Mathematics - 2017-01-23 (page 2 of 4)

It should be a bunch of 1 dimensional affine subspaces, not Riemann spheres

@Krijn An angry one? ... Try a screwdriver?

@TedShifrin I will. Is coconut water nice?

Tobias Kildetoft

4:32 PM

@TedShifrin So your recommended drink to make the anger go away is a screwdriver?

@DHMO But 12 is a multiple of 4, so y could be also 2

@Semiclassic: One node, possibly, yes.

Hush, @Tobias. :)

@Twink we want numbers of the form x12y

Yes, @Krijn :)

Hm. I guess I should reserve 'Riemann Sphere' for the projectivized curve.

4:33 PM

@Twink since it is divisible by 4, the last two digits, which is "2y", should also be divisible by 4

Right.

On CP^2 you get a bunch of spheres

Yeah.

@Semiclassic: If this is projective, then of course any pair of lines in $\Bbb P^2$ intersect.

I think I do want to be projective here.

@Twink Any number of the form xz00 is divisible by 4 so the last two digits should also be divisible by 4

4:34 PM

@TedShifrin Right.

Hi @Twink. Long time no see.

@DHMO and why just the last two digits? we could get 3 digits, for example 36*5=180

Not sure why you say "one node", though. If I've got $n$ linear factors, then there should be $\binom{n}{2}$ intersections and as many nodes.

hi @TedShifrin

Sorry, @Semiclassic. I thought we started with just two lines. Sure.

4:35 PM

@Semiclassical at most

@Twink if a number is divisible 4, then its last two digits are divisible by 4

we observe that 100 is a multiple by 4

@Balarka: Why at most? If there are no triple points?

@BalarkaSen Yeah. I'm assuming singular points of at most multiplicity 1.

@Twink so if a number is divisible by 4

it will still be divisible by 4 if we subtract 100

so we continue subtracting 100

until there are no more 100 left to subtract

and we would obtain the last two digits

oh :D

4:36 PM

@TedShifrin @Semiclassical Yeah, I didn't notice that triple points were not allowed

which is divisible by 4

@Balarka: Should I remind you about the leak at the back of your head?

I actually do want to consider triple points eventually, so I don't object to them in principle. But for now I want to avoid that complication.

Now, am I correct in thinking that a 'generic' perturbation will resolve all nodes?

hey @TedShifrin

Yes, @Semiclassic. Intersection of generic is generic :P

4:38 PM

By looking at $f + \epsilon \cdot g = 0$, yeah, I think so

I have a linear algebra question @TedShifrin

Yeah.

@TedShifrin I was about to ask you a geometry question. Continuing our earlier discussion, can we see geometrically what $d\omega_1 = 0$ means? From the formulas that means $\omega_{12} = f\omega_2$... which means... $d\omega_{13} = 0$ in a principal moving frame... no idea what all this means geometrically.

All this falls into my "stupid question streak"

@DHMO and you used that since it's divisible by 9 the digits must sum a multiple of 9?

@Twink yes

4:40 PM

Okay. Then (continuing my logic) a generic perturbation will replace each node with a handle. Since each sphere (again, over CP^2) touches at one point, I'll get some smooth Riemann surface of higher genus.

@DHMO @Krijn thank you :)

@Semiclassical Yup

mmkay.

@Balarka. not stupid questions, so shaddup. ... $\omega_{12} = 0 \pmod{\omega_2}$ means that the $e_1$ curves are all geodesics. If they're also lines of curvature, then what do you know about the principal curvatures of those lines of curvature?

Ahh, $\omega_{12}(e_1) = \omega_2(e_1) = 0$ in a principal frame, so $e_1$-curves are geodesics. Very nice.

4:44 PM

Amusingly, I can hear a lecture across the hall here talking about (spatial) curvature of (magnetic) field lines.

And I'm pretty sure they have to choose particular (reference) frames in order to make the computations possible.

Hmm, follow-up question.

Suppose my curve is specifically of the form $f(x,y)=(y-x)\prod_{k=1}^n (y-y_k)$ with distinct $y_k$.

@TedShifrin Well, the $e_1$-curves in a principal frame are lines of curvature aren't they? So that curve is planar.

So the real slice of $f(x,y)=0$ is a bunch of horizontal lines and a slant line crossing through all of them.

Why planar? I certainly don't see that, @Balarka. I was asking what you knew about $k_1$.

You want only one $x$, @Semiclassic?

Lines of curvatures which are geodesic are planar.

Right.

4:51 PM

I think $k_1 = 0$.

Oh, right, @Balarka. No, Lots of counterexamples to that.

Hm

If I projectivize, I've got spheres $y-x=0$ and $y-y_k z=0$ for $k=1,\ldots n$.

(I'm sounding out my thoughts as I go, so sorry if I ramble)

You have some bad intersections going on at infinity. All those $\Bbb P^1$'s (except the first) intersect in one point.

That's what I thought.

This is a highly degenerate case.

So, could I describe the situation as such? I've got $n$ spheres which touch at a common point, and which each touch another sphere ($y=x$) at a distinct point.

4:55 PM

You can draw a (schematic) picture with a bunch of lines passing through a fixed point and a last line crossing them all transversely.

Right.

@TedShifrin Oh I guess a longitude in the standard torus which is not "on the top" works as a counterexample. Certainly $k_1 \neq 0$

Longitude meaning the "long curves", not meridians.

@Balarka: Remember we're not talking about a single curve. We're talking about a whole local picture.

The question which I think I can't answer without giving more details is what will happen to that picture upon perturbation.

I don't follow your example, @Balarka. The "fat" circles are almost never geodesics.

4:57 PM

Certainly I expect the nodes which touch the y=x sphere to resolve into handles.

But the singularity at infinity is a lot less obvious.

It's no different. Nah, the multiple point breaks up into lots of nodes, and then you resolve those.

Hmm.

Oh duh. It's geodesic only when it's on the top. That was rubbish.

@Balarka: That's totally rubbish.

You know full well the top curve is not a geodesic.

@TedShifrin do you use Grindr?

4:58 PM

LOL, @Twink. This is not the place for such discussions.

I'll take that as a yes

You'd be wrong, but ... I really do not want to discuss this here.

then give me your email :)

sorry, sorry. not top. the shortest and the longest of those.

But it does work with the other family of circles, @Balarka. However, in that case, $k_1$ is totally constant. It does not have to be. You might check out whether $d\omega_1=0$ in that case. ....

5:03 PM

u.u

Ugh, campus wireless is down

What I was about to say: What makes things complicated in my particular case is that the perturbations aren't generic

So I can't take "all singularities resolve" for granted.

Is point slope form of a line in an oblique axis $${y - a \over x- b} = m = {\sin \theta \over \sin(\omega - \theta) }$$

I think it is but I can't find a reliable source stating this.

@Semiclassic: True enough.

@A---B: What do you mean by "a line in an oblique axis"?

Hi.

@TedShifrin A coordinate system in which the angle between x and y axis is $\omega$.

5:15 PM

Oh ... Then that's a funny notion of slope.

@TedShifrin Literally, I can't find a page on this thing.

@TedShifrin You were absolutely right :D

You should trust me, @Mahmoud. I taught that stuff more than 13 times.

@A---B: I'm still not sure I understand your definition. Are you still measuring with a unit distance along each axis?

@TedShifrin Sorry, had to step away for a while. What's an example where $k_1$ is not constant?

I'd elaborate on what I mean by non-generic, but right now I'm having to chat on my phone because of the aforementioned outage

5:17 PM

I'm telling you that $d\omega_{13}=0$ tells you something else about $k_1$. Think about other surfaces of revolution.

So, no-go on that for now. :/

Hey @Ted

Hi @Danu

I have a textbook here saying that a Lie subalgebra of the Lie algebra $\mathfrak g$ of some Lie group $G$ corresponds uniquely to a connected Lie subgroup.

Doesn't $SO(3)$, $SU(2)$ show that that's wrong?

@TedShifrin http://www.mysearch.org.uk/website1/images/pictures/296.1.jpg

I need to find point slope form of a line in that green coordinate system.

5:20 PM

Did they forget simply connected?

and there, it's working again. :/

@TedShifrin Well, $d\omega_{13}=0$ can happen in a principal frame if $k_2 = 0$, so $k_1$ might as well be nonzero there

@Balarka: Try to be as general as possible.

No takers?

Also hi Balarka

Isn't SU(2) a double-cover of SO(3)?

5:22 PM

Yup.

What's your $G$ here, @Danu?

@TedShifrin Is my equation correct ?

@TedShifrin Any Lie group

I don't think so, @A---B.

I get the right hand side $\cos\omega + \sin\theta\sin\omega$.

Hi @Danu

5:23 PM

I mean for your example, @Danu.

$SU(2)$, for instance

Just take the trivial (total) subalgebra

$SO(3)$ is not a subgroup.

Ah, subgroup

great

:D

No

No way

What about $GL(3,\Bbb C)$

Abstractly, given any Lie algebra, there's a unique simply connected Lie group with that as its Lie algebra.

$SU(2)$ is not likely a subgroup.

Oh.

$SU(2)$ and $SO(3)$ should embed into that both, no?

yeah, definitely should

5:25 PM

@TedShifrin :S

@TedShifrin I did not follow how ?

I have too many things to think about, @Danu, so probably you're right.

lol

@A---B: Assuming $a=b=0$, a point on the line is of the form $x(1,\sin\theta)$ in usual coordinates. But we get your $y$ by dotting with the unit vector $(\cos\omega,\sin\omega)$.

$$\lim_{x\to 1} (x^2+x+1)=3 \iff (\forall \epsilon \gt 0)(\exists \delta=\operatorname{min}(1,\frac{\epsilon}{4}) \gt 0)(\forall x\in \Bbb R): |x-1|\lt \delta \implies |f(x)-3|\lt \epsilon$$ Does this seem to be right ?

5:30 PM

Yes, @Mahmoud. Keep doing more practice.

I wouldn't write it that way, though. I prefer sentences with words.

@TedShifrin I find symbols more space consuming, and less chaotic.

@TedShifrin So it should be $${y - a \over x- b} = m = \cos\omega + \sin\theta\sin\omega$$

I believe so, @A---B. You checked my argument, right?

@TedShifrin: I don't have a solid understanding of what the proofs are about tough.

@Mahmoud: You really should write. Given $\epsilon>0$, choose $\delta = ...$. Then show it works. You need to learn to write good proofs. These do not consist of lots of symbols.

5:34 PM

@TedShifrin : Okay, I will.

I gave you lots of model proofs in those notes.

@TedShifrin Yes. But don't trust me cause I don't have much idea of vectors.

Vectors are your friend, @A---B.

@TedShifrin Yes. (Hopefully)

So slope intercept form will be $$y = (\cos\omega + \sin\theta\sin\omega)x + c$$

Yup.

5:40 PM

Thank you @TedShifrin

Sure :)

@TedShifrin Do you know any use of these coordinate system ??

A true physical application? No.

Mathematically such things arise ... You see general coordinate systems in linear algebra.

@TedShifrin Thanks.

@A---B One place you see oblique coordinate systems in physics is solid state, specifically crystal structures.

5:47 PM

hey @TedShifrin

hi Karim

A tangential vector field on $\mathbb{S}^n$ is a continous map $\sigma : \mathbb{S}^n \rightarrow \mathbb{R}^{n + 1}$ such that for all $x \in \mathbb{S}^n$, $\sigma(x)$ is perpindicular to x. Show that if n is odd then $\mathbb{S}^n$ has non-vanishing vector field.

Here is my argument

Just write one down.

The reason being is that the planes of atoms need not be orthogonal to one another. They may prefer to order in a way that involves an oblique angle.

Consider $\sigma$ given by $(x_1,...,x_{2n}) \mapsto (-x_2,x_1,-x_4,...,-x_{2n},x_{2n - 1})$

5:49 PM

Right, Karim.

This is continous as it is combination of sum and cosine

But would you look at a line in that configuration, @Semiclassic?

Well, you'd certainly look at planes in that configuration. That'd tell you about different scattering directions.

Cool.

Some details here: users.aber.ac.uk/ruw/teach/334/bravais.php

5:51 PM

@TedShifrin I am gonna prove the following: If $\mathbb{S}^n$ has a non-vanishing tangential vector field, then the antipodal map is null homotpic.

Right. Standard result. It's even in my book :)

cool

What's also neat is that, because you are doing scattering experiments in that context, you end up talking about the Fourier transform of such lattices.

So you end up doing stuff with lattice duality in crystallography.

(indeed, it's in the momentum space (the dual space) that one formulates Bragg's law for scattering most appropriately)

Liad

Hi, i got a question regarding "the fundamental theorem of calculus" . is it true that if F(g(x)) ' = f(g(x)) *g'(x) ? (assuming F = $\int f $ and x is fixed).

@MikeMiller I just got this joke like ten minutes ago while I was cooking

Liad

5:58 PM

i think my claim is true, but how is this derived from the fundamental theorem of calculus ?

I want to calculate the degree $[\mathbb{Q}[\sqrt{3}, \rho] :\mathbb{Q}]$.

It holds that $$[\mathbb{Q}[\sqrt{3}, \rho] :\mathbb{Q}]=[\mathbb{Q}[\sqrt{3}, \rho] :\mathbb{Q}[\sqrt{3}][\mathbb{Q}[\sqrt{3}] :\mathbb{Q}]=[\mathbb{Q}[\sqrt{3}, \rho] :\mathbb{Q}[\sqrt{3}]]\cdot 2$$ and $$[\mathbb{Q}[\sqrt{3}, \rho] :\mathbb{Q}]=[\mathbb{Q}[\sqrt{3}, \rho] :\mathbb{Q}[\rho]][\mathbb{Q}[\rho] :\mathbb{Q}]=[\mathbb{Q}[\sqrt{3}, \rho] :\mathbb{Q}[\rho]]\cdot 4$$ since we know that the polynomial $f=x^4-2x^2-1\in \mathbb{Q}[x]$ is irreducible, and $\rho\in \mathbb{C}$ is a root of $f$.

Hi @iwriteonbananas

Hi Nlalarkan

I mean Balarka

lol

@BalarkaSen Whatchu up to

@Danu Do you know model categories?

6:02 PM

didn't really get the joke.

been fiddling with forms

@MaryStar You can check if $\sqrt{3} \in \mathbb{Q}(p)$ which solves your question

I want a new phone

But I don't want to spend any money

What are my options?

@iwriteonbananas No.

@iwriteonbananas There are very good Asian brands that aren't as expensive.

Otay otay

pickpocket one

6:04 PM

Like Xiaomi?

Yeah fuck it, i'll just steal a phone

@Krijn How could we do that?

@MaryStar I didn't think of an elegant way, but probably squaring a number in $\mathbb{Q}(p)$ and check if that can equal 3

@iwriteonbananas I've been using xiaomi phones for almost 4 years and they've always been great

@Krijn Ok, I will try.

They don't support a certain bandwith that my provider uses tho

And I don't feel like getting a new provider

6:09 PM

@MaryStar Or try it the other way, check if $\rho$ is a root of a degree 2 irreducible polynomial over $\mathbb{Q}(\sqrt{3})$

@Krijn Why of degree 2 ?

Thanks for the input however

@MaryStar Well, you know it's at most degree 4

It can't be degree 3 or 1 (do you know why)

Astyx

Hi chat

Hi @Astyx

6:12 PM

@Krijn Because $\sqrt{3}$ is a root of $x^2-3$, so the degree $[\mathbb{Q}(\sqrt{3}:\mathbb{Q}]$ is $2$.

@MaryStar No, this is not it.

If it was degree 1, then $p \in \mathbb{Q}(\sqrt{3})$, so it could not be a root of an irreducible polynomial of degree 4 over $\mathbb{Q}$

The total degree is 4, or 8, as you can find from the first formula you gave

I have a sequence easy to enumerate (tuples (a,b,c) with 0<a<b<c that satisfy an equation, so I can iterate and test each candidate). Is there a standard method/approach to get an indexable sequence without having to store/compute the entire sequence (eg you can get tuple #100 by simply asking for index 100, and you'll get the 100th lexicographically least tuple without computing the previous 99)?

So either $\rho$ is degree $2$ over $\mathbb{Q}(\sqrt{3})$ giving total degree 4

Or$\rho$ is degree 4 over $\mathbb{Q}(\sqrt{3})$ giving total degree 8

Hello, is there a better way to determine if this matrix wolframalpha.com/input/… is diagonalizable without finding it's eigenvalues or without analyzing it's characteristics polynomial monotonicity?

@user379685 Calculate its determinant?

6:16 PM

What would that tell me?

@A---B: Just realized I had a typo in what I wrote. The $\sin\theta$ should have been $\tan\theta$. So it's very close to your original formula. It's now $$\cos\omega+\tan\theta\sin\omega = \frac{\cos(\theta-\omega)}{\cos\theta}.$$

@user379685 What would the eigenvalues tell you?

Also if it doesn't have zero as an eigenvalue

The connection is that the determinant is the product of all eigenvalues

could you link me the theorem, proof?

No wait sorry, I'm mixing things up

Is there a continuous surjection from $\mathbb R_{≥0}\to\mathbb R$?

6:19 PM

@A---B: It still bothers me. We should get "infinite slope" when $\theta=\omega$.

2

Oh, no, this is all wrong.

Correct answer: The slope in the new coordinates is $\dfrac{\tan\theta}{\sin\omega-\cos\omega\tan\theta} = \dfrac{\sin\theta}{\sin(\omega-\theta)}$. This is pretty close to what you had originally. Sorry. I had to use a bit of linear algebra to get it right.

@s.harp I'd expect a "snake shape" to work.

The local extrema of $y=\sin(x)+cx$ all lie on $y=\sin(x)-x\cos(x)$, even as you vary $c$.

@s.harp Try $x\sin x$.

Map $[0,1]$ to $[0,1]$, $[1,3]$ to $[-1,1]$ (going from 1 to -1), then $[3,6]$ to $[-1,2]$ etc @s.harp

"back and forth"

It's cool to see the phenomenon I just mentioned on Desmos (since they have sliders)

@TedShifrin No problem, We all make mistakes. I also made the mistake by saying your calculations are correct, my fault.

6:29 PM

@responses yeah, thanks^ I figured out something like that too. Was looking for a counter example to math.stackexchange.com/questions/tagged/functional-analysis which now has two nice answers

We have that $$[\mathbb{Q}[\sqrt{3}, \rho] :\mathbb{Q}]=[\mathbb{Q}[\sqrt{3}, \rho] :\mathbb{Q}[\sqrt{3}][\mathbb{Q}[\sqrt{3}] :\mathbb{Q}]=[\mathbb{Q}[\sqrt{3}, \rho] :\mathbb{Q}[\sqrt{3}]]\cdot 2$$ and $$[\mathbb{Q}[\sqrt{3}, \rho] :\mathbb{Q}]=[\mathbb{Q}[\sqrt{3}, \rho] :\mathbb{Q}[\rho]][\mathbb{Q}[\rho] :\mathbb{Q}]=[\mathbb{Q}[\sqrt{3}, \rho] :\mathbb{Q}[\rho]]\cdot 4$$

When we divide these two relations we get $[\mathbb{Q}[\sqrt{3}, \rho] :\mathbb{Q}[\sqrt{3}]]=2[\mathbb{Q}[\sqrt{3}, \rho] :\mathbb{Q}[\rho]]$.

@Semiclassical Oh I see the link. Thanks for it.

@AkivaWeinberger Heh, I essentially "discretely" did that :P

@A---B: To be more reassured with the correct answer, note that when $\omega=\pi/2$, it gives the usual answer, and that when $\omega=\theta$, you get infinity for the slope. It's right :)

This is what I get for trying to do three things at once. My apologies.

Oh, hey bananas @iwriteonbananas :)

@Balarka: You should have gotten that $k_1$ doesn't change as you move in the $e_2$ direction (so it sounds very much like it must be like the surface of revolution case).

@Danu They both embed into that, but they correspond to different subalgebras.

Just because subalgebras are isomorphic doesnt mean they're the same subspace!

6:38 PM

Okay.

So it's important to fix the "ambient group".

Yes

@Danu Piecewise linear category :P

We have that $[\mathbb{Q}[\sqrt{3}, \rho]:\mathbb{Q}]$ is $\leq [\mathbb{Q}[\sqrt{3}] :\mathbb{Q}]\cdot [\mathbb{Q}[\rho] :\mathbb{Q}]$. Therefore, we get that
$[\mathbb{Q}[\sqrt{3}, \rho]:\mathbb{Q}]\leq 8$.

From $$[\mathbb{Q}[\sqrt{3}, \rho] :\mathbb{Q}]=[\mathbb{Q}[\sqrt{3}, \rho] :\mathbb{Q}[\rho]][\mathbb{Q}[\rho] :\mathbb{Q}]=[\mathbb{Q}[\sqrt{3}, \rho] :\mathbb{Q}[\rho]]\cdot 4$$
we get that $[\mathbb{Q}[\sqrt{3}, \rho] :\mathbb{Q}[\rho]]$ is either $1$ or $2$. So $[\mathbb{Q}[\sqrt{3}, \rho] :\mathbb{Q}[\sqrt{3}]]$ is either $2$ or $4$.

if i know the determinant of A, can i get det(A+I)?

Astyx

Hi chat again

6:45 PM

@user379685 No. Try two examples.

Is there a way to give hints to mathjax on where to line-break in an equation? I've tried googling but nothing coming up is helpful in my case

After an operator like $+$ usually, I think

so that they don't think the first line is the entire equation

@user379685 for $2\times 2$ matrices you have $\det(1+x)=1+\mathrm{Tr}(x)+\det(x)$, for higher dimensional matrices you can find a construction in this paper: igm.univ-mlv.fr/~berstel/Mps/Travaux/A/… :)

@AkivaWeinberger : The problem for me is that I have {a : b | c} and I'd like it to break on : or |, but it is breaking in the middle of c (which has a couple of long expressions joined with $\land$

@Extrarius you can force a linebreak with double backslash

6:52 PM

@s.harp : I was hoping for something that says "break here if you need to break, before you fall back to your usual heuristic", but if I cant find such a thing, I'll break it manually like that

Q: Relation between Ihara's $\zeta$ and knot polynomials

Hi there...

any experts for graph AND knot polynomials around:

0

$\hskip2.3in$ has $A=\pmatrix{0&3\\3&0}$. It has three faces, so the rank of $G$ is $\chi(G)=2$. The reciprocal of Ihara's $\zeta$ function can be evaluated $$ \frac{1}{\zeta_G(u)}={(1-u^2)^{\chi(G)-1}\det(I - Au + (k-1)u^2I)}\\ =(1-u^2) (4u^4-5u^2+1) $$ Looking at the orientation of the edges...

@draks... Why do you expect there to be a relationship?

7:10 PM

@MikeMiller because the trefoil knot can be derived from the given 2-vertex cubic graph...

So? The trefoil knot can be derived from the trefoil knot, but I don't expect the Jones and Alexander polynomials to be related.

What? No I build my graph, calculate Ihara's zeta and then blow up the graph to get the knot and find several polynomials. Now I wonder if there is a relation. If it is Jones, Alexander or another I don't know...

So my graph is still hidden in the knot. Is one of the knot polynomials hidden in Ihara's zeta?

given the characteristic polynomial of A -λ^3 -3λ^2 + λ +2 how can i calculate the determinant of A^3+A^2-A-I?

7:27 PM

@user379685 Are you referring to my question?

no

phew...

Do you know the answer to my question?

did you talk about the Cayley-Hamilton theorem in class? @user379685

yes

7:30 PM

ok so use the fact that a matrix satisfies its own characteristic polynomial

you know that $-A^3-3A^2+A+2=0$

A^3+A^2-A-I=-2A^2+1

but what's det(-2A^2+I)?

hello?

@draks... All I'm saying is you should have a reason to believe there's a relationship before asking for/looking for one.

(One reason to think there shouldn't be is that the graph you use to construct the knot is hardly an invariant of the knot.)

Woops I left to make dinner, hmmm, you're right, I forgot a $3$ in my calculations..m.

Woops I left to make dinner, hmmm, you're right, I forgot a $3$ in my calculations...

7:50 PM

@MikeMiller the graph is kinda image of a map from knots to graphs, since by combining certain strands of the knot we get the graph...

@Alessandro, @user379685: If you let $B=A^3+A^2-A$, you want to evaluate the characteristic polynomial of $B$ at $t=1$. Can you find the coefficients of that polynomial in terms of eigenvalues of $A$?

@ted shifrin what's t?

Oh, I'm thinking of the characteristic polynomial as a polynomial in $t$. Sorry :)

hmmm i don't quite get what you mean:c

$\det(B-tI) = p_B(t)$ is the characteristic polynomial of $B$.

7:57 PM

yes

Therefore $\det(B-I) = p_B(1)$.

Ah, that's clever

There's still work to do, of course.

So you know the coefficients of the characteristic polynomial in terms of the eigenvalues of $B$ (sum, product, etc.) ... Can you relate those to the eigenvalues of $A$?

hmmm