Mathematics - 2024-10-24 (page 1 of 2)

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Xander Henderson

00:10

The monolith aliens said stay away!

Obliv

00:21

Can someone help me understand these lecture notes

user image

is that "given $a \in \mathbb{R}$ and $\epsilon >0$" part meaning to quantify for all $\epsilon>0$ or just fixing some $a,\epsilon$

I'm guessing it's the latter, otherwise $V_{\epsilon}(a)$ would just be all of $\mathbb{R}$

and example 1. how would $\varnothing$ be an open set if there are no points for which the definition applies to?

Sine of the Time

@Obliv suppose $\emptyset$ is not open, what can you say?

leslie townes

01:27

obliv: do you mean, "how can emptyset fail to satisfy a definition that requires absolutely nothing in the case of the emptyset"? or as sine has phrased it, how could emptyset fail to be empty? (it would, among other things, have to have a point in it, to fail to be open)

6 hours later…

Pizza

07:51

@SineoftheTime So at 4:30 I have the exam, I'll let you know how it goes :)

Sine of the Time

08:06

@Pizza :)

Soumik Mukherjee

08:19

All the best @Pizza

1 hour later…

Pizza

09:37

@SoumikMukherjee A thousand thanks !

2 hours later…

psie

11:50

I feel silly since I'm almost 100% sure it's a typo, but in this answer, the author writes $\|T^{-1}(y)\| \leq \frac{1}{\alpha}\|T(x)\|=\alpha\|y\|$ where $y=T(x)$, and one of the comments also says that we should be able to deduce $\|T^{-1}(y)\| \leq \alpha\|y\|$, but shouldn't the inequality read $\|T^{-1}(y)\| \leq \frac{1}{\alpha}\|y\|$?

Then it makes sense too that $\alpha$ is some number $>0$.

Otherwise the condition $\alpha>0$ doesn't really have any effect.

Jakobian

12:10

@psie answer?

psie

Question, I mean.

Jakobian

well, what do you think?

psie

I think I'm right :)

Jakobian

right. You don't need anyone to confirm you on such triviality I'm sure

psie

the comments made me insecure

Jakobian

12:14

no need to be so easily swayed by other people

trust your logic

psie

working on it

but thanks

Jakobian

good luck

2 hours later…

Jakobian

13:59

When can a Tychonoff space be densely embedded in a non-normal space?

I think I might have an idea

Adil Mohammed

0

A: Properties of Cross Product to infer the null space of this transformation

You are right; the negative of a linear transformation is negative. Indeed, since $T(x)=-C(x)$ and $C(x)$ is linear we have $$T(x+y)=-C(x+y)=-[C(x)+C(y)]=-C(x)-C(y)=T(x)+T(y)$$ and $$T(cx)=-C(cx)=-cC(x)=cT(x)$$ The null space is the set of vectors whose cross product with $v$ is $\vec{0}$. You m...

is this answer correct?

Vladimir Lysikov

@AdilMohammed Yes.

Adil Mohammed

my issue is probably with some sites saying its cross product and others saying its dot product (like wikipedia i think). i assume thats probably because of different definitions

Vladimir Lysikov

14:38

cross product and dot product are two different things, they are both bilinear
The map sending $x$ to $v \times x$ is a linear map $\mathbb{R}^3 \to \mathbb{R}^3$, its kernel is the line spanned by $v$ and its range is the plane orthogonal to $v$ (for nonzero $v$)
The map sending $x$ to $v \cdot x$ is a linear map $\mathbb{R}^3 \to \mathbb{R}$, its kernel is the plane orthogonal to $v$ and its range is $\mathbb{R}$ (again, for nonzero $v$)

ModularMindset

14:59

Can a graph be augmented by a line bundle?

Xander Henderson

15:31

@ModularMindset What does that even mean?

Soham Saha

Hi everyone. Why does the Wikipedia article for Holmium-magnesium-zinc quasicrystal say that the faces are true regular pentagons? In what sense are they “true”?

Holmium–magnesium–zinc quasicrystal

A holmium–magnesium–zinc (Ho–Mg–Zn) quasicrystal is a quasicrystal made of an alloy of the three metals holmium, magnesium and zinc that has the shape of a regular dodecahedron, a Platonic solid with 12 five-sided faces. Unlike the similar pyritohedron shape of some cubic-system crystals such as pyrite, this quasicrystal has faces that are true regular pentagons. The crystal is part of the R–Mg–Zn family of crystals, where R=Y, Gd, Tb, Dy, Ho or Er. They were first discovered in 1994. These form quasicrystals in the stoichiometry around R9Mg34Zn57. Magnetically, they form a spin glass at cryogenic...

Looks pretty beautiful in any case though…

Jakobian

@AlessandroCodenotti for Linelof spaces, I believe I've proved its equivalent to compactness

Compact and zero-dimensional, of course

Maybe I should write a paper on this

Ryder Rude

16:08

hi

what do u think about the many worlds interpretation of quantum mechanics

Xander Henderson

16:22

@SohamSaha The relevant context is in the opening paragraph of the article you cite:

> Unlike the similar pyritohedron shape of some cubic-system crystals such as pyrite, this quasicrystal has faces that are true regular pentagons.

The word "pyritohedron" is a link to en.wikipedia.org/wiki/Dodecahedron#Pyritohedron .

> A pyritohedron is a dodecahedron with pyritohedral (Th) symmetry. Like the regular dodecahedron, it has twelve identical pentagonal faces, with three meeting in each of the 20 vertices (see figure).[3] However, the pentagons are not constrained to be regular, and the underlying atomic arrangement has no true fivefold symmetry axis.

Vladimir Lysikov

And this is important because crystals (structures with translation symmetry) cannot have order 5 axes

आर्यभट्ट

17:11

I everyone I was watching how Fabinnoci number and music is related but video took some new turn creating different mathematical expression from the same numbers can anyone help me in getting crux of the video?

I am not a math major but read number theory in my high school and university

Jakobian

17:29

2

Q: Do Stone-Čech compactifications have property that disjoint closed subsets of $X$ have disjoint closures in $\beta X$?

I came across this in Van Douwen´s paper, Characterizations of $\beta \mathbb{Q}$ and $\beta \mathbb{R}$, Proposition 4. Van Douwen writes: "We show that $\gamma H$ = $\beta H$ by showing that disjoint closed subsets of $H$have also disjoint closures in $\gamma H$." ($\gamma H$ is just an arbitra...

general-topology proof-explanation compactness compactification

a misleading answer (and comment) about Stone-Cech compactifications and their properties corrected

nickbros123

we know closure as the smallest closed set containing a set, and interior as the largest open set within a set, but is there a notion for smallest open set that contains a set?

im not sure if this thing always makes sense

Ben Steffan

such a thing generally doesn't exist

take a point in $\mathbb{R}^n$, say

nickbros123

yeah I was thinking when two open sets dont contain one other

@BenSteffan makes sense

I wanted to prove this result: If $X$ contains two disjoint closed sets $F_1$ and $F_2$, then there exists open sets $G_1$ and $G_2$ such that $F_1\subseteq G_1$ and $F_2 \subseteq G_2$ with $G_1$ and $G_2$ disjoint themselves

ModularMindset

Is anyone aware of anyone who has generalized Ihara zeta functions on k-regular graphs?

Ben Steffan

@nickbros123 that's false generally

nickbros123

17:42

I shouldve added: in metric space $X$

sorry

Ben Steffan

a space that satisfies that condition is called normal or T4

nickbros123

I guess Ill have to use some property of distances here. I have been proving most theorems I came across with just complements, closures and interiors etc that I forgot I was doing metric spacess :)

Ben Steffan

distances should be useful here, yes ;)

Jakobian

a hint is to use distance from a set functions

The point is that a closed set in a metric space is a zero-set (using those functions), and for two disjoint zero-sets $Z_1, Z_2$ there exists a function $f:X\to [0, 1]$ such that $f$ is $0$ on $Z_1$ and $1$ on $Z_2$.

ModularMindset

I suppose those Fat Graphs could provide a little generalization

Pizza

17:48

@SineoftheTime

Anyway I finished, I just didn't manage to do everything unfortunately, I mean I did everything but I didn't finish everything

However, I should have done something well, that is, the exercise procedures are all correct, I think.

psie

I'm working an exercise where the aim is to prove all norms in $\mathbb R^n$ are equivalent. I need the following fact; the unit sphere in the $1$-norm is compact. I guess I need to show boundedness and that it is closed. Or is there a faster way to go about this?

I have no idea how the unit sphere looks like in the $1$-norm.

Sine of the Time

@Pizza glad to hear that

Was it hard?

@psie it pretty fast to show it's bounded and closed, isn't?

Pizza

@SineoftheTime The differential equation was easy, I just forgot how to solve the integral, so I left it half done, I'm sad about that because it was easy, now I'll show you the exam

user image

Ben Steffan

@SineoftheTime but that's not enough to show it's compact :)

unless the question is whether the unit sphere in the 1-norm is compact under the usual norm

Pizza

Exercise 2 had 2 separate requests

That is, the exercises are not connected

I don't know how much was feasible in 2 hours, the double integral if you try to find what theta varies between is a bit strange but it should be 5π/4 and 3π/2

The first one was easy

The 5, is it done with Stokes' formulas?

I'm sad about the differential equation, the integral was easy but during the test I had a blank

I used the parameter variation method

Thorgott

18:00

I sat in a lecture just earlier today where we discussed the equivalence of norms in a finite-dimensional vector space over any complete valued field

Pizza

@SineoftheTime What do you think about the exam ?

Ah, for the theory questions I managed to write the Poincaré theorem, instead for the differentiability questions I wrote the theorem and the proof

psie

@BenSteffan hmm, I've already established that the identity map from $(\mathbb R^n,\|\cdot\|_1)$ to $(\mathbb R^n,\|\cdot\|)$ is continuous. Then I'd like to conclude that since the unit sphere in the $1$-norm is compact, and continuous functions map compact sets to compact sets, it is compact in $(\mathbb R^n,\|\cdot\|)$.

Then I'd like to say that the norm function is continuous, and hence attains a minimum on the compact set. This minimum is $\inf\{\|x\|:\|x\|_1=1\}$ and it is greater than $0$, which implies continuity of the inverse of the identity.

I'm using here the equivalence; identity bicontinuous $\iff$ norms are equivalent.

But I'm not sure about the statement "the unit sphere in the $1$-norm is compact".

Sine of the Time

@Pizza pretty standard

Pizza

@SineoftheTime Exercise 5, is it done with Stokes' formulas?

Sine of the Time

@BenSteffan right, I was thinking about the $2$-norm

@Pizza yes

Pizza

18:08

Oh ok I did it like this

I don't know if you want to try to find out what theta varies between in the integral.

Sine of the Time

we discussed question 6 a couple of days ago

Pizza

yes, I wrote the Poincaré theorem there, I did the proof of the differential theorem

Sine of the Time

I misread

Pizza

Could you do this by making the system where the circle and the line intersect?

Sine of the Time

$\frac54 \pi \le \theta \le \frac 32\pi$

Pizza

18:12

I mean, I got confused here because I was getting strange coordinates,

@SineoftheTime Yes I wrote it like that

Sine of the Time

you used shifter polar coordinates right?

Pizza

Yes

Sine of the Time

in which point?

Pizza

-2 for y

0 for x

Sine of the Time

nice

Pizza

18:14

How did you find them so quickly?

I mean theta

Sine of the Time

if you draw a picture you don't need to do computations

Pizza

Yes but how do you recognize 5π/4?

Sine of the Time

note that $y=2-x$ has slope $-1$

and the slope is $m=\tan \alpha$

Pizza

user image

I had made this system

Sine of the Time

it's easy to see that the line itercepts the $y-$axis in $-2$ and the $x-$axis in $2$, creating a $45-90-45$ triangle

Pizza

18:19

That's why I didn't understand, but seeing -√2 and having radius 2 then I understood

Sine of the Time

geometry is important in double integrals, no need to solve the system in my opinion

if you draw the line and circle, you see easily the angle

no need to waste time solving a system

Pizza

and yes ... I was used to doing it like that

Sine of the Time

though the procedure is correct and it's used in the general cases

Pizza

@SineoftheTime yes but look at the points that come out of the system

That is, y I mean

Anyway, luckily I did well.

Sine of the Time

@Pizza yes, the points are correct

but you don't need them

Obliv

18:22

How could "the intersection of a finite collection of open sets is open"

Ben Steffan

by definition :)

Pizza

@SineoftheTime yes, but how did I figure out that it was 5π/4 from those points of y?

Wait

Obliv

can I get an example? I'm thinking this isn't open: $(0,2)\cap(.5,1.5) = [.5,1.5]$ not open

Ben Steffan

recalculate the intersection, carefully

what you wrote is not correct

Obliv

oh right

$(.5,1.5)$ doesn't contain the endpoints so it wouldn't be in the intersection

Sine of the Time

18:24

@Pizza do you mean: from the points you found, how to find the angle?

Pizza

Yes

Sine of the Time

everyone uses $(a,b)$, I'm the only one in chat who use $]a,b[$ :((

@Pizza you can find theta such that $(\rho \cos \theta,\rho \sin \theta)=$your point

but this is strategy is not the optimal one

Joako

Hi everyone, I got a question about the expected value of the hyperbolic cosine of the standard Brownian motion, which has an answer here, but I would like to someone else to review it since following what I saw in a YT video it could be mistaken (and I aiming to use it).. I left a comment by the way, with the mentioned video

Sine of the Time

the (second) best way is to use the relations which define $D$

Obliv

yes nvm

Soumik Mukherjee

18:31

Will be facing a cyclone today

Sine of the Time

@SoumikMukherjee In my region there are floods since last week

Soumik Mukherjee

I hope you are okay

Sine of the Time

I'm ok, but the area near me is under the water

Soumik Mukherjee

:(

Sine of the Time

user image

where you see water, there are usually crops

Soumik Mukherjee

18:48

Terrible situation

Sine of the Time

every year is the same unfortunately

Obliv

is the closure of any open subset of $\mathbb{R}$ just $\mathbb{R}$

why wouldn't it be just the sup/inf of the subset..

Ben Steffan

@Obliv most certainly not

Obliv

that's how we're defining it for some reason.

Ben Steffan

?

you're not defining the closure to be $\mathbb{R}$

Obliv

18:52

user image

a limit point is any point outside of the set so the set of all limit points of an open subset of $\mathbb{R}$ is its complement

so the closure of this subset is $\mathbb{R}$?

Ben Steffan

@Obliv no

that's not what a limit point is

Obliv

user image

wait

it's the point in which ALL epsilon neighborhoods of that point intersects A at some point other than it

Ben Steffan

this does not say "any point outside the set," now does it? :)

Obliv

I thought it was for some epsilon neighborhood

ok yes that makes more sense

Thorgott

@Ben TIL: the simplicial category $\mathfrak{C}(\Delta^n)$ considered as a simplicial object in $\mathbf{Cat}$ can be described as a 'comonad resolution' (don't ask me what that means in general) of $[n]$, more precisely its given via $k\mapsto(FU)^{k+1}([n])$, where $F\dashv U$ is the canonical adjunction between $\mathbf{Cat}$ and $\mathbf{Graph}$

Ben Steffan

18:58

!?

Thorgott

my reaction, too

nickbros123

@Jakobian I think this is one of the criterion for disconnectedness no?

oh never mind i didnt read that properly

ModularMindset

@XanderHenderson Take a topological graph embedded on a surface and endow it with a line bundle i.e. a vector bundle

ModularMindset

19:25

Let $S^2$ be the 1-dimensional unit circle, and let $G$ be a 4-regular graph embedded into $S^2$. Assume that $G$ is the union of three distinct geodesic loops $\gamma_1, \gamma_2, \gamma_3 \subset S^2$, such that the vertices of $G$ correspond to the intersection points of these geodesic loops. Let the set of vertices of $G$ be denoted by $V(G)$, and the set of edges by $E(G)$.

We endow $G$ with a smooth line bundle $L \to G$, and assume that this line bundle is compatible with the Levi-Civita connection $\nabla$ induced by the Riemannian metric on $S^2$. This means that for each edge $e …

more details on that

i made a trivial notational errror. S^1 should have been S^2. I tried to fix it

anyway that is an example I have been thinking about

ModularMindset

19:43

This basically gives you electromagnetism

Pizza

@SineoftheTime In your opinion, which exercise is the most difficult?

Anyway I'll let you know as soon as I know the results

Sine of the Time

@Pizza maybe the particular solution of the DE

but I should do the computations to judge

Pizza

@SineoftheTime 2 easy integrals came out, only one for a void I didn't do it and I moved on, maybe the part after that the derivative with arctan etc came out

the derivative had to be done, something quite long came out

Sine of the Time

what did you use? Variation of parameters?

Pizza

Yes

Sine of the Time

19:52

did you do 4 with Gauss Green?

Pizza

In the end i have to find y'

@SineoftheTime yes, but I was not able to write the result of the last integral, but I did all the parameterizations

Sine of the Time

did you manage to conclude ex.5 ?

Pizza

Yes

I did it all

Sine of the Time

that's good news. What is the minimum to pass? 18?

Pizza

Yes

I hope I passed, even if I could have done more

Sine of the Time

19:55

I don't know how the points are distributed, but it seems you did a good job

I assume you did 1 and 2 perfectly

Pizza

The first one yes, the second one only at the absolute extremes I did all the parameterizations but I was not able to insert all the points due to lack of time

but it shouldn't be a serious mistake, in the end I just had to calculate the points

I did the parameterization part well, then I also calculated the vertices

psie

I'm considering asking my question on the main site, maybe someone there knows what the authors of my book meant when they said "the unit sphere in the $1$-norm is compact." Note, this is on an exercise on equivalence of norms in $\mathbb R^n$, so I can't use that fact.

Sine of the Time

the parametrizations on the square are straightforward

Pizza

yes indeed

but I wanted to make sure I wasn't making any mistakes

Sine of the Time

are you planning to do the oral part?

Ben Steffan

20:02

@psie what do you mean "what they meant"

the statement is not ambiguous

psie

@BenSteffan well, you were saying earlier checking closedness and boundedness here is not enough

Ben Steffan

indeed

do you know what the word "compact" means?

it's not closed + bounded

Pizza

for now I have learned the proof of Schwarz's theorem, differential, Fermat, the whole part on Stokes and divergence (with Gauss Green), and one on Taylor series

But the one about the differential has already been written in the exam, I don't think he Will ask for it

psie

@BenSteffan well, in $\mathbb R^n$ it always is, right? Otherwise, totally bounded and complete.

Ben Steffan

In $\mathbb{R}^n$, yes

usually the word means "every open cover has a finite subcover," but perhaps your analysis texts take a different view

psie

20:06

Right, that has been mentioned too. But to prove the unit sphere is compact in the $1$-norm, does this argument work? It is bounded, i.e. $\|x\|_1=1<2$ for example and that the preimage of the continuous function $x\mapsto\|x\|_1$ of the closed set $\{1\}$ is closed.

Gian's Pizzeria

guys

Ben Steffan

you just acknowledged that closedness + boundedness is not enough

Sine of the Time

closed+bounded=compact in $\Bbb R^n$ with the usual topology (i.e. the one induced by the $2$ norm)

correct me if I'm wrong @Ben

Ben Steffan

you're right :)

that's precisely the point: you don't know that the 1-norm generates the usual topology yet, so you cannot use this

but you have another criterion, as you say

psie

Ok.

Sine of the Time

20:09

Since the equivalence of norm is in fact an equivalence relation, I'd use the $2-$norm instead of the $1-$norm

@Gian'sPizzeria girls

psie

It's really strange, because in my book the Heine Borel theorem has already been proved. So can I simply not use it?

Ben Steffan

It doesn't apply

Gian's Pizzeria

Today in the test $y''+2y'+9y=1/sin(x)$ came out, I applied the similarity method with [1/(a sin(x) + b cos(x))], can you confirm that it's right?

Ben Steffan

Heine-Borel applies to $(\mathbb{R}^n, \lVert \cdot \rVert_2)$

Sine of the Time

compactness depens on the topology, HB works for the topology induced by the 2 norm.

psie

20:11

oh, I have never realized that. I thought HB works for $\mathbb R^n$ (and any norm). Full stop.

Sine of the Time

@Gian'sPizzeria are you sure about the text of the problem?

Thorgott

@psie it does, you just might not know it yet

psie

4 mins ago, by Ben Steffan

Heine-Borel applies to $(\mathbb{R}^n, \lVert \cdot \rVert_2)$

sku

have trouble understanding limit points. A point x is a limit point of a set A if every epsilon-neighborhood V(x) of x intersects the set A in some point other than x. According to this definition, all points of (0,1) besides 0 and 1 are limit points? can someone give some intuition on limit points?

Sine of the Time

Since all norms are equivalent, then they induce the same topology. But you can't use it because that's what you're trying to prove

Xander Henderson

20:16

@sku 0 and 1 are also limit points.

leslie townes

well, they might be. 'limit point' is relative to the ambient metric space implicit in "V(x)"

i don't mean to confuse the matter. but the x in "a point x is a limit point..." isn't just out there flapping in the breeze

Sine of the Time

leslie 1 Xander 0

sku

what would an example of not a limit point in the set (0,1)?

Xander Henderson

@leslietownes Oh, sure. I was assuming that we are in $\mathbb{R}$ with the usual topology. That feels implicit in the notation, i.e. the interval $(0,1)$.

@sku 7.

$\pi$.

leslie townes

between this and the different norms on R^n, lots of metric space particulars today

Xander Henderson

20:20

$-10^{100}$.

leslie townes

sku: some subsets of metric spaces contain all of their limit points, or have all of their points being limit points. you are not going to find a point in (0,1) that is not a limit point of (0,1). this example or set of examples is likely just testing how you map definitions onto relatively 'familiar' subsets of R. it is not trying to suggest that there is something particularly interesting about the limit point concept for an open interval in R

Xander Henderson

@sku Assuming that $(0,1)$ denotes the open interval from zero to one, and that the metric defining your $\varepsilon$-neighborhoods is the usual one, anything in the set $\mathbb{R}\setminus[0,1]$ is not a limit point of $(0,1)$.

sku

I guess I am trying to understand why we are learning this concept. It has to be teaching something or a stepping stone to something else...

Xander Henderson

@sku What class are you taking?

Is this a real analysis class? A topology class?

Something else?

sku

Real Analysis

Gian's Pizzeria

20:23

@SineoftheTime y''+9y=1/sin(x)

I remembered badly

sku

chapter 3 of Abbotts book

Xander Henderson

Okay, so do you remember calculus? And how you had to compute a lot of limits there?

sku

yes

Xander Henderson

And how you dealt with notions like continuity and differentiability?

sku

yes

Sine of the Time

20:24

@Gian'sPizzeria I'd use variation of parameters

Xander Henderson

In a more general setting, these things are all defined topologically, i.e. in terms of open sets.

Gian's Pizzeria

@SineoftheTime I applied the similarity method with [1/(a sin(x) + b cos(x))], can you confirm that it's right?

Sine of the Time

It does not look right

Gian's Pizzeria

...

Why?

Xander Henderson

So you are building the tools which are going to eventually allow you to push through all of those nice calculus theorems.

Sine of the Time

20:25

you use asin x+ bcos x for a particual solution of y''+9y=sin x

Gian's Pizzeria

Yes

Sine of the Time

for 1/sin x, it's not enough to search a soln in the form 1/(a sin x+bcos x)

sku

I see. one qq. in [0,1], is 1 not considered a limit point?

Gian's Pizzeria

@SineoftheTime So if I wrote like this I automatically got everything else wrong?

Sine of the Time

I'd say so. Did you manage to find a particular solution?

Gian's Pizzeria

20:30

Yes, but I'm getting something very long

Sine of the Time

why did you not use variation of parameters?

Gian's Pizzeria

Because I saw the sin and the similarity method immediately came to mind

leslie townes

sku: does every interval of the form (1 - e, 1 + e), e positive, intersect [0,1] in some point other than 1? that's the question

sku

Clearly the answer is yes. since 0<1-e<1

I mean it intersects wirh 1 also. I guess we mean intersect a point not including 1. Then no. I think I got it

I was confused with language. Thank you all

Sine of the Time

@Gian'sPizzeria this would have worked if it was =sin(x)

Gian's Pizzeria

20:39

Why?

Sine of the Time

because this is not how differentiation works

Alessandro Codenotti

@Jakobian interestingly there is an answer from 3 years ago claiming that it is a characterization of beta X for normal X but it has been deleted

imbAF

Hello, I have a question about eigenvalues and eigenvecttors

user image

Is there an intuitive way to prove the first statement

without calculating the eigenvalues and then the eigenvectors and then comparing them

?

Could I argue that if the scalar product of the two is zero than they are eigenvectors that correspond to differen eigenvalues

hence proving the statement

Jakobian

@AlessandroCodenotti it is

Ah yeah

leslie townes

imbaf: a very low tech approach would just be to write out what Hx = lambda x (where x is your (u, v*)^T vector) would mean in terms of a system of two scalar equations in u and v. it turns out to imply, without too much rewriting or effort, the system of scalar equations that tells you Hy = -lambda y (where y is that other vector)

imbAF

20:53

But how is implied that

-v , u* is the eigenvector to -lambda ?

y is the other vector?

@leslietownes I thought of considering the complex conjugate in both sides

leslie townes

yes but to make sense of this, if you are stuck, it may help to write this out as a system of scalar equations that you complex conjugate

if rewriting the vector equation as a vector equation isn't seeming to lead you anywhere

imbAF

I did write them as two equations

$\epsilon u + \Delta v*=\lambda u$
$\Delta* u -\epsilonv*=\lambda v*$

Is there a math problem solving channel ?

Xander Henderson

21:13

@sku 1 is a limit point of $[0,1]$.

The idea is that if $x$ is a limit point of $A$, then any punctured neighborhood of $x$ must intersect $A$. In the language developed for you thus far, $x$ is a limit point of $A$ if for any $\varepsilon > 0$, the set $$ \left[ (x-\varepsilon, x) \cup (x,x+\varepsilon) \right] \cap A$$ is nonemepty.

Every neighborhood of $x$ intersects $A$ at a point other than $x$ (it might also intersect at $x$---it certainly will if $x \in A$, but that doesn't matter).

sku

Thank you so much @XanderHenderson and @leslietownes. I can see how this would relate to continuity in a future chapter.

leslie townes

imbaf: if you label those as (1) and (2), if you conjugate both sides of (1) and multiply through by -1 you get - delta' v - eu' = -lambda' u' which i will call (3), and if you conjugate both sides of (2) you get delta u' - v epsilon = lambda' v, which i will call (4). (4) and (3) are respectively the first and second entries of the vector equation Hy = -lambda' y, where y = (-v, u')

i'm writing ' for conjugate because asterisk sometimes triggers emphasis formatting in chat

imbaf: so Hy = -lambda' y. and since H is equal to its hermitian transpose it has real eigenvalues, so lambda' = lambda.

there might be some sign errors or misplaced ' in my calculation, but something like that is the idea. you won't need to evaluate what the eigenvalue lambda actually is, but you might need to know that it's real

(implicit in all of this, i think, is that epsilon is real)

imbAF

21:31

@leslietownes I might need a moment to understand this

But you are complex conjugating both equations in both sides

leslie townes

its tempting to want to do this all in terms of matrix algebra and not write components, but you then need to 'matrix-ify' the operation of producing (-v, u') from (u,v'), which is still going to make express use of the fact that we have only two components

so it sort of amount to a system of two things anyway

imbaf: yes, if a = b (as complex numbers) then a' = b'

imbAF

the prime is representing the conjugation?

leslie townes

yes, as explained previously i am not writing * for conjugation because repeated stars in chat sometimes cause things to be emphasized instead of to appear as stars

imbAF

Let me try it

leslie townes

and rather than keep track of what will or won't set off people's chatjax formatting i often choose to just avoid triggering it :)

if i don't type $ or * the evil formatting can't get me

imbAF

21:38

Ok so I tried it and this is where I am

So i had:
$\epsilon u + \Delta v'=\lambda u$
$\Delta' u -\epsilon v'=\lambda v'$

I did:
$-\epsilon u' - \Delta' v=-\lambda u'$
$-\Delta u' +\epsilon v=-\lambda v$

I see, so you are trying to recreate the unchanged components of the matrix, while complex conjugating the expression where u is?

@leslietownes but shouldn't you only complex conjugate:$\epsilon u + \Delta v'=\lambda u$ without changing sign

and complex conjugate and change sing for this $\Delta' u -\epsilon v'=\lambda v'$ ?

leslie townes

writing x = (u, v') and y= (-v, u'), if i had to describe it in short words, i wrote Hx = lambda x as a system of scalar equations in u and v'. i then complex conjugated both of those equations to get a system of scalar equations in u' and v. i then "noticed" (by just rearranging those equations) that this turns out to be, or give, a system of equations equivalent to the vector equation Hy = -lambda y

some of the way you're talking about it suggests that maybe you're focusing in on pieces of these equations and somehow "transforming" them individually, or that you think i might be doing that, and i'm not

i'm using tools like "if a and b are complex numbers and a = b, then a' = b'" and "if a and b are complex numbers and a = b, then -a = -b"

and other stuff like "if a and b are complex numbers, then a + b = -(-a) + b"

imbAF

If i do complex conjugation on both and sign change on the first, I end up with one extra minus

leslie townes

there are a lot of operations going on here where there is an opportunity to make something resembling a sign error. note that the matrix H itself has one minus sign in it, and the eigenvalue the question is asking us to derive things about also has a minus sign in it

imbAF

Since H is hermitian I can consider $\lambda$ not to change , it's real

but I don't get the expression

But how can something so trivial be so difficult

I don't think you can show what the exercise is asking that way

@leslietownes I dd it

Thanks

leslie townes

22:02

hooray :)

imbAF

And if I need to argue what i did

would it be ok to say that I Tried to recreate the matrix on the LHS

and that led to an eigenvalue expression that showcases the other eigenvalue and its corresponding eigenvector

leslie townes

well, yes, the idea is definitely to rewrite the complex conjugate of H (u, v')^T = lambda (u, v')^T as another matrix equation that expressly involves the matrix H (rather than, for example, the entrywise complex conjugate of the matrix H)

and if you're careful, you find that you get what the exercise is asking for

imbAF

I understand

leslie townes

the key is that whether we refer to it as 'rewriting' or 'recreating' or whatever else, the process is not one of just rearranging symbols to get the desired result, but actually deducing new true equations from old true equations using things like "if a = b then a' = b'"

imbAF

but that holds only if a and b are the same in value and both real

no?

leslie townes

22:15

they can be complex too. the only use of anything being real in the above was in using things like epsilon' = epsilon and lambda' = lambda above, i.e. that epsilon was real and that lambda was therefore also going to be real

psie

I've struggled with understanding why the unit sphere in the $1$-norm is compact. I turned to the main site, but without success so far.

leslie townes

you can get a matrix-ification of the calculation by doing something like letting e.g. K = [[0,-1],[1,0]] and checking that K^2 = -I and (from the way your particular H was defined) KHK = H', where H' is entrywise complex conjugate. the matrix calculation is then, from Hx = lambda x, conjugate both sides, use H' = KHK and then left multiply by K to deduce H Kx' = -lambda' Kx', so that if x is an eigenvector for lambda, Kx' is an eigenvector for -lambda'

a slight generalization of the exercise. where the opportunity for all the sign errors is baked into the computation that KHK = H'

Sine of the Time

22:55

@psie before this exercise, have you proven that if $\|\cdot\|_a$ dominates $\| \cdot \|_b$, then the open sets of $(V,\|\cdot\|_b)$ are open in $(V,\|\cdot\|_a)$ ?

Obliv

I want to show $(E^{\circ})^c = \overline{E^c}$ so I started by analyzing the set definitions: $(E^{\circ})$ is the set $\{x\in E: \exists V_{\varepsilon}(x)\subseteq E\}$, so the complement is the set $\{y: y\notin E^{\circ}\}$ but idk how I'd define the complement of the interior of $E$.

psie

@SineoftheTime hmm, I don't think so :( what were you thinking?

by dominate you mean $\|\cdot\|_b\leq c\|\cdot\|_a$ for $c>0$ I assume

Obliv

I think in negations and logic statements so when I see complement I think of negations of set definitions, that's why I'm doing it this way

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