ M admits an open neighborhood U and a homeomorphism ϕ : U → V
to an open subset V ⊂ R
n. We call (U, ϕ) a chart of M."
Does the part of "Such that, ... " follow from the previous attributes,or is it part of the definition?

Koro

@ParamanandSingh Hi. Do you know about SL Loney's trigonometry book? At my school library long ago, I found the book in hindi. There I learned that स्पर्शज्या (अ)= ज्या(अ)/कोज्या(अ)

And numbers 1,2,3,... were also written in hindi.

I couldn't find any x, y etc. in the book.

in hindi books also, there write sin x, cos x etc. but in that book everything was in hindi.

Mad

Apparently it does not follow, this is a different requirement.

Koro

@Mad it's a part of the definition.

Also, your definition as written is wrong.

Mad

@Koro how come?

Koro

you should say clearly where V lies.

Mad

12:33

R^n?

Koro

exactly.

Mad

its just copy paste error.

Koro

for every p, there is an open set U in M and an open set V in R^n such that...

ncert.nic.in/textbook.php?lhmh1=6-6 also uses x,y,z etc.

@ParamanandSingh I vaguely remember that very old Maths books from author Dr. Harswaroop Sharma were completely in hindi.

I also want to learn topological terminologies in hindi but I couldn't find any topology book in hindi. I know it exists because I saw one mathematics question paper in which translation of 'compact set, closed set' etc. was written. This means there are some people who write that exam in hindi.

sadly, I don't remember the name of the exam.

PNDas

14:23

I have that $u_m\in H^1(U)$, $||u_m||_{L^2(B_1)}=1$ and $||\nabla u_m||_{L^2(B_1)}\to0$. The books says that there exists a constant $u_0\in H^1(B_1)$ such that $u_m\rightharpoonup u_0$ in $H^1(B_1)$ and $u_m\to u_0$ in $L^2(B_1)$. But I am only getting subsequence. Can you please check my ideas?

First, the sequence is bounded in $H^1$ so it has a weakly convergent subsequence in $H^1$. Now Rellich Kondrashov says that this subsequence has another subsequence which converges in $L^2$ (it'll converge to the same function). Now using the definition of weak derivative we get that weak derivative of the limit function is $0$ which means the limit is constant everywhere.

ephe

15:12

How can I show that if $B$ is open then $B\cap\overline{A}\subset \overline{B\cap A}$? I've seen this proof: https://math.stackexchange.com/a/3027499/967766 but I don't think it's correct since $s$ isn't arbitrary. I've tried the following:

Let $x\in B\cap\overline{A}$. Then since $x\in\overline{A}$, we have that $\forall r>0 \ B(x;r)\cap A\neq\varnothing$. But $x\in B$ also so $B(x;r)\cap A\cap B\neq \varnothing$. Since $r$ was arbitrary, $x\in\overline{A\cap B}$.

But I suppose it's wrong since I didn't use that $B$ was open.

Edit: I can understand the proof using just open sets but I can't figure out how it works using balls in $\mathbb{R}^n$

Alessandro Codenotti

You know that $B(x,r)\cap A\neq\varnothing$ since $x\in \overline{A}$, and you know that $B(x,r)\cap B\neq\varnothing$, since $x\in B$, but how are you jumping to $B(x,r)\cap A\cap B\neq\varnothing$?

ephe

@AlessandroCodenotti Oh okay. For some reason I thought that the intersection should have contained the center but that's clearly not true. Thank you.

sunny

15:52

I'm reading a soft introduction to complex analysis and I have a basic question. What does it mean for a (line) integral to be "locally independent of the path"? In particular, what does the word "locally" mean in this context?

ephe

16:17

I still can't figure out why this proof works:

Let $x\in {\overline A}\cap B$. Since $B$ is open, we can chose $r > 0$ such that $B_{r}(x)\subset B$. For some $0 < s < r$ we still have $B_s(x)\cap B\neq\emptyset$ and since $x\in\overline A$, then $B_s(x)\cap A\neq\emptyset$. Thus $B_s(x)\cap(A\cap B)\neq\emptyset$.

We have just shown $x\in \overline{A \cap B}.$

What happens if $s\geq r$? Couldn't the intersection be empty then? I'd really appreciate some help.

Jakobian

@sunny it means that the line integral over two homotopic paths is the same

that's what they mean at least

Koro

7

Q: If $0<A\le x_n\le B~~ (\forall n)$, and $\lim_{n\to\infty}\sqrt[n]{x_1x_2\cdots x_n}=A$. Prove $\lim_{n\to\infty}\frac{x_1+x_2+\cdots+x_n}{n}=A.$

Problem Assume $ 0<A \le x_n\le B~~ (\forall n)$, and $\displaystyle \lim_{n \to \infty}\sqrt[n]{x_1x_2\cdots x_n}=A$. Prove $\displaystyle\lim_{n \to \infty}\frac{x_1+x_2+\cdots+x_n}{n}=A.$ Is it true or not ? Probably it holds, but seems to be difficult to prove. If we consider using the squee...

real-analysis calculus limits

Ted Shifrin

16:48

@sunny Locally means locally: If you wiggle the path a little bit, the integral won't change. But consider, for example, integrating $\int dz/z$. If you change your path from $A$ to $B$ so that now it winds an extra time around the origin, the integral will change.

Sine of the Time

hi, what are some techniques to prove that a space (of function) is closed?

sunny

makes sense 🤸‍♂️

Ted Shifrin

@Sine Too broad a question. Take a convergent sequence?

Sine of the Time

I'm struggling to prove that $V=\{f\in L^2 | \, xf \in L^2 \}$ with the norm $\|f \|^2=\int (1+x^2)|f|^2 $ is a Banach space

the integral is over $\Bbb R$

Ted Shifrin

17:03

If $f_n\in L^2$ and $f_n\to f$ in $L^2$, then we know $f\in L^2$. Note that convergence in your norm implies convergence in the $L^2$ norm.

Sine of the Time

but should't I take $f_n \in V$?

Ted Shifrin

Of course, you take $f_n\in V$.

I need to leave now. Good luck :)

Sine of the Time

Ok, I'll wait for your help :)

Jakobian

@SineoftheTime I thought you did that already

Sine of the Time

I didn't manage to solve it

Jakobian

17:11

I think you want to follow Ted's advice. Given a Cauchy sequence $f_n$ in $V$, take limit of $f_n$ in $L^2$, call it $f$. Then show $\|f_n-f\|\to 0$

This is the "obvious" candidate for the limit

Sine of the Time

Ok, I'm trying to do it. I'll let you know

Jakobian

you also have some limit $g$ of $xf_n$ (in $L^2$), might be good thing to consider

We then show that $g = xf$ or something

Yeah I'll let you do this one

Sine of the Time

17:27

@Jakobian can you check if my proof is correct? It'll take my some time to write it

Jakobian

sure

Sine of the Time

$f=\lim_{n\to \infty}f_n$ in $L^2$ and $g=\lim_{n\to \infty} xf_n=xf$. That means $\|f-f_n \|_2<\epsilon_1$ and $\| xf_n-xf\|_2<\epsilon_2$. Now $\|f-f_n \|_V=(\int |f-f_n|^2 +x^2|f-f_n|^2 )^{1/2}\le \|f-f_n \|_2+\|xf-xf_n \|_2<\epsilon_1+\epsilon_2$

Jakobian

Justify that $g = xf$

Sine of the Time

$\lim_{n\to \infty} xf_n=x\lim_{n\to \infty}f_n=xf$ (?)

Jakobian

The limit is in $L^2$. What makes you think you can just take $x$ out of the limit

Sine of the Time

17:37

oh right, I'm dealing with convergence in norm

So I should prove that there's also pointwise convergence?

Jakobian

There will be no pointwise convergence

we're not even dealing with functions, but equivalence classes of them

Sine of the Time

you're right

Jakobian

however, you can use that there is a subsequence which almost surely converges

Sine of the Time

what do you mean by almost surely

Jakobian

This will mean there is a subsequence $n_k$ such that $f_{n_k}\to f$ and $xf_{n_k}\to g$ almost surely

well, I'm usually just saying almost surely, its just from probability

almost everywhere

Sine of the Time

17:43

ah ok

hence $xf=g$ almost surely (?)

Jakobian

anyway, you can see this means that $xf = g$ almost everywhere, that is as equivalence classes

$xf = g$ in $L^2$ lets say

Sine of the Time

ok

the final part of the proof is right?

Jakobian

Did you know that lemma we used about existence of subsequence which converges almost surely?

Sine of the Time

yes

Jakobian

@SineoftheTime well its certainly not written like an analysis proof

What are epsilons, what are $n$

I know what you're trying to do but you're not doing it right

I'd really wouldn't even bother

We've proven $f\in V$ already so we might write $\|f-f_n\|_V \leq \|f-f_n\|_2 + \|xf-xf_n\|_2\to 0$

just arrow that it converges to $0$

Sine of the Time

17:48

So the epsilon are not important

Jakobian

You can do without them, yes

Sine of the Time

ok makes sense

X4J

I saw this question:
Suppose $T: \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$ is a linear transformation with $0 < dimKer(T)$.
Is it possible that there exists $u_1, u_2 \in \mathbb{R}^{n}$ s.t $u_1$ is the only vector that satisfies $T(u_1) = u_2$ ?

Trying to answer this, my only intuition was that if by the way of contradiction the argument is true then $(T(e_1), \dots, T(e_n))$ are independent since otherwise we would have two different representations in respect of $(e_1, \dots, e_n)$ for $u_2$ and so for $u_1$ and that would imply that there are two vectors that satisfy this, which is a…

Jakobian

18:05

@X4J if $T(x) = u$ has solutions in $x$, then fixing one such solution say $u_0$, gives us that the set of al solutions is $u_0+\text{ker}(T)$

this of course means, there will always be more than two

X4J

18:22

@Jakobian This is very obvious when you represent it as solutions to an equation, thanks. Thing is, I am not sure if the intuition\idea I described above is actually true (this is also a different perspective than the one you represent I guess)

Jakobian

18:36

The thing with intution is that its yours and not mine

Ted Shifrin

19:03

@X4J I didn't read your whole argument. It seems so convoluted and complicated. Just use the meaning of $\ker T$. If $u_1$ satisfies $Tu_1=u_2$, then so does $u_1+v$ for $v\in \ker T$. Also, a bit of advice. Do NOT use the same letter for elements of domain and range. That leads to confusion (for you and for the reader). Use $u$'s for elements of the domain and some other letter, say $z$, for an element of the range.

Yeah, see, when I look at your argument, you're using the same basis for domain and range. The argument should work for $T\colon\Bbb R^n\to\Bbb R^m$ with $m$ different from $n$ and then you'd have to take different bases for domain and range. It isn't about bases and representations of vectors with respect to bases. It's purely about the linear map $T$, which you are ignoring.

I don't see what your intuition is intuition about here, at all.

amWhy

19:42

Alas! @TedShifrin, where is the Great Pumpkin/Great Gourd, with the mean face?

Ted Shifrin

@amWhy robjohn had a wonderful Halloween nom-de-plume, but I haven't seen him today.

amWhy

I missed my chance, yesterday. Looks like he was shaped like a pumpkin pie?

Horiatiki

@copper.hat Can you explain yourself better please? Explain yourself with a practical example relating to goals. Anyway I saw your reply. I left you a comment because something isn't right. Thank you

Is there anyone who can help me? BOUNTY+50: Gauss-Newton algorithm to solve a non-linear system?

X4J