linear-algebra

[haughty berry]

did i lol

[lavish jewel]

yes

you mean A(5A + 3I) = 4I

[haughty berry]

Yeah

[lavish jewel]

you're still doing something wrong

[haughty berry]

Yeah forgot to take out the square

and the A

ugh sorry

[lavish jewel]

looks better

[haughty berry]

My bad

I should not be making mistakes like this, sorry lol

[lavish jewel]

don't sweat it, was just pointing it out 😛 everyone makes typos

[subtle gust]

From 2nd to 3rd step

We multiplied both sides by the inverse

And took another A as a common factor

[haughty berry]

idk 4 typos in a text so short... 😅

[quaint pond]

What’s up guys

[haughty berry]

yeah

[quaint pond]

I’m learning linear algebra at college this semester

I was wondering if this holds true

I was just thinking about it

R^2 to R^3 transformation can never be onto right?

[haughty berry]

yep

[quaint pond]

Because we are increasing the row by one

[subtle gust]

I'm still confused :)

What did we fo from 3rd step to last

[haughty berry]

Multiply by A^-1 and divide by 4

[subtle gust]

Oh

[haughty berry]

idk what you mean by this, but a linear transformation from V to W where dim V < dim W cannot be surjective in general
the proof is pretty simple. Suppose {v1...vn} is a basis of V, then {Tv1...Tvn} is a spanning set of ImT. Since n<dim W, this cannot be a basis, so dim Im T != dim W meaning Im T doesn't equal W.

A similar thing holds for transformations between W and V. If dim W > dim V, then T can never be injective (the proof is similar)

[lavish jewel]

i'd just be a little more careful with calling {Tv_i} a basis a priori

spanning set, sure

[haughty berry]

oh yeah my b

[grave garden]

Hiii guys

Why $\dim S_{**} = n-1$ ?

[stoic python]

Potato

[zealous flame]

Is there a way to represent λ_max as a function of a symmetric matrix S?
λ_max is the maximum eigen value of the matrix S.

[zinc timber]

what a weird notation

[grave garden]

[zinc timber]

like u want a notation for λ_max?

[zealous flame]

Actually, I am trying to do b part here

https://cdn.discordapp.com/attachments/423244559682764800/946071773730992159/Screenshot_20220223_212118.png

[zinc timber]

not like there's an explicit function where you can put in S and expect to get the λ_max

$\lambda_{\text{max}} = \sup_{x\neq 0} \frac{x^TSx}{x^Tx}$

[zinc timber]

called the Rayleigh quotient

idk if that can help

[zinc timber]

what's S**?

[grave garden]

The set of solution $X\cdot A=0$

[stoic python]

Potato

[zinc timber]

,tex \begin{gather*}
X\cdot A = P\cdot A \
\implies (X-P)\cdot A = 0 \
\implies X-P \perp A
\end{gather*}

[zinc timber]

then dim S** is the dim of the orthogonal subspace of A

which has dim = n-1 (if A is non zero)

[grave garden]

You use $\dim A+\dim A^\perp=n$ right ?

[stoic python]

Potato

[zinc timber]

yes

[grave garden]

I see

Thanks mate !

[spare widget]

The set of solutions of Ax = 0 is made from the vectors that A sends to 0, i.e. the kernel

Idk what the X dot A is supposed to be though

[zinc timber]

(although technically A is a vector so dim A is non sensical, use ⟨A⟩ instead )

[spare widget]

I thought A is a matrix here, npw X dot A makes sense

[grave garden]

⟨A⟩, is this inner product or sth ?

[spare widget]

who uses capital letters for vectors though

I though it was a matrix

[zinc timber]

ya bad notation grr

lmao

[zinc timber]

no subspace generated by A, span{A}

[grave garden]

I see

( This book if from 1987 or sth, people usually say it used a bad notation, but hmmm this book is easy to read for me )

[zinc timber]

small letters weren't invented back then

[grave garden]

[spare widget]

What's the book

[zinc timber]

[grave garden]

Serge Lang, linear algebra

[zinc timber]

it's Lang's book

[grave garden]

I love his book

[compact tartan]

just a quick subspaces question, If the vector space is all continuous functions on the reals,

would the class of functions 2f(x)=f(2x) be a subspace

I say yes because

2f(0)=f(2*0)=0 so 0 is contained
(f+g)(2x)=f(2x)+g(2x)=2f(x)+2g(x)=2(f+g)(x) so addition is closed,
a(f(2x))=a(2f(x))=2(af(x)) so scalar multiplication is closed

but im not sure

[zinc timber]

like you already proved it and still not sure

[compact tartan]

5 ppl in my group disagreed w me

😭

ty!!!

[zinc timber]

reason for their disagreement?

[compact tartan]

said scalar multiplication wasnt actually closed

told me to try a=1/2

which still works

so all dubs today

[zinc timber]

[lusty falcon]

Hi

guys, could someone help me with this exercise

Point 3

[zinc timber]

translate

[outer goblet]

i define {x,y} as V and the other one as W

and then have to show that V is a part of W and vice versa

but im kinda stuck on how to show that W is a part of V?

any help?

[spare widget]

Demonstrate that the eigenvalues of alphaA are ...

[night wren]

Can you always decompose a codomain into the image of a linear transformation direct sum some other subspace? i.e. T(V)+W'

[spare widget]

write the eigendecomposition for A

[spare widget]

The complement

[zinc timber]

not needed also may not be possible

[spare widget]

I am assuming he has something additional not in the picture

Because they are asking him to prove alpha lambda_1, ...

[zinc timber]

try to show that αA-αI is singular

no I mean there may not even be enough eigen values to write an eigen value decomposition

[outer goblet]

can some1 help me

[wintry steppe]

"V is a part of W" is just not true, unless you're referring to their spans. but proving that isn't enough to show that the second set is also a basis

[spare widget]

So a Jordan normal form, I assumed A not being defective yeah

[outer goblet]

is this not what symbol C means

that long c

like this is the solution but i dont get it

[wintry steppe]

this shows they span the same space. so your new set spans. maybe a word or two on why it's linearly independent is in order

remember a basis is a spanning set that's linearly independent

[outer goblet]

ah so long c meants span sry i forgot

but still how do those two euqation prove that they sapc

span

[lavish jewel]

it means subset

[wintry steppe]

C means the set of complex numbers.

oh, "long" C

[outer goblet]

no not this

yeah

[lavish jewel]

$\subset$

[wintry steppe]

{x, y} is a basis, so its span is the entire space

[stoic python]

Edd

[outer goblet]

yeah i get this one

[outer goblet]

but not the other way around

[zinc timber]

lol long C

[outer goblet]

🙃

but idk how

means that V is a subset of W

[wintry steppe]

these equalities show that a linear combination of x and y (so an element of V) is also a linear combination of 1/2(x+y) and 1/2i(x-y) (so an element of W)

[outer goblet]

ah so bascially i can use scalar multiplication and addition on the other vectors to go to the first ones

ait thx

[spare widget]

I am trying to interpret a system that I get from the KKT conditions

$\begin{bmatrix} A & B^T & C^T \ B & 0 & 0 \ C & 0 & 0\end{bmatrix} \begin{bmatrix} \boldsymbol{x} \ \boldsymbol{\lambda} \ \boldsymbol{\mu}\end{bmatrix} = \begin{bmatrix} \boldsymbol{0} \ \boldsymbol{b} \ \boldsymbol{c}\end{bmatrix}$

[stoic python]

criver

[spare widget]

Where A is symmetric

As I understand the above Bx =b and Cx = c are enforced exactly, while Ax = 0 (what would have been the original equation) is enforced weakly

I am trying to understand how this would compare to doing something like

$\min_{\boldsymbol{x}}|A\boldsymbol{x}|^2, \quad \text{s.t.} , B\boldsymbol{x} = \boldsymbol{b} ,, C \boldsymbol{x} = \boldsymbol{c}$

[stoic python]

criver

[spare widget]

Would this result in the same solution?

Note that even though I wrote mu, both constraints are equality ones

i.e.Bx = b, Cx = c

And I know that thdy are linearly indepebdent so KKT is applicable

[zinc timber]

I don't under how you get the matrix

[spare widget]

I have an original energy

[zinc timber]

i get A^TA instead of A

[spare widget]

Yes, let me elaborate

The original formulation is:

[zinc timber]

ooh

[spare widget]

$\min E(x) \implies \nabla E = Ax = 0$

[stoic python]

criver

[spare widget]

Now I add constraints Bx = b, and Cx = c and add them to the energy

[spare widget]

So then I get this

Now the question is, is this the same as solving

[somber loom]

If a linear transformation is 1:1, is it automatically onto?

[spare widget]

$\min_{\boldsymbol{x}} |A\boldsymbol{x}|^2, \quad \text{s.t.} , Bx=b, , C x = c$

[zinc timber]

that's what I was saying actually that if you go from the optimization to the matrix one you get A^tA instead of A

in that matrix

[stoic python]

criver

[zinc timber]

no, if they have same dim and are finite then yes

[somber loom]

Thank you

[zinc timber]

I assume A is +ve definite

?

[spare widget]

yes symmetric positive definite

[zinc timber]

because |Ax|²=x'A'Ax and grad is 2A'Ax

[spare widget]

yes the grad is A^TAx

[zinc timber]

so you can use cholesky decomposition

[spare widget]

I am trying ti understand whether the kkt system gives the same solution

[zinc timber]

$A=L^TL$ then $\min_{x}\norm{Lx}^2$

[zinc timber]

probably no

[spare widget]

in both Bx = b and Cx = c are enforced explicitly

[zinc timber]

what are the conditions you get from here?

[spare widget]

in the L2 formulation one finds the x such that Ax is closestto 0

[spare widget]

That I can solve explicitly

I can enforce them explicitly

Think removing dimensions of x

That get fully determined by these

[zinc timber]

no I mean to say that you can check if both of them give the same optimization problem

[spare widget]

well one involves Ax + B^T\lambda+ C^T\mu = 0 while the other A^TAx = 0

[zinc timber]

I'm kinda rusty on my diffs but 2A'Ax+ λ B'+ μ C' =0 is one comdition

[spare widget]

So one makes up for Ax!=0 through the lagrange multipliers

[lavish jewel]

hmm?

[spare widget]

While the other makes up for the strong constraints through an L2 minimisation

[zinc timber]

idk I'm not seeing how those two are eqv

@lavish jewel

[spare widget]

Bx = b, Cx = c basically fix some parts of x

[lavish jewel]

the what

[spare widget]

Then we want to get Ax = 0, but this is impossible with the constraints, xo the lagrange multiplier approach just gives: Ax +B^T\lambda + C^T\mu = 0

So lambda and mu supposedly can make up for Ax != 0

[lavish jewel]

what's the original problem?

because the way you're saying it it just sounds like the feasible set is empty

[spare widget]

$\min_{\boldsymbol{x}}E(\boldsymbol{x}) \implies \nabla E = A\boldsymbol{x} = \boldsymbol{0}$

[stoic python]

criver

[spare widget]

Thebn I add extra constraints

Bx =b, Cx = c that I want to enforce strongly

[lavish jewel]

then yes, one need not have Ax = 0

[spare widget]

Yes it's essentially overdetermined at this point

But the KKT approach handlex this through

Ax + B^T\lambda + C^T\mu = 0

[lavish jewel]

mhm

[spare widget]

While an L2 approach would habfle it through

$\min_{\boldsymbol{x}}|A\boldsymbol{x}|$

Under the aforementioned constraints Bx = b, Cx= c

[stoic python]

criver

[spare widget]

So I am trying to figure out what the difference is

Because it's clear that both enforce Ax = 0 weakly

[lavish jewel]

which two?

[spare widget]

kkt vs the L2

[lavish jewel]

also the kkt one is E(x) + B^T lambda + C^T mu

[spare widget]

nabla E(x) is Ax

[lavish jewel]

i know

the loss function in kkt includes the original loss, not its grad

you then differentiate all the terms

not just E(x)

[spare widget]

Yes what I wrote is nabla Lagrange = 0

That's what the Ax +B^T\lambda + C^T\mu = 0is

[lavish jewel]

oof yes, sorry, i forgot what the original constraints were

ok

[spare widget]

question is is there actually a difference

Both enforce Ax = 0 weakly

So I would expect both to result in the same solution

A, B,C are linear operators here

[lavish jewel]

can you write the full L2 problem you're talking about

you mean minimize the 2-norm of Ax with the given constraints?

[spare widget]

It's a discretisation of $\min_x\int|\nabla x|^2$ leading to $\Delta x = 0$, so $A$ is a discretisation of $\Delta$

The strong constraints are Dirichlet data

[stoic python]

criver

[spare widget]

That's the full problem

[lavish jewel]

i meant more generally, it's just you keep saying kkt and "the L2"

i don't need to know the full problem, just what you meant by L2 in that context

because you keep throwing E(x) and Ax around and idk which one you mean when

[spare widget]

Then then kkt results in the system I mentioned, while the alternative being the Lw thing

The original is basically

[lavish jewel]

i guess i don't mean the original either

you said the kkt is Ax + B^T ... = 0

what's the other

[spare widget]

the other is

$\min_x|Ax|^2, \quad \text{s.t.} ,Bx = b,, Cx= c$

[stoic python]

criver

[spare widget]

The original being

$\min_x E(x) = \min_x||\sum_iD_ix||^2, \quad \text{s.t.} Bx = b, Cx =c, , \nabla E(x) = Ax = 0$

[stoic python]

criver

[spare widget]

The kkt minimize the original

Whkle the other thing is

$\min_x |\nabla E(x)|^2 = \min_x|Ax|^2, \quad Bx =b, , Cx = c$

[lavish jewel]

the $\Vert Ax\Vert_2^2$ has as a first term $2A^TAx$

[stoic python]

Edd

[lavish jewel]

the original has as first term $Ax$

[stoic python]

Edd

[lavish jewel]

the other terms are the same

[stoic python]

criver

[lavish jewel]

in kkt form

now, regarding what that means for the minimizer

the question is whether 2A^TA x has the same minimizer as Ax

what do you know about A's rank

[spare widget]

under the constraints Bx =b, Cx = c

A has 1 zero eigenvalue, but is symmetric positive semi-definite, the constraints Bx=b, Cx=c guarantee that there is a unique solution

The logic behind the |Ax| minimisation

[lavish jewel]

i don't think in general these have the same minimizer

[spare widget]

Is that in both cases we want to enforce Ax= 0 weakly

It's clear that the latter enforces this in an L2 sense, but in what sense does kkt enforce this?

It'sstill in a weak sense, but is it L2 or something else

[lavish jewel]

i honestly don't see the Ax minimization here, not with the extra constraints

[spare widget]

what do you mean?

[lavish jewel]

i don't see why minimizing Ax should work if you have extra constraints

[spare widget]

I still don't get what you mean

[lavish jewel]

in kkt, you are minimizing the 2-norm of the whole thing

Ax + C^Tmu + B^T lambda, or whatever the order was

you can minimize the two norm of all of this together to look for a saddle point of the lagrangian

[spare widget]

that's the gradient of the lagrangian

The Ax + C^Tmu + B^T lambda isthe gradient of the lagrangian

[lavish jewel]

yes

and that's what kkt cares about

[spare widget]

I guess in the l2 case I would instead get

A^TAx +C^Tmu + B^T lambda = 0

[lavish jewel]

2A^TAx, but yes

that's what i said before

[wintry steppe]

does this line go thru both x0 and x1?

[lavish jewel]

yes, xoxo

[wintry steppe]

how do u know that though?

[lavish jewel]

substitute t = 0 and t = 1

[wintry steppe]

but the bounds arent known right?

[spare widget]

R

[wintry steppe]

like patr a of the problem was

[spare widget]

t is in R

[wintry steppe]

how does that affect it though?

[spare widget]

I think I am starting to grasp it. Generally A^TAx = A^Tb would have the same solution as Ax = b for a consistent system with a nonsingular matrix, but here I have the extra constraint terms, which would make this different

[lavish jewel]

it can be ANY real number

right

[spare widget]

You can pick any t in (-infinity, infinity)

And since you get x0 for t =0 and x1 for t =1 the line passes through those

[lavish jewel]

still, rn i'm tired and i can't think of a way to prove it, criver. but you would have to show the solutions to Ax +C^Tmu + B^T lambda = 0 and 2A^TAx +C^Tmu + B^T lambda = 0 are the same

[wintry steppe]

what if u had x_0 + tx_1

[lavish jewel]

then it doesn't pass thru x_1 in general

[spare widget]

Yeah, I am not sure they are at this point, one minimises E and the other |\nabla E|

[wintry steppe]

if t was 1 wouldnt it?

[spare widget]

Then it's x_0 +x_1

If there is a t for which it is x_1

Then x_0 + t * x_1 = x_1

this does not have to hold though

It's an overdetermined system in anything higher than 1d

[lavish jewel]

this would boil down to x_0 = (1-t)x_1 -> x_0 = cx_1, meaning the two vectors are actually scaled versions of each other, xoxo

[spare widget]

And even then a 0 singularity

[lavish jewel]

so whenever x0 and x1 are not linearly dependent, x0 + tx1 does not pass through x1

[spare widget]

You'd basically need to be able to subtract out x0, which is what Edd mentions with the two being collinear

Either way, thank you for the help. That did help. I finally realized that at least in the general case those minimize L2, but wrt a different quantity. And even if the two minima are equal without constraints, that shouldn't necessarily be the case in general, since the level lines of E and |\nabla E| do not have to agree.

[lavish jewel]

without constraints already Ax and A^TAx have different minima

A^TAx has minima at x = 0 + nullspace component

while Ax goes off to -inf

[spare widget]

Ax = 0, A^TAx =0 would have the same minimum I think for A non-singular

[wintry steppe]

ok cool so x_0 + tv would be through x_0 and parallel to v

[spare widget]

The pseudo-inverse matches the inverse for invertible stuff

So they should agree as long as I modify A to make it non-singular (in my case I can do that by incorporating Bx=b into A, then I do know that the modified A is nonsingular, it's the Cx=c constraint that I needed the kkt for)

[wintry steppe]

why is this parallel to v but the other one goes thru x_0 and x_1?

[wintry steppe]

so when x0 and x1 are linearly independent, it isnt going thru x1 but how do u know if they're linearly independent or not?

[lavish jewel]

you test the definition

[wintry steppe]

i think its just when they can equal to zero?

[lavish jewel]

when c1v1 + c2v2 = 0 for nonzero ci

[wintry steppe]

so here , if there is a t such that (1-t)x0 + tx1 = 0

then it wouldnt go thru x1

but idt there is such t so therefore, thats why it goes thru x1

[spare widget]

(1-t) x0 + t x1 = x0 + t (x1-x0)

when does this pass through some point p?

when there is a t such that x0 + t(x1-x0) = p

or

t(x1-x0) = (p-x0)

now if you plug in p = x1, it's clear that t = 1 works

or if p = x0, then it's clear that 0 works

if you plug in p = (x1+x0)/2 -> you get t (x1-x0) = (x1-x0)/2

so t=1/2 works

and so on

in practice

if you have 2 vectors

u and v

and you want to check

t u = v

then you would check that v[i]/u[i] = v[j]/u[j] for all i,j

and the t would be

t = v[i]/u[i]

v[i] denoting the i-th component of v

[wintry steppe]

hmm ok that somewhat makes sense

i feel this is some easy concept that is just completely going over my head haha 😭

[spare widget]

it's membership of a point from a parametric curve

let's say you have the curve

r(t)

and you want to check that the point p is from it

then there must exist a t

such that r(t) = p

for linear stuff this is easy, because you can invert along each coordinate

e.g.

(x0,y0) + t * (x1,y1) = (px, py) <-> t = (px-x0)/x1 = (py-y0)/y1, provided that (x1,y1) != (0,0), if the equalities don't hold, e.g. (px-x0)/x1!=(py-y0)/y1 then the point is not from the line

if (x1,y1) = (0,0) then if (x0,y0) = (px,py) it;s from the set, otherwise not

[wintry steppe]

ah ok that makes a lot of sense

tysm!

[halcyon spindle]

Let M and M' be $m \times n$ and $n \times p$ matrices respectively and let r be a integer less than n. Suppose we have decompose the two matrices into these following blocks $M = [A | B]$ and $M' = [\frac{A'}{B'}]$ where A, A', B, B' are $m \times r, r \times n, m \times (n-r), (n-r) \times p$ matrices respectively. \I am having trouble showing MM' = AA' + BB'.\ \ Letting $w_{ij}$, $w'{ij}$, $p{ij}$, and $s_{ij}$ denote the entries for M, M', MM', and AA' + BB' respectively. We have \ $p_{ij} = \sum_{v=1}^{n}w_{iv}w'{vj}$, \ $s{ij} = \sum_{v=1}^{r}a_{iv}a'{vk} + \sum{k = r+1}^{n-r}b_{ik}b'_{kj}$

[stoic python]

Plegasus

[halcyon spindle]

I am now having troubling showing p_{ij} = s_{ij}.

expressing that form p_ij using A, A', B, and B' instead of M and M' is what I am struggling with it right now.

[spare widget]

you could try

[halcyon spindle]

Note I know s_ij = p_ij since that is how I computed s_ij. Any entry p_ij in MM' is just the sum of the corresponding entry in AA' and BB'. But I don't know if that rigorous enough for the proof.

Wait, I did it wrong.

[spare widget]

$M = \begin{bmatrix} C_{11} & C_{12} \ C_{21} & C_{22} \end{bmatrix}$

[stoic python]

criver

[spare widget]

Then split A = [c11; c21], B = [c12; c22]

etc

actually nevermind, you don't need that

I misread your question

what you wrote is trivial afaik

$(MM^T){ij} = (A_i, B_i) \cdot (A_j,B_j) = A_i\cdot A_j + B_i\cdot B_j = (AA^T){ij} + (BB^T)_{ij}$

did I write this out correctly

I am currently multitasking so it's a bit hard

point is element ij is row_i dot col_j

but row_i is two parts and col_j is two parts

then split it into (v1,v2)

I think it should read A_j, B_j in the second

since it's columns of the tranpose

[halcyon spindle]

Sorry not following what your trying to do here but I have the right argument in my head now. Thanks anyway.

[spare widget]

A_i denotes row i of A

$(MM^T){ij} = M_i\cdot M_j = (A_i, B_i) \cdot (A_j,B_j) = A_i\cdot A_j + B_i\cdot Bj = (AA^T){ij} + (BB^T){ij}$

[stoic python]

criver

[chilly hazel]

can someone tell me what the hell I'm looking at in (5)

how did they get that?

what do you even call that

[spare widget]

which part

[chilly hazel]

(5)

why would x = Gamma^-(1/2)z minimize v(x)

and why is z the smalles eigenvector of (5)

how on earth did they get that stuff

[spare widget]

seems like some generalization to

$\inf_x \frac{x^TA^TAx}{x^Tx}$

which you know is minimized for lambda_min * z_min

[stoic python]

criver

[spare widget]

indeed decompose x

as a linear combination of the eigenvectors

$x = \sum_i \alpha_i z_i \implies |Ax|^2 = |A(\sum_i \alpha_i z_i)|^2 = |\sum_i \lambda_i \alpha_i z_i|^2$

[stoic python]

criver

[spare widget]

then if you choose a x that has a component only in the direction z_min

then you get the minimum

the above seems to be something like that

but with respect to an inner product with metric tensor Gamma

so

$\langle u, v\rangle_{\Gamma} = u^T\Gamma v \implies |u|^2 = u^T\Gamma u$

[stoic python]

criver

[spare widget]

ok I think I got it

[chilly hazel]

someone told me to look up "Schur decomposition"

that's what I've been doing so far lol

[spare widget]

I think the idea with the generalized eigenvalue

is that you have A^TGAy = \lambda G

if y is a generalized eigenvector A^TGA wrt to G or something like that

then decompose x

$x = \sum_i \alpha_i y_i \implies \frac{x^TA^T\Gamma Ax}{x^T\Gamma x} = \frac{(\sum_i\alpha_i y_i)^T\Gamma(\sum_i\lambda_i\alpha_i y_i)}{x^T\Gamma x}$

[chilly hazel]

this is the part before that btw

so you got a VAR(1) model with assumption E[S] = 0

[stoic python]

criver

[spare widget]

nvm

it's actually simpler

they do the substitution

$x = \Gamma^{-\frac{1}{2}}y$

[stoic python]

criver

[spare widget]

then rewrite v

[chilly hazel]

[spare widget]

wrt y

[chilly hazel]

this is a little above my head lol

[spare widget]

it's simple you'll see

[chilly hazel]

I managed to get all the way here

but I'm completely stuck now

x is a nx1 matrix of portfolio weights btw

gamma the covariance matrix

[spare widget]

$v(y) = \frac{x^TA^T\Gamma A x}{x^T\Gamma x} = \frac{y^T\Gamma^{-1/2}A^T\Gamma A \Gamma^{-1/2} y}{y^Ty}$

[stoic python]

criver

[spare widget]

you follow until here?

[chilly hazel]

A just some factors

[chilly hazel]

is this "Schur decomposition"

[spare widget]

I just substituted x = Gamma^{-1/2} y

[spare widget]

it is not

but do you follow until this point?

[chilly hazel]

where does that ^(-1/2) come from

ah wait nvm yeah

[spare widget]

it's just a substitution

to get rid of the Gamma in the denominator

[chilly hazel]

mhm

[spare widget]

Now you apply the same trick I mentioned before

Denote

$C = \Gamma^{-1/2}A^T\Gamma A\Gamma^{-1/2}$

[stoic python]

criver

[spare widget]

and let $z_i$ are the eigenvectors of $C$ and $\lambda_i$ be the corresponding eigenvalues:
$Cz_i = \lambda z_i$

[stoic python]

criver

[spare widget]

Now decompose y as a linear combination of the eigenvectors

[chilly hazel]

which one do you mean exactly?

[spare widget]

the idea that

$z_{\min} \in \arg\min_y \frac{y^TCy}{y^Ty}$

[stoic python]

criver

[chilly hazel]

oooooh yeah I got it

if you choose C to be that

finding the eigenvector corresponding to the smallest eigenvalue will minimize v(x)

[spare widget]

yes that'

is the idea

[chilly hazel]

that was a lot simpler than I thought

thanks man!

[spare widget]

the only trick was getting rid of the denominator

which they do through the substitution

idk why you'd need schur here

[chilly hazel]

he said you'd need that to get to the gamma^(-1/2)

[spare widget]

Gamma is symmetric psd

[chilly hazel]

it is

[spare widget]

then \Gamma^{-1/2} is just taking Q^T \Lambda^{-1/2} Q

i.e. do the eigendecomposition and then take the square root and reciprocal

[chilly hazel]

so I got the smallest eigenvector z of C now, let me see if I can figure out why x = \Gamma^(-1/2)z will minimize v(x)

really appreciate the help!

[chilly hazel]

actually, what exactly is y in this case?

just some undefined vector?

[spare widget]

https://math.stackexchange.com/questions/2345363/finding-vector-for-supremum-norm-of-matrix

[chilly hazel]

$v(x) = \frac{z^T\Gamma^{-1/2}A^T\Gamma A \Gamma^{-1/2} z}{z^Tz}$

[stoic python]

Vertox

[spare widget]

x = \Gamma^{-1/2}y

I reserved z for the eigenvectors

[chilly hazel]

yeah it's from that formula but what does it stand for

[spare widget]

you can use the opposite of this, i.e. for the minimum

[spare widget]

it's just a substitution required to get rid of the Gamma in the denominator

[chilly hazel]

okay I why x = Gamma^(-1/2)z minimizes v(x)

denominator cancels out

numerator is minimum

[spare widget]

see this

https://math.stackexchange.com/questions/2345363/finding-vector-for-supremum-norm-of-matrix

[chilly hazel]

so you get the smallest possible v(x)

[spare widget]

and set A = \Gamma^{1/2}A\Gamma{-1/2}

[chilly hazel]

yep got it open

[spare widget]

the A from there corresponds to \Gamma^{1/2}A\Gamma{-1/2}

then you can check that

$|A^{theirs}z|^2 = z^T\Gamma^{-1/2}A^T\Gamma A\Gamma^{-1/2}z$

[stoic python]

criver

[spare widget]

then by the logic there, but applied for the minimum

you get that it is minimized when you choose z_{min}

[chilly hazel]

mhm

that was so much simpler than I imagined

I thought they were doing some absolutely fancy shit

[spare widget]

basically if you pick a vector in the direction that the matrix scales the least

then you minimize the L2

the only trick is the substitution to get rid of the Gamma in the denominator

I learned something new

[snow plinth]

would this be valid

[spare widget]

That's fine yes, though it depends on your tutor whether they'll accept it

You can trivially show T [x, ...] + T[z, ...] = [..., x-1] + [..., z-1] != [..., x + z - 1] = T [x+z,...]

[chilly hazel]

ahhh I get where schur get's into play @spare widget

you use it to actually compute Gamma^{-1/2}

when you implement it

[lapis osprey]

If I have the set {0, 3, 6, 9, 12}

how do I write it in builder set notation?

I know that each time it goes up by 3.

{0, 3, 6, 9, 12}

{n | n ≤ 0}

{2n | n = 1, 2, 3, 4}

{3n | n = 0, 1, 2, 3, 4}

{3n | n ≤ 2}

{3n | n ∉ Z}

[spare widget]

I also noticed that the proof I linked you is only applicable to the Gamma^{1/2}A Gamma^{-1/2} matrix, although you can probably show it for the whole thing, but not in the way using the derivative and lagrange multiplier

[chilly hazel]

I actually found a presentation for this paper where they do it a little bit differently

I think they cancel out the numerator here

and then maximize the denominator

to minimize v(x)

a lot more elegant

don't need to do Schur or anything

and you need to figure out A in the other method anyway

I'll see if that cancels out the numerator tomorrow lol

I'm so tired

ah nvm

they get rid of the x^Tx

$v(x) = \frac{A^TΓA}{Γ} = Γ^{-1}A^TΓA$

[stoic python]

Vertox

[chilly hazel]

getting rid of x is bad tho

[spare widget]

That looks like nonsense

[chilly hazel]

yeah I know you can't divide by a matrix

so it becomes the inverse

[north steeple]

Not sure if this is right channel for this

anyone able to help me understand this better

i dont really understand the examples it gives either

[cold flint]

Would someone be able to walk me through how to obtain the Laplace transform of 2e^-4t*u(t-1)? I'm not sure how to go about it and I believe that solving it has something to do with shifting but I wasn't quite sure but I could also show what I have

[still lodge]

det(A) = 1 means that the basis vectors dont flip after the linear transformation described by A is applied right

[zinc timber]

yes

[still lodge]

epic

[wintry steppe]

with regards to this problem, how would i find something parallel to this line?

https://cdn.discordapp.com/attachments/540211747613704221/946112052789932172/Screen_Shot_2022-02-23_at_12.31.06_PM.png

[halcyon spindle]

Can you clarify exactly what you mean by “something”?

[wintry steppe]

a vector parallel to the line

[halcyon spindle]

v = x_1 - x_0 is one, you can take any two point on that line, take the difference to get a vector parallel to that line.

[wintry steppe]

wait

v = x1-x1?

do u mean x1-x0

how do u know that quantity is 1

[halcyon spindle]

Yeah sorry.

wdym?

[wintry steppe]

how do u know x1-x0 is 1?

[halcyon spindle]

I meant v = x_1 - x_0 is one vector parallel to that line.

[wintry steppe]

ah ok

cool

ty

[spare widget]

You can't get rid of x as you did

[spare widget]

Do (1-t)x0 + t x1 = x0 -t x0 + t x1 = x0 + t (x1-x0)

[grave garden]

Hii guys

Why $^t(^tYAX)=^tYAX$ ?

[stoic python]

Potato

[spare widget]

Because Y^TAX is a scalar

A scalar transposed is still the same scalar

Though this notation keeps getting worse

[lavish jewel]

what book is this

[spare widget]

I hope I don't get to see

[lavish jewel]

wth

[spare widget]

$_{-1}Z$ next

[stoic python]

criver

[zinc timber]

lol

[spare widget]

Supposedly Serge Lang

[grave garden]

Yes, Serge Lang

Normally what is the transpose symbol?

[gray dust]

using t is common but writing it to the left isnt

$A^t$

[stoic python]

RokabeJintaro

[spare widget]

Or capital T, also vectors are typically lowercase

[grave garden]

I see

$^T_tA_t^T$

[stoic python]

Potato

[grave garden]

Beautiful

[lavish jewel]

$\tau^{-1}{M}$ represents a matrix that has not been transposed

[stoic python]

Edd

[spare widget]

Is that actual notation?

[lavish jewel]

no, i'm joking

[grave garden]

[spare widget]

[true jacinth]

how can i express a cartesian product $A_1\times A_2\times\dots A_n$ as a set of n-tuples ? i understand with examples but not how to write the generalized notation

[stoic python]

DonutsenPLS

[vivid barn]

you mean like $A_1 \cross A_2 = {(a_1, a_2)|a_1 \in A_1, a_2 \in A_2}$

[stoic python]

worm

[true jacinth]

yes

but with n-tuples

[vivid barn]

i think you can use the big pi for this

i've seen textbooks do it

[true jacinth]

like $A_1\times A_2 \times\dots A_n ={(a_1,(a_2,\dots,a_n))|a_i\in A_i, 1\leq i\ leq n}$

[vivid barn]

that would work i think

[stoic python]

DonutsenPLS

[lavish jewel]

why the parentheses?

[spare widget]

$\prod_{i=1}^n A_i = {(a_1, \ldots, a_n) ,:, a_i \in A_i}$

[vivid barn]

probably a typo i'm assuming

[stoic python]

criver

[lavish jewel]

what criver wrote now is succinct

[true jacinth]

yes that's what i mean

[lavish jewel]

if you wish you can add a $,,, 1 \leq i \leq n, i \in \mathbb{N}$

[stoic python]

Edd

[lavish jewel]

or i \in {1,2,...,n}

[spare widget]

By the way, I did some numerical experiments and the L^TL vs L kkt formulations are indeed not equivalent.

[lavish jewel]

not surprised

[wintry steppe]

What book?

[grave garden]

Serge Lang's linear algebra

Im sad now that this is not always true

[zinc timber]

wait what

WTF

WHAT BOOK IS THAT

[grave garden]

This again -,-

[zinc timber]

BY DEFINITION OF INNER PRODUCT, $\ip{v, v}$ IS NEVER ZERO UNLESS $v$ ITSELF IS 0

[zinc timber]

IDK WHAT NOTION HE IS USING

[grave garden]

It could exist too, like how he provide the example ?

[spare widget]

Tis magic

[grave garden]

If you have the book too, it is at page 135

[spare widget]

You sure you don't have some fake pirated copy

[zinc timber]

he's using the definition $\ip{v, w} = v\cdot w$ but according to (modern) definition, $\ip{v, w} = v\cdot\overline{w}$

[spare widget]

That's for C though

[languid sphinx]

https://math.stackexchange.com/questions/295588/what-is-the-intuition-behind-hermitian-product

[zinc timber]

(1, i) is element of C

[spare widget]

hadn't read that part mb

[wintry steppe]

its still an inner product though

oh yeah where v is important

[spare widget]

Guess Lang's a fan of nonstandard stuff

[grave garden]

Im a bit confused rn 👁️

[zinc timber]

?

[wintry steppe]

v,v never equals zero for some people because they want inducing norm to be easy

[zinc timber]

@grave garden follow other book, fr now

[wintry steppe]

lol

the v*w bar is standard

[zinc timber]

why do you think that condition is enforced then

things are much more nicer

[grave garden]

What easy book do you have in mind ?

[zinc timber]

LADR/LADW

[spare widget]

physicists and engineers would be very confused if you feed them this definition of inner product

[wintry steppe]

vector spaces over finite fields are cool

[spare widget]

I would guess eveybody would be actually

[languid sphinx]

LADR does not mesh well with determinants

Did not hear anything too bad about W so far

[wintry steppe]

and inner product spaces over them are also

[spare widget]

You typically learn standard stuff before nonstandard stuff

[grave garden]

heyyyy im an engineer

[wintry steppe]

its whatever small detail

[spare widget]

Find a better book

[languid sphinx]

Did you get confused

[spare widget]

There's a huge choice

[languid sphinx]

I thought engineers would begin with Strang's instead

[zinc timber]

there's also one by AMS society that deals with advanced matrices

[lavish jewel]

i'm also an engineer, this book sux

[zinc timber]

apes strong together

[lavish jewel]

🐒

[grave garden]

Well i have no knowledge, so it feels okay wkwk

[wintry steppe]

Dont worry

[spare widget]

Do worry

[wintry steppe]

its all just convention

[zinc timber]

wym changes the whole subject

[wintry steppe]

just dont call the given v,w an inner product

Not really

[zinc timber]

from LA to abstract

[spare widget]

Nobody's going to be explaining the standard conventions in an engineering paper

[wintry steppe]

Ok true

[grave garden]

I prefer sth straight forward, not like manual book

[spare widget]

They assume you know those conventions

Same with physics

[wintry steppe]

But they dont need to bring up norms if they dont want lol

[languid sphinx]

I thought Strang would be straight forward

[zinc timber]

how else are you gonna prove Reisz Presentation theorem?

[spare widget]

Axler, hoffman, insel, the done wrong one, halmos, there's plenty

[zinc timber]

v*w\bar is not a convention, but that's what it should be

(well std one at least)

[spare widget]

It is convention in physics and engineering to assume one takes this inner product when it is not specified

though I've seen physicists swap the conjugation argument

[zinc timber]

?

[spare widget]

As in bar{u} \cdot v = <u,v>

[zinc timber]

No you can't do that

it's no longer liner in the first component

[spare widget]

They do it sometimes

[zinc timber]

[spare widget]

It's some convention so gotta be careful

[zinc timber]

if you mean $\ip{u, v} = v^*u$ then it's ok

[zinc timber]

but not u^*v

[spare widget]

ik what you mean, but there are still people that flip it, so gotta be careful

[zinc timber]

Idk I'm good with stf stuffs

[spare widget]

https://physics.stackexchange.com/questions/482105/question-on-notation-for-the-inner-product-of-complex-vectors

[zinc timber]

oof

that's why mathematicians and physicists can never be friends

[spare widget]

I like physicists better, don't shoot me

[zinc timber]

[spare widget]

I definitely regret picking CS over physics

[zinc timber]

well I like all three of these

[spare widget]

also in physics they seem to cover a broader range of mathematics than mathematicians in bsc in the unis I have been in

[zinc timber]

except these notational inconsistency

[spare widget]

granted, not in the same depth

[zinc timber]

ya true, I've seen my friends cover more syllb in 1 sem than our entire course

I'm mostly interested in QM and Cosmology stuffs tho

would love to do my phd on these

[spare widget]

they also do mathematically "illegal" stuff that "just works"

[zinc timber]

that's why they keep a mathematician in the premises

to figure out the details

[spare widget]

https://en.wikipedia.org/wiki/Constructive_quantum_field_theory

[spare widget]

we do that in CS also

keep a mathematician to fit a model to the results 😂

[zinc timber]

[wintry steppe]

are there infinite dimensional vector spaces which arent complex?

or does it always have to be a hilbert space or banach space

[zinc timber]

there are infinite dimensional spaces that are neither Hilbert nor Banach

[wintry steppe]

ah

[zinc timber]

ex C[0,1] under L¹ norm

[wintry steppe]

but are they complex vector spaces?

[zinc timber]

it's real

L² norm also works

[wintry steppe]

wait can you explain what C[0.1] is ive never seen this before

[zinc timber]

set of all continuous functions from [0,1] to ℝ

check it's a vector space

and a normed space under L² norm given by || f || = (∫ f² dx )^½

[wintry steppe]

ohh, but is there a vector space that is just a set of vectors from [0,1] to R and is L^2 but not complex?

[zinc timber]

?

[wintry steppe]

so not a set of all continuous functions

[zinc timber]

what? wym?

[wintry steppe]

i mean this but instead of continuous functions you just have normal vectors

[zinc timber]

??

what is a normal vector?

these are perfectly normal to me

[wintry steppe]

continuous functions are not vectors

[zinc timber]

who said that

[wintry steppe]

yea so im asking if you have such space with vectors instead of continuous functions

[zinc timber]

they are more that vectors, they form an R algebra

[spare widget]

in linear algebra the elements of a vector space are termed vectors

he means finite dimensional vectors, e.g. R^n

[wintry steppe]

ohh

lol

ok

[zinc timber]

is Rⁿ the only vector space you are aware of?

[wintry steppe]

nono

thats not my question

then im good

[zinc timber]

,w vector space

[stoic python]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Ryuzaki#5323

[zinc timber]

if you are thinking vectors as arrows, don't.

[spare widget]

it's fine if you're analytic geometry class

it's not enough in linear algebra

[spare widget]

I have a matrix of the form

$M = \begin{bmatrix} L & A_1^T & \ldots & A_m^T \ A_1 & 0 & \boldsymbol{0}^T & 0 \ \ldots & \boldsymbol{0} & \boldsymbol{0}\boldsymbol{0}^T & \boldsymbol{0} \ A_m & 0 & \boldsymbol{0}^T & 0 \end{bmatrix}$

[stoic python]

criver

[spare widget]

And I want to find the first k rows of M^{-1} where k is the number of rows of L

Then I believe I get the equations

$\begin{bmatrix} I_{k\times k} & O_{k\times (n-k)} \end{bmatrix} M^{-1}M = \begin{bmatrix} I_{k\times k} & O_{k\times (n-k)}\end{bmatrix}$

[stoic python]

criver

[spare widget]

L in the above is not full rank, but M is full rank

Then my guess is that this should have the form: [I O]M^{-1} = [L^{-1} O]

where L^{-1} is the specific pseudo-inverse, such that L^{-1}L = I, and L^{-1}A^T_k = O for all k

Is this correct for the [I O]M^{-1} = [L^{-1} O] or am I missing something

Yeah nvm, I think I am wrong

the general solution just requires that

$\begin{bmatrix} I_{k\times k} & O_{k\times (n-k)} \end{bmatrix}M^{-1} = \begin{bmatrix} R_0 & R_1 & \ldots & R_m\end{bmatrix}$

[stoic python]

criver

[spare widget]

such that

$R_0L + \sum_i R_iA_i = I, \quad R_0A_i^T = O$

[stoic python]

criver

[spare widget]

show that the transpose is equal to the matrix (use the property of transposition of matrix product and that A is symmetric)

then try to show that x^TMx > 0

well write out the symmetry first

that is

(V^TAV)^T =?= V^TAV <- prove this first

yes, I would use a decomposition since V is full rank

note

(AB)^T = B^TA^T

order changes

yes

also for inverse if the matrices are invertible

you usually have to be more detailed, e.g.

(V^TAV)^T = V^TA^T(V^T)^T = V^TAV

where A^T=A since A is symmetric

I am guessing you would do something like y = Vx

then y^TAy>0

the tricky part is showing that y^TAy=0 only for x = 0

so shouldn't it be m>=n?

i.e. V needs to compress x to a smaller space

ah, maybe I am wrong

V is n rows and m columns

nvm, I am probably just wrong

I have to think about it when I am not in a meeting

[lavish jewel]

how about doing an EVD of A

since it's symmetric pos def, it's diagonalizable in some orthonormal basis

[vale herald]

someone germam here?

[lavish jewel]

so that you get V^T Q^T D Q V, and you can simply make a matrix D^(1/2) by taking the elementwise square root, as D is diagonal

this means V^T A V is of the form W^T W, which is symmetric positive semidefinite

then all that is left is to show that V^T Q^T D^(1/2) has rank m

what you can do is, knowing that Q is a full rank orthogonal matrix, the image of a set of linearly independent vectors is linearly independent

and so the rank of the m x m matrix is m (by grouping Q with V, and V^T with Q^T), so that the symmetric pos semi def matrix is symmetric definite

@slender frost

[zealous flame]

https://cdn.discordapp.com/attachments/903481101744484362/946448752418390036/Screenshot_20220224_221905.png

How do I proceed with this one?

[lavish jewel]

you can use what i wrote above

the direction BB^T -> M is SPD is easy. in the backwards direction, if you can use the existence of M's eigenvalue decomp, then you're just about done

[zealous flame]

So, M = X U X^T

Can I say B = (U^0.5) X

Then multiplying B B^T

oh no

am i going in the wrong direction?

[lavish jewel]

you need to do both directions

[spare widget]

Don't you mean D^{1/2}?

[lavish jewel]

hmm?

[spare widget]

Q is the matrix of the eigenvectors

[lavish jewel]

i meant D, yes, oops

thanks

[spare widget]

I have a question related to that

In the above you have a map from a smaller space (m)

To a larger one (n)

doesn't this mean that I can take a vector that is nonzero from R^m

And Vx= 0?

[lavish jewel]

no, because they state V has full column rank

this means its kernel is trivial

[spare widget]

Ok, so the opposite will break things then I guess

[lavish jewel]

yes

[spare widget]

Since if we go from R^n to R^m the kernel will be at least n-m

[lavish jewel]

if m > n, then the null space is always nontrivial

hmm that's not what i mean

or is it

[spare widget]

I just used V^T

Which goes from R^n to R^m

[lavish jewel]

ah it was correct, cuz V is n x m, not m x n

mhm

[spare widget]

so with that only semidefinitness would be provable

[lavish jewel]

right

[spare widget]

Also if the rank of V was not full, I could nevertheless get a 0

[lavish jewel]

yep

[spare widget]

Since that would imply that the columns are linearly dependent

And I would just need to solve Vx = 0 in that case to find such a vector

[lavish jewel]

yes

[spare widget]

Great, thank you

[zinc timber]

you mean m<n for R^n -> R^m?

[lavish jewel]

backwards

for R^m -> R^n

[zinc timber]

ok then it's fine

[lavish jewel]

(it's the notation in the image)

(i know it's backwards)

[zinc timber]

Nah I was referring to criver's message

[wintry steppe]

If you have a training set of five ordered pairs of xi,yi

And you’re told to find the k=1,2,3 nearest neighbors for x =1

Then you just wanna find the one closest x value ordered pair, then two closest x value ordered pairs, then three

Right?

[lavish jewel]

sounds about right

[wintry steppe]

Ok amazing

Tyy

[wintry steppe]

I have the following augmented matrix

Which is from the following system of equations: ```
z = 30
2x+5y+3z = 121
4x+3y+6z = 207

I constructed the equations above and the answers are z = 30, x = 3 and y = 5

However, solving the above matrix gives z = 30, y = 5 and x = 12

I even checked the solution for the augmented matrix with a calculator, same invalid answer

I checked the solutions for the system of equaitons, which turned out to be the same as intended

Why is this happening?

nvm

[spare widget]

It's x = 30 not z = 30

[molten pilot]

A is a square matrix

I don't understand the last equivalence

[nocturne jewel]

The transformation associated with A is surjective

[molten pilot]

yeah, that's the definition of being onto

but why

[wintry steppe]

Yeah, realized my mistake...... after 3 damn hours trying to finish my guass-jordan elimination assigment....

thank you

[molten pilot]

I just reasoned that after $[A|b]$ is reduced to $[\tilde{A}|\tilde{b}]$

[stoic python]

DarQ

[molten pilot]

you can reverse the steps to get to the original

[nocturne jewel]

If A has all pivot columns, then A is invertible

so T is bijective

where [T]=A

[molten pilot]

this is part of the proof of that statement lol

[nocturne jewel]

Alternatively just apply rank-nullity

[molten pilot]

but the equation $A\vec{x}=b$ only has solutions if $\tilde{b}$ has non-pivotal 1s

[molten pilot]

haven't reached that yet

[stoic python]

DarQ

[molten pilot]

is there a typo here?

[stoic python]

DarQ

DarQ

[nocturne jewel]

given they're talking about inverses, I doubt it

[slow scroll]

They were just giving a counter example if injectivity did not hold

[molten pilot]

well

[slow scroll]

Oh no wait

Yea it looks like a typo

I agree with your correction

[molten pilot]

yeah

coz I can think of a non injective square matrix where the only vector such that Ax=x is the 0 vector

[gaunt grail]

where is a good place to learn linear algebra

other than like khan academy

any good textbooks?

[nocturne jewel]

#book-recommendations

[wintry steppe]

can someone help me out with this prob? Im confused

[gaunt grail]

Oh sorry, I didn't see that channel.

[wintry steppe]

can anyone help me for that prob?

[zinc timber]

?

[wintry steppe]

the pic I just sent

[zinc timber]

what is confusing you

[wintry steppe]

like what the true statements are

I know it cant be a cuz matrices with an inverse cannot be communative

but other than that idk

[fickle patrol]

Can someone help me understand this question?

[wintry steppe]

can someone help me for this question

is everyone sleeping?

[quartz compass]

confused about what

[wintry steppe]

like im not sure on how to answer it @quartz compass

[quartz compass]

you put a check in the boxes where the statement is true

[wintry steppe]

well yes ik that

but idk which statements are true

[quartz compass]

go through them one at a time

[wintry steppe]

I did

I dont understand

[quartz compass]

tell me what you think about every single one of them

why you think they might be true or false, give a reason

give your best guess

and best reasons you can think of why

[wintry steppe]

ok so the first one its false

because it must be associative and not communative

[quartz compass]

sounds good

[wintry steppe]

and from there on im not sure

[quartz compass]

try looking at a special case when A is a 1x1 matrix

in other words, pretend they're just regular numbers

[wintry steppe]

im not understanding how that coorelates to this prob

what do you think are the true statements because like this involves understanding the rulings regarding inverses

I dont think D is true

@quartz compass

can you just give me the answer and then explaii because this is due in 10 mins

[quartz compass]

no

[wintry steppe]

bruh

[wintry musk]

I do some matrix math at work, and one thing that thought I proved to myself over the years is this (which might not actually be true in general):
Lets say you have 3 Euclidean coordinate systems: L1 (Local 1), L2 (Local 2), and W (World) (lets assume that all the matricies are 3x3 and are orthonormal)
You are given 2 transformation matrices. One that takes you from L1 -> W (lets call this matrix A) and the other takes you from L2 -> W (Lets call this matrix B).
Lets define T to be the total transform that goes from L1 -> L2: T = (B)^(-1) A (Im assuming column major matricies here.)

So my assumption is that R (which I will define here as the Rotation from L1 to L2) would be: R = (T)^(-1) = (A)^(-1) B
So my questions are:

1.) in general is the rotation L1 to L2 equal to the inverse of the transformation from L1 to L2
and if so
2.) under what contexts would this rotation work "as expected"... ie would this rotation only work in L1 space? or maybe L2? or maybe even W?

Any thoughts on this would be very helpful! Thanks all.

[zinc timber]

answer is 7

[spare widget]

T is the rotation from L1 to L2 not T^{-1}

If what you are talking about is related to a CG API like opengl or directx be aware that they have a broken understanding of what column majority means and what an mxn matrix means.

To be precise I understood your formulation as

$[v]{W} = A [v]{L_1}, , [v]{W} = B [v]{L_2} \implies [v]{L_2} = B^{-1} [v]{W} = B^{-1}A [v]_{L_1}$

[stoic python]

criver

[spare widget]

Note that this matrix is a rotation only if A and B are rotations

[wintry musk]

To answer your question, this isnt related to opengl or directx (so i dont think I have learned there broken understanding)

[spare widget]

one could be a rotation + reflection for instance, in which case you will not get a rotation

[wintry musk]

The TeXit you have seems accurate to how I was defining A and B

[spare widget]

So what is it related to that involves column majority

[wintry musk]

I just mentioned column major as opposed to row major (in which the code I work in uses)

Back to the REAL details tho! Im kinda confused how my defined T is actually the rotation (as opposed to the inverse of T)

[spare widget]

but as I said, provided all 3 are orthonormal bases that do not change handedness/orientation, the B^{-1}A is the rotation matrix

Both T and T^{-1} are rotation matrices provided the bases are orthonormal and don't change handedness

But T by definition takes a vector from L1 to L2

T^{-1} by definition takes a vector from L2 to L1

[wintry musk]

Right, so T would show how the components of a vector in L1 would "look" in L2

[spare widget]

The code you're working with could have the same issue, gotta be careful when someone mentions column/row majority, it is not a mathematical concept, it is a memory layout one

[wintry musk]

But if I wanted a rotation vector (which rotated all of the basis vectors of L1 to match the basis vectors of L2), wouldnt that rotations vector be T^(-1)?

[spare widget]

A matrix is not a vector

[wintry musk]

I will definitely double check this @spare widget, you could be onto the source of my issue.

[spare widget]

Also the basis vectors transform differently than contravariant vectors

Your matrices transform contravariant vectors here

[wintry musk]

covariant vs contravariant isnt something ive ever been able to nail down 😦 ... someday tho!

[spare widget]

basis vectors are covariant and transform in the opposite way

[wintry musk]

it sounds like i need to explore how/why basis vectors would transform differently than just "regular" vectors

[spare widget]

https://youtu.be/A1h_eucHFW4

[wintry musk]

cool, I'll look into this.

But to recap, your answer is that T would be the rotation matrix that rotates the basis vectors of the L1 to L2, correct?

[spare widget]

Also relevant: https://en.m.wikipedia.org/wiki/Active_and_passive_transformation

[spare widget]

Yes

Both T and T^{-1}=transpose(T) are rotation matrices

But T by definition transforms from L1 to L2

Whether this matrix rotates counterclockwise is irrelevant

So for a specific choice of L1 and L2 it may turn out that for your "setting" it looks like it rotates in the "negative" direction, for other choices it may rotate in the "positive" direction

[wintry musk]

In the general cases I was looking at, when applying my rotation matrix to L1, it either perfectly matched L2's basis, or it was completely wrong

(In my case it turned out that the inverse of the rotation matrix yielded my desired result... which might be explained by the difference between covariant and contravariant)

[spare widget]

Because basis vectors transform in the opposite way

[wintry musk]

Thanks so much for your time @spare widget! I really appreciate your help

[tribal smelt]

A, B are vectors. If |A + B|= |A- B| , then vector A,B are

1. AxB
2. A.B
3. A=B

Can anyone help me understand this ??

[lavish jewel]

try expanding the expressions using dot products

[spare widget]

Try 3) then |A+B| = |A-B| = |0| then it holds only for A=B = 0