@JunaidRehman I don't know, but this discussion on the Wiki page re: SVD seems relevant: en.wikipedia.org/wiki/…

@BAYMAX lol, so for eigenvalue $\lambda_1=1$ I get the eigenvector $\left(t,0,\cdots,0\right)^T$, where $t\in\mathbb{R}$

@NaCl Which you can normalize to $e_1$ since constant multiples of eigenvectors are also eigenvectors.

Yeah

@JunaidRehman In particular, it contains the remark: "...while related, the eigenvalue decomposition and SVD differ except for positive semi-definite normal matrices"

Thanks @Semiclassical. Thanks for the link. I read it earlier but I am still confused about the relation between $U$, $V$, and $W$ for a square matrix case.

nah, better not do that. Our course didn't talk about that

stare

Your course doesn't talk about the fact that if $\vec{v}$ is an eigenvector with value $\lambda$, then so is $c\vec{v}$?

Yes. We didn't even talk about determinants, which is just ridiculous

(Also, this only works if $c=0$. But then again, if $t=0$ then what you wrote is a zero vector and so isn't an eigenvector.)

Oh, good point, I have to exclude 0

yes as eigen vectors cannot be a zero vector!

So I can have infinitely many eigenvectors to one eigenvalue?

astonishing

Yes. However, one typically only worries about eigenvectors up to equivalence.

So we'd take $e_1$ as representative of that entire class of eigenvectors.

So thos "normalized" eigenvectors?

Right.

Aaah, it's an equivalence class

Okay

There are various ways to normalize, though.

One could take the norm to be 1 (the most literal normalization).

Or one could require that the first nonzero entry be 1.

Both of those would uniquely determine a specific eigenvector, though usually not the same one.

@JunaidRehman Well, if the matrix $M$ of interest is a square matrix and unitary, then $M^\dagger M=MM^\dagger =I$.

Right

Yeah, it's just about having a handy representative of our class

Now for $\lambda_2=\mu$ I get the eigenvector (for even $n$, where $n$ describes the dimension of the given matrix) $\left(0,t,-t,t,-t,\cdots,t\right)^T$ and (for odd $n$) $\left(0,-t,t,-t,\cdots,t\right)^T$, where $t\in\mathbb{R}\setminus\{0\}$, as before

Is there a formally better way to describe it?

You're basically looking for eigenvectors of $\begin{pmatrix} 0 & 1 & 0\cdots & 0 \\ 0 & 0 & 1 \cdots\\ \vdots\end{pmatrix}$

except, y'know, not terribly written. (i've excluded the upper-left zero because it's irrelevant.)

oops, I kinda forgot $\mu-\mu=0$, lol

In component form, that's the matrix $S$ with $S_{n,n+1}=1$ and zero otherwise.

@NaCl Since it's up to a constant multiple, you could always start with a plus sign

and have the final sign be $(-1)^n$

$(0,t,-t,t,-t,\dots,(-1)^n t)^\intercal$

@Semiclassical Then it should be $\left(0,t,0,\cdots,0,t\right)^T$

@SteamyRoot That looks nice

Something seems off. But I think I'm just not awake right now.

Actually, yes, something is definitely off.

first $x_1(1-\mu)=0$, second $x_2(\mu-\mu)+x_3=0$, third $x_3(\mu-\mu)+x_4=0$,$\cdots$, $n$-th $x_n(\mu-\mu)=0$

What you've got in the lower-right corner is a Jordan block.

Yes, I'm aware of that

As such there should only be one eigenvector associated to that block, and it should just be $e_2=(0,1,0,\cdots,0)^T$.

(up to equivalence blah blah blah)

So the only eigenvectors are $e_1,e_2$.

But why is $x_n=0$? Doing $(M-E\cdot\lambda_i)\cdot \vec{x}=0$ yields for $\lambda_2=\mu$ the above system of equations

I just got the results for eligibility to the ENS. Happiest day of my life yet

@NaCl That's... a good point.

So I'm quite sure it should be $(0,1,0,\cdots,0,1)^T$

Well, I'll admit I'm looking at Mathematica here.

And it just gives zeros past the second element.

humm, ok

I'll ask my tutor

thanks!

Actually, I see it

oops, I do as well

You listed off a few of the equations. But if you look at the (n-1)th case, that's $x_{n-1}(\mu-\mu)+x_n=0$.

The $n-1$-th equation yields $x_n=0$

Exactly.

haha

So that's consistent.

Thanks!!

np

@Semiclassical So, all uitary matrices admit the same SVD? Based on the fact that $A^{\dagger} A = A A^{\dagger} = I$

Not sure. It seems wrong on the face of it.

Yes.

I think that section may be misleading us, though. That's saying that if we manage to find the SVD, then it'll satisfy blah blah blah

It's not telling us how one finds the SVD.

Let me get it from another angle. We know the eigenvalue relation $A \times u = \lambda u$ where $u$ being the eigenvector and $\lambda$, the eigenvalue.

Going back to first principles, the right-singular vectors will satisfy $Av=\sigma u$ and $A^\dagger u = \sigma v$.

Aah right.

So therefore $A^\dagger Av = A^\dagger (\sigma u)=\sigma^2 v$.

But $A^\dagger A=I$ for a unitary matrix, so $v=\sigma^2 v$.

Yes.

If $\sigma=1$, then literally any nonzero $v$ will be an eigenvector.

So we can take $V=I$ as the basis.

And similarly for $U=I$. But that would seem to assert that $A=I\Sigma I^\dagger $ is already an SVD, which seems wrong.

Yes.

Though, these are all degenerate eigenvalues.

That may be the escape clause here.

The Wiki page doesn't seem to support that interpretation, though.

There is something wrong.

Yeah.

The bigger picture of my problem is, I am trying to find the eigenvectors of a unitary matrix using Matlab

A simple test case may be illuminating. Suppose $A=\text{diag}(i,-i)$.

for the repeated eigenvalues, the eigenvectors are not orthonormal.

at mtlabcentral someone suggested to use SVD in place of eig because it is more robust.

I will test the case you gave above.

What Mathematica gives for that is illuminating.

Hi @Krijn

Hey @Balarka

I just asked my first question on overflow

In that case, let $U=\begin{pmatrix} 0 & i \\ -i & 0\end{pmatrix}$ and $V=\begin{pmatrix} 0 & 1 \\ 1 & 0\end{pmatrix}$

Haha

I have been thinking about asking a question on MO but don't know if I should

Question on?

Hi chat

Heyo

some foliation shit

(you might recognize these as Pauli matrices if you've done any quantum physics)

Yes.

@BalarkaSen Foliation is almost the same as stratification, which is what I'm looking at :D

These are both unitary, but $UIV^\dagger = UV = A$.

Oh, indeed?

So what Mathematica does is take the unitary matrix, factorize it into two other unitary matrices, and declare that the SVD with $\Sigma=I$.

@BalarkaSen No, only in geological terms, I think

I am sorry, I am not following the notation. Eigenvalues of pauli matrices are 1, and -1. What is $U$ and $V$ here?

$\sigma_2$ and $\sigma_1$.

Whereas the $A$ I gave earlier was $A=i\sigma_3$.

(I may have a minus sign wrong somewhere, probably on the definition of U. I can never remember where the signs go on $\sigma_2$.)

But the statement should go like this. Note that $i\sigma_3$ is a unitary matrix, and that $i\sigma_3 = \sigma_1 \sigma_2 = \sigma_1 I \sigma_2^\dagger$ since the Pauli matrices are Hermitean.

Yes.

My point is that this is an SVD, since $I$ is obviously diagonal with nonnegative real entries and $\sigma_1,\sigma_2$ are unitary.

If that doesn't seem useful...well. I basically agree.

This seems to suggest to me that the SVD is really not going to be useful for a unitary matrix.

I have the similar feeling.

SVD might be useful only in the case of rectangular matrices

or perhaps in case of square matrices as well when you are interested in the magnitude of eigenvalues

Right.

For unitary matrices, though, it seems pretty useless.

Yes.

What are you wanting to do with these unitary matrices again?

Actually I am looking to get their orthogonal eigenvectors. Some relation like $A x = c x$

Hmm.

but the eigenvectors I am finding are not orthogonal

in some cases

I tried the svd and it gave the orthogonal basis

but since we do not have a relation of above kind, it is not useful at all.

Yeah.

@JunaidRehman Within the same eigenspaces, I take it?

Yes, within the same eigenspace.

You could just apply Gram-Schmidt to make them orthogonal, then...?

Actually some eigenvectors have no non-zero entries and interestingly they are orthogonal

Well, one has the spectral theorem for unitary matrices.

1) All eigenvectors have magnitude 1. 2) Eigenvectors corresponding to distinct eigenvalues are orthogonal. 3) The eigenvectors comprise an orthogonal basis.

2) distinct

if they are repeated, we can apply Gram-schmidt to get them orthogonal?

Wait. Eigenvectors with "no non-zero entries"?

Yup.

Yes

Wouldn't that just be a zero vector?

I imagine that was supposed to be all non-zero entries

I am sorry

no worries

all non-zero

Presumably the point is then that those eigenvectors without no zero entries all correspond to distinct eigenvalues. Which seems plausible on the face of it.

If there's no obvious reason for them to represent the same eigenvalue, they probably don't.

(weasel words out the wazoo right there, though)

so if I have a matrix $V$ of eigenvectors. Columns are not orthogonal. Though I have the relation $A v_i = \lambda v_i$. If I apply Gram-Schmidt on the matrix and obtain the orthogonal columns, does the relation $A x_i = \lambda x_i$ still remains?

where $x_i$ is the $i$th column after the Gram-Schmidt

Sure. All Gram-Schmidt does is take linear combinations.

You should only apply GS within the same eigenspace

^

Right, corresponding to same eigenvalue?

Right.

Yup

I see.

Since each vector in the subspace is an eigenvector with that eigenvalue, all linear combinations of them will also be eigenvectors.

So yeah, that should work.

That said, I'm not sure how numerically stable doing Gram-Schmidt is.

So there may yet be a smarter way to proceed.

I see.

Thank you very much @Semiclassical and @SteamyRoot. This discussion was really refreshing and useful for me.

glad to be of help

Hmm, here's a thought actually.

Yes

Blarg, no. Silly thought, doesn't work, ignooooore

You can share it for the sake of it. Maybe to give some insight, if you want.

Okay, slightly less silly thought. I wonder if there's a way to modify your matrix $A$ in such a way that the matrix becomes non-degenerate.

And then infer the original eigenvectors/eigenvalues.

Like $A + I$?

That's what I was thinking originally. But it doesn't lift the degeneracy.

Why do you call this matrix degenerate, because of repeated eigenvalues?

Right.

Yes, then this does not solve the issue.

For me, Gram-Schmidt sounds like the way to adopt.

Yeah, start there.

How large are your matrices?

$3\times 3$

Okay, that's not bad at all.

If it was a large matrix this could be an issue, but 3-by-3 is small enough that GS shouldn't be an issue.

Actually the size is not an issue. I wanted to clarify my concept of it.

Sure, I meant with regards to numerical stability

Yes. I can actually do the GS by hand.

Right.

Had a homework assignment. So plenty of practice ;)

How do you type math (Latex) in the chat? Manually or using some tool?

Your typing is fast.

Manual.

I just have a lot of practice.

Nice.

@Astyx alors vous avez prévu un truc avec Ted ?

How is this even possible

Well, it's impossible to do with 4 or less triangles.

And, among the numbers that can be expressed as a sum of squares, it's more likely that larger numbers can we written as a sum of squares in more ways.

(once you have 5 triangles, you can just subdivide a triangle in two smaller ones, though you may have to multiply all lengths with a common factor...)

or bruteforce it in Python

@JunaidRehman Detexify can be of help

Your brain does the translating. I don't even see the Latex code. All I see is blonde, brunette, redhead. Hey uh, you want a drink?

Hey @N3buchadnezzar

Ahoy

Your integral book is turning out to be helpful.

For learning Norwegian? :p

hey anyone familiar with universe ?

haha, no, for computing integrals. have to do that a lot for the various exams i have this year

universe is too friggin big man

Heh great. It covers atleast most of the standard integrals

No I mean universe in the context of Grothiendieck @BalarkaSen

lol

Just turned in my thesis, so maybe I have some time this summer to start reworking it into English.

@N3buchadnezzar Yeah, it has a lot of material in it.

Oh, congrats.

hi everyone btw

missed you guys :)

Thanks =)

I'd love an English version, but the Norwegian isn't being much of a trouble actually.

Big scary integral in one of my PSE question under preparation:

$$\lvert x_1y_1\rangle \otimes \lvert x_2y_2\rangle = \frac{1}{N^2}\int_{\Bbb{R}}^{(4)}e^{(x'+d)^2+(y'+d)^2}e^{(x'-d)^2+(y'-d)^2} \lvert x'y'\rangle\otimes \lvert x''y''\rangle d^4\mathbf{x}$$

how are you @BalarkaSen ?

@N3buchadnezzar Thanks. I will look into it. I am pretty new (almost 2 years new) with Latex stuff ;) .

i'm good

@Secret Did you invent that notation for multiple integration yourself?

@BalarkaSen Yeah, its easy to read integrals in another language ;)

learning anything new lately ?

@SteamyRoot yeah, I invented it (and used that in one of my MSE questions)

mostly foliations and things

cool

As it was mostly written 5 years ago a lot of the explenations and stuff is just plain wrong. So woulød be nice to make a second edition.

I've seen people use $d^n x$ before, but a $^{(4)}$... meh

I am learning algebraic geometry, homological algebra, and some advanced commutative algebra.

neat

@SteamyRoot Used that in my thesis to denotote integrals over the polydisc lol :p

alternately, using existing notation, another way to write multiple iterated integrals is simply use $(stuff)^{(-4)}$. This notation is used in differentiation context such as some integral differential equations

Afaik, standard notation for this would just be $\int_{\mathbb{R}^4} d^4 \mathbf{x}$

Ah I see

I dunno if there is a standard notation for the volume form on R^n

$$ \int_0^{2\pi} \cdots \int_0^{2\pi} f(e^{i x_1}, \ldots , e^{x_n}) \frac{\mathrm{d} x_1 \cdots \mathrm{d} x_n}{(2\pi)^n} = \int_{\mathbb{T}^n} f(z) \mathrm{d}m_n(z)$$

What was your thesis on?

Oh hey, brillouin zone

Hey everyone!

@BalarkaSen Hankel forms and Neharis theorem. Basically assiocating functions on the infinite polydisc with Dirichlet Series. Nothing revolutionary. Just really compact notation ^^

@Semiclassical Hey I suspect an error in this paper on gravitational waves tat.physik.uni-tuebingen.de/~kokkotas/Teaching/NS.BH.GW_files/…

Dimensional analysis on eq. 2.40

@BalarkaSen omega?

Uh oh @Arctic... Normally I'd write omega as a generic form

I get $[TM^{-2}]$ when it should just be T

@Secret Well, I don't know what the ket's contain, but the exponentials can be simplified alot. It smells like substitution to polar coordinates.

yeah, probably a jacobian of $r'^2$ and $r''^2$ will pop up somewhere after the substitution

@Balarka in the manifolds chat we're discussing whether eating a donut is cannibalism up to diffeo...

Yeah, I agree. @Lozansky

Well, it depends on what happens to the ket's, but you'd have just $r'$ normally. Not sure if a substitution for the $x''$ and $y''$ is necessary.

@Lozansky Equation 2.39 looks to be right dimensionally.

Yes

What's the condition for coalescence?

I presume it's something like "a is sufficiently small" but I don't know how small.

Merging together

Hmm.

It happens very rapidly

Okay. So there's some finite time for $a\to 0$.

Yeah

Incidentally, one way to solve that equation is probably to write it in the form $\frac{da}{dt}=-\frac{1}{a^3}\frac{a_0^4}{t_0}$

Where $t_0$ is defined implicitly by comparing with what they wrote

@N3buchadnezzar Ah, analytic stuff I know nothing about :P

@arctictern Interesting, never seen that one.

If I multiply $a^3$ over to the left, I can rewrite that as $\frac{d}{dt}a^4 = 4a^3\frac{da}{dt}=-4\frac{a_0^4}{t_0}\implies a^4 = (-4 a_0^4/t_0)t+C$

@Daminark Hmm, interesting question (:P)

@BalarkaSen I mean, it's on the "volume form" wikipedia page all over. it's the usual letter I've seen (although I've only read riemannian stuff peripherally.)

Since $a(0)=a_0$, this simplifies to $a^4=a_0^4(1-4t/t_0).$

So the time to go to zero is $t_0/4$.

@Semiclassical I think it should be $M^2$ not $M^4$

I have never read the wikipedia page on volume forms. You might be right, I am generally bad at remembering notation.

Comparing with what they had, one finds $1/t_0=64 G^3 \mu M^2/5c^5 a_0^4$

and therefore $\tau = t_0/4 = \dfrac{5c^5 a_0^4}{256 G^3 \mu M^2}$

Which, yes, has $M^2$ not $M^4$.

So yeah, good attention to detail.

A way one could have caught this is by working out what the period of rotation of a planet around the Sun would be. (The masses would be different, but the dimensions would still have to match.)

And that presumably would behave again as $c^5 a_0^4 /G^3 M^3$.

Which sounds very Kepler's third law - ish.

It is derived using Kepler's third law

Right.

Oh, of course. Because they initially had it in terms of $\Omega$.

Yeah

\begin{equation} \begin{cases} Q_{xx} = - Q_{yy} = \dfrac{1}{2} \mu a^2 \cos(2 \Omega t) \\ Q_{xy} = Q_{yx} = \dfrac{1}{2} \mu a^2 \sin(2 \Omega t) \end{cases} \end{equation}

Right-o.

Anyways, sounds like you've got it covered.

Is the quadrupole tensor $Q^{\mu \nu}_{TT}$ with the transverse-traceless gauge applied to a rotating binary system

Is that a question, or a statement?

It was a statement, supposed to follow right after the previous message...

ahh

This gives me way more reasonable values

@Astyx @Hippa on va me laisser savoir la décision?

Hi @Semiclassic @Lozansky

@TedShifrin Hi Ted! How's Europe?

It's fine.

@LeGrandDODOM @Ted Le plan serait d'aller à un crêperie rue Montparnasse. On a dit qu'on verrait sur place les disponibilités (il y en a suffisamment pour qu'on ne risque pas que tout soit complet). Il restait à s'accorder sur le lieu et l'heure de rdv

Et j'ai eu mes résultats d'admissibilité aux ENS et à Polytechnique

hi! @TedShifrin

someone can take alook? math.stackexchange.com/questions/2314637/…

Et les résultats sont bons, @Astyx?

Formidable! Félicitations!

@Liad Have you checked the answer here? Maybe it contains what you need?

Astyx, envoie-moi un email lorsq'on saura.

@SteamyRoot my question is exactly about part 2

Hi @Ted @Liad and everybody else

hi

Hi @Alessand

liad, I'm confused. The closure of a finite union is the union of the closures.

@Ted C'est également à toi de nous dire. Enfin si j'ai bien compris tu n'as pas de préférences pour l'heure ?

Really? i tried to prove it and got stuck on something so i thought it isn't true.

Totally true, Liad.

Hi there

I have a question about MSE

im gonna try it again.

math.stackexchange.com/q/2313791/348589 This question is what I asked.

That is put on hold.

Saying : "This question is missing context or other details"

But it is well answered and nothing missed.

@Astyx: 12 ou 12:30, si ça convient. Le difficulté sera de se reconnaître so on ne dit pas exactement où.

@kayak: That's the usual explanation when there's a homework-looking question with minimal effort.

Is there a simple way to see that a matrix of the form $\begin{pmatrix} 0 & A\\ A^T& 0\end{pmatrix}$ (where the blocks are submatrices) always has signature $0$?

@TedShifrin Ohh..

@s.harp what is signature?

See the eigenvalues come in plus/minus pairs, @s.harp?

ok the direction $\cup_{i=1}^l \overline{ V_{x_i} } \subset \overline{\cup_{i=1}^l V_{x_i}} $ is easy. the other direction does not work for me @TedShifrin

@TedShifrin Definitely its not related to my HW. It comes from my head and trial and error.

@TedShifrin Moi ça me convient en tout cas, je verrai avec les autres et je te tiens au courant par mail ! Je pense que nous te reconnaitrons assez facilement (et on dit "la difficulté")

Explain that and give more concrete thoughts on what you've tried, in the future, @kayak.

@Ted I don't know how to see that right now

are the tricks with the char. polynomial?

if $x \in \overline{ \cup_{i=1 }^lV_{x_i} }$ then for each nhbd $U$ of $x$ we know $U$ intersects $\cup_{i=1 } ^l V_{x_i}$ , so it is true for some $i$ , but it can be different for different $U$s doesn't it?

is it true that $\det\begin{pmatrix}\lambda\Bbb1& A\\ A^T&\lambda\Bbb1\end{pmatrix} = \lambda^{2n}- \det(A)^2$?

There are known things about det of block matrices with certain hypotheses, @s.harp

@TedShifrin Thanks for your suggestion. Is it not allowed or not suggested to ask true or false questions? It could be a big deal to solve something because it can give someone whether this direction is right.

Yes, @s.harp.

alright thanks ;), wiki has a general formula for this kind of thing

@arctictern I have been thinking about the following thing: say $\text{Map}(S^2, S^2)$ be the connected component at $\text{id}$ of space of self-maps of $S^2$ (there are Z-many connected components of Map(S2, S2)), and $\text{Map}_*(S^2, S^2)$ be the space of based maps.

Then you get a fibration $\text{Map}_* \to \text{Map} \to S^2$ given by evaluating a map $S^2 \to S^2$ at $p$. This gives a homotopy exact sequence $\pi_2(S^2) \stackrel{\partial}{\to} \pi_1(\text{Map}_p) \to \pi_1(\text{Map}) \to \pi_1(S^2) = 0$. Now $\pi_1(\text{Map}_*) = \pi_3(S^2)$, so now I want to understand what that map $\pi_2(S^2) \to \pi_3(S^2)$ is. If it's anything other than the zero map, $\text{Map}$ has to be simply connected which I don't think I believe.

Meh, it won't accept a * so I used Map_p up there at some point

@Liad: Is $\bar A \cup \bar B$ a closed set contaning $A\cup B$?

yes

Ah, chatjax no longer works on my ipad.

you write that fast on ipad?

Is it the smallest, Liad?

math.stackexchange.com/q/2314906/348589 This question looks more HW but I got +1 lol

@TedShifrin i can use this def. in a general topological space? i have the def. $x\in \overline A $ iff $x \in U$ then $U$ intersects $A$

$U$ is nhbd of $x$

That's not stated carefully.

from what you wrote we have $\overline{A\cup B} \subset \overline A \cup \overline B$

why not?

For all $U$?

yea

But that gets you stuck for the reason you said.

Actually Map is not simply connected so that is going to be the zero map. Hm.

OK, I need to go.

Bye, all.

See you.

bye.

Bye Ted

does there exists dirac deta function with 3 variables e.g. $\delta (a,b,c)$?

so basically I am trying to construct an integral requiring 2 dummy variables to become equivalent to a specified dummy variable

(Replace id by constant map up there by the way)

@TedShifrin I just finished watching comey's testimony

Could someone please explain the 5th step of the wavy curve method here: brilliant.org/wiki/wavy-curve-method

WHat's the easiest/ best way to solve such inequalities: wolframalpha.com/input/?i=((x-1)(x%2B2)%5E2)%2F+(-1-x)%3C+0

Sign of how utterly weird I can be: I just went and made a paper + tape model of the beaches of the Riemann surface I've been looking at

@Semiclassical One man's weird is another's "frickin' sweet"

Lol

I think I did it a bit backwards, alas

Backwards in which way?

Like, det = -1?

Well, do you know what the Riemann surface for log looks like?

Like an infinite staircase

Usually, that winds CCW as you move up the staircase. Mine winds CW

Ah.

I guess that would make it the surface for -log?

Yeah