Mathematics - 2015-11-11 (page 4 of 4)

Ainz Ooal Goal

20:03

@TedShifrin What's the set of all orthogonal, diagonalizable matrices on $\Bbb{R}$ ?

Ted Shifrin

on $\Bbb R$???

Ainz Ooal Goal

yes

Ted Shifrin

$\pm 1$.

Do you mean $n\times n$ real matrices?

Ainz Ooal Goal

Ah, by 'on R' I meant 'on $\mathcal{M}_n(\Bbb{R})$'

Ted Shifrin

So what do you know about the eigenvalues of an orthogonal matrix? (We'll work only with $\Bbb R$.)

Ainz Ooal Goal

20:05

$\pm1$

Ted Shifrin

Allow eigenvalues to be complex.

I meant we'll restrict to real matrices.

Ainz Ooal Goal

Their module is 1

Ted Shifrin

Right. So which will be diagonalizable over $\Bbb R$?

Ainz Ooal Goal

Well, they'll need to have eigenvalues $\pm1$

Ted Shifrin

So you've answered your question.

Use the spectral theorem, of course.

What is the geometric interpretation of all these linear maps?

Ainz Ooal Goal

20:11

Uh ? The spectral theorem ? Why would orthogonal matrices be symmetric ?

And I'm not sure what the geometric interpretation is, other that it's the map in an orthogonal basis

Mike Miller

They're not, but they're certainly normal.

Ted Shifrin

well, there's a more general spectral theorem, but don't worry about that. Suppose you had eigenvalues $\pm 1$ in the standard basis. What is the map?

Ainz Ooal Goal

Well, that depends of the matrices $P$ used to get $M=PDP^{-1}$ doesn't it ?

Ted Shifrin

That isn't answering my question. Answer it first. Then we'll discuss $P$.

Ainz Ooal Goal

The map corresponding to $D$ is just $u(e_i)=\pm e_i$, but I don't know for M. So it's a symmetry

Ted Shifrin

20:16

Yes, it's an orthogonal reflection in (or across) a subspace.

Prepositions are always difficult.

OK, so the hermitian version of the spectral theorem will tell you that there is a unitary basis (working over $\Bbb C$). However, when the eigenvalues are real, that unitary basis will in fact be an orthogonal basis. You should figure out what the hermitian spectral theorem tells you about the general orthogonal matrix.

Balarka Sen

@TedShifrin I think I have an idea, but it's pretty quirky. You need a map from a nonHausdorff space to a Hausdorff space or something like that. If $f : X \to Y$ is such a map, and every closed set appears as preimage of a point, then for any two closed $A, B \subset X$, $f(A) = y_0, f(B) = y_1$. But as $Y$ is Hausdorff you can take two nbhd of $y_0$ and $y_1$ which are disjoint, hence pullbacking would give you disjoint nbhds of $A$ and $B$.

You can make this fail by some nonHausdorff assumption.

Ted Shifrin

No, I'm happy to stay in Hausdorff spaces. Just not metrizable.

But be careful. When you use Urysohn, you get a function that's zero on $A$. But when can you guarantee it's zero precisely on $A$? (So, start with a normal space. I'm fine with that.)

OK, lunchtime for me. Back later.

Balarka Sen

Yeah, well, you can find non-normal spaces which are Hausdorff.

Ainz Ooal Goal

@TedShifrin From what I've gathered, the hermitian spectral theorem tells us that for any hermitian matrix M, we can build an orthogonal basis of R^n from eigenvectors of M. But I don't see how to apply that here, since a random matrix with eigenvalues +-1 isn't hermitian ..

Mitch

advice sought:

I've asked a question on math.SE that has gotten very little attention:

-2

Q: Assumptions needed for proof of the Pythagorean Theorem from examples

There are lots of geometric dissection and reassembling proofs of the Pythagorean Theorem (PT). I want to know what are the necessary ideas to allow a proof using discrete instances. For example, we can, not too terribly painstakingly, show that a 3,4 rt triangle has 5 as its hypotenuse, and lik...

(and two downvotes)

Ainz Ooal Goal

20:30

54 views isn't 'very little attention'

Mitch

Euclidean geometry (and really geometry from euclid) is not particularly popular, expected to be pretty elementary.

@AinzOoalGoal (things fall off the top of the activity list so quickly)

BUt my question is expected to be not so terribly simple.

Ainz Ooal Goal

The activity list goes so fast on mse :o I only browse the 'Newest' list

Mitch

So how do you get people to look at a question? I've already put a 50 bounty on it.

Ainz Ooal Goal

What's " rt triangle" ?

Mitch

right triangle

Ainz Ooal Goal

20:32

Oh ok.

Mitch

is that too abbreviated?

Ainz Ooal Goal

Well, it's not obvious for me.

Mitch

rt

Ainz Ooal Goal

I don't get the two downvotes though

Mitch

ha ha. right

Ainz Ooal Goal

20:34

@Mitch how do you even know it's polynomial ? (using purely mathematical reasonings, not physics)

Mitch

I added the tag 'algebraic-geometry' because I hoped that people with experience with higher order multinomials would be able to say 'oh yeah the form of the solution should be this because of this' allowing interpolation (in some lagrange generalization)

but that was removed.

@AinzOoalGoal that's just an analogy.

Ainz Ooal Goal

@Mitch Then, how can knowing the relation for a finite number of points help at all ? There are always infinitely many curves going through a finite set of points

Balarka Sen

It doesn't seem like algebraic geometry to me.

Mitch

The linked question was posted in math education because I expected them to have some ideas about quick ways to show particular examples.

@AinzOoalGoal sure. I'm hoping one of the asumptions (or observations) would be that the realtion should be of some particular form. I guess I'm not getting that across.

(to me, I think that's the whole point, I don't know the form, how do you know the form)

spexel

Is there a way to increase the number of questions per day?

i really need some help on a problem and i can't post it :(

Ainz Ooal Goal

20:38

As a physician, $d a^2 + e b^2 = f c^2$ seems obvious. Each time you increase the area of both squares linked to two sides, you increase linearly the area taken by the 3rd square @Mitch

Mitch

@BalarkaSen is there a better way to explain it? Do you have any suggestions on how to make it better?

Ainz Ooal Goal

@spexel If you're reached the cap, then you're probably asking a bit too much :O Let others ask question too ! Anyway, why don't you ask here ? :-)

Balarka Sen

I don't have any suggestions. I am just saying the removal of the algebraic geometry tag was perfectly fine, as the question has nothing whatsoever to do with algebraic geometry.

Mitch

@AinzOoalGoal That's maybe the thing that is not obvious to me (I'm not thinking hard). I vaguely remember something like that, but I don't see how to justify that. There's also thedesired strategy of not wanting to immediately prove PT. I want to get close enough to use examples as the last step.

@BalarkaSen sure. a suggestion for a better tag?

Ainz Ooal Goal

The main problem is that, after all, proving the PT is very easy, and hence by proving that the formula is of a certain form, you'd probably have done more than enough to prove it at once @Mitch

Balarka Sen

20:41

as I said, I have no suggestion whatsoever.

Mitch

@AinzOoalGoal I don't see that (unless I go ahead and see the butterfly proof)

@AinzOoalGoal Yes, that is a problem. but it is sort of a game I'm playing, trying to use other tools.

PVAL

This really depends on how you define "length" and "angle"...

Mitch

In what way?

PVAL

Well if you define the distance between two points using the usual square root formula, PT is trivial.

Mike Miller

@Ainz: He's invoking the spectral theorem for normal operators.

Mitch

20:46

@PVAL yes, those are logically interdependent

PVAL

If you define say distance from zero as the unique function with d((x,0),0)= |x| that is rotation-invariant then other things happen.

Ainz Ooal Goal

@MikeMiller I only know it for symmetric matrices (and now for hermitian ones)

Mike Miller

Which is why he said "Don't worry about it".

Ainz Ooal Goal

But then, how do I get my answer :(

Mike Miller

Ok, so here's the spectral theorem for normal operators: "The spectral theorem is also true for normal operators, though the eigenvalues needn't be real". Now invoke it.

Ainz Ooal Goal

20:52

I don't get how we can apply it here. We were saying that our matrices necessarily have eigenvalues 1 or -1. the spectral theorem just tells us that any (symmetric / hermitian / normal) can be used to create an orthogonal basis from its eigenvectors.

Mike Miller

That's usually known as a diagonalization.

Ainz Ooal Goal

oooh

Mike Miller

What are you looking for? I thought you wanted to know it was diagonalizable.

Ainz Ooal Goal

I was searching for the set of orthogonal, diagonalizable matrices

We concluded that it was necessary that their eigenvalues be 1 or -1

And from the spectral theorem, any orthogonal matrix with eigenvalues 1,-1 is diagonalizable ?

Mike Miller

Aye.

Ainz Ooal Goal

20:58

Thanks :-)

Anonymous

Hello everyone.

Anonymous

I have a really basic question about percentages.

Anonymous

I have 3897 students, each have a rank from 1 - 3897.

Anonymous

How do I give them a rank based on percentages?

Ainz Ooal Goal

@samayo What do you mean 'based on percentages' ?

Anonymous

21:05

Like the way Stack Oveflow rates profiles, example top 0.1%

Ainz Ooal Goal

rank in percent = rank/3897*100

Anonymous

perfect!

Anonymous

@AinzOoalGoal Do you know any good sites where I can learn some basic Maths stuff?

Anonymous

Interactive site would be nice, if you know any.

Ainz Ooal Goal

@samayo 'maths stuff' is vague

@samayo But, do look at setosa.io, it's well made. Not very advanced stuff, but interesting

Anonymous

21:11

Anything aside or better yet, the very next step after +, -, x, /

Anonymous

At first glance, it looks terrifying :)

PVAL

Theres these things that have paper in them that are often covered partially by a harder type of paper or some cardboardish thing.

Anonymous

@AinzOoalGoal Definitely not what I am looking for.

Anonymous

It is some sort of advanced physics stuff

Ainz Ooal Goal

@samayo The problem is that 'the very next step' isn't well defined

Anonymous

21:14

I quit school very early, so aside from addition, mult.., sub.. and division I literally know nothing.

Anonymous

all the things I hear like, Algebra, Geometry, Sin, Cosin are foreign to me

Ainz Ooal Goal

I'm just a student so I don't know any good source to start :(

Anonymous

ehh, no problem

Anonymous

But, what in your opinion is the next topic to learn?

Ted Shifrin

M le méchant: HINT: If $A$ is orthogonal, $iA$ is hermitian.

@Samayo: You need to learn some more, and what you learn depends on what you're interested in pursuing. You should look at Khan Academy.

Ainz Ooal Goal

21:21

@TedShifrin @MikeMiller helped me for the end :-) so basically, the set we were searching for is the whole set of orthogonal matrices with eigenvalues $\pm1$

Ted Shifrin

Yes, M le méchant, but what I just told you is more basic than learning about normal operators. But, no matter.

Anonymous

@TedShifrin I tried Khan Academy once, but they are mostly video tutorials, I just wanted to read. But I didn't know where to start :\

Ted Shifrin

Oh, @samayo. What are you interested in learning?

MickLH

I am working with "fractional calculus" but my orders are real. Can I call it "real ordered calculus" or will someone burn me at the stake?

Ted Shifrin

I have no unearthly idea, @MickLH.

That's what Fourier transforms allow you to do when you're fancy and grown up.

Anonymous

21:23

@TedShifrin Anything beside addition, subtraction, multiplication and division.

MickLH

@TedShifrin Good enough for me! You both are a math teacher and didn't react with anger! :D

Anonymous

I think they are called "Arithmetic operations"

Ted Shifrin

@samayo: But what is your aim? Physics? Economics? Or just more math for math?

Anonymous

@TedShifrin Ah, no. I don't go to school. I just wanted to learn more about the basics of Maths.

Ainz Ooal Goal

@TedShifrin Uh ? Why is that ? Why would $A^tA=I_n$ imply $<iAX,Y>=-<X,^tAY>$ ?

Ted Shifrin

21:25

@Samayo: This is a strange suggestion, but I highly recommend this book by my former colleague. It is meant for people who want to teach elementary and middle school, but it is full of fascinating stuff explaining arithmetic and then building algebra and geometry.

Anonymous

@TedShifrin Thanks, bookmarked.

Ted Shifrin

Le méchant: Sorry, I screwed up. I was thinking skew-symmetric. My apologies. Yes, you do need normal.

I really should retire for good. :)

Ainz Ooal Goal

Noo :(

Anonymous

In addition, is there a website that lists mathematical steps/topics to learn from scratch based on their difficulty or grades?

MickLH

@samayo I'm not so sure any given concept has an intrinsic difficulty value that can be ordered.

Anonymous

21:31

Maybe by grade? From grade 1 - high school.

MickLH

@samayo Nobody can decide for you, but maybe it will help to know a few examples: In electrical engineering I make heavy use of differential equations and optimization problems. In programming I make heavy use of linear algebra and recurrence relations.

Anonymous

Hm, so it's harder than I thought. You know when you start learning English, you get started with the basics like .. alphabet, vocabulary, parts of speech, tenses ...

Anonymous

I thought there would be something like that for learning Maths

MickLH

You already pointed out that you're murky on trig, I guess it's safe to say get comfortable with that... You'll end up using it so often that people don't even consider sine a special function.

Mary Star

When we have a system of two differential inequations $$L_1 y \neq f_1 \\ L_2 y \neq f_2$$ where $L$ is a differential operator, can we write it in the form $Ly\neq f$ ? For example, when we add the two inequations? Or can we not add two ineqautions?

Antonio Vargas

22:23

[Nevermind, I am an egg.]

Américo Tavares

36

Q: What is the purpose of this site?

What is (and what should be) the purpose of math.SE? As far as I can say, various users have different views on this question. Some users view it as a repository of knowledge. Some users approach it as teaching opportunity. Similar to previous, but slightly different: It could be understoo...

discussion

Alec Teal

@AméricoTavares what was the question? Things that matter not?

skull petrol

Is that what a purpose means?

Agawa001

i saw this link posted often here

skull petrol

22:39

@AméricoTavares you should inline a link and then it may get starred.

Agawa001

i m curious to know 'what is the purpose of this chatroom'

skull petrol

Ask on meta :)

What is the purpose of this site?

10

:-)

Agawa001

@skullpetrol i cant understand , why cant mods pin anything on the starboard ?

skull petrol

Dunno, I think a room owner can do that.

Loong

@Agawa001 they can

Agawa001

22:47

@Loong can it not be buried again after pinning it

skull petrol

:-)

Agawa001

@skullpetrol i see it pinned now :)

Anthony

Is there a good way (read: an algorithm) to find polynomials satisfying sums of algebraic numbers?

For instance, how would I go about finding the polynomial such that $P(\sqrt{2}+\sqrt{3}) = 0$?

Ainz Ooal Goal

@Anthony P(X)=X(2^.5+3^.5) ?

Anthony

@AinzOoalGoal I think I don't understand what you're saying- $(\sqrt{2}+\sqrt{3})^2\neq 0$

Ainz Ooal Goal

22:57

@Anthony ._. sorry I meant P=X-2^.5-3^.5

Anthony

Oh, and sorry. The polynomial needs to be over $\mathbb{Q}$. That's important.

:P

Balarka Sen

Since Hippa has pointed out the loophole : Yes, there is an algorithm, @Anthony. You need to find all the Galois conjugates of that algebraic number in the Galois closure of the extension of $\Bbb Q$ obtained from adjoining that number, and then compute the product of linear factors of the form $X - a$ where $a$ is a Galois conjugate of the algebraic number. It's a bit sophisticated, but that's the idea.

Anthony

Seeing as we are heading into Galois theory, then, I'll assume that this problem is meant to not be algorithmic.

Thanks, @BalarkaSen.

Balarka Sen

For example, Galois conjugates of $\sqrt{2} + \sqrt{3}$ are $\pm \sqrt{2} \pm \sqrt{3}$. Now you multiply factors of the form $(X \pm \sqrt{2} \pm \sqrt{3})$.

You'll get your desired polynomial.

@Anthony It's an algorithmic problem, just that you need to use Galois theory in the algorithm :P

Anthony

Yeah. Thanks @BalarkaSen!

Balarka Sen

23:01

Sure.

Karim Mansour

Hey I came to ask a question

@BalarkaSen in the box topology is it true that if we have cartesian product of connected space will it be connected?

Balarka Sen

I don't think so, no. I don't know of an example off the top of my head, although I should.

Mike Miller

@Anthony: alternatively, "resultant polynomial"

which I think is the name of what Balarka is describing above

Balarka Sen

I have forgotten what the name of this construct is :P But I guess, yeah.

OK, box topology on countably many product of R is not connected is it?

Mike Miller

any CAS should have something that computes resolvents.

resultants

Karim Mansour

23:12

I found the following example but I don't understand it

Consider the cartesian product $R^{\omega}$ in the bxo topology. We can write $R^{\omega}$ as union of the sets A consisting of all bounded sequences of real numbers, and the set B of all unbounded sequences.

why is the following sets disjoint ?

Mike Miller

I'm sure you can see why they're disjoint if you think about it a bit.

Balarka Sen

eh, I am not really sure about bounded sequences, but if I look at the subset of convergent sequences converging to a point, I think that gives you a separation.

Karim Mansour

I see

Balarka Sen

Because if you look at a convergent sequence $(x_n)$ converging to $a$, you can look at all the sequences $(y_n)$ such that $y_n \in (x_n - 1/n, x_n + 1/n)$. These guys also converge to $a$ by squeeze principle, hence forms an open neighbprhood of $(x_n)$ inside the subset. So, the set is open.

Karim Mansour

I see

Balarka Sen

23:19

Now, for proving that complement is open, that is clear because if you have a sequence which converges to some other number $b$ you can perturb similarly to make nbhds around it completely contained in the interior. If it does not converge at all, perturb anyway and none of the things inside the nbhd will converge to $a$.

So, I think that works. Box topology is not my favorite topology anyway.

Karim Mansour

I am just gaining more intuition about box and product top

Balarka Sen

Product topology is nice.

Karim Mansour

yes

Mary Star

@AntonioVargas The inequations $L_1 y \neq f_1 , L_2 y \neq f_2$. are equivalent to $L_1y = f_1+a , L_2y \neq f_2+b$, where $a, b \neq 0$.
We can add these two equations and we get $Ly=L_1y+L_2 y=f_1+a+f_2+b, a, b \neq 0$.
Then we have $L \neq f_1+f_2$.

Is this correct? But if $a=-b$? Do we maybe have to suppose that $a,b >0$ ?

Karim Mansour

I am currently reading section of connectedness and compactness

then I will do my topology assignment

Ainz Ooal Goal

23:21

@Chris'ssistheartist

Chris's sis the artist

@AinzOoalGoal ?

Balarka Sen

Let me know when you prove that (finite, to begin with) product of compact spaces is compact. Beautiful application of the tube lemma.

Ainz Ooal Goal

@Chris'ssistheartist Do you know any special results about the inverse gamma function (for x>1) ? Or somewhere I could find some ?

Karim Mansour

alright

Chris's sis the artist

@AinzOoalGoal You should look at the graph of gamma function before thinking of its inverse. I suppose you have in mind some restrictions.

Ainz Ooal Goal

23:23

Sorry, I mean x>2 (forget the x>1 nonsense)

Karim Mansour

hm

why is it true that when some topological space is connnected then its closure is also connected?

@BalarkaSen

Balarka Sen

not sure what closure of a space means.

you take closure of a subset in the ambient topological space

Chris's sis the artist

30

Q: Inverse gamma function?

This is an analysis question I remember thinking about in high school. Reading some of the other topics here reminded me of this, and I'd like to hear other people's solutions to this. We have the gamma function, which has a fairly elementary form as we all know, $\Gamma(z) = \int_0^\infty e^{-...

ca.analysis-and-odes special-functions

Américo Tavares

@skullpetrol Could you please explain what does it mean?

Ainz Ooal Goal

I've seen that MO link, but all it really does it point to a dead link to an approximation, or this paper ams.org/journals/proc/2012-140-04/S0002-9939-2011-11023-2/…

Chris's sis the artist

23:26

@AinzOoalGoal that might help. Unfortunately these days I'm mainly focused on some specific corner of research for my book.

Karim Mansour

in munkres they are trying to show that $R^{\omega}$ is connected in the product topology, so they showed that $R^{\infty}$ is connected which I understand

Ainz Ooal Goal

Well, ok :-) thanks @Chris'ssistheartist

Chris's sis the artist

@AinzOoalGoal I find anything else helpful I let you know. :-)

Karim Mansour

they mention that they need to show that closure of $R^{\infty}$ equals $R^\omega$ from which it follows that $R^{\omega}$ is connected

why does it follow that $R^{\omega}$ is connected from this ?

Américo Tavares

@skullpetrol The question is recent.

Mike Miller

23:29

@KarimMansour: Try to prove it yourself. It's a good exercise with the definitions.

Karim Mansour

alright

Balarka Sen

I was writing the proof :P

skull petrol

@AméricoTavares a link to your question has been pinned on the star board :-)

Balarka Sen

Try to draw a picture. It's visually clear.

Américo Tavares

@AlecTeal The question is recent.

Mike Miller

23:30

@BalarkaSen If I've taught you anything, it should be that the best way to teach is not to.

Balarka Sen

I agree. But then I probably shouldn't teach at the crack of a dawn.

What are you working on? (I don't expect I'll get an answer, but worth a try)

skull petrol

@AméricoTavares more people will see it with it pinned.

Mike Miller

Ask another day and I'll answer. You remind me that I should be doing that instead of this, anyway.

Américo Tavares

@skullpetrol I see, thanks!

skull petrol

np pal :-)

Balarka Sen

23:35

Paul Erdös tells a joke

►

re "ask another day".

FizzledOut

23:47

Can anybody quickly tell me why the derivative of $2x$ is $2$, but the derivative of $2x$ in $y^2 = 2x + 1$ becomes $2dx/dz$ (if the function is with respect to z)?

(Assuming x is an independent variable)

Ted Shifrin

@Balarka @Karim: Neither of you is supposed to be here!

@FizzledOut: Well, if $z$ is the independent variable, now $x$ is dependent ...

@Karim: The relevant proposition is proved in Munkres, btw.

FizzledOut

@TedShifrin Don't mind me, I'm confusing myself. In any case, this is interesting. Wolfram Alpha will spit out that the derivative of 2x as 2, which I know, but if the derivative of the equation gives 2*(dx/d__), then isn't the derivative of 2x alone assuming that dx/d__ = 1?

skull petrol

Professor @TedShifrin are you ignoring @SohamChowdhury?

Ted Shifrin

I already answered him earlier, @skull.

Karim Mansour

yeah I noticed it @TedShifrin yeah I came to ask a question and leave

haha

Ted Shifrin

23:54

But I don't necessarily respond to every summons, no.

@Fizzled: I don't know about Wolfram Alpha, but Mathematica forces you to say what variable you're differentiating with respect to.

skull petrol

He wanted me to ask you :-)

Ted Shifrin

You do what everyone wants, @skull?

2

You know what comes next, right?

skull petrol

Slavery?

Ted Shifrin

Yes, that's one option ... or leaping unceremoniously off a bridge.

Ainz Ooal Goal

Or, getting punched by @Ted

skull petrol

23:56

Good point.

Ted Shifrin

I don't punch, Le méchant, although you definitely deserved a number. I have only smacked Balarka a few times. Smack \ne punch.

Ainz Ooal Goal

Ted Shifrin

LOL ... Is that my being unable to solve math problems? I told you so.

Mary Star

We have the inequations $L_1 y \neq f_1 , L_2 y \neq f_2$, where $L_1$ and $L_2$ are differential operators.
They are equivalent to $L_1y = f_1+a , L_2y \neq f_2+b$, where $a, b \neq 0$, right?
We can add these two equations and we get $Ly=L_1y+L_2 y=f_1+a+f_2+b, a, b \neq 0$.
Then we have $L \neq f_1+f_2$.

Is this correct?

Ted Shifrin

It makes no sense at all @MaryStar.

And, by the way, independent of all that, the sum of two inequalities could be an equality.

$3\ne 5$, $4\ne 2$, and yet $3+4=5+2$. Amazing stuff.

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$Mathematics$ Mathematics