@Cookie: OK, it works out again by a symmetry argument (but way less tricky than yours). Try $e^u = \tan x$ and see what you can do with the integral.

@ted Hey I gotta run, but thanks for inspiring the problem again :) I'll be thinking about it for a while!

Can anyone help me invert a matrix in GF(2)?

OK, we can continue the discussion later. It works.

how is your christmas Prof.Ted ?

What is the problem, @Startec?

Merry christmas btw if I didn't say it

Fine, Karim. I hosted a large dinner party — still have leftovers :(

@TedShifrin this one: math.stackexchange.com/questions/2582865/…

nice @TedShifrin

I didn't specify that this was in a binary number field

so I am not sure if his answer is correct.

You certainly don't mean to invert ... that's not a square matrix.

@Ted have you ever looked at Huybrechts' book on complex geo?

Good point. I want to find B which is: (ATA)^−1 * AT

Only a little bit, Eric, as Danu was studying it. So I answered some questions for him. I learned/taught from Chern, Wells, and G/H.

ah ok, i managed to acquire copies of huybrechts' book and G/H

I don't know what you're writing. So what is $A^T A$, precisely? It's got $1$'s on the diagonal.

G/H is a p hard read

Yup, Eric, but it's got great stuff in it.

@TedShifrin sorry - how can I be more clear? I really just want to invert that matrix that I posted

Huybrechts leaves out most of the geometry :)

@Startec: So $A^T A$ is not invertible (even mod 2).

It has three rows the same.

Concentrate on rows 3,5,6,7 of your matrix. What are they?

there's soooo much stuff in it

Right, they are the generators!

Hi @Ted and everyone

Now you can only go backwards @Startec if you start with a vector that's in the column space of your matrix. The column space is 4-dimensional, sitting inside 7-space.

Hi @Alessandro

@TedShifrin Just want to make sure I understand this. So in constant sheaf a section over a connected space is just an assignment over an element g of G. So the morphisms over sections are just maps between singletons ?

What do you mean by morphisms over sections?

I've never heard of such a thing.

Is that a Hamming (7,4) code?

@AlessandroCodenotti yes it is

@TedShifrin By that I mean to say just the restriction maps.

So you just restrict a constant function to a smaller domain to be ...

hi @AlessandroCodenotti

@Alessandro: Maybe you should help, since you recognize this stuff. I only know linear algebra :P

ohh ok yeah sorry for the confusion @TedShifrin yeah it is clear now.

@TedShifrin this is Linear algebra, and you are helping me. This is exactly what I am trying to figure out here, how to work backwards.

So here's a way to create a left-inverse, @Startec. Have you learned elementary matrices?

Yes.

Cool. So you know that the reduced echelon form will be $\begin{bmatrix} I \\ \hline O \end{bmatrix}$.

one second. For some reason your markup is not rendering in the chat.

(See LaTeX in chat up there >>>>^^^^^.)

@TedShifrin linear error correcting codes are linear algebra (and I forgot the little I knew about them I'm afraid)

@Alessandro: I know, but I don't know the code theory. :P

I should have taught it in linear algebra, but there was never time.

@TedShifrin the text I am using hasn't covered reduced echelon form yet, but I do know the concept behind elementary matrices, so I think we can proceed. (got the latex working thanks)

@Startec: Figure out the product of the (7 by 7) elementary matrices $E$ that will give you reduced echelon form and take the top 4 rows. That'll be your left inverse.

We had a really quick introduction in the abstract algebra course, there is a master program in cryptpgraphy and codes theory here so they try to fit a bit of this stuff in the undergrad curriculum

I've taught a tiny bit of codes in abstract algebra, but only a tiny bit, Alessandro.

i remember taking two semesters of cryptography

@Startec: Because of rows 3,5,6,7, this is going to be really simple.

You can move those rows to make the top 4 rows of the reduced echelon form.

Excellent. So @TedShifrin you suggested starting with a vecotr that is in the column space of the matrix, for instance, could that be, [1, 1, 0, 1, 0, 0, 1]? Could we start with that?

You can, but I'm telling you how to construct the $B$ so that $BA = I$ ...

Right, okay, I'm following, but when you say "move those rose to make the top 4 rows", how do I do that?

What matrix do you multiply by $A$ that will move its 3rd row to the 1st row, move its 5th row to the 2nd row, etc.?

@AlessandroCodenotti all the math you learned, you've forgotten?

I'm shocked!

He's only forgotten a little bit, Demonark.

@TedShifrin Ah, right okay. This ^ technique has also not been covered yet in this text (which is rough because obviously I need to know it). Can you guide me?

:P

Actually huh I'm wondering, is it efficient to compute a matrix inverse using the characteristic polynomial?

@Startec: Suppose you have a smaller matrix $A$ (say 2x3). What 2x2 matrix $E$ can you put in front of it so that when you multiply $EA$ you switch the rows of $A$?

I think it is a [1] in each position corresponding to those rows....

Hell no, Demonark.

Damn

Think about the difficulty of computing the char poly in the first place :P

@TedShifrin give me a second. I will figure this out

(You can always read my linear algebra book instead, @Startec ... :D)

Well, I'm not sure how efficiently you can triangularize a matrix

is anything involving matrix inverses efficient

You're gonna use row operations, so you might as well use row operations to find the inverse, Demonark. It's basically something like $O(n^3)$, I think.

So basically the difficulty will be dependent on how efficiently you can do triangularization, and then matrix powers. The latter could be nasty

And true

I guess that doesn't sound horrible

@TedShifrin I am not sure. My understanding of elementary matrices is to think of "adding one row to another row". I don't know how the relate to swapping

That's why I told you to try a 2x3 example. Do it with your fingers and figure out where to put the 1s and 0s. :)

Once you see the pattern, it'll be easy to understand/generalize, @Startec.

Right.

What's the 2x2 matrix you put in front to do that?

No. It won't be that!

Use your fingers and what makes it work?

Don't you multiply matrices by moving your fingers? Left hand for left matrix, right hand for right matrix? :)

There are three kinds of elementary matrices, @Startec. You're talking about only one kind.

I'm curious about this multiplying with fingers.

It's how I always demonstrated it on the blackboard in classes, Salt.

Dotting a row vector with a column vector.

i see

i was hoping for a more elaborate trick

LOL, nope.

It's just a visual for teaching.

But it helps.

Nowadays, I often multiply matrices by treating each column as a vector, then distribute each component to the column vectors of the matric on the left, then add them up

i.e.:

That's the first way of defining matrix times vector, Secret, yes. Take the linear combination of the columns of your matrix given by the entries of the vector. Not useful for this, however

Fingers, huh

well matrix times matrix is basically the process of matrix times vector repeated twice (for 2x2 matrices) since each column acts independently, thus demonstrating the linear wrt arguments of matrices and linear operators in general

But we're trying to invent the matrix on the left.

as for that..., I have not read the full transcript yet, sorry if missing the point

lol

inb4 the matrix is [0 e2 e1] so our matrix is [e2 e1]^T

so that [e2.0 e2.e2 e2.e1; e1.0 e1.e1 e1.e2] becomes [0 1 0; 0 0 1]

@Secret is that what you're getting at?

@TedShifrin sorry to be so literal, but could you give me more instruction on this using my fingers part?

A has three columns

Just multiply matrices as you are used to doing, @Startec. I find it helpful to move the left hand across a row while moving the right hand down the column, multiplying and adding as we go.

and I am not sure exactly how I can move rows like this

Well, you want 0 times the entry in the first row plus 1 times the entry in the second row, right?

If you guys are being so stubborn about this, you can dig out my YouTube lecture where I demonstrated matrix multiplication. I'm sure it's in there somewhere :P

@Secret that's the point

don't use complicated machinery to do simple things :P

@Startec how do you multiply matrices?

your left finger traces through 1 2 3 from left to right

at the same time, your right finger traces through 7 9 11 from top to bottom

so when your left finger is at 1, your right finger is at 7

Thanks to Leaky for the lesson :P

when your left finger is at 2, your right finger is at 9

when your left finger is at 3, your right finger is at 11

But you can.

You can ask yourself, "Self. What will make the answer I want come out?"

> Self. What will make the answer I want come out?

> Self.

$\Huge{\rm Self.}$

where are you, self?

That should have been ... Self: What will .... ? :P

@Startec "if I do this technique to this number, I will get this number." that's the easy part

"which number can I do this technique to, if I want to get this number?" that's the part that needs more thinking about

@Startec: As I said above: chat.stackexchange.com/transcript/36?m=41986630#41986630.

but "needs more thinking about" doesn't mean "impossible"

lmao

LOL. When you guys get obsessed, you guys get obsessed.

for every pair of number pointed by fingers, multiply them. For every row and column of numbers traces through, add those products together to get the corresponding matrix entry

Getting here....

(26 letters in this language is simply not enough to describe that can be elegantly described using 256x256 pixels in a 256 color space)

The 1 in the first row picks up the second row of the original matrix; the 1 in the second row picks up the first row of the original matrix.

👉🏻👇🏻

Now what's going to do it with larger matrices?

Okay, let's see if I can generalize back to the original question....

Right.

larger martices

So, to shift the rows in G

         👇🏻
👉🏻[? ?] [0 0 1]
   [? ?] [0 1 0]

Stop it, Leaky.

lol

@Secret pictures hardly compare to unicode

@Salt NB Inspired by einstein summation notation, I have developed a way to multiply matrices as the sum of outer product of its corresponding row and column vectors. It is equally complex thus no less time consuming, but manually it is faster for me because I am more used to vectorised operations (and visually there are only n terms in the sum). Will tell you more about that later cause I don't want to confuse the current discussion

for some reason i was reminded of reading about how eigenvalues of random matrices repel each other years ago and i've been off trying to find something about it

maybe it's only true for random unitary matrices

@Salt what do you mean by repel?

@TedShifrin you are actually being very helpful. I'm going to get through this.

You'll get it any moment now, Startec.

@LeakyNun archive.ymsc.tsinghua.edu.cn/pacm_download/21/… the introduction covers it

probability density is smaller when two eigenvalues are closer together

Yes. That's what you want to end up with.

> First, the eigenvalues tend to repel each other; the probability density for unitary eigenvalues $e^{i\theta_j}$ $$f_2(\theta_1,\ldots,\theta_n) = \frac 1 {(2\pi)^n n!} \prod_{j<k} \| e^{i\theta_j} - e^{i\theta_k} \|^2$$ is smaller when two eigenvalues are close together on the unit circle.

i seem to recall watching someone smoothly change these random matrices and watching the eigenvalues move towards each other and then away, never touching or crossing

hmm

probably by raising to a power?

@TedShifrin any idea?

Not even thinking about it.

what is this picture of hands and matrices that's starred

hi GFaux

good evening Ted

the one posted 26m ago by Secret

Yup. There's your $B$ !!! Congrats, Startec.

IT took a while, but you understand things a lot better. Check it out.

Okay, but there is still a question which is that, how did you know that you want the generators there, in the first four rows? Like what if we had [0, 1, 0, 0] in the third row instead of the 2nd?

> one

1. It's 1.

Remember you want $BA = I$ ... $I$ is the identity matrix. Check it out.

Oh right okay so, knowing that, how did you know that if the generators were in those rows than BA would be the identity?

Do the multiplication and you'll understand for yourself.

oh right, of course. There it is

@Salt that’s a pretty typical result, in fact: if there’s no symmetry which permits a level crossing as a parameter is varied, you’ll instead end up an avoided crossing

There are assumptions for that, of course

i remember it surprising me anyways

but i'm not clear on the details, to be fair

are the diagonalizable matrices dense? @TedShifrin

Fair enough. I mostly have vague recollections of quantum mechanics

Something something Wigner

yeah, wigner matrices

you did type out 1 on the left

why do you insist on having “one” inside?

Um, you aren't multiplying correctly. ...

@LeakyNun they are

@Daminark why?

@salt there’s an animation like that here: en.m.wikipedia.org/wiki/Avoided_crossing

Because I am doing this in python and one is different than the integer value 1.

So, take a complex matrix A

I'm no python.

you’re doing multiplication in python??

You can write $A = P^{-1}BP$ where B is triangular

you can override int.__add__ i think

but you need to be able to do it yourself

or just have a class that inherits int

Now, if all the entries of the diagonal of B are distinct, it's diagonalizable (and so is A). If not, you add a tiny diagonal matrix D to B so that it is

correct, Demonark.

interesting, but i knew there were connections of random matrix theory to quantum mechanics

Huh? I don't get that at all. You can get nontrivial Jordan canonical forms in this scenario.

And then you just look at the norm of P^{-1}DP, which is controlled

Demonark, you're talking about perturbing to make the eigenvalues distinct. OK. So? You're arguing that diagonalizables are dense.

@TedShifrin yes, which is coincidentally what i asked, thank you very much

I didn't notice you asking anything, but OK.

The idea is that $P^{-1}(B+D)P$ should be a matrix whose norm is close to A

@TedShifrin Ah, I see. I flipped some things in my original matrix. Thanks

Damned irresponsible of you, @Startec!

Well technically it's because my book was using a different diagram than the one I posted

Even more irresponsible :P

I never saw that, Leaky.

Correct argument, Demonark.

$\|A - P^{-1}(B+D)P\| = \|P^{-1}DP\| \le \|P^{-1}\|\|D\|\|P\|$, so you just make $\|D\| < \frac{\epsilon}{\|P\|\|P^{-1}\|}$ and you're set to go

So how would you do it with Jordan form?

I misunderstood what you were doing, originally.

It's the same idea.

Ah I see

@Daminark yes

Aight, sick

it is non-diagonalizable only finitely often

But yeah that's a proof of Cayley Hamilton actually

@Daminark :o

really?

that's a neat proof

I didn't know there's a proof that I can understand :P

The idea is that you show that if $A_n \to A$, then $p_{A_n}(A_n) \to p_A(A)$

how do you guys think of elements in $V_1 \otimes \cdots \otimes V_n$?

@Daminark ya, I get it

As $v_1\otimes\dots\otimes v_n$?

like span

@TedShifrin I don't think that's enough

no, not span

you want lin.combs of those guys

Yes, I do need linear combs of those guys.

But it was sorta a stooopid question.

It's true that not all elements are of that form but they're a good prototype

then it's not very helpful I think

You can often define/prove things on those guys and extend linearly

multilinearly

Lol, I mean linearly within the tensor product

I don't think that's really what you mean, but OK.

i know it isn't span, but that's how i often intuit it

You're thinking $\oplus$, not $\otimes$, @Salt.

oh

And, actually, you're thinking $+$ rather than $\oplus$.

no, i don't think i was

lol!

Well, with span there's no requirement that the subspaces intersect only in the 0-vector.

i thought V cross W is spanned by a basis consisting of pairwise products of elements of V's basis and W's basis

No.

That's tensor product, not direct product.

So that is what I wrote originally. But nothing to do with "span."

Belated Xmas, Semiclassic.

I dunno,I thought we were talking about tensors

same to you, Ted

I was talking about tensors originally, Salt. And then you said "span." Which really made no sense.

but that's how i think of a tensor product

I give up.

Are you thinking of tensor product in the quantum context?

i'm not

kk

@Semi what's the quantum context of the tensor product?

Suppose you've got an electron. It's a spin-1/2 particle, which means that the spin part of the wavefunction is represented by a complex column 2-vector (normalized to have length 1)

the basis states are just $e_1,e_2$ which conventionally correspond to the spin z-component being $\pm \hbar/2$

now suppose you've got two such electrons, and you want to be able to talk about them at the same time.

To characterize a given pair of electrons, we can measure the z-components of their spin angular momentum separately. You can either have them be up-up (both +hbar/2), down-down (both -hbar/2), down-up, or up-down.

So you go from needing a 2-component vector for one electron to needing a 4-component vector for two electrons.

And the usual way you do that is to take the up-up combination to be $e_1\otimes e_1=\begin{pmatrix} 1e_1 \\ 0 e_1\end{pmatrix}=\begin{pmatrix} 1 \\ 0 \\ 0 \\ 0\end{pmatrix}$

and similarly $e_1\otimes e_2,e_2\otimes e_1, e_2\otimes e_2$ for the up-down, down-up, and down-down combinations

formally I'm not sure that's how tensor product is defined, but it is how the kronecker product is defined and that notation uses $\otimes$ as well.

Similarly, you can now talk about linear maps on these vector spaces

and the Kronecker product $A\otimes B$, with $A,B$ being 2-by-2 matrices, acts on a state $u\otimes v$ as $(A\otimes B)(u\otimes v)=(Au)\otimes (Bv)$

so the Kronecker product is the right matrix representation of the tensor product of $A,B$ as maps on these vector spaces.

Hi all; I'm really confused on a Number Theory concept; It would be great if anyone could clarify it.

So the question is: Find the product of all of the positive divisors of 450 that are multiples of 3.

nvm, I'll post it on the website

why? what part are you having difficulty with?

So the part I'm having trouble with is this:

My method to solve the above question was to first generalize the prime factorization of the factors of 450 that are multiples of 3, as so:

your bounds are wrong

^

Yeah, I just saw

that's still wrong