Question

How to multiply surfaces as matrices intro:

Answer 1

btw, can this be thought of as a version of convolutions then

Answer 2

We can choose a matrix to be any place on a surface, it doesn't matter where. Just as long as they're evenly placed it will work out. So we are going to define our matrix \[\LARGE A=\left[\begin{matrix}-1 & -2 \\ 2& 8\end{matrix}\right]\]\[\LARGE A(x,y)=(-y)^xx\] So here I'm saying we're taking samples at steps of 1 and going from 1 to 2 on the x and y axes.

We can sample the same function in different areas different ways and get rectangular matrices or whatever, and it's quite easy to show that \[\LARGE A^T(x,y)=A(y,x)\] Now for multiplication: \[\LARGE \sum_{n=1}^2A(x,n)B(n,y)\] We are simply going over the 2x2 matrix over the bounds. If B is really A on the same surface, we should scale it so that at n=1 and n=2 we hit the right spots. This is just like how with matrices we have: \[\LARGE A_{ij}B_{jk}=C_{ik}\] So let's let B be the identity function, \[\LARGE \delta (x,y)\] when x=y it is 1, otherwise it is zero.

\[\LARGE A=AI=\sum_{n=1}^2A(x,n) \delta(n,y)\]\[\LARGE A(x,1) \delta(1,y)+A(x,2) \delta(2,y)=A(x,y)\] How do I know this statement is true? Just evaluate it at y=1 or y=2 and see that you will drop off all the other terms.

Whether you left or right multiply simply sorts it out into its rows or columns.

Answer 3

Notice though, all of these functions are secretly not discrete though. So the product AB is generally another surface that we can sample from. And we also have the ability to sorta just sample infinitely between points and turn this into an integral. Or collapse our matrices into single vectors and it's just the regular old dot product we had with integrals in the past.

Answer 4

does it even matter how many points u can select for A, or if it has to be symmetric or anything, is it fully dependant on just the functions and knroicker

Answer 5

You can select as many points as you like. It just has to obey the regular matrix multiplication rules that the inside parts of their indices line up, you can multiply a 3x7 times a 7x5 matrix just as easily and your summation will run from 1 to 7.

Answer 6

oh okay so like u did
x=1,y=1
then
x=1,y=2
2 1
2 2

Answer 7

ok i can see why the transform is true now

Answer 8

If you recall watching this video, these Fourier Matrices are quite nicely expressed as functions. https://www.youtube.com/watch?v=umt6BB1nJ4w&list=PL49CF3715CB9EF31D&index=26 and their inverse is quite easy to compute and see that by following the rules I wrote above will give us the identity surface as well.

Answer 9

Every surface is the sum of a symmetric and skew symmetric matrix \[\LARGE A(x,y)=B(x,y)+C(x,y)\]
This isn't very surprising but hey it's useful since it's not very clear how to actually represent an arbitrary matrix with functions.\[\LARGE B(x,y)=\frac{A(x,y)+A(y,x)}{2}\] \[\LARGE C(x,y)=\frac{A(x,y)-A(y,x)}{2}\]

Answer 10

But really just like Fourier series it seems like it shouldn't be too big of a deal.

Answer 11

ok i got what u said so far

Answer 12

what do u want me to watch in this lecture there no fseries

Answer 13

https://www.youtube.com/watch?v=M0Sa8fLOajA&list=PL49CF3715CB9EF31D#t=1148

There, sorry I think I sent you the wrong thing.

Answer 14

Essentially here is the formula in my scheme\[\LARGE (F_n)_{ab}=e^{i \frac{2\pi}{n}ab}\] and like he says earlier, we're starting this at k=0 and go up to k=n-1 as the dimensions of the matrix, so we're creating an nxn matrix this way.

Answer 15

wait im rewatching from hemetian matrices part in this lecture

Answer 16

he just started F_n

Answer 17

The inverse is just the hermitian of this matrix, which is quite obvious to see the conjugate transpose is really just 1/Fn really convenient haha.

Answer 18

thats cool u can also tell those columns had to be orthogonal because of that inner product of and sin and cos functions showing up in the inner products

Answer 19

and u know that since its different roots m=/=n in the sin(mx)*cos(nx) or cos(mx)*cos(nx) represenation of their innerproducts

Answer 20

u know m =/=n because theyre all for a different set of roots along each column or row

Answer 21

So like multiplication wise we have: \[\LARGE \sum_{k=0}^{n-1}e^{i \frac{2\pi}{n}xk}e^{-i \frac{2\pi}{n}ky}=\delta(x,y)\]

Answer 22

right okay thats what u mean u had complex representaion of this identity

Answer 23

hmm id have to see this thru tho

Answer 24

ok wait i see it

Answer 25

hmmm wait

Answer 26

how is it zero for x=/= y

Answer 27

F_n^H*F_n=I

Answer 28

Try simple cases like n=2, n=3, n=4.

Answer 29

actually there's a 1/sqrt(n) factor out front of each of these matrices I forgot

Answer 30

ya ya np

Answer 31

The only thing is it will give you instead of the identity will be a scalar matrix with that n value.

Answer 32

ur function is the identiy if x ,y we chose are integers

Answer 33

are u tryna now say we can continue this to non discrete

Answer 34

sure thing d000d

Answer 35

lol

Answer 36

tell me

Answer 37

i think we are starting fourier transforms again soon, gotta revisit this then

Answer 38

now imma start my exam-.-

Answer 39

ok have fun I'm playing around right now trying to figure stuff out. Keep in mind this means we can also do stuff like tensors and have A(x,y,z) and then by permuting the variables we have higher order transposes.

Answer 40

Here's a fun question, what's the transpose of a matrix in polar form?

Answer 41

the identiy wud be equivalent to e^itehta

Answer 42

well e^i*2pi

Answer 43

ud have to get a matrix in its polar orthonormal form and maybe the transfmore of that wud be its inverse in polar too

Answer 44

transpose*

Answer 45

hmm theres gotta be some way to extract a general transpose in polar form from these properties

Answer 46

\[\LARGE A^T(r(x,y),\theta(x,y))=A(r(y,x),\theta(y,x))\] \[\LARGE r=\sqrt{x^2+y^2}=r^T\]\\[\LARGE \theta^T=\tan^{-1} \frac{x}{y}=\tan^{-1}\cot \theta\]

Answer 47

oh cool!

Answer 48

A^T=A(r^T,theta^T) O_O

Answer 49

what can we do with this

Answer 50

feel like we are just saying something with matrices what we can say anyway by switching axis

Answer 51

Yeah that's all we're doing really. It's just sorta a lame playing around of stuff to see what it looks like for fun.

Answer 52

ahh ok vcarry on
http://dmpeli.math.mcmaster.ca/TeachProjects/Math3C03_14/home4.pdf

Answer 53

see if anything peaks ur interest like the last question perhaps -.-

Answer 54

lmao the most random first comment here