@Trey Really? It's just $\frac12(u+1)-1 = \frac12(u+1-2) = \frac12(u-1)$.

is $\frac12$ valid LaTex?

Proving Euler's theorem and Fermat's little theorem without explicit mention of group theory feels...dirty.

what do you mean by dirty?

Sacrilege, maybe? I don't know. It feels wrong in a way I can't explain.

well isn't Euler's theorem older than group theory?

@Fargle: I did them in the first chapter of my algebra book, and the 6th chapter is group theory.

It is, as is FLT. I don't know. I just know that they're almost immediate in a group-theoretic context, but then again, maybe their immediacy follows exactly from the work I'm doing developing the properties of the integers.

It's easy induction with mod p stuff, @Fargle.

I just want to be able to say, "This is a corollary to Lagrange's Theorem," rather than, "Well, note that if $x_1,\dots,x_{\phi(n)}$ are the naturals less than $n$ which are coprime to $n$, then $(ax_1)\cdots(ax_{\phi(n)}) \equiv x_1\cdots x_{\phi(n)}\;(\text{mod }n)$ and then cancellation implies the result."

start ChatJax

Usually it's good to understand things different ways.

That is a good point.

I make good points twice a year.

I dunno---I guess I'm glad to have seen these properties borne out not just from an abstract angle, but from just slowly building up definitions and theorems about the integers themselves. But still.

Group theory would kill, like, half of this class.

Well, mr group theory show-off ... do that damn problem I gave you.

Allow me to elaborate: rudimentary chapters-1-and-2 group theory up to Lagrange's theorem kills half the class.

And all you need for me is the Fundamental Homomorphism Theorem.

For the polynomial factorization problem?

Yup. From group theory, that is. The rest is high school algebra.

It was $x^4 - 10x^2 + 1$, yes? Give me a second, I just finished my homework.

LOL, yes.

hi Eric

hlo how r things

doing ok, I guess, and you?

I'll be honest, I don't see how you wouldn't need at least the ring-theoretic version.

Because we were talking about the basic notion of reciprocity, @Fargle. That's just very basic group theory.

@TedShifrin p good, i’m basically finished with apps now except i still wanna tweak my statements a bit

It'll be next quarter before you do math, Eric :P

But my point, long ago, when you gave me the problem, was that I don't even know the statement of quadratic reciprocity as it relates to number theory, much less which fact or facts constitute its translation into group theory

i’ve been doing loads of math! but also i do just have so much other stuff going on for apps so that’s true

i’ll at least be continuing w sid and taking fewer classes total

So don't tell me---I'll do the research right now---but that was my overall point back then.

@Fargle: It's about squares mod $p$. Given $a$, $b$, if they are or aren't squares, what about $ab$?

Hi Ted!

I already told you that ages ago, too, I'm pretty sure.

hi Kasmir

i got this cool Q

I think you did. And you told me my proof was overkill.

I believe you on that, but I do not understand it yet.

we want to extend it to a map from V-->W

by T(v) = Sv if v in U

T(v) = 0 if v is not in U

need to show that T is not linear

my attempt was , we pick an element u in U

st S(u) is not 0

and we consider the element u+v

this element is clearly in V because it is a vs, and it is not in U

Yeah, that's not a good way to do linear algebra, @Kasmir.

hmm why not

am not done yet BTW !

Because you work with subspaces, not with "in" or "not in".

so what should i say ?

is it a style thing ? or totally wrong?

Isn't the point to define $T$ so that it is linear?

no to show that T is not linear

That's an actual homework problem to show you what I said — that this is not the way to do linear algebra?

Yes, if you add something nonzero in $U$ to something not in $U$, then of course it's still not in $U$.

hmm i dont get it :D

@KasmirKhaan the question is wrong

He's saying he was given the question.

let me post the Q again

and lets regroup !

Hi leaky btw

well, when I say "the question is wrong", I mean the longer sentence "either the question is wrong or you rephrased it incorrectly" @Ted

he said "need to show $T$ is not linear" ... so that is correct.

look, you added one more sentence

i only added S not being 0

the rest was the same

That's needed, or Leaky wins.

it is needed ofc

anyway !

we pick v in V that is not in U

My point is: Do you know how to do it correctly so as to obtain a linear extension?

That's what is more interesting to me.

Yes TED!

we take a basis of U

As long as you pick $u$ so that $Su\ne 0$. Of course, for that you need $S$ to be not the zero map.

Yes let me answer the first Q

then ill do yours!

OK, u+v now is in V

and not in U , it should get mapped to 0

if T is linear

You shouldn't be asking this ... seriously. You should know whether what you have is correct.

but T(u+v ) = T(u) +T(v) = S(u) +0 = S(u)

but S(u) was picked to be not zero

Yeah, yeah.

I know it is correct

but i need to see all details

Then don't waste our time?

TED! am redoing LA

Fine. Redo it.

to make my proofs look better

@KasmirKhaan what is v?

not just answer the Q

v is an elemnt in V but not in U

so the thing that is important here , just saying that u is differnt from 0 would be wrong

Hey everyone!

If you're really being careful, you need to justify why $u+v$ is not in $U$.

because it might be in kernel of S, and hence i made an error

Yes, that's correct.

hi Demonark

okay thanks Ted ! this is a good point

hi @Daminark

if u+v were in U

you never said that

He did before you showed up, Leaky.

v = u+(v-u)

u in U , and v-u in U

@Ted no he didn't

so then my assumption that v in not in U is wrong

contradiction

Not the way you've written it, @Kasmir.

@LeakyNun I did damn it leaky !!!

Why is $v-u\in U$?

@KasmirKhaan show me the message

Leaky, enough.

This is tiresome enough.

I will put you in timeout for being too obnoxious.

i mean v = (u+v) -v

two elements in U would be in U because it is a subspace

Typo

what typo !

Read what you just typed.

yes yes v = (u+v) -u

thanks Ted!

So stuff like this you should get right, and immediately, without needing us.

okay sorry , i just wanted help to make it regorous

Did you learn quotient vector spaces in your course or in algebra?

doing them soon

we did not cover anything good in LA

The best way to do the extension, if you know quotients, is to use them.

But you need not.

okay thanks! iwant to learn more stuff, i noticed that LA is the key for many courses in math, being bad at LA is a death sentence :D

but i only appriciate it when i did abstract algebra first, then one can see how pure LA is

so to answer your Q Ted

to make this map T : V---> W , as an extension

You never watched my linear algebra lectures, either :D

i did

but you do things different in US

:0

Plenty of rigor in them ... Well, of course we do. And my course is different from 99% of the courses in the US.

i used ur videos the most when I did analysis , the course was almost the same as US

no, @Kasmir, my course is very unusual for the US.

@Fargle: I answered.

But, @Kasmir, I meant the linear algebra in particular, to get used to proofs.

aha did not know that ! but greens theorem and line integral and double integral were your master piece imo :D

i shall take a look at them again then ! ._.

so to make thsi map T linear

we get two basis one for U and one for V

vectors in V we can map to 0

and vectors in U we can map as S

this is surpriensly linear ._.'

No, no.

Slow down.

okay

Way too sloppy.

$U\subset V$.

we can start with a basis of U

U= u_1 ,..., u_n

OK

then we extend it to a basis for V

OK

u_1,...,u_n, v_1,....,v_k, we map u_i as S

and make v_j to 0

this is a basis for V now

OK. Notice that what you're doing is inducing a linear map on $V$ by sending $V/U$ to $0$ and using $S$ on $U$.

ahaaaaaaaa

Neat thanks Ted ! did not think of it like that

But it totally make sense now ._.

You lift a basis for $V/U$ to vectors $v_1,\dots,v_k$ as you suggested. Alternative viewpoint.

alright coolio indeed :D ill keep doing more exercices and come back to ya :D

still here @TedShifrin ?

Of course I'm still here.

i meant for how long :D

Well, I dunno.

Ill text ya when am done with other exercices :D

Try to do more self-criticism on your own.

well do !

I see I effectively shut up Leaky with the threat. :) Mike left the chat because I wasn't threatening people.

@TedShifrin but your threat can only shut up people who respect you

Mike was referring to other people

Well, the others I can actually implement the threat. I just don't need the stress, but I can do it.

Yes, I know he was.

Not that he treated me with a whole lot of respect, but I get it that he's anxious too.

Hey, Ted, I ordered one of the books you've recommended in here and it arrived yesterday: Dan Pedoe's "Geometry A Comprehensive Course."

Though, there are a few things that confused me as I was reading through the first bit.

Thinking out loud for a second here, let's say $\alpha$ is some algebraic integer, and let's look at its Galois conjugates. I think a Galois automorphism should have operator norm at most 1 because it's finite order, so if $|\alpha| < 1$, then any Galois conjugate should also have absolute value $< 1$, but that should be impossible because the norm has to be a non-zero integer

Like, he says "The points A and C together with all points B between A and B form the segment AC" and I'm pretty sure the second B is a typo and should be C. Am I correct?

@Rithaniel: It is not a super-easy book. But it has amazing stuff in it.

Yup.

Wow, what page?

4

Wait, no, not 4

2

Right, I found it on 2. Yup.

Well, as I learned with all my books, even the most fastidious authors end up with mistakes in their books.

@Daminark what do you mean by $|\alpha|$?

Absolute value, as a complex number

Alright, thought I might be missing something important there. Glad my interpretation of it as a typo was correct.

That sounds right, Demonark.

Also, I'd never seen the area of a triangle given by half the determinant of a 3x3 matrix before. Still don't have a good mental grasp on why that's the case.

I guess that's the issue: I don't actually know that, off-hand.

Determinants, in any dimension, give the (signed) (hyper)volume of a parallelepiped with the column vectors as its determining edges.

Alright now I'm finally cool with the proof of Burnside's theorem

That does sound familiar.

@Daminark is that relevant?

So, we go from volume of a 3d object to the area of a triangle.

Think about the 2D case for a second, because the visualization is easier. The determinant of a matrix like $\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ gives you the volume of the parallelogram, two of whose sides are given by the vectors $[1,0]^T$ and $[1,1]^T$.

Yeah, basically during the proof you end up with an average of roots of unity being equal to an algebraic integer

Suppose V is finite dim and dim V > 0 , W is infinite dim, how to show that L(V,W) is inifite dim ?@TedShifrin

I mean i should somehow show that there is no spanning set

but how would one start this ?

Can you give an example of the kind of 3x3 matrix you're talking about, @Rith? I'm not familiar with this specific construction (personally I think you'd be able to do the whole thing with just 2d determinants if you're in the plane, but I can see how you'd need it if you were in space).

So it has absolute value at most 1, and by the argument equal to 1

Hold on, I'll get the Latex worked up

But then if an average of roots of unity has absolute value 1, then they're just the same

@Daminark oops, I thought burnside lemma

Oh no I meant the business about groups of order $p^aq^b$ being solvable

Burnside's lemma is real easy

matrices are usually on a planar grid

is there a notion of matrices that aren't planar rectangles of numbers

but instead quadrilaterals on some surface geometry?

@Kasmir: Show how to give infinitely many linearly independent elements.

@Rithaniel: Draw a picture and think about the picture.

it seems obvious Ted but donno how to start this

i mean since W is infite dim

Do it first when $\dim V = 1$.

chess anyone? lichess.org/xG5EZakd

mumbles something about universal property of product and coproduct

well if Dim V = 1 , then Dim W = 1,2.....n....

Huh? Don't write nonsense.

lol

no i mean

@Rithaniel Okay, so I said stuff about the columns earlier, but it works just as easily with the rows (because determinant is the same for transpose).

$\{\partial R(x,t)/\partial t\}*\{\partial R(x,s)/\partial s\}=\{\partial R(x,m)/\partial m\}$ can someone help me classify this differential equation?

arent the linear maps from V--> W infinite dim ?

Do you see how the row vectors of the matrix correspond to your original problem?

even tho V and W both finite Dim ?

@Kasmir: Most definitely NO. You need to understand that case first.

is there a basis that spann all linear maps from one vector space to another?

Original problem? You mean how they correspond to the vertices? Yes, of course.

mumbles something about dual and tensor product

@Rithaniel Okay, so the determinant itself corresponds to the volume of a parallelepiped in 3-space whose edges are determined by the three row vectors of your matrix.

@TedShifrin am really stuck

@LeakyNun is that mumble direct to me ?

Figure out the dimension of $L(\Bbb R^n,\Bbb R^m)$.

Now what happens if you cut that parallelepiped in half with the plane $z = 1$?

So I know it's a partial differential equation

and on the LHS there is a product

@KasmirKhaan no, I'm mumbling to myself. that's what "mumble" means

okay i shall !

but the product is confusing to me

(And more to the point: can you see, visually---in either a diagram or your mind's eye---why that plane cuts the parallelepiped exactly in half?)

because it's two partials of $R$ multplied together..

Well, I can see why it cuts it in half because each z is 1, that's plain. Though, trying to think of how to describe the shape I see in my mind's eye.

A distorted tetrahedron with one vertex at the origin and the other three at points $(x_i ,y_i ,1)$

By "distorted" do you just mean "not regular"?

Yes.

But this implies that the volume of this tetrahedron is equal to the area of it's base.

any body have a clue?

Unfortunately, I am clueless, ultradark. (Sorry, had to make a pun)

@Rithaniel That's the part where I'm starting to get more confused myself. Where did you see this?

Dan Pedoe's book "Geometry: A Comprehensive Course"

Which I picked up on Ted's recommendation.

Ahh.

It's in other places as well. I've watched some youtube videos and read some stackexchange answers talking about this. None have quite illuminated the situation. The reminder that determinant corresponds to volume was a useful push.

BTW @Fargle @Rithaniel: That area problem is in my linear algebra book :P ... You need to just see volume = area of base times height. You've lifted the triangle up to the plane $z=1$.

Yes, determinant = signed volume.

@TedShifrin I thought for tetrahedra it was 1/3 of that, which is why I was confused.

Yes, you're right.

But you don't have a parallelepiped, either.

another chess match anyone? lichess.org/gKJURUfo

@TedShifrin Why not?

Because you're chopping with the plane. You have to add vectors to make a parallelepiped.

But, both you guys, you don't need all this. Just use properties of determinants to reduce to a $2\times 2$ determinant, that will give the area of the triangle.

Oh, yeah, good point. Move one of the coordinates to the origin and then do the easy thing.

Right.

Well, I've ran the numbers on paper several different ways, I'm just trying to get a solid visualization of things.

@Rith To be more explicit, take the cofactor expansion of the determinant along the row of zeroes to get the 3x3 determinant as a sum of 2x2 determinants, and then take the row vectors and "shift" them all by, say, the vector $[-x_3, -y_3]$, so that $(x_3,y_3)$ becomes the origin.

Then you end up with an expression that "obviously" gives the area of the shifted triangle, and because shifting preserves area, you're done.

Oh, don't do cofactors. Just do column operations.

Ah, fair enough. Listen to the actual mathematician, he knows what he's doing :P

Well, of course, it all comes out the same.

Did the solution I sent you earlier look correct, by the way, @Ted?

Oh, I haven't looked at my email. I've been busy all over the place.

Yup. Now all you need is the "simple" group theory proof about the squares. As I said, FHT. Isn't the problem cool, now that you get it?

The misconception I had had is that I thought you meant "three different ways" as in "do the factoring after mod p, and then find three categories of factorization that will work for all mod p" which is why I was barking up the "1 mod 4 vs 3 mod 4 vs 2" wrong tree. Your more specific hint really helped crack it.

No, what you just said is way too complicated!

Yes.

tED

Ted !

sdfjhsadfkasdf

But that was the angle I thought I was meant to take all that time, which is why the FHT and reciprocity stuff really threw me.

Still dont see it

OMG, it's a Meow.

Ted Factorial

@Kasmir: How do you usually represent linear maps from $\Bbb R^n$ to $\Bbb R^m$?

I remember that meow fellow

@TedShifrin with matrix

Heya @MeowMix

Now figure out the dimension of the space of linear maps.

mby n?

mn

ih fargle

and ted

Done.

but hmm, the idea of finding a spanning list

of linear functions from V--> W

You can give a basis easily.

Using the matrices.

aha ._.'

aha

our elements are the matrices in this case

and each matrix is in 1-1 correspandance with a linear map

so we got them all covered into the matrix :o

wow we never did this in LA ._.

this is so good :D

How do you give the basis, though?

Which $mn$ matrices?

1 in ij and rest 0

There you go.

hmm it quite handsome :D

because if we think of non linear maps

Back to @Rith's problem---I agree that after the cut, we don't have a parallelepiped, but before the cut it seems like we do. I'm confused precisely because it seems like, based on the fact that a tetrahedron's volume is 1/3 base area times height, we shouldn't have that the tetrahedron's volume is equal to the triangle's area, but it is.

this cannot be done so clean

@Fargle: To quote from the solution I wrote to the problem in my book :)

The original determinant is the (signed) volume of the parallelepiped. The tetrahedron has 1/6 that volume.

But the volume of the tetrahedron is 1/3 the area of the base times the height, so the area of the triangle is 1/2 the determinant.

The crucial thing is getting that 1/6.

Ah. I had the mental picture wrong.

(Always draw diagrams, kids.)

That was the first thing I said to Rithaniel :)

I'm trying to think about why 1/6.

btw Ted

Ah, you want to think about a different base.

in rep theory

one does things like f : G --> GL(V)

what is GL (V)

Invertible linear maps from $V$ to $V$.

hmm

that is why we use matrices?

Sure.

@Fargle: Yeah, use the triangle spanned by two of the vectors as your base.

Ahh. That makes a lot of sense then.

My problem is that I'm bad at drawing parallelepipeds, so I ended up just drawing a doubled tetrahedron, which is very not a parallelepiped. And you did say as much...

Clearly I need another cup of joe.

Yeah, but since when do you listen to me?

Therein lies the problem: the inherent gap between doing, and knowing consciously that you ought to.

Ah, the indignance of youth.

"marginal" means minority right

then how come "marginal distribution" is a sum