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5:24 AM
Given a matrix $M$ and its characteristic polynomial $p(x)$, can we find a matrix $N$ whose characteristic polynomial is $q(x):=p'(x)$?
 
 
2 hours later…
7:10 AM
Morning all!
 
7:43 AM
Morning
 
Hey @Rithaniel :)
 
So, I have to make a video for this English course I need to complete before I can get my undergrad. Been working on that all morning.
 
Sounds fun.. haha
 
Decided to explain the concept of the cyclic group of two elements, because anyone could understand it.
Also, yeah, I wish I could focus down entirely on math, but there are other things I need to be concerned about.
 
That's the college system in the US right?
 
7:50 AM
Yeah, they want to ensure that people get "well-rounded" educations instead of more specialized educations.
 
I quite like that tbh, but you're right it's nice to focus on something you enjoy more
 
Indeed, it's just stressful because I'm not good at social stuff. At least, I haven't been, in the past.
What are you up to, this morning?
 
Well that'd improve with practice :)
I'm at work, my penultimate day here
and then I have 2 weeks of free time before I move to Deutschland
 
Bah, practice!
Also, woo, big transition time!
 
Yeah :) It really can't come soon enough
 
7:56 AM
Well, you were at that job for a while. I'm sure there's stuff you'll miss, too.
 
 
2 hours later…
9:27 AM
@AkivaWeinberger companion matrix is the keyword
 
 
3 hours later…
12:04 PM
Do you think a simple algorithm for intelligence exists?
 
 
2 hours later…
2:01 PM
How would one evaluate the limit $\frac{(\sin{(x)})^4}{2(\sin{(x^2)})^2}$ as $x\to0$ with l'hospital rule?
 
 
1 hour later…
3:06 PM
Corollary 2.2.4: If $Y$ is a closed subspace of a Hilbert space $H$ and $X \subseteq H$, then $(Y^{\perp})^{\perp}$ and $(X^{\perp})^{\perp} = [X]$. Proof: Since $[X]$ is a closed subspace of $H$, and $(X^{\perp})^{\perp} = ([X]^{\perp})^{\perp}$, it suffices to prove only the result concerning $Y$...
Why is $(X^{\perp})^{\perp} = ([X]^{\perp})^{\perp}$?
I've tried to derive this, but I could only derive tautologies.
 
can an algebraic variety be orthographically projected to $\Bbb R^2?$ If so how much information is preserved after the transformation?
 
3:47 PM
what is the collection of some algebraic varieties called?
 
4:00 PM
@AkivaWeinberger your link is a nice one, I read it all a year ago and it give a good overview of neural network, but deep learning is far from being intelligent.
The thing I keep from it is the proof that "neural network can approximate any non-linear function"
Is that intelligence? to be able to find a non-linear function for every task you encounter? I am not sure about that
 
@user193319 it suffices to prove $X^{\perp}=[X]^{\perp}$. If $A \subset B$, then $B^\perp \subset A^\perp$, so $[X]^{\perp} \subset X^{\perp}$. The other inclusion follows directly from linearity and continuity of the inner product
 
4:16 PM
Howdy @Mathein
 
hi @Ted
 
@AkivaWeinberger Wissner-Gross's equation : alexwg.org/publications/PhysRevLett_110-168702.pdf can be conceptually interesting, I did not dig far in though
 
Hey @Ted @Mathein
 
Hi @ÍgjøgnumMeg. Happy penultimate day!
 
Hi @ÍgjøgnumMeg @TedShifrin @MatheinBoulomenos
 
4:19 PM
Thanks so much @Ted ! :)
Hey @Alessandro
 
Hi demonic @Alessandro.
 
@Ted my boss took the team out for an unexpected meal at lunch time and the team had bought me leaving presents, which was really nice :)
 
Aww, that's actually quite unexpected and sweet.
You should be wracked with guilt :P
 
Hi @Alessandro @ÍgjøgnumMeg
 
@Ted haha, well they've known since January that I'm leaving so it's fine hehe
 
4:27 PM
Foucault pendulums demonstrate parallel transport around a line of latitude of a sphere?
 
See my diff geo notes, yes, DogAteMy.
 
No I wanna independently think of all the things in the book first, like that Borges story with Cervantes
 
Well, then you can't ask questions if you're going to be independent.
Warning: The physical situation is, of course, just a good approximation. It is not literally parallel transport.
 
Yeah 'cause there's centrifugal force isn't there?
 
That's fictitious. But, yes, there's a small but nonzero tangential component to the acceleration of the bob.
 
4:36 PM
what is parallel transport around a line of latitude?
 
How much differential geometry do you know? I only have one minute.
 
Parallel transport is when you move a vector along a surface such that it remains tangent to the surface
 
heya guys
does anyone do anything with logic here?
cos I'm looking for a recommendation for an introductory book on modal logic
 
You do it in such a way that the rate of change of the vector remains perpendicular to the surface (so in a sense, from the surface's perspective there is no rate of change)
The GIF shows parallel transport of two vectors around a loop
 
not rate of change, DogAteMy. That's not vectorial.
Anyhow, I need to get going. Hi/bye Sha.
 
4:39 PM
As you can see, they don't necessarily end in the same orientation they started with - the curvature of the sphere messes with it
On a plane it would just be simple translation
@TedShifrin I'm thinking of it being embedded in $\Bbb R^3$
rather than an intrinsic surface
@rapasite And a line of latitude is a horizontal circle on the sphere (so parallel to the equator, or circling a pole)
 
@TedShifrin hi/bye Ted!
 
@ShaVuklia Dunno, sorry
♫ Hello and goodbye / I just unemployed you ♫
 
ah okay, thanks yea it's hard to find logicians
 
(That's like one of two things I remember from Evita)
(Also "Don't Cry For Me, Argentina")
 
so your vector will only change orientation when you pass in the pole right?
 
4:44 PM
This doesn't have to do with the pole
If you chose any other triangle of that size the same thing would happen
 
5
Q: Books about modal logic?

RobboI've just approached modal logic reading "An Introduction to Non-Classical Logic" of Graham Priest. I am looking for some books that treat this argument in a more extensive way than the book I am reading. I am especially interested in the philosophical side of modal logic. Can you suggest me s...

 
The offset of where it ends up at the end of the loop versus where it started, is proportional to the area of the loop
 
but I mean the change is when you switch meridian
 
Pretend the North Pole is at the center of that triangle
and we're just seeing the sphere at an angle
The same thing happens
 
ok my bad, I will read about that to try to understand, you mean any close path will have this effect and it is because the curvature of the sphere right?
 
4:48 PM
Yeah
And the amount it rotates by is proportional to the area of the loop
or more precisely the amount of curvature contained in the loop
 
what about the equator?
aa ok the amount of curvature is 0 for the equator
cause it follow a geodesic right?
 
Better GIF
Here's another GIF
It starts with a flat piece of a circle
and then is folded into a cone
 
I mean if you choose the hole equator as the blue path
 
If you chose the whole equator, the region it contains is a hemisphere
and the hemisphere essentially has 360 degrees worth of curvature in it
which is why the vector ends the way it started
If you cut a hemisphere in half
so you have a quarter of a sphere
 
that is cool
 
4:57 PM
then it ends up facing exactly 180 degrees away from where it started
You can also imagine this
I think it's called a "(spherical) lune"?
with angle θ
Then parallel transport around its boundary rotated the vector by 2θ
If you cut it in half, so you have a triangle that starts at the North Pole, goes to the equator, goes sideways, and goes back to the North Pole
parallel transport around that rotates the vector by θ
--- With a cone
almost all of a cone has zero curvature
(because you can cover it with a piece of paper without deforming the paper)
but there's a lot of curvature at the vertex of the cone
so on a cone, this would depend entirely on whether or not your loop goes around the vertex or not
If it doesn't then the vector would end up exactly where it started
@rapasite This is called holonomy by the way
 
you just blow my mind
 
Yay
Arright I have to go to class
 
thx
 
5:42 PM
I still don`t see the vector turning 360° in the case where the vector is orthogonal to "the plan which contain the equator" and the path is the equator. Am I wrong?
 
6:17 PM
Look from the perspective of someone standing at the North Pole @rapasite
or from someone standing anywhere not at the equator really
 
6:38 PM
@rapasite You might want to take a look at my (free) differential geometry text. The link is in my profile. :)
 
Anyone not busy enough to help with a linear algebra problem?
 
Me. I am not busy enough to help with it :)
 
Why do you need to be sufficiently busy to help with a linear algebra problem?
 
I just meant in case you're busy with other things.
 
@MatheinBoulomenos Because procrastination is all we do
 
6:43 PM
true
 
Such a cynic, demonic @Alessandro.
What's the question, @kyle?
 
@TedShifrin it seems accurate based on personal experience
@Mathei do you have some intuition on nilpotent groups and central series to share? Like why are they natural things to consider
 
Assume you are given a pair of matrices A, B which satisfy AB = BA. Show that if
we set $C = A^@ + 2A$ and $D= B^3 + 5I$, then $CD = DC.$ Then try to generalize this in some interesting way, namely find a property for matrices C,D with that certain property, then CD = DC. For example, $C = A^2 +6A$ and $D = 3B^3 - 2I$ will also have CD = DC.
 
OK, @kyle. So did you get the first part? You just wrote it down and checked it, right?
 
I'm having trouble even showing the first claim. I don't see how the commutativity of A and B implies $A^2B^3 = B^3A^2,$ for example (when you expand it out).
 
6:46 PM
Well, $AABBB = ABABB$ for starters.
To write it down explicitly takes a bunch of steps, but you can get there.
 
@AlessandroCodenotti As far as I recall, the study of nilpotency actually originated in Lie algebras
where it is a more common property
 
Okay, should I try to get there before I proceed?
 
You should certainly convince yourself that you could :)
 
@AlessandroCodenotti not really, no
 
I don't know if you know some formal proof procedures like mathematical induction.
 
6:47 PM
I'm familiar with induction, yeah.
I just suck at linear algebra.
 
This is not interesting linear algebra, trust me.
 
For a finite group, they are just direct products of $p$-groups and have some useful properties like the normalizer property
 
@kyle: So if you think inductively, you should see that if $AB=BA$, then $A^2B=BA^2$ and $AB^2=B^2A$ can be checked easily. Then you should see that if you've proved $A^jB^k=B^kA^j$ holds when $j,k\le m$, then you can see that it holds when $j,k\le m+1$.
 
@TobiasKildetoft ah, unfortunately I know nothing about Lie algebras, they're one of those things I keep hearing about and I'll need to learn about at some point
 
Usually people would do this by working with $AB^k$ first, and then do powers of $A$ after that.
 
6:50 PM
In some sense, being nilpotent is as close as you get to being abelian without actually being so
 
In some sense for a finite group at least, being nilpotent is also about having a lot of normal subgroups
 
Hmmm makes sense, Abelian means that the lower central series has two terms, nilpotent that it might be longer but still becomes trivial
@MatheinBoulomenos I see, I'm mostly interested in infinite finitely generated groups though
 
@TedShifrin OK I got it - how do I proceed from here?
 
You're done by the distributive property.
 
The more general question now - how do I find this interesting property that C and D have?
 
7:05 PM
Well, what do you guess from the discussion you and I have just been having?
 
I started drinking this coffee five minutes ago. How is it that I feel -more- tired now?
(That’s a rhetorical question: coffee takes time to have an effect. But stiiiill)
 
Actually, after the caffeine kick, you do feel more tired.
 
7:24 PM
I think anything that ends up yielding a bunch of terms with the form $A^{n}B^{m}$ that we can legally swap around by that induction proof (the rest we can deal with by associativity of addition which is easy).
So C and D have to be polynomials in A and B?
Doesn’t seem like a very specific property.
 
There you go.
It's very specific.
 
That’s it?
 
Yup.
They're not asking you to prove that this is necessary, but it's certainly sufficient.
 
Ah I see. That was really helpful!
 
Glad you feel better :)
 
7:28 PM
Thanks.
 
hi @ted
 
Hey @Eric
 
how’s it going
 
hi @Ted
 
Hi @Leaky
Other than my degenerating vertebrae/disks in my neck, I'm doing peachy. :)
 
7:37 PM
well i guess that’s good-ish
 
Everyone should feel free to upvote my comment here :P
What have you learned so far, @Eric?
 
@TedShifrin speaking about that, do you think $\frac{\partial z}{\partial t}$ is bad notation?
 
im in a gr lecture atm and am very bored
 
awww
 
my pde lecture talked about frobenius in weird language without ever saying frobenius
 
7:41 PM
I am fine with partial derivative notation. When there's any ambiguity, then the thermodynamics notation $(\partial f/\partial x)_{y,z}$ is suggested.
@Eric: I suspect non-geometric analysts don't think about Frobenius :P
 
@AkivaWeinberger when you say "perspective of someone" what does that mean, I think my problem is that my view is stuck with the euclidean frame of reference
 
do pure pde analysts really think about general first order pdes that often
 
@Leaky: I'm assuming that you're talking about $z=z(x,t)$, for example. I do not like it when there is a composition involved. That leads to all the Leibniz headaches and misconceptions.
 
outside of like specific examples
 
@Eric: I doubt it, but I'm the last person you should ask.
 
7:43 PM
Hello
 
ryan is also in this lecture
 
@Ted so like $\frac{\partial z}{\partial t} = \frac{\partial z}{\partial x} \frac{\mathrm dx}{\mathrm dt} + \frac{\partial z}{\partial y} \frac{\mathrm dy}{\mathrm dt}$
 
We should start coming again in a month
 
No, I do not approve of it with compositions, because $z$ represents two totally different functions on the two sides of the equation.
Of course, I teach it that way in standard multivariable courses, but never in my own.
@Ryan @Eric: You never know when you might miss a 20-second nugget of beauty.
 
7:45 PM
@rapasite ^This is the world from the North Pole's perspective
 
yeah i might keep coming unless i’d rather go to department tea
 
@AkivaWeinberger ooh a part of USA is covered in white
 
Inconsiderate to schedule a grad course during tea hour.
 
@LeakyNun ?
Oh, Alaskan panhandle?
I guess the Rockies continue up through there
 
@TedShifrin it’s in the physics dept
 
7:47 PM
He’s actually doing what Fernando did in his class
 
Aha. Of course.
 
But with a physicist twist
 
yeah except fernando never wrote down christoffel symbols
 
@AkivaWeinberger ^
 
he wrote the geodesic eq in its full glory
 
7:48 PM
Lol he just left a derivative and didn’t do the product rule
 
But if you ever want to compute something ... ?
I think I usually convinced my undergrad diff geo students that Christoffel symbols weren't really Christ-awful.
 
@RyanUnger didn’t he go from the divergence form to the normal form in one step
idt thats that big a jump
 
I think everyone in the Riemannian geometry class knows Riemannian geometry
 
maybe not the undergrads
 
Well, that's a vague statement. If you truly know everything there is to know, you shouldn't be in grad school.
You mean everyone knows the beginnings of the course?
 
7:50 PM
@TedShifrin he means the basics, the class seems like it’s going to go to cool places
 
I’m sure they can write down the Christoffel symbols
 
Eric, you still have to learn complex geometry.
 
Yeah Eric
 
@LeakyNun Ugh right forgot about that
 
im being ganged up on
 
7:51 PM
also alaska is closer to russia than i thought
 
There’s no complex geometry here
 
There's another part of the USA that's far too orange
 
Someone is talking next week about Calabi-Yau degeneration
 
@AkivaWeinberger ok so the vector is gonna point toward the north pole and thus it will rotate while moving on the equator, but still in 3d Cartesian Euclidean space it does not rotate, I need to investigate ^^
 
@Leaky: That's how Sarah Palin could see Russia from her kitchen window.
 
7:52 PM
@rapasite True
 
is this a meme @Ted
 
Yeah think about other examples of holonomy on the sphere and on other surfaces (like cones) and get back to me @rapasite
 
@LeakyNun it was back in the day
 
No, it's a true story that most American citizens who aren't babies will remember.
 
@RyanUnger how is this a meme
 
7:53 PM
What happens when you parallel transport along a line of latitude (not the equator)? Hint: draw a cone that's tangent to the sphere along the line of latitude @rapasite
 
It was made into memes
 
a meme is just anything people won’t shut up about
 
The humidity in this room is too much
 
@rapasite Alternate hint: this discussion started with Foucault pendulums, which measure latitude
@LeakyNun People keep on proposing proposals for bridges across the Bering strait from America to Russia
It's not gonna happen, 'cause, not worth it
Both coasts are really far from any roads
 
some commentors on reddit said that if usa helps hk it would be WWIII
what do yall think
 
7:57 PM
I'm afraid that the deluded narcissist-in-chief will start WWIII (and IV?) without even realizing what he's doing. That's how everything goes now.
 
america wouldn’t openly intervene unless we’re really dumb
 
let's hope he doesn't get re-elected
 
Small 'unless'
 
Well, we are, @Eric. I speak of the "we" who are running the country.
 
Big unless? I dunno what the right phrasing would be, the point is we're pretty dumb
 
7:58 PM
@AkivaWeinberger My students typically had a hard time understanding why parallel translation in the cone was the same. Of course, I assigned this as a required exercise every year, and did talk about it in class.
 
@TedShifrin i mean u know i don’t disagree that our president is incredibly dumb but the forces that decide whether or not we want to go to war for whatever incentive is way more than what goes on in the white house
 
Well, the biggest war-monger just "quit." But I don't trust any of 'em.
Anyhow, back to math(s).
 
i just mean there are pretty good reasons we don’t openly antagonize china like we do much of the world
 
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