yup

Hi all. Does anyone have a recommendation for an undergraduate level textbook for component free methods for solving algebraic and differential vector equations?

ok :]

either, or. It doesn't have to be both

can you be more specific in terms of what problem(s) you're trying to solve, @prokaryoticeukaryote?

@TedShifrin "Undergraduate" ça veut dire avant le bac non ?

non, @Astyx, ça veut dire qu'on est étudiant à l'université, en commençant.

@TedShifrin and how do we write the dual of point $P$?

Ah, je me disais aussi ... merci !

@meow: That doesn't make sense unless you mean the corresponding pencil $\lambda_P$.

There is no mapping between $\Bbb P^2$ and $\Bbb P^2{}^*$, only a correspondence.

(I was disputing this terminology with Alessandro a few days ago.)

a correspondence from points to points?

no, point <----> line and line <----> point

so wait

a correspondence from lines in $\mathbb{P}^2$ to points in $\mathbb{P}^{2*}$?

and vice versa ...

lines in $\Bbb P^2{}^*$ correspond to points in $\Bbb P^2$ ($\lambda_P$ <---> $P$)

can't you correspond points in $\mathbb{P}^2$ to points in $\mathbb{P}^{2*}$ though?

I can't. How can you?

so a point in P^2 is a line through the origin excluding the origin right?

in $\Bbb R^3$, yes

What is $\Bbb P^{2*}$ ? Is it the dual space of $\Bbb P^2$ ?

why can't you just correspond it to the plane that it's normal to? (which would be a line in $\mathbb{P}^2$, and thus we could correspond it to a point in $\mathbb{P}^{2*}$)

No, it's the projective space built on the dual vector space, @Astyx

Right

no, that's a line in $\Bbb P^2{}^*$, @meow

what is?

@TedShifrin For example, if you know $U \times V$ and $U \cdot W$, you can solve for U, as long as V and W are not perpendicular, using the BAC-CAB rule. As for problems I am trying to solve, an example of an equation I am trying to understand is $\nabla \left( U V^T\right) + \nabla (V U^T) = F$, where V and F are known, and U is unknown.

I want to know if this equation has a well defined solutoin, or if it requires extra information. But really, it would just help to have an idea of general coordinate free equation solving methods, if they exist

@prokaryoticeukaryote: I'll have to think about this in a bit.

a line in $\mathbb{P}^{2*}$ is a pencil, i.e. a set of planes in $\Bbb R^3$

so a single plane, is a point in $\Bbb P^{2*}$

Reread p. 341, @meow. You're getting all confused.

@TedShifrin sure. Thanks for responding

@meow: It's important to keep the Latin letters and greek letters different. You're confusing the actual spaces $\Bbb P^2$ and $\Bbb P^2{}^*$.

which one of my statements are false?

Vectors live in $\Bbb R^3$. Normal vectors to planes coordinatize $\Bbb R^3{}^*$.

ok

You're trying to just work with one vector space. The dot product allows you to do so, but it's really better to keep $\Bbb R^3$ and $\Bbb R^3{}^*$ (the set of linear maps $\Bbb R^3\to\Bbb R$) distinct.

You want something that is geometrically meaningful without depending on actual coordinates (that's the abstract explanation).

i'm confused, what is $\mathbb{R}^{3*}$

lol forgot the asterisk

The set of linear maps $\Bbb R^3\to\Bbb R$. It's called the dual vector space.

I did not discuss it that way in the book (but if you learn about multivariable calculus and tensors and differential forms right, you'll see it there).

@meow-mix The point, basically, is that projective space comes from the vector space structure on $\Bbb R^3$. Using further structure (like the dot product) is unnatural.

You can identify those two spaces using it. But these violent desires have violent ends.

You can give an isomorphism $\Bbb R^3{}^*\to\Bbb R^3$ by either choosing a basis or using a dot product. But one doesn't need to have this isomorphism for all the geometry to work out.

I think MikeM and I just said basically the same thing.

I'm back

@TedShifrin $ | a_n | < 1 $ for large n because that's a requisite to converge right ?

Yes, @Maks.

The terms must go to 0 eventually.

i realized it very late, but before you ask a question, ask yourself why you can't answer it yourself

Good, @Null.

@Null was that directed to me ?

no, it was just something I realized

It was directed at Null :) But it's not a bad comment.

Oh ok, I always ask here because I'm not 100% sure of things haha

Where's Balarka actually?

Balarka's swamped with exams in his school.

like a very late insight

@TedShifrin yay me too !

Well, Balarka's still in high school :P

Oh, where is he from ? here we've finished high school

India.

Good evening everyone

Just a question, on mathematical analysis II, how many different topics you see ?

Hi @Alessandro

@TedShifrin is it ok to say "i don't understand"?

Here I have an exam on the 23rd of December...

it is somehow disrespectful to ask questions that are answerable by yourself, if not for some wiki

Ugh @Alessandro

@TedShifrin So hopefully, re: the comment I sent you, you're less ashamed.

At this rate I'll have christmas dinner with my prof next year

LOL @MikeM ... was it good?

Hi @ted!

The question "how are you" is viewed in a different light that way

@meow: Are you trying to work out exercise #5?

no :P

It might help solidify things.

Quite.

The arctic char would have cost twice the amount, so I'm glad it was what the waitress suggested (the lamb soup, that is).

i'm just still confused on why we can't relate points to lines in $\mathbb{P}^2$; but I'll take your word for it

like, if I really care about how anyone is, I would observe him, and answer that myself

@meow: The point is to keep the two different projective planes separate, or else it gets totally confusing.

You're trying to combine everything into one picture.

yes, ok that i understand

As I've said, you're not wrong, but then you'll get far more confused in the end.

ok then.

How much linear algebra do you know?

I predict that if you keep working with stuff, at some point in a few hours you'll realize for yourself what I've been saying.

@MikeM: Not that much :P

OK. I was just going to make the point that it would be foolish to think that there was only one vector space of dimension $n$.

P.S. @MikeM: I love arctic char :D

The MSE user or the fish?

LOL ... the fish.

@MikeMiller but arent all vectorspaces of dimension n isomorphic to each other?

@Null umm i dont think so

Over the same field, yes @null

maybe so

Hi @Alessandro

@Null Yes, but not naturally so. There's no canonical isomorphism between them. And that matters sometimes.

Of course they are. But you need coordinates to understand how.

Hi @Astyx

@Ted It was 4200 isk, so I passed.

i'm feeling stupid lol

I understand, @MikeM. Of course, I'm in a different stage of life, but when I travel food is one of the main things I want to experience and enjoy.

Get used to it, @meow, and keep working/learning.

We all feel stoooopid from time to time.

My professor quickly showed us last year that there is a canonical isomorphism between a (finite dimensional) vector space and its dual's dual (bidual?), but I failed to see the significance of this fact then (and I still do :/ )

Perhaps you appreciate it more if you realize that in infinite dimensions it may fail, @Alessandro.

@Ted Especially I

Because I am

@MikeMiller ah, so (1,0) might be isomorph to (1,0,0), but uncountable more ways are there to assign a "rule". What is a canonical isomorphism? (i know the meaning of both words, just can't say one)

Ta guelle, toi, @Astyx.

Doesn't it always fail in infinite dimension? @ted

:(

Nope @Alessandro

Oh, man, I feel so stupid when doing maths

Certainly not in a Hilbert space.

@Alessandro: Do you know about $\mathscr L^p$?

hi chat

rehi @Semiclassic

no @ted

a canonical isomorphism between A and A exists right?

OK, @Alessandro, it's in general a Banach space (Hilbert for $p=2$). It is "reflexive" (equals its double-dual) for $1<p<\infty$.

@Null Like, the identity?

Hi @AndrewT

@Krijn yep

Hellu @TedShifrin!

How are you?

Doing well, @AndrewT, and you?

Quite well. Using the fact that I just finished one exam to procrastinate practicing for the next (and more important) one.

isn't it true that for an infinite dimensional space $\dim V^*>\dim V$? @Ted (we never really talked about infinite dimensional spaces yet in a class so I'm not very familiar with them)

Imma go, good day/night everyone

Not always, @Alessandro. Consider the facts I just told you above :)

@AndrewT: Most people here seem to specialize in procrastination.

Wouldn't that imply $\text{dim }(V^*)^*>\text{dim }V^*>\text{dim }V?$

@TedShifrin He's thinking algebraically, not analytically.

@TedShifrin to the point I procrastinate procrastination

@null Wouldn't that just be productivity?

Ah, good point, @MikeM.

@TedShifrin Yeah, a chatroom for nonresearch level maths might not be the best idea for math students.

@MikeMiller What do you mean by this?

@AndrewT: Yes, too many people waste too much time here. It's OK for me to waste time if I so wish, though.

If $V^*$ is the vector space of all functionals, then he's right (where by dimension we mean cardinality of a Hamel basis). Ted's talking about continuous functionals and Schauder bases.

What's an example of an infinite commutatibe ring that isn't a field?

@MikeMiller Apropos of your starred quote, how is Westworld going? :)

Ahhhh. This cleared things up a bit for me.

@TedShifrin one of the profs at my university is hardcore. He means every minute which is not spent on math, is indeed a wasted one.

That's excessive, @Null.

^

I didn't see the last episode. I mostly like it.

I might start watching Westworld to upgrade my procrastinating

@fluffy_muffin Well, what examples of infinite commutative rings do you know?

Now we know Semiclassic is procrastinating when he engages in algebra discussions :P

point.

Z, but integers aren't a field but is an I.D, C, R, Q, etc. Should correct myself looking for such a ring that is neither an I.D. nor a field.

You can think about other integral domains.

Think back to algebra in high school.

yeah, if you just didn't want a field you could just do R or Q.

i banned diablo 2 from my pc because i wasted 10 years on it. so from that point of view anything i do now is productive in some way

What's that thingy about finite commutative rings are always field or something? Hard proof to go along with it

No, easy proof. But you need integral domain.

in $\mathbb{P}^{2*}$, the vector $\zeta \in \Bbb R^3$ has homogeneous coordinates $[\zeta]$ which is a plane perpendicular to $\zeta$ (in $\Bbb R^3$), right?

I'm assuming you mean the ring of polynomials? But, is R is an integral domain then so is R[x]. So, that doesn't work either.

@meow-mix What mapping from R^3 to P^2* do you have in mind?

No, @meow, $\zeta$ is a vector, which you interpret as the normal vector to the plane.

@TedShifrin This one is easy indeed, but I recall a similar stronger statement with a harder proof

Or maybe I was just silly a few years back

Good luck proving $\Bbb Z/6$ is a field, @Krijn.

@TedShifrin should have said "with normal $\zeta$"

If you're thinking [x,y,z] <-> (x,y,z), i'm fine with that.

@meow, so, yes, you're right, sorry ... $[\zeta]\in\Bbb P^2{}^*$ represents the line in $\Bbb P^2$ corresponding to the plane in $\Bbb R^3$ with normal vector $\zeta$. Precisely.

@TedShifrin Ahhh, I found what I was looking for I think: Every finite division ring is commutative

okay. sorry, i'm going to take a break, and think about all this.

Don't know why I was thinking of that, though

and i'll do some more exercises in a bit

$n!=1\cdot 2\cdot\cdots \cdot\frac{n}{2}\cdot\cdots(n-1)n$ how can i rigorously notate the half of n (it is not known wether it is even or odd)

what do you mean, @Null

@meow-mix one factor is very close to the half of n, for big n

Oh, so you don't want ID either, @fluffy. I missed that.

OK @meow

@fluffy_muffin Can't you just take an infinite product of a suitable ring?

Nevermind. We can take Z x Z.

Yup.

just wondering, is the center of $GL$ just the scalar matrices?

:33999247 No, your solution is much better

Yes, @Alessandro. Good exercise.

scalar matrix meaning $\lambda I$?

Yup.

right right

Hm, I see, I'll think about a proof

I think I know what I'd try, but I'm hitting my afternoon crash.

plus, y'know, procrastination

@fluffy_muffin Only if $n$ is a prime

its characteristic is n, and it's an integral domain iff n is prime

heya tern

heya

I see Mike and three others are enjoying westworld

@arctictern Westworld is great

Four, if my download speed won't let me down

Hi chat

@TedShifrin Bonsoir Ted

Tomsoir

Bonsoir, @JeSuis.

@TedShifrinDifferential Equations, Dynamical Systems, and an Introduction to Chaos: c'est un super bouquin!

DogAteMy: To take a break from analysis for a bit. Suppose I hand you a (smooth) convex plane curve. Then in every direction there will be precisely two points with tangent lines in that direction. Show that the distance between the planes is always at least $2/\kappa_{\text{max}}$, where $\kappa_{\text{max}}$ is the maximum curvature of the curve.

Eh bien, ça te plaît, @JeSuis? :)

Guys, I have a little problem at home, how much do you know of physics ?

Oui, c'est un livre qui donne beaucoup d'intuitions

Nous américains aiment bien donner des intuitions de temps en temps :P

hehe

DogAteMy: Of course, I meant "distance between the lines" ... not planes.

Speaking of intuition, @JeSuis, I think the exercise I just gave DogAteMy is some nice intuition.

@TedShifrin does curvature means courbure in french ?

Yes.

never met :p

Tu ne blagues pas?

Never met even for a curve in $\Bbb R^2$?

Je n'ai jamais travaillé sur la "courbure"

OK.

Je ne dirai plus rien.

pourquoi ?

@TedShifrin Bonne nuit

what would $z\to\infty$ mean if $z$ is a complex number?

or does it simply make no sense?

better asked: what would $z=a+bi$ represent for $a,b\to\infty$?

It's usually $|z| \rightarrow \infty$ when it comes up

well unreleated to this. If i want to show the limit of $\frac{1}{\sqrt[n]{n!}}$, with the hint to use $k\geq \frac{n}{2}$ for half of the terms as an approximation, do I approximate the other half with $k\leq \frac{n}{2}$?

(half of the terms of the factorial of course!)

@KajHansen also hi ;)

Wait, that's not helpful lol

@KajHansen i thought about something like this too!

(and saw the flaw)

That does work a decent bit though

let's do it!

^tomorrow lol

Due tomorrow do tomorrow (morning)

:-D

i just want to leanr for my exam, so i'm really not sure wether I should make my excercises when I still have knowledge holes from the past excercises

how would you tackle that?

my exam is in like 2-3 months

I'm not sure tbh

@Null Make the past exercises, then the new ones?

You seem to have enough time

it's like doing butterflies to get huge shoulders, when you really need to first learn how to do butterflies properly

Participating in as much math as possible as it pops up naturally is a decent strategy long-term, like encountering questions IRL and on MSE, but idk for 2-3 months

maybe downloading as many exams available on the internet and trying to solve at the end one in the timespan given?

Old exams are a decent way of studying

@KajHansen I probably did 40% of my bachelor this way

Then again, only for courses I wasn't interested in