linear-algebra

[old flame]

Same case for kernel and image I guess right

[spiral star]

yea

im(f) just applies to any function

and ker(f) is defined for homomorphisms

(and can be generalized in category theory)

[old flame]

Damn, it extends quite far then

[spiral star]

its just nice to have the same terminology for different things

especially when those things overlap

linear algebra will overlap a bit with group theory once you get to determinants

which is probably very late since you read axler lol

with his irrational fear of determinants

[old flame]

Yeah that's true LOL

Determinants is the last chapter

Oh well I guess I just have to wait a while

So when it overlaps things gets more interesting ?

[spiral star]

uh

idk

determinants are super useful for doing all the eigenvalue theory stuff

characterizing similarity

but yea...

i dont really remember how axler does all that

[gray dust]

a tiny addition, an isomorphism is a bijective homomorphism, note domain/codomain need not be the same set (tack on domain=codomain & you got a bijective endomorphism hence an automorphism). and related is that in the context of linalg it's a somewhat big thing that same dim vector spaces (over same field) are isomorphic, ie there exists an isomorphism between em

[old flame]

Oh I haven't reached there yet, sounds exciting 👍

Do these notation simplify proofs ? Or when do we use them ? As I haven't encountered a problem requiring them yet

[spiral star]

@gray dust is it just me or does your name evolve over time lmao

[gray dust]

very slowly if at all, rokabe was maybe 6mo ago

[spiral star]

so it is just me

lol

[gray dust]

i'll change it more often, just for you

[spiral star]

oh noes

[gray dust]

too late

[spiral star]

[gray dust]

Do these notation simplify proofs ? Or when do we use them ? As I haven't encountered a problem requiring them yet
fancy speak doesn't affect how good your proof writing is, this is just the vocab flow has accrued over time having experience beyond linalg

[spiral star]

(my LA prof actually did topological algebra)

but yea, it's just notation

i think it's good to learn the different notation

i dont see where using the more general notation would make anything easier in LA

only very minor things

truth is, im just too lazy to change my ways

[gray dust]

not that you need to

[spiral star]

i will stick to what i learned in universal algebra and category theory :p

[zinc copper]

for a set V to be a vector space, must the additive inverse of an element v be unique or can there be multiple?

[marble lance]

@zinc copper It is not an axiom that it must be, but it follows from the other axioms. So yes, "there can be only one"

[zinc copper]

how does it follow?

[spiral star]

gwoup theowy

[zinc copper]

ah anyways ive figure out why my example cant be a vector space, but now im curious

[marble lance]

Suppose y and z are both additive inverses of x

Can you show that y = z?

[zinc copper]

x+y = 0 = x+z. subtract x to get y =-x = z

oh nvm

i cant just subtract

hmm

[marble lance]

Yeah, start with x + y = 0, then z + (x + y) = z

[zinc copper]

right

and with associativity

[marble lance]

Yes

[zinc copper]

(z+x)+y = y

so z = y

aight

thanks

[marble lance]

Np

[errant wyvern]

Hi, when is there some dedicated way to finding all the invariant subspaces of a linear opertaor give by its matrix?

[rose coral]

Ok so I'm a bit rusty, but I believe what you do is find the Jordan Canonical Form

Which is like a generalization of finding eigenvalues

You should always be able to find the Jordan Canonical Form if you are willing to deal with complex numbers

Wait, actually I don't think you need to go through the whole process if you just want the invariant subspaces

Are you asking out of curiosity or because you want to know the algorithm for finding the invariant subspaces?

[errant wyvern]

Well i have these exercieses on my exams which requier that from me, was wandering if there is an algorithm for it or if its nuanced?

So yeah if theres an algorithm that would help out immensly

[rose coral]

Ok so just finding the invariant subspaces isn't tooo terrible

Do you know of the characteristic polynomial?

[errant wyvern]

I'm generaly given the matrix so i can find it if thats what you are asking?

[rose coral]

I'm just not familiar with what people are taught so I wanted to know if you knew of it

So first thing you need to do it factor the characteristic polynomial

As you might know, the roots tell you the eigenvalues

Well, for the invariant subspaces you also care about the multiplicity of the roots

[errant wyvern]

could i do the same with det(a-xI) to find the eigenvalues and its multiplicity?

[rose coral]

Yeah

So say that the char poly is $\det(A - xI) = (x - 1)(x + 2)^2(x^2 + 1)$

[stoic python]

The_Vman:

[rose coral]

Then for each factor you get an invariant subspace

For 1 you get the eigenspace $\ker(A - I)$

[stoic python]

The_Vman:

[rose coral]

Maybe you use nullspace instead of kernel

For the factor $(x + 2)^2$ you get what is called a generalized eigenspace, in this case corresponding to the eigenvalue $2$

[stoic python]

The_Vman:

[rose coral]

Which is $\ker\Big((A + 2I)^2\Big)$

[stoic python]

The_Vman:

[rose coral]

And finally you get one for $x^2 + 1$, which is $\ker(A^2 + I)$

[stoic python]

The_Vman:

[errant wyvern]

Maybe you use nullspace instead of kernel
@rose coral myb im lost in translation, but doesnt in english litterature kernel and nullspace mean the same thing?

[rose coral]

Yeah they do, just wanted to include both terms just in case

[errant wyvern]

allright, also in all of theese "A" reperesents the matrix of the linear operator?

just checking for confidence sake

[rose coral]

Yup, and you find the kernel by row reduction or whatever you prefer

And so the space decomposes as $\bR^5 = \ker(A - I) \oplus \ker(A + 2)^2 \oplus \ker(A^2 + I)$

[stoic python]

The_Vman:

[rose coral]

Where each one is an invariant subspace

It's also important to note that the invariant subspace corresponding to some $(x - \lambda)^n$ will be $n$-dimensional

[stoic python]

The_Vman:

[rose coral]

And that's it

Well

Actually each $\ker(A - \lambda)^n$ might have smaller invariant subspaces in it

[stoic python]

The_Vman:

[rose coral]

And the sum of any invariant subspaces is again invariant

[errant wyvern]

allright so if i have like this example

this would lead me to belive that invariant subspace is 3dimensional, based on eigenvalue 1

i miswrote on the pic, eigenvalue is obviously 1

[rose coral]

Sorry we're talking about the $\begin{bmatrix}1 & -1 & 0 \ 1 & 1 & 0 \ 0 & 0 & 0\end{bmatrix}$?

[stoic python]

The_Vman:

[errant wyvern]

yes, in the example we can see the linear opertator by being given its efect on the vectors {x,y,z}, we extrapolate the matrix which then eqiuals to what you have written here

The_Vman:
@stoic python

and then we find the eigenvalue, which in this case is 1, of multiplicity 3, correct?

[rose coral]

I don't think so, you should get an eigenvalue of 0 with multiplicity 1

Not sure about the other two factors

[errant wyvern]

yes

i made a mistake my bad sorry for wasting time

im hurrying and making stupid mistakes

[rose coral]

No worries, happens to everyone

Ok so in this case it looks like the characteristic polynomial is, up to a sign, $x((x - 1)^2 + 1)$

[stoic python]

The_Vman:

[rose coral]

Where the quadratic doesn't have real roots

Oh btw if you have the matrix in blocks like this one, you can calculate the characteristic polynomial for each block and then multiply those together

So here the (x - 1)^2 + 1 comes from the top left, 2x2 block

But in terms of the invariant subspaces, here there isn't much to compute as the blocks give the subspaces. The only thing that could potentially happen is that the 2x2 block broke down further into invariant subspaces

But the polynomial doesn't have real roots (it is irreducible), so it doesn't

[errant wyvern]

allright so what do i do here to find invariant subspaces?

[rose coral]

You could go through finding the kernels of corresponding to the factors

But the factor corresponding to 0, which corresponds to the bottom right 1x1 block, will have invariant subspace spanned by (0 0 1), the kernel of the original map

And the 2x2 block doesn't further break up into invariant subspaces, so its invariant subspace is simply given by the span of (1 0 0) and (0 1 0)

[errant wyvern]

i see

thank you

really appreciated the help

[rose coral]

No problem

[errant wyvern]

i also now noticed a patern that my professor tends to give on this sort of exercise

its either a projection operator or something of the sort in the example above

[rose coral]

Oh cool, that makes it easier

[errant wyvern]

yeah, but i really appreciate your help, it may not help too much here but it definitly will in the future

[rose coral]

Anytime

[errant wyvern]

here is another example of the exercise, hopefully i did it well here

here is a linear operator of projection in R^3 to R^3, where it maps {x,y,z}^t to {0,y,z}^t. Extract its matrix, then found eigenvalues. Eigenvalues are 0 and 1, 1 having multiplictiy of 2. Found kernel of (P-1 I) and it is spanned by vectors {0,1,0} and {0,0,1}

[rose coral]

Yup that works

Now here the invariant subspace spanned by {0, 1, 0} and {0, 0, 1} decomposes into the 1d subspaces

So in this case the subspace spanned by {0, 1, 0} and another spanned by {0, 0, 1}

[errant wyvern]

I see

[rose coral]

It could have been that it doesn't decompose

[errant wyvern]

yeah but here it does i just skimmed past it

thanks for pointing it out

[rose coral]

No problem

Also if the matrix is diagonal, it fully decomposes into the 1d subspaces given by the standard basis vectors

And as an example $\begin{bmatrix}2 & 0 & 0 \ 0 & 1 & 0 \ 0 & 1 & 1\end{bmatrix}$

[stoic python]

The_Vman:

[rose coral]

has it so that the generalized eigenspace corresponding to 1 doesn't decompose, because of this 1 off the diagonal in the bottom left 2x2 block

Here the invariant subspace isn't that hard to see, but if the matrix wasn't this nice, you would have to compute $\ker(A - I)^2$ not just $\ker(A - I)$ to get the full 2d invariant subspace

[stoic python]

The_Vman:

[errant wyvern]

I see

Will be on the lookout for that in the examples im given

Another question, if i have a list of matrices with the task of checking which ones are simmilar, all i need to do is find their eigenvalues right? And if i have the same, but instead of checking it they are simmilar i need to check if they are equivalent matrices all i need to do is check their ranks?

[spiral star]

@errant wyvern yes, for equivalence you only have to compare their ranks.
and no, for similarity its not enough to find eigenvalues.

if two matrices are similar, they have the same eigenvalues. but not all matrices with the same eigenvalues are similar. you need a stronger condition

so if they have different eigenvalues, you immediately know they cant be similar

but if they have the same, it doesnt tell you anything

the usual trick for similarity is to show that both matrices are similar to the same normal form

for example, if both matrices are similar to the same diagonal matrix, then they are similar

and if they are not diagonalizable, then you can compare their jordan normal forms

[errant wyvern]

I see

Allright thanks

[spiral star]

i guess if you want some steps for checking whether matrices A and B are similar: 1) find their characteristic polynomials. if they are different, then A and B cant be similar. 2) check geometric multiplicities. if A and B are similar to the same diagonal matrix, they are similar. 3) if neither matrix is similar to a diagonal matrix, find their jordan normal forms and compare those

i hope you must never resort to finding the JNF lol

[errant wyvern]

yeah i see

finding the jnf is annoying and gruntwork but it aint hard

just would rather skip it lul

[spiral star]

yea xd

but comparing JNF will always tell you if they are similar

but best case scenario is you find your answer before you have to go through that

you can use a few facts about the characteristic polynomials as well

if you get distinct linear factors, like (x - a)(x - b)(x - c)... where a,b,c are different

then you have similarity to a diagonal matrix

and you dont have to take any extra steps

because then the geometric and algebraic multiplicities already match up

so for those you can read off the answer pretty much

basically, try to avoid unnecessary computation

[rose coral]

@glad current does this make sense?

$\begin{bmatrix} 1 & 2 & -3 \ 5 & 0 & 1\end{bmatrix} \begin{bmatrix} a \ b \ c \end{bmatrix} = a\begin{bmatrix} 1 \ 5\end{bmatrix} + b\begin{bmatrix} 2 \ 0\end{bmatrix} + c\begin{bmatrix} -3 \ 1\end{bmatrix}$

[stoic python]

The_Vman:

[glad current]

yeah

(that's how 3brown1blue explains it)

[rose coral]

Ok awesome

I think its the best way of remembering it

[hexed mural]

Question

can you parameterize a vector Space?

for a Basis vector and a coordinate vector?

[half storm]

What do you mean parametrize?

[hexed mural]

a line

the vector space

parametrize it

So for example

[half storm]

Can you parametrize a line in a vector space?

[hexed mural]

yes

[half storm]

If depends on what the vector space is.

[hexed mural]

This can be explained by example

[half storm]

Not all vector spaces are $\mathbb{R}^n$ n $\in \mathbb{N} \cup { 0 }$

[stoic python]

JohntheDon:

[hexed mural]

@half storm

instead of typing this can I link the paper?

and we can discuss it?

it's probably easier

https://peroxide.dk/papers/collision/collision.pdf

What I am trying to figure out is. How he parameterized the ellipsoid vector space

page 7-11

[half storm]

eh I'm not familiar with this. Might be better to ask someone else.

[hexed mural]

what section

[half storm]

<@&286206848099549185>

[hexed mural]

the question section?

alpha to theta

thanks god father John

[half storm]

Yea questions section. lol no problem.

[little frigate]

Is there a website I can use to calculate matrices based on a position like

say I have a formula like c_ij = i + j

I'd like to know if there's a website that can calculate this

[quartz compass]

whatcha mean calculate

like fill out the matrix that looks like that with those as entries?

[little frigate]

Yea

[quartz compass]

I'd use a good enough calculator or programming language like matlab or octave

[little frigate]

Dang

[quartz compass]

ti-nspire can make matrices that way for instance

[little frigate]

Alright thanks

[olive kraken]

anyone know how i can find the OG matrix from this

i have to figuure out if the determinant is 0

im not sure if there is a way to do that other than working backwards to find the matrix

oh wait i think nonsingular just means that it can be put into the RREF where theres a 1 for each row

[quartz compass]

one way is to write out a 2x2 matrix with entries a,b,c,d then plug in everything to X'=AX

[surreal thistle]

can component of a along b be a negative value?

nevermind i guess it can

[gray dust]

scalar projection, yes

[olive kraken]

im stuck on this

even for a 2x2 matrix i don't really know how to prove that without using numbers

let alone an nxn matrix

[rose coral]

What does it mean for lambda to be an eigenvalue of a matrix A?

[olive kraken]

it makes it a zero vector?

or something like that

[rose coral]

Maybe you are thinking of the matrix A - (lambda)*I

[olive kraken]

yea

[rose coral]

A scalar lambda is an eigenvalue of a matrix A if there exists a non-zero vector v such that Av = lambda * v

[olive kraken]

so the direction is preserved but the magnitude changes

[rose coral]

Pretty much, if lambda is 1 or -1 then the magnitude doesn't change

But yeah the direction is preserved

Ok so what would it mean for $\lambda^2$ to be an eigenvalue of $A^2$?

[stoic python]

The_Vman:

[olive kraken]

theres a non zero vector Av = ((lamda)^2)+v

?

[rose coral]

$A^2v = \lambda^2 v$, yeah

[stoic python]

The_Vman:

[olive kraken]

oh yea

[rose coral]

Ok so to prove the statement, you need to use the assumption that $\lambda$ is an eigenvalue of $A$ to find such a $v$ for $A^2$

[stoic python]

The_Vman:

[rose coral]

Right?

[olive kraken]

yes

i dont even know how to move foward from that though

[rose coral]

Ok so here's what my thought process is when I'm given a problem like this

First I understand what I need to do, I think we've done that part

Now I look at what I can use to prove the statement I want

So, I need to find a special vector v that relates to A^2. And since lambda is an eigenvalue of A, I'm given a special vector for A

My gut instinct is to see if the same vector works

I'll set it up

[olive kraken]

ok

[rose coral]

I'm given that $\lambda$ is an eigenvalue of $A$. Unravelling that definition, I'm given some non-zero vector $v$ such that $Av = \lambda v$

[stoic python]

The_Vman:

[rose coral]

Now I want to try out this vector with A^2, see if it works

How can I tell if it works?

[olive kraken]

i guess if it equals the lamda squared thing ?

[rose coral]

Yeah, I want to start out with $A^2v$ and see if I can turn that into $\lambda^2 v$

[stoic python]

The_Vman:

[rose coral]

So, I start with A^2v. I want to get some lambda stuff, so I use the assumption

A^2v just means apply A to v, and then apply A again

So $A^2v = A(Av)$

[stoic python]

The_Vman:

[olive kraken]

A is a matrix and so what is lambda again

[rose coral]

It's a scalar, so probably a real number

[olive kraken]

oh ok

[rose coral]

But the point is, now I can use my assumption that $Av = \lambda v$

[stoic python]

The_Vman:

[rose coral]

If I plug that in, I get $A^2v = A(Av) = A(\lambda v)$

[stoic python]

The_Vman:

[rose coral]

Do you think you can continue?

[olive kraken]

maybe

so then you apply the lamba to both sides?

[rose coral]

Not quite

I'm guessing you want to get another lambda so you have lambda^2

The idea is that the A already is like a lambda, when A acts on this vector its the same as scaling by lambda

[olive kraken]

ohh

[rose coral]

So you should be able to get the other factor of lambda from the A that is left in $A(\lambda v)$

[stoic python]

The_Vman:

[olive kraken]

A = lambda?

is Av and lambda v just multiplication?

[rose coral]

Yeah the A and lambda are multiplying the vector v

But the tricky thing is that A doesn't necessarily scale everything by lambda, only this specific v

(Well actually it should do the same thing to whole span of v)

Ok so the fact I wanted you to use was $A(\lambda v) = \lambda * Av$

[stoic python]

The_Vman:

[rose coral]

And then you can replace $Av$ again by $\lambda v$

[stoic python]

The_Vman:

[rose coral]

The idea is that whenever you have a matrix A, you can pull out scalars. So A(5x) = 5A(x)

Ok I hope that wasn't too confusing, its just how I think about doing proofs

[olive kraken]

i mean its a good explanation im just super unfamiliar with this stuff so its taking some work to digest but i think im following

[rose coral]

Ok I'm glad you think it was a good explanation/you were able to follow

What you should take away is that a matrix A does the same sort of thing to vectors in the same span

Which is captured by A(3x) = 3Ax

So if A rotates a vector v by 90 degrees, it also rotates 3v by 90 degrees. If it skews v onto the y axis, it also skews 9v onto the y axis (and also 0.2*v and 12v etc.)

[olive kraken]

ohh okay

[rose coral]

And if A scales v by lambda, it also scales 3v or 5v or lambda*v by lambda

So is scales the entire line spanned by v, by a factor of lambda

So this whole thing was saying, if A scales some line (say spanned by a vector v) by a factor of lambda, then applying A^2, meaning applying A twice, should be like scaling that line by lambda twice, meaning it scaling by lambda^2

Personally I think that the best way to understand linear algebra is geometrically

Have you watched 3Blue1Brown's essence of linear algebra on youtube?

[olive kraken]

no i need to watch that i havnt looked into the geometric aspect of linalg at all

but its super confusing as of now so that hopefully will help

[rose coral]

I would recommend it. It explains everything very visually and is a great way to build intuition, and understand what is going on

Other than that just keep asking questions

[olive kraken]

okay i will i hope i can understand this stuff eventually

as of now i literally have no idea why i do any of the operations that i do which sucks

also can i just send a summed up version of what we discussed to make sure i didnt make any errors?

[rose coral]

Yeah sure

[rose coral]

Yeah that looks good

When you write a proof though its useful to have words explaining what each step is

So you can start by something like "Since A has eigenvalue lambda there is a vector v such that (first equation in black)"

And then you can say stuff like "by (the equation in black) we can plug in lambda v for Av" ...

But it just takes practice

And btw if you find the videos confusing, or something you find something you've read/learnt in class confusing, just ask. People are also here to help with those types of questions

[olive kraken]

wow

thank you so much for the help

[rose coral]

Anytime

[olive kraken]

this honestly gives me confidence in my math skills, i was beginning to think that I was in my final days of math because i was getting so lost

but i will keep pushing and hopefully understand this stuff

[rose coral]

I believe in you

Yeah I would say the way linear algebra is usually taught isn't great

[olive kraken]

Thank goodness the internet is here to help lol

[rose coral]

Lol yeah its great

[hexed mural]

Hello I have question regarding this derivation

What do they mean by Orthonormal base for the plane how are these definitions derived? for the vectors axes

[paper egret]

my lin alg is incredibly rust

any1 got any starting points?

[wintry steppe]

a common technique for showing that some vector x has a certain representation with respect to a basis {u_1, ..., u_n} is to apply your inner product to both sides of x = sum x_i u_i

the question is worded a bit weirdly

but i gather from it that your inner product is just x^T y. try using this inner product along with the equation x = sum x_i u_i

[paper egret]

i wasnt too sure if x_i is an element of vector x

[wintry steppe]

x_i is a scalar here

[paper egret]

o

h

so its not

in the vector x

thats confusing

nice

[wintry steppe]

it's a bit deceptive since x is an element of R^n

so you'd be inclined to represent it in the standard basis

[paper egret]

damn shudve used a different letter instead of x_i then, maybe c_i woulda been better

[wintry steppe]

maybe

i like how it says the n-dimensional vector space as if R^n is the only one

[paper egret]

so we know that

x = c1u1+c2u2 + ... + cnun is true

set every c to 0, except c_i = 1

x = ciui

uh oof my proof is garbage

[wintry steppe]

you can't change the values of the c_i's

[paper egret]

shit

[wintry steppe]

if you're writing x = sum c_i u_i, then your goal is to show that c_k = x^T u_k for each k

[paper egret]

we know that there is a set of c_i's that make the vector x

uhh

hmm

[wintry steppe]

here's something super straightforward you can do: try subsituting your expression for x into x^T u_k

(which is equivalent to what i said at the start and probably less confusing)

[paper egret]

if i transpose x

from the definition

x = c1u1 + c2u2 + ... + cnun

transpose this

what would the right side look like?

[wintry steppe]

transpose is linear

that might help

[paper egret]

lemme think it thru a tad bit more

uhhh

[wintry steppe]

just keep in mind your c_1, ..., c_n are scalars and u_1, ..., u_n are elements of R^n

[paper egret]

so the transpose applies to all vectors

or my mistake

applies to each term

[wintry steppe]

mmhm

[paper egret]

xT = c1u1T + c2u2T + ... + cnunT

sorry for the crappy notation

[wintry steppe]

nah it's fine i get what you mean

that's correct

[paper egret]

if i multiply that by ui

i dont know what happens LMAO

xTu_i = (c1u1T + c2u2T + ... + cnunT)(u_i)

[wintry steppe]

you forgot the left side of the equation

that is important

;)

[paper egret]

that

[wintry steppe]

right

[paper egret]

but how does the right side

give c_i

h u h

[wintry steppe]

tell me what property {u_1, ..., u_n} satisfies

this is where it should come in

[paper egret]

its a basis

i never studied orthonormal

so maybe it might be that

[wintry steppe]

do you know what orthonormal means at least?

[paper egret]

lemme wikipedia real quick

oh it just means

the vectors are orthogonal to each other

[gray dust]

and?

[paper egret]

wait no

[wintry steppe]

hi roketto

[paper egret]

wait thats correct?

[gray dust]

there's more. hi tera

[paper egret]

all unit vectors

and orthogonal

o

that means any dot product between two vectors from the basis is equal to 0

but

i dont see how that applies here

[wintry steppe]

simplify (c1u1T + c2u2T + ... + cnunT)(u_i) some

[paper egret]

lol i dont know how

FUCK

WAIt

IM RETARDED

GImme a sec

distribution duh

oh

thank

you

so much

you furry god

[wintry steppe]

do you see the final result?

[paper egret]

yessir

uh wait real quick though

ui^T dot with ui is equal to 1 right?

[wintry steppe]

yes

why?

[paper egret]

uh idk

tbh

[wintry steppe]

orthonormal

the normal part means that any vector dot itself is 1

[paper egret]

o h

[gray dust]

hence unit norm

[paper egret]

damn so it was all in the orthonormal

yea

orthogonal + normal = orthonormal

i shudve paid attention to the orthonormal more

[wintry steppe]

slight nitpick: "u_i^T dot u_i" should just be "u_i dot u_i" or "u_i^T u_i"

[paper egret]

fuck me

thank you again

learned something new

and a nice lin alg review ig

[gray dust]

tera strikes again

[wintry steppe]

now go do 20 problems of the exact same flavor

jk

you can get a slightly more general expression for x with respect to an orthogonal basis by following the exact same method

which is sometimes useful

e.g. in the proof of gram-schmidt iirc

fun fact that might be worth keeping in mind

[pastel kettle]

Is this where also i can ask questions?

[gray dust]

if linear algebra related

[pastel kettle]

I have no idea what inner product im supposed to use and dunno what im supposed to doooo

[rose coral]

Ok so they've given you the norm associated with an inner product

And you want to know what the formula for the inner product is so that you can find the space perpendicular to E

Right?

[pastel kettle]

Yes

And also further

And projection as well

[rose coral]

Yeah

[pastel kettle]

I feel like im losing hair bc of this

[rose coral]

Ok so Theorem 2.1.13 lets you get the formula for the inner product from the norm

So I would plug in the equation you're given for the norm to get what the inner product is

[pastel kettle]

Yes so instead of y do i put x? So it can be $llx+xll$

[stoic python]

im_molee:

[rose coral]

Well I think you want to know what <x, y> is

If you replace y by x then you'll just get the norm back

I think you need to know what the inner product between two arbitrary vectors is

[pastel kettle]

I meanim confused cuz from the question y is not given

Yes very true

[rose coral]

Oh ok

[pastel kettle]

Thats why im stuck from beginning

[rose coral]

So there is this inner product <x, y> that you are not given

[pastel kettle]

They not given

[rose coral]

Yeah

But you can find what it is

In Theorem 2.1.13 the x and y are just variables

[pastel kettle]

Oh is it $(1,2)^T$

[rose coral]

No

[stoic python]

im_molee:

[rose coral]

You don't want to compute the inner product of a specific pair of vectors just yet

You just want to know the formula

[pastel kettle]

Oh yes

Truee

So far i know $llxll^2 = <x,x>$

[stoic python]

im_molee:

[pastel kettle]

Am i right?

[rose coral]

Yes

But you also know that <x, y> = ||x + y||/4 - ||x - y||/4

If x and y are any two vectors

So use the formula you are given for the norm to compute ||x + y|| and ||x - y||

[pastel kettle]

So do i need to compute <1,2>?

[rose coral]

No

[pastel kettle]

Ugh

Just with letters then?

[rose coral]

Also <1, 2> doesn't make too much sense because you need to input vectors into < , >

Yeah with letters

[pastel kettle]

So there, $x = x_1$ and $y=x_2$?

[stoic python]

im_molee:

[rose coral]

No

[pastel kettle]

Oh yeah bold x is vector

Y= (y1,y2)

[rose coral]

So you'll have two vectors x and y, and you'll get a formula for <x, y> in terms of x1, x2, y1, y2

[pastel kettle]

Yes

[rose coral]

Yeah exactly

[pastel kettle]

So i need to right down the root thing as in terms of x and y

[rose coral]

Yeah

[pastel kettle]

$(x_1 - x_2)^2 + 2(x_2)^2 + (y_1 -y_2)^2 + 2(y_2)^2$

[stoic python]

im_molee:

[pastel kettle]

This??

[rose coral]

Not exactly

[pastel kettle]

Times 1/4

[rose coral]

That would be ||x||^2 + ||y||^2

You want to look at ||x + y||, so in the root formula you'll get something like $\sqrt{\Big((x_1 + y_1 )- (x_2 + y_2)\Big)^2 + 2(x_2 + y_2)^2}$

[pastel kettle]

Ohh

And we can also change + to minus

[rose coral]

You want to look at $||x + y||$, so in the root formula you'll get something like $\sqrt{\Big((x_1 + y_1 )- (x_2 + y_2)\Big)^2 + 2(x_2 + y_2)^2}$

[stoic python]

The_Vman:

[rose coral]

Yeah for the other term you'll have -y1 and -y2 instead of the y1 and y2

But the root goes away like you wrote above

[pastel kettle]

And we times 1/4

[rose coral]

Yes

[pastel kettle]

So if we do that we get <x,y>

[rose coral]

Yeah that gives <x, y>

[pastel kettle]

Then we move to Projection formula?

[rose coral]

Yeah, you look at the linear map which is the projection

I would hope that the formula for <x, y> simplifies some amount

[pastel kettle]

I am writing down

Waitt

Oh no

Made mistake

[rose coral]

I think you are missing some brackets

[rose coral]

Yeah that looks good

[pastel kettle]

I will try to substitute them to A B c d

[rose coral]

Be careful, you got rid of some minus signs

[pastel kettle]

Is it correct??

Ohh

The second term

Last term

[rose coral]

You should have (...)^2 - (...)^2 + 2(...)^2 - 2(...)^2

Yeah

[pastel kettle]

Shall i subsitute

[rose coral]

I think it continues to simplify

[pastel kettle]

Oh with squared term?

[rose coral]

Yeah

[pastel kettle]

Hmm

Let me think

[rose coral]

I think you should maybe deal with the terms in pairs instead of substituting for them

[pastel kettle]

Hmm

[rose coral]

For example, the last two terms simplify to $8x_2y_2$

[stoic python]

The_Vman:

[rose coral]

There should also be a lot of canceling in the first two terms

[pastel kettle]

How did u get 8

Oh expand

[rose coral]

Yeah

Every other term in those last two should cancel

[pastel kettle]

Shall i just expand all

[hexed mural]

Is anyone familiar with Linear Combination of Vector Spaces?

[pastel kettle]

(X1+y1)^2 -2(..)

[rose coral]

Yeah I think you should expand it all

Is anyone familiar with Linear Combination of Vector Spaces?
@hexed mural Sorry we are still using this chat, but you can ask your question in one of the other chats and I'll help you as soon as I can

[hexed mural]

Sounds good

I'll wait

[rose coral]

Or someone else will help you

👍

[hexed mural]

it's cool I've been kinda waiting all day for someone with some familiarity

[pastel kettle]

So just open all brackets

[rose coral]

Yeah I think so

[pastel kettle]

Damn give me a sec

[rose coral]

It looks like the majority of the terms will cancel

it's cool I've been kinda waiting all day for someone with some familiarity
@hexed mural If you ask your question in #help-5 I possibly help you in between helping im_molee

[hexed mural]

No worries I can wait

[rose coral]

Ok

[rose coral]

You should have some minus signs

I think you forgot that the second term had minus signs

[pastel kettle]

Yeah its kinda cancelling out

-2 there!

[rose coral]

Yes but the $x_1^2$ should cancel

[stoic python]

The_Vman:

[pastel kettle]

Hmm i dont see posibility hmm

Ugh

Oh wait

[rose coral]

Ok so on the second to last line

[pastel kettle]

Right??

That whole minus is missing

[rose coral]

Where you were going to substitute c and d

in front there should have been a minus

[pastel kettle]

Yess boss

[rose coral]

That should let you cancel the $x_1^2$ at some point

[stoic python]

The_Vman:

[rose coral]

In the end all of the terms should be mixed like the $8x_2y_2$, nothing squared

[stoic python]

The_Vman:

[pastel kettle]

Okay still doing itt

[rose coral]

Cool

[pastel kettle]

I got $2x_2y_2$

[stoic python]

im_molee:

[pastel kettle]

After times 1/4

[rose coral]

Yeah that's good

[pastel kettle]

$<x,y> = 2x_2y_2$

[stoic python]

im_molee:

[rose coral]

Wait you didn't get any other terms?

[pastel kettle]

No i didnt

So i calculated is (1)

[rose coral]

Ok, I got $<x, y> = x_1y_1 - x_1y_2 - x_2y_1 + 3x_2y_2$

[stoic python]

The_Vman:

[rose coral]

Yeah, also using (1)

[pastel kettle]

Last 3 lines

[rose coral]

Wait you got exactly the same thing I did

[pastel kettle]

Lol

[rose coral]

Lol

You just forgot them after dividing by 4

Anyways

[pastel kettle]

Hmm?

Oh

Wait no

Oh i see

[rose coral]

The terms $\frac{1}{4}(4x_1y_1 - 4x_1y_2 + \cdots)$ you wrote down were correct

[stoic python]

The_Vman:

[pastel kettle]

I got confused with the 1 and 2

And cancelled out

[rose coral]

Ah cool ok

So you get $<x, y> = x_1y_1 - x_1y_2 - x_2y_1 + 3x_2y_2$

[stoic python]

The_Vman:

[pastel kettle]

Yess

[rose coral]

Anyways, now we can compute $P_E$

[stoic python]

The_Vman:

[pastel kettle]

So what we got is the top part of fraction

[rose coral]

Yeah now we'll plug in values

So $P_E$ is a linear map, and we need to find the matrix associated with it

[stoic python]

The_Vman:

[rose coral]

Do you know how to do that?

[pastel kettle]

Wait let me try

Is our xj

1,2

[rose coral]

Yes

We only have one vector

[rose coral]

Well actually we need to make sure (1, 2) has norm 1

[rose coral]

That's almost right

[pastel kettle]

Can u see

[rose coral]

Sorry, I got confused with the norm 1 thing

[rose coral]

Yeah, but now there is only one variable

Ok, first for the denominator you got 5? I got 9

[pastel kettle]

(1,2) so 1+5

1+4

[rose coral]

But we are not using that norm

[pastel kettle]

Ohhh

Yes

Ur right

1/9

[rose coral]

Awesome, and then on the numerator we have $<x, (1, 2)^T>$

[stoic python]

The_Vman:

[rose coral]

So we use $y_1 = 1$ and $y_2 = 2$ in the formula we found

[stoic python]

The_Vman:

[pastel kettle]

What do you mean by we have nummerator of that

[rose coral]

Because the expression is $\frac{<x, x_j>}{||x_j||^2}x_j$ right?

[stoic python]

The_Vman:

[pastel kettle]

Oh yes

Oh so yes y1 and y2 will be 1 and 2

[rose coral]

Which simplifies to $\frac{1}{9}\begin{bmatrix}-x_1 + 5x_2 \ -2x_1 + 10x_2\end{bmatrix}$

[stoic python]

The_Vman:

[pastel kettle]

Oh fuck am i dumb why didnt i simplify

[rose coral]

It's ok

[pastel kettle]

So what does this tell us? The Pex = to that

[rose coral]

Yes

So what is the matrix associated with $P_E$?

[stoic python]

The_Vman:

[pastel kettle]

It gives me (-1,5)^t

Hmm confusing

1,2?

[rose coral]

You are looking for a, b, c, d so that $\begin{bmatrix}a & b \ c & d\end{bmatrix}\begin{bmatrix}x_1 \ x_2\end{bmatrix}$

[stoic python]

The_Vman:

[rose coral]

is Pex

[pastel kettle]

Oh can u do voice channel?

If u dont mind??

I think last bit is most confusing

[rose coral]

Sure give me a second

[pastel kettle]

Aite

Do u hear me haha

[rose coral]

Yes, let me check my settings

[pastel kettle]

Aite

[rose coral]

This is $\frac{1}{9}\begin{bmatrix}-1 & 5 \ -2 & 10\end{bmatrix}\begin{bmatrix}x_1 \ x_2\end{bmatrix} = \begin{bmatrix}-1/9 & 5/9 \ -2/9 & 10/9\end{bmatrix}\mathbf{x}$

[stoic python]

The_Vman:

[native rampart]

Just find the inner product and check

[rose coral]

$\begin{bmatrix}-1/9 & 5/9 \ -2/9 & 10/9\end{bmatrix}$

[stoic python]

The_Vman:

[rose coral]

\begin{bmatrix}1 & -5 \ 2 & -10\end{bmatrix}

[stoic python]

The_Vman:
Compile Error! Click the reaction for details. (You may edit your message)

[rose coral]

$\begin{bmatrix}1 & -5 \ 0 & 0\end{bmatrix}$

[stoic python]

The_Vman:

[rose coral]

$\begin{bmatrix}1 & -5 \ 0 & 0\end{bmatrix}\begin{bmatrix}x_1 \ x_2\end{bmatrix} = \begin{bmatrix}0 \ 0\end{bmatrix}$

[stoic python]

The_Vman:

[rose coral]

$\begin{bmatrix}x_1 - 5x_2 \ 0 \end{bmatrix} = \begin{bmatrix}0 \ 0\end{bmatrix}$

[stoic python]

The_Vman:

[rose coral]

So $x_1 = 5$ and $x_2=1$ works

[stoic python]

The_Vman:

[pastel kettle]

Thabk u so muchhh

[rose coral]

No problem!

@hexed mural Hi, I finished helping out im_molee if you want to ask your question now

[hexed mural]

alright I hope it's not something out your expertise, I've spent some time reading and brushing up

The question is as follows

[rose coral]

Ok I hope I can answer it too

[hexed mural]

So I hope it's not a bother

http://www.peroxide.dk/papers/collision/collision.pdf

in this document page 7-11

What I am interested in his how he derives a Vector Space as a Linear Combination

for the ellipsoid

[rose coral]

Ok

[hexed mural]

then if if we have time I was also wondering how he parameterizes it

along the velocity vector

[rose coral]

Ok I'm kinda getting the ellipsoid thing

[hexed mural]

what I think

is he does a linear combination

of the Radius vector

(1,1,1)

and (x,y,z)

then creates that matrix

CBM

which he uses to convert location velocity

then parameterizes it

that's what I understand but the math is abit above me

[rose coral]

Ok so personally I believe the "radius vector" doesn't make much sense

But I think I know what he means

[hexed mural]

that's what I thought too

LOL

he's implying that

[rose coral]

Really you want a set of three vectors to define the ellipsoid

[hexed mural]

the (x,y,z) vector is (1,1,1) in elipse space

I think that's what he's saying

[rose coral]

Well more than that I think

He's saying that if you take the unit sphere

And then stretch by X in the x direction, by Y in the y direction, and Z in the z direction, then you get the ellipsoid he is describing

[hexed mural]

That's a bit confusing

but that stretching has to be that of an ellispse

the values

for X,Y,Z

[rose coral]

Sorry I don't get which part you are confused by

[hexed mural]

so

[rose coral]

Are you confused at how stretching like that gives an ellipsoid?

[hexed mural]

yea

that's some transform

right?

applied to the unit sphere to give the Ellipse

[rose coral]

Yeah

[hexed mural]

is that transform

CBM?

[rose coral]

Not quite

So I was just describing how you can get an ellipsoid

CBM is more like a change in measurement

So is this for like programming a game?

[hexed mural]

when you say change in measurement you mean from one Vector Space to Another

Yea

I am learning

how collison works

[rose coral]

Not exactly from one vector space to another

[hexed mural]

been digging deep into it for a while

[rose coral]

That's cool!

Ok so

Lets imagine this ellipsoid exists in the game world

Not sure that's how its used but

Let's just go with it for now

[hexed mural]

The ellipsoid is used as a collision detection

it's sweeps across a velocity vector

then once detection is detected you do some collision resolution

anyway as you were saying

[rose coral]

Ok, but the ellipsoid is like "physically" in the game world, even if is just used to detect something

So in the game world you've conveniently labelled the points in space with 3 coordinates

Right?

Each point in the space is has like some x, y, and z components

[hexed mural]

The object will have a transform

yea

World coordinates

[rose coral]

Cool

But the point is that if you want to talk about an ellipsoid, or maybe the camera is suddenly rotating, you might want more convenient coordinates than the world coordinates

[hexed mural]

I don't think there should be any rotation

[rose coral]

You want to represent the same points in the "physical" space in game, but maybe while running some code it's simpler if you changed how you label the points

[hexed mural]

for the collison

[rose coral]

Yeah in this case there might not be

But if you need to pick different coordinates due to rotation, the same concepts apply

[hexed mural]

you mean to go from two different coordinate systems right

[rose coral]

My point is just that we want to talk about the same space but with different coordinates yeah

[hexed mural]

for now we can just care about it as something in world space

We can just care about Object to World Collision

So an object moving in world space has its world coordinates

and handling collision with walls

floors ceilings

etc

[rose coral]

That makes sense

And so the author of this thinks that to do the collision detection its easier to use "ellipsoid" coordinates instead of the world coordinates which are probably pretty arbitrary when it comes to collision detection

[hexed mural]

yea

I mean

I don't understand why he does this in the first place

Ellipsoid space

I am curious if this is how all Primitives are done

so

what I mean as primitives

are different types of geometry styles

you can have

Capsule, Box, Ellipsoid, Sphere

but at the end of the day

you're gonna have a plane touching some other plane

[rose coral]

Oh cool

[hexed mural]

if this interests you

Here was something else I was trying to figure out

it might be a bit simpler than what i asked earlier

ok so in this one

this is a top down view

so let's suppose we have a collision detection in a BOX we check 4 corners

once we have a collision detected

we find the path shortest from the center to the wall center and compute a slide location

what this is doing is sliding the object away from the wall

[rose coral]

Right

[hexed mural]

the question I am curious about is how they compute that Slide Location

apparently

it's done something like this

https://www.youtube.com/watch?v=oom6R-M2lvQ

[rose coral]

I would imagine its that you decompose the vector into components

[hexed mural]

how so

how is that equation derived for DP

[rose coral]

DP is?

[hexed mural]

delta position

in the previous image

its the start -> goal

[rose coral]

Oh so how you would have moved if there were no wall

[hexed mural]

yea how they get that goal point I think is the shortest distance

to the center of the wall

from the center of your object

As soon as collision is detected

this stuff is not easy btw

so don't be discouraged haha

[rose coral]

Wait don't you start out with the goal position?

[hexed mural]

no

the goal i think is computed

[rose coral]

Then what information do you start with?

[hexed mural]

Delta

[rose coral]

And the start position

[hexed mural]

which will be your

position yea

then you also

know

the Normal for that wall

[rose coral]

Oh ok then it shouldn't be bad

[hexed mural]

i assume that subtraction

is between two vectors

A + -B

head to tail

to get the slide vector

I think

or it's a projection

[rose coral]

That would make sense

Ok so the goal position should be start + delta

[hexed mural]

yea

[rose coral]

And if you never detect a collision that's where you end up

[hexed mural]

yes

[rose coral]

Now

Say you detect a collision at some intermediate point C

for collision location

[hexed mural]

yea

box collison

so one of the corners

[rose coral]

Yeah

And let, idk, r be the amount you would have continued going to the goal position if for no wall

Sorry r is a vector, some amount of the delta that's left over

[hexed mural]

I think

collision location is

halfway

always

wait no

Goal = Delta + Start where

[rose coral]

If $n$ is a unit normal to the wall, then $(n \cdot r)n$ should be the component of $r$ that is going straight into the wall

[stoic python]

The_Vman:

[rose coral]

So you should slide according to $r - (n \cdot r)n$

[stoic python]

The_Vman:

[hexed mural]

yes

how did you find that

[rose coral]

So the dot product lets you project vectors

[hexed mural]

that equation

would be the slide vector

you solved for it

assuming it's Unknown right

you projected the Start to Goal Vector with the normal right?

[rose coral]

Not exactly, only the part that was left over from the collision detection

So however much you still have left to move

[hexed mural]

you mean collision to goal

[rose coral]

Yeah

[hexed mural]

so that's why you subracted R

so you'd get the distance from collision to goal

right?

[rose coral]

No

I can write it in terms of delta and the collision location if you want

But it would just be plugging in delta - (collision location - start location) = r

I subtracted because the term $(n \cdot r)n$ corresponds to the component going straight into the wall

[stoic python]

The_Vman:

[rose coral]

In your diagram, you have a vector from collision to goal and from collision to slide location

[hexed mural]

Yeap

[rose coral]

The $(n \cdot r)n$ represents the other side in that right triangle

[stoic python]

The_Vman:

[rose coral]

Pointing straight into the wall from the slide location to the goal

Because it is some scalar (n dot r), times the vector n whose direction is straight out of/into the wall

[hexed mural]

ah

[rose coral]

The scalar just makes sure its the right length

[hexed mural]

yeah ok we scale down it down with dot product yes of cource

N is the Normal

out the wall

from goal

[rose coral]

Yeah

For this thing to work it should be length 1, otherwise you need an extra factor

[hexed mural]

computing in my head

one second

[rose coral]

sure

[hexed mural]

k

did you

draw a right triangle

from start point

with goal normal

?

to find r

like a larger right triangle

[rose coral]

To find r, I just took the goal position - collision position

[hexed mural]

So R is the vector from the center of the object to the goal positon

just to be sure

we are on the same page

right?

[rose coral]

If the object is like beside the wall at the collision position yes

Not at the start position

[hexed mural]

ah

so you mean

r = Vector from Start -> Collision location

[rose coral]

No, from collision to goal

[hexed mural]

k

ok

so it's as follows

1. Collision Detected with World Object.. a wall.
2. Find R vector from Collision Location to Goal with the Delta
3. Take the Dot Product with R and the Normal of the Goal Position of the Wall

4. Subtract R - Result of Part 3. to Get the slide component Vector

[rose coral]

Yeah exactly

[hexed mural]

the two vectors

R->N

it's head to tail

[rose coral]

Well, in part 3 you get a number

[hexed mural]

so you have to subtract

[rose coral]

You need to subtract N, the normal, times the number from part 3

[hexed mural]

yea the Normal is unit vector

it's what's left over

from the projection

[rose coral]

Right but the dot product gives a single number

[hexed mural]

yeah

a scalar

[rose coral]

You want to subtract a vector

So I'm saying that you subtract the normal scaled by that number from the dot product

Which is a vector

[hexed mural]

yeah

[rose coral]

Ok cool

[hexed mural]

this will give us the final point

[rose coral]

Yeah

[hexed mural]

some point

yea

that's cool

is there a way to get that slide vector another way without doing a subtraction

or you have to do vector subtraction

[rose coral]

I don't know of a way

[hexed mural]

man that is wild

makes sense

[rose coral]

Yup, vectors