linear-algebra

[lavish jewel]

then what matters is if the vector containing the 3 rgb values at one point in the image is a linear combination of the rgb values at other points in the image

not that rgb is uncorrelated with height and width

but rather how rgb varies w.r.t. itself as the other 2 values changes

(assuming you care about rgb space. other matrix shapes can have different ranks.)

[neat relic]

oh ok..I think I understand clearer now, like what matters is the linear combination of 1 feature (i.e. height, width, rgb) to itself at a point. So what happens if it is a linear combination of it?
I know we learnt in linear algebra that it means it is not linearly independent, but am not sure how it relates to image similarity and collinearity haha..

[lavish jewel]

i guess it would be what you call collinearity

means the matrix is rank defficient

[neat relic]

oh ok..I think I have enough to work with, learnt lots today from you, thank you for your time and help.

I'll look into the different distances like Euclidean distance, Cosine distance, etc and figure out the best one to use.

[lavish jewel]

"best" is really subjective

but yeah, go give that a read

[neat relic]

cool cool, yeh lucky there's like youtube tutorials around and stuff haha. But yeh will check those out, and yeh thanks heaps for your help.

[lavish jewel]

aight

[wind pasture]

this statement is false right?

[dire thunder]

A^TA would be n x n matrix

[wind pasture]

yeah so theres no way

well there is a way

but i mean generally

[dire thunder]

ye

prolly

[wind pasture]

how about this statement?

[dire thunder]

https://math.stackexchange.com/questions/3573576/given-an-n-times-m-matrix-a-prove-that-colata-colat for first one

what is P_3, set of polyomials of degree up to 3?

[wind pasture]

yep

[dire thunder]

yes then one-to-one is equivalent to being onto

[wind pasture]

your link says something different than the Q

[dire thunder]

i mean dim P_3 = 4 and dim R^4 = 4

[dire thunder]

it was for question with col spaces

[wind pasture]

right but it has col(A^T) on the RHS

[lilac stratus]

if dim E = dim F, then any onto linear transfo between E and F is one-to-one

[wind pasture]

col(A^T A) might equal col(A^T)

[dire thunder]

yes which shows that your statement was false

[wind pasture]

oh ok

[lilac stratus]

uh ?

[dire thunder]

they had statement Col(A^TA)=Col(A)

[lilac stratus]

oh, yeah

[wind pasture]

i guess this is true too then

[dire thunder]

welll

dim V = dim null S + dim ran S

and dim V >= dim ran S since S is onto

moreover, dim ran S = dim W in that case

thus you get dim V >= dim W

@wind pasture can you see how to complete this argument?

[wicked lion]

How can I calculate the 3d collision of two spheres and whether or not they are intersecting?

[wind pasture]

uh dimW >= dim ran T

[dire thunder]

yes, and since T is onto, dim ran T is?

[wind pasture]

dim V?

[dire thunder]

yep

[wind pasture]

dimV = dimW = isomorphic?

[dire thunder]

yes, any two finite-dimensional spaces with the same dimension are isomorphic

[wind pasture]

great thanks

[dire thunder]

yw

[wintry steppe]

you can prove that by constructing the isomorphism by sending each element of one basis to the other

[lilac stratus]

what do you know about the spheres ?

[wicked lion]

@lilac stratus absolutely nothing.

😄

i mean i know like

they have a radius

lmao

[boreal crane]

bijektively

[lilac stratus]

Ok so you have radius and centers, right ?

[dire thunder]

||is that coincidence that your name contains russian vulgarism, shika ||

[wicked lion]

Uh, I don't know how to calculate it. I just know they exist. @lilac stratus

[lilac stratus]

how do you went to calculate the intersection of two spheres if you only know they exist but not where

[lilac stratus]

||no, initially that was a private joke with a friend, and then that lasted and here I am 2 years later with the same nick||

[lilac stratus]

(i'm probably misunderstanding the question )

[wicked lion]

uh

[dire thunder]

Hint: point is on intersection of two curves/surfaces iff it satisfies equations of both these curves/surfaces

[wicked lion]

so like, you know how in games, there are spheres over different areas, and when a player overlaps with the sphere it causes something to happen

i essentially want to calculate whether or not something overlaps something else in 3d space

[lilac stratus]

You can't speak about sphere equation if you don't have the radius/the center of the spheres, though, can you ?

[dire thunder]

how then they would speak about any spheres, though?

[lilac stratus]

Well that's why I am confused lol

[wicked lion]

Does that make sense

[lilac stratus]

Well yes, but you need informations on that something to do so

[wicked lion]

idk how to help with that honestly

[lilac stratus]

If you're programming, and the sphere is a data structure like this:

struct Sphere {
  radius: real
  center: 3DPoint
}

then yes, you can compute the intersection of two spheres

[wicked lion]

Well yeah

but i think thats irrelevant to the formula isn't it?

the formula would be the same for each and every single situation where you have to do this

i would assume at least

[lavish jewel]

is it for sure 2 spheres?

[lilac stratus]

it is irrelevant, indeed, but I was kinda confused cause you told me you had no information on the spheres

[lavish jewel]

or generally 2 geometric objects

[lilac stratus]

but you actually have the radius and the center

But like, if you have two spheres S1 and S2 the first one being of radius r1 with a center (x1, y1, z1) and the second one with a radius r2 and a center (x2, y2, z2), then the points of S1 are the points (x, y, z) satisfying the equation:
(x - x1)² + (y-y1)² + (z-z1)² = r1²
and the points of S2 are the points satisfying the equation
(x - x2)² + (y - y2)² + (z-z2)² = r2².
You're looking for points satisfying both equations

[wicked lion]

oh i'm more trying to learn 3d math

for games

stuff like dot product, cross product, intersections of collision, etc

i've been interviewing with some major game s tudios and realized i don't know enough about 3d math.

I know a lot about programming in general though.

Which is kind of interesting when you think about it lol.

[lilac stratus]

Well, I'm not sure dot product or cross product is relevant here, I'm fairly sure the best way to compute it programming-wise is really to solve equations (not necessarily the ones I've given tho)

[wicked lion]

Yeah I know it's not relevant here. I was just saying those are vector math concepts which I've found to be very common on interviews lol.

Honestly, I assume calculating intersecting circles is very similar to intersecting spheres but probably an added term or something

[lilac stratus]

yeah

[lilac stratus]

oh, yeah they are, since this kind of linear algebra is used a lot in 3D programming

Actually, if you want to learn this stuff

I'd recommend you to try doing like a small ray-tracer

if you're a programmer that's probably one of the best way of learning about it (and a funny one)

[wicked lion]

Where could I read up on that?

[lilac stratus]

2secs

https://www.realtimerendering.com/raytracing/Ray Tracing in a Weekend.pdf is really really good

[wicked lion]

o good lord thats a lot of reading XD

[lilac stratus]

there's a lot of code too

and you get funny results quite fast

(like after reading ten pages you already have some code you can play around with)

[wicked lion]

Now here's a interesting question I have seen: "Given thousands of spheres inside of a cube, that are floating/moving... how can you optimize collision checks"

[lilac stratus]

Also, although it does not only speak about raytracing, this website is nice https://www.scratchapixel.com/index.php?redirect

[wicked lion]

Now I assume this question requires me to have a deep understanding of why the collision works

soo that i could potentially, remove terms or something

so it's less calculations

honestly though, i can't imagine collision checks to be very expensive...

[lilac stratus]

that's one of the most expensive part (in computing time) of the 3D maths

[wicked lion]

really

[lilac stratus]

a good method would be something AABB inspired

I'm only know a little bit of 3D programming, so there may be a lot better, but AABB would be decent, I think

[lilac stratus]

probably adapted for spheres, like not just looking for intersections of the bounding box but also trying to vaguely guess when the bounding box intersects but not the actual underlying spheres

(not sure how one would do that though)

[wicked lion]

i gotta figure out like

in which order to learn stuff lol

i have so much stuff i want to learn

and very little time

during the day tbh

i have to work from 7 to 3 pm every day

so its hard to figure out how to prioritize this stuff

[lilac stratus]

I'd recommend you to like, just code, and learn the maths while doing it

that actually works quite well, if you already know some very basic linear algebra

[wicked lion]

Honestly, what's embarassing is that the last math course I took was basically algebra 2 lol

i never took calculus

[zealous junco]

spheres usually is hard to compute though i'd typically approximate it using a square or something XD

[wicked lion]

😭

[lilac stratus]

and if you don't, start with the first chapters of scratchapixel (the website I linked earlier), and then, well.. code (a small raytracer, or even just some rotating cube using an already existing graphic library)

[coral skiff]

hey, got a linear algebra question in #help-0 about iteratively solving linear systems with elements on the diagonal of A that aren't all non-zero if any of you think you can help

[lilac stratus]

like these chapters @wicked lion

[wicked lion]

yeah i saw those

they look pretty impressive

where'd u find this lol

[lilac stratus]

I used to spend a lot of time programming, and 3D programming was something I wanted to learn, so I've spent a lot of time looking for learning resources

It was kinda fun, a few days ago I wanted to restart doing some 3D programming related stuff so I coded a small 3D ascii renderer lol (really small, like only 200 lines)

anyways

[wind pasture]

i got (1, 0, 0), (0, 1, 0), (0, 0, 1)

am i tripping or is this right?

[wicked lion]

sounds like fun

im not very confident in my math skills honestly

my education in mathematics has been godly awful

[lilac stratus]

Well, I kinda dropped out in like 8th grade, and didn't restart actually going to school since then, so I'd say my education wasn't great either lol. Luckily, internet is full of nice resources to learn by yourself

[wicked lion]

Yea lol

So little time in the day to actually sit down and learn tho lol

[lilac stratus]

[wicked lion]

whats that mean lol

[wintry steppe]

Anyone good with vectors, i can pay

[dire thunder]

#❓how-to-get-help

[wind pasture]

if i have a vector A

does A^x conserve eigenvalues?

how about x*A

[stable kindle]

A^x?

[dire thunder]

xA conserves since its char polynom preserves roots

[stable kindle]

what, like (a1^x a2^x a3^x)

[wind pasture]

like A^2 or A^5

[dire thunder]

it also conserves afaik

[stable kindle]

what's the square of a vector again

[dire thunder]

they prolly meant matrices

[stable kindle]

hmmmm

[wind pasture]

my bad

this is what im trying to solve exactly

i found the eigenvalues of A

now im trying to do part b

[stable kindle]

so i think A^2 will have the square of the eigenvalues of A, tbh

it'll conserve the same eigenvectors

but it'll scale by the eigenvalues twice?

i mean maybe the eigenvalues are just 1 and 0 and -1 or something neat

[dire thunder]

alright so for xA we have

let T be associated linear transformation

xT(v)=x\lambda v where lambda is eigenvalues for T

thus x\lambda v is eigenvalue of xT and thus of xA

https://proofwiki.org/wiki/Eigenvalue_of_Matrix_Powers

and for A^x

[stable kindle]

yeah, as i said

[wind pasture]

so xA conserves?

[stable kindle]

yes

easy to see if you think about it

[wind pasture]

but how would i solve part b?

[dire thunder]

not exactly conserves, but eigenvalues of xA are x times eigenvalues of A

[wind pasture]

actually i think i have a way of solving it

[stable kindle]

well first what are the eigenvalues

of A

[wind pasture]

2, -3, -1

[stable kindle]

huh

[wind pasture]

can't i just say that eigenvalues are the diagonal entries

[stable kindle]

lol, only if it's a diagonal matrix

what are the actual eigenvalues

wait

i might be bad at linear algebra lemme check something rq

[wind pasture]

i think the condition is triangular matrix

[stable kindle]

yes i'm bad

you are right

ok but wait since it's diagonal you can basically just know the diagonals of A^5 and A^2

and so you can get the diagonal of B really easily

[wind pasture]

yeah exactly

[stable kindle]

ok that's how you're meant to do it

[wind pasture]

i added them and got a 0

[stable kindle]

lol

[wind pasture]

in the last row at least

[stable kindle]

i get the diagonal of B as ||(32, -243, -1) + (8, 18, 2) + (2, -3, -1)||

[wind pasture]

you mean 2 not 20 right

[stable kindle]

hrnnn

yes

[wind pasture]

i got that too

how would i solve part c though?

if i show that A is or isn't diagonalizable, does that mean B is or isn't?

nvm, B is diagonalizable because its an nxn matrix with n distinct eigenvalues

[boreal crane]

Yeah but in this Case i guess B is diagonalizable exactly when A is because the Eigenvectors dont change just the eigenvalues

and if there is a base of eigenvectors of A there is also one for B

[native rampart]

No

[boreal crane]

Nope x^3 +(-x^3) is not in the set

[quick oak]

oh

what about mult closedness

and one item showing

[boreal crane]

wait i just need to check what you mean cuz i study math not in english

[wintry steppe]

0*(x³)

[quick oak]

oh...

what should i do for this one?

[wintry steppe]

view matrices as arrays

[boreal crane]

it is

[wintry steppe]

it's basically just saying that some entries are 0 and others are arbitrary

[quick oak]

Aij means determinant right

but im still stuck

[wintry steppe]

i think it means the i,j entry of A

[boreal crane]

yeah if you want i-j-1 to be divisible by 2 you just have the possiblity that one is 1 and the other is 0 so this is just the space of diagonal matrices

[wintry steppe]

you can ignore that condition it doesn't really matter

Hello! I have a small question about this exercice, I have to find the value of alpha such that these two linear applications have the same reduced form (don't know what the term is in english but it's basically almost the diagonal form, so it's upper triangular with the eigenvalues on the diagonal and a 1 on the upper left). So when the discriminant of the characteristic polynomial is 0 or smaller than 0, we can find this reduced form. And my idea was to calculate the reduced forms of the two linear applications and separate among the case of delta=0 or delta<0 but it seems to be way too intense on the calculation and I asked myself if there is another way of solving this problem that is more efficient 🙂

[boreal crane]

its important because it tells you that A1,0 and A0,1 are always 0

[quick oak]

ohhhh nice

[wintry steppe]

Any subset of matrices where some entries are 0 and the others are arbitrary is a subspace

[boreal crane]

jup

[wintry steppe]

so it doesn't really matter which ones are 0 and which aren't

[quick oak]

so answer is yes i guess huh?

[boreal crane]

yes

[quick oak]

what can I do here?

[stable kindle]

determinant

[quick oak]

determinant is zero?

@stable kindle

[stable kindle]

??

yes

[quick oak]

what do you mean determinant?

[stable kindle]

if x isn't 0

wait

non-trivial

so yeah determinant of a must be 0

[strange delta]

what does it mean by describe span{v_1,v_2,v_3} ?

[quick oak]

you sure

i am little confused

-101a^2-210ab-34a-109b^2+442b determinant is this

[stable kindle]

lol

oh jesus christ

[quick oak]

i multiplied the matrixes

this

[stable kindle]

no, i'm a fool

wait

ok so you literally gave us AX

not A

wait

no

argh

[quick oak]

soooo....

it is not hard i believe

[stable kindle]

ok so the determinant of a product of matrices is 0 iff at least one of the matrices' determinants is 0

so that's easier now probably

[quick oak]

i need to find determinant of each matrix

then ensure at least one of them is zero

[stable kindle]

yeah

it's a lot easier since they're simple diagonal matrices

and simple numbers

[dire thunder]

why you assume X is matrix

A = product of these two matrices

X is presumably vector in F^3

so do det of A which should be zero

[quick oak]

one is -a+b

other is -a-b

[stable kindle]

right, so at least one of those is 0

[quick oak]

sooo

what can i do from here

b-a = 0

or

-b - a = 0

is that the realtion?

[hard vault]

Man I feel so stupid... How do you prove that if W is a vector subspace of V with the same dimension then W = V?

[dire thunder]

finite dimensional?

anyway

just take basis for W and try to extend it to basis of V

[hard vault]

Well, you can't I guess because they have the same dimension. Right?

[wintry steppe]

if you consider adding nothing to it an extension then you can

[dire thunder]

any basis of W has the same length as basis of V thus spans V (by lin. independence)

and by definition it spans W

[hard vault]

Wait, how does it span V?

[dire thunder]

it is linearly independent and has length dim V

any such set of vectors spans V

[hard vault]

Can you argue like this: if it doesn't span V then I could always add elements to the basis of W. But they have the same dimension so it must span V?

Wait, doesn't the proof you "posted" use the same argument as mine?

[wintry steppe]

yes it kinda does

[hard vault]

Okay! I guess it's valid then. Btw I like you pfp, Carla

[wintry steppe]

thanks i like yours too

[zealous junco]

So for

can someone give an example where say $W_i, i \in I$ are subspace of $V$ and $W_i \cap W_j = 0$ pairwise but then $W_i \cap \sum_{j \in I\backslash {i}}W_j > 0$?

[stoic python]

Anticipation

[lilac stratus]

yeah

actually I can give you an idea on how to construct such an example

If you take two supplementary subspaces F and G

and H = Vect(f+g) for some nonzero f and g in F and G respectively

then F inter H = {0}, G inter H = {0} but F inter G+H = F + V = F

so like to give a concrete example

W1 = (0,1)R², W2 = (1,0)R², W3 = (1,1)R² and i would be 3 here

does that help @zealous junco ?

[wintry steppe]

Anyone good with vectors, i can pay

[lilac stratus]

(i'm actually not sure I understood the question right, since the notation saying that "a subspace being > 0" is kinda weird)

[native rampart]

Take W_1=span{e_1} ,W_2=span{e_1+e_2}, W_3=span{e_1-e_2}

mb

didn't see someone answered already

[lilac stratus]

(no prob)

[wind pasture]

is there such a W?

[wintry steppe]

count dimensions

if W and W^\perp were isomorphic, what could you say about their dimensions?

[wind pasture]

they are equal?

[wintry steppe]

right

is that possible?

[wind pasture]

why wouldn't it be?

[wintry steppe]

you tell me

R^5 is the direct sum of W and W^\perp

[wind pasture]

5 = dim(w) + dim(w perp) ?

[wintry steppe]

yeah

[wind pasture]

but since 5 is odd

its not possible?

[wintry steppe]

yup

[zealous junco]

oh yeaa ops thanks

3 vectors in R^2 would work haha i was being dumb

[dire thunder]

how you order subspaces tho

[lilac stratus]

well you need to chose them carefully, to make sure that atleast 2 of the intersections are equals to {0}

[lilac stratus]

did he speak about any order ?

[zealous junco]

its just an index set

[dire thunder]

0

[zealous junco]

oh proper subset

[lilac stratus]

oh yeah, that notation confused me too

he meant nonzero subspace ig

[zealous junco]

anyway does it make sense to say $\sum_{i \in I}W_i$ if I is not finite?

[stoic python]

Anticipation

[dire thunder]

well how would u define infinite sum of vector spaces?

[zealous junco]

ok right

[lilac stratus]

it may make sense in some topological context (not even sure), but it definitely doesn't make sense in the general case

like, $\sum_{n \in \bN}{\bR}$ wouldn't make anysense

[stoic python]

Shika-Blyat

[lilac stratus]

($\bR$ as a subspace of itself)

[stoic python]

Shika-Blyat

[zealous junco]

So if I want to write V = direct sum of W_i but I can have infinite many W_i such that this holds

[dire thunder]

wym

[zealous junco]

like lets say V is infinite dimensional

then I could find infinite index I so that W_i nontrivial and direct sum of W_i is V

i think?

[dire thunder]

[brittle juniper]

i don't see what prevents you from defining the sum of arbitrary many subspaces to be the subspace generated of the union of the subspaces

[dire thunder]

well assume we speaking abt P -- vector space of all polynomials

ye

then W_i = span(x^i)

[lilac stratus]

well actually that would make sense

[dire thunder]

and P = sum W_i

[zealous junco]

this is my prof writing

idk what I is here

that index has to be finite?

[dire thunder]

i do not think so

well

yes, seems it is not really important finite or infinte dimension (and thus index set) is

[zealous junco]

Then ok i guess the infinite sum does make sense for subspaces

its just that if w in that summed subspace then w = lin combo of finitely many of vectors in W_i

like it would be if $w \in \sum_{i \in I}W_i$ then $w = \sum_{i \in I}w_i$ where almost all $w_i = 0$.

[stoic python]

Anticipation

[lilac stratus]

(sorry ig what I was saying was totally wrong )

[zealous junco]

all good math is hard

quotient space is hard

[native rampart]

inb4 you learn all algebra is quotienting

[lilac stratus]

^ lol

[strange delta]

for part d) when showing the lhs is a subset of the rhs, is it ok to just say that linear combination of v1,v2,v3 is a subset of the complex field so there is no need to show anything

[zealous junco]

yea just say C^3 over C has dimension 3

[strange delta]

did i do this correctly? pretty sure it's wrong

this is my answer to part d

[hoary osprey]

you havent really show anything

[strange delta]

what should i do

[boreal crane]

do you need to show d)? from the picture you sent?

[strange delta]

yeah

[boreal crane]

You need to write those vectors in a 3x3 Matrix and do the Gauss algorithm if the rank of the Matrix is 3 your vectors are linear independent and every vector can be expressed by linear combination of them so the vectors are a base of C^3 @strange delta

[zealous junco]

well i mean if a) is true in your answer then so is d) if not then d is false

if your working in field of complex number

[strange delta]

the mark scheme says this but it just assumed it had a solution and went ahead

[lavish jewel]

it didn't assume anything, they showed it was invertible

[wintry steppe]

guys. I need help about the logic of linear-algebra-multiplication.

We're just multiplying the matrices the way it's introduced.

But just memorizing.

I've seen some 3B1B videos.

[lavish jewel]

have you learned what a dot product is and what it means?

[wintry steppe]

I know dot product from physics.

[lavish jewel]

all right. the definition of matrix multiplication is taking dot products between the rows of the matrix on the left and the columns of the matrix on the right

[wintry steppe]

yeap. I also know that.

[lavish jewel]

you can interpret it as a bunch of projections, then

[wintry steppe]

but it's interpreted this way.

[strange delta]

so is question is essentially just asking if theres a possible solution, not if the system system can satisfy everything in the complex field

[wicked lion]

can someone explain to me what the length of a vector is actually representing?

[lavish jewel]

no. it is asking if there is a unique solution for any x

[wicked lion]

like, ifth e length of the 2d vector (9, 2) is 9

what does 9 mean?

and how is that different from distance

[strange delta]

ah ok

[wicked lion]

im kind of stuck on the difference between length vs distance

[strange delta]

i guess i was overthinking it

[tame mural]

I would say length=norm

To be clear with language

and distance is a metric involving 2 vectors

So at least on a basic level

Norm is a function that needs only 1 vector

And distance requires 2

[wicked lion]

norm = ?

norm as in normalization?

[tame mural]

Norm = the "length" of a single vector

[lavish jewel]

norm is a type of length

[wicked lion]

how can a single point have any length though

🤔

[lavish jewel]

you represent the point by a vector that points from the origin to the point

[wicked lion]

a dot has no length

[lavish jewel]

the length of that vector is the distance of the point to the origin

[wicked lion]

its a dot

lol

[tame mural]

Yeah but it has a length to the origin

[wicked lion]

well that brings me to the next question: what is the origin?

the origin of what?

[lavish jewel]

of your coordinate system

[tame mural]

it's the identity element in your system

[lavish jewel]

additive identity, anyway

one refers to the origin as the 0,0,...,0 coordinate

[wicked lion]

ok so

the length is

the distance between the starting point of ur coordinate system?

i dont get it still

let me give an example

[tame mural]

Yes, but let's call that the norm to be clear

[wicked lion]

len(9, 2) = 9

[lavish jewel]

between the starting point and another point

distance requires 2 points

[wicked lion]

so if we know the length of the vector 9,2 is 9

what is 9?

9 movements to the right, on the x axis?

if its that, then why?

why isnt 9 elements up on the y?

[lavish jewel]

because you made that up

lol

[wicked lion]

hmm?

[lavish jewel]

the length of a vector is usually its euclidean distance to the origin

using the pythagorean theorem

you decided to make up a length that only takes the x coordinate

that's something different

not wrong, but if you made it up, you also have to check what properties it has

the traditional length of a vector [9,2] would be sqrt(9^2 + 2^2)

[wicked lion]

yea

[lavish jewel]

in general though, you can make up any kind of "distance" that is a "metric"

[wicked lion]

but like in this case sqrt(9^2 + 2^2) does = 9

[lavish jewel]

the interpretation of those, like the one you presented, is not necessarily straightforward nor meaningful

,w sqrt(9^2 + 2^2)

[stoic python]

Results provided by WolframAlpha

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

Requested by Edd#5182

[lavish jewel]

wolfram says you're wrong about that

[wicked lion]

probably just have my setup wrong then

in my code lol

probably an order of operations issue lol

ohhh wait

no i know why

[lavish jewel]

my guess is you mixed ints with floats

[wicked lion]

its coz im using int

yep

well no

its all ints

so it will floor it

9.2 becomes 9

[lavish jewel]

classic

[wicked lion]

but

in this case it's fine

it's just an example

and what i want to know is more just about

how can i use length, what is it, and how is it useful?

like

distance between vectors = USEFUL!

length of a vector = ?

like i dont see how getting the length of a vector is practically useful for anything

in the real world

I'm completely ignorant here on the use cases of it

[wicked lion]

the length of 0,0 is 0

[lavish jewel]

the length of a vector can be used to represent speed without a direction, electric field intensity, radiated power intensity, force

[wicked lion]

we know that much lol

the length of 9,0 i assume is 9?

[lavish jewel]

mhm

[wicked lion]

in your example i mean

ok so

0,0 length is 0 because we are at the origin

[lavish jewel]

you ever done trigonometry?

[wicked lion]

i did, a long time ago

lol

very long time ago

[lavish jewel]

you'll have to review it

[wicked lion]

i bet

[lavish jewel]

you're missing middle school level stuff here

[wicked lion]

ye

i've been afraid of math honestly

for a long time lol

not until recently did i decide i wanna try and get better xD

so its gonna be a steep learning curve probably

for a while

but uhm

let me make a video

of something real quick

https://youtu.be/kolmyllGQ-Q

@lavish jewel

[lavish jewel]

@wicked lion the second point is the origin

as you say, the distance from any point to itself is always 0, there is no distance

but vectors are usually built from a "head" and a "tail" as "head-tail", and if nothing is specified, you can just place the tail at the origin and the head at the point, so you have those 2p oints

[pallid rampart]

Consider the linear transformation $(\alpha_1,\alpha_2,\alpha_3,\dots)\mapsto\sum_{i=1}^\infty\alpha_i$

[stoic python]

Whoever

[pallid rampart]

Well why isn't it one?

Yes

So I think it's pretty clear that this is linear

And it's the sum is always finite since all but finitely many alpha are nonzero

No it's because it's a finite sum

[wicked lion]

you see, i'm coming at this from a very different background honestly. lol

but i think i get it now

the idea here would be like

[pallid rampart]

yes

[edgy kelp]

https://i.imgur.com/oY25H8J.png

This would be B

Not enough info

right?

[wicked lion]

alright so i finally understand size i think lol

https://www.youtube.com/watch?v=lCtH7fYwn9c made a video, hopefully my understanding makes sense 😄

[lavish jewel]

looks ok

[rare spade]

How is $||1||^2=2\pi$ ?

[stoic python]

Timur

Timur

[wintry steppe]

could you define what these mean?

[rare spade]

It is the 1-norm squared I think

Like $$||v||_2=\sqrt(x1^2+y1^2)$$

[stoic python]

Timur

[rare spade]

This is confusing asf

However, sqrt(1^2)^2=1 ?

[lavish jewel]

sure

[stable kindle]

i also do not understand where 2pi is coming from

[lavish jewel]

$\Vert 1 \Vert_2^2 = 1$

[stoic python]

Edd

[rare spade]

I don't get this then

[stable kindle]

what's the context for your original thing

[wintry steppe]

they probably mean the norm where you integrate a function's square on [0, 2pi]

norm on L^2[0, 2pi]

[stable kindle]

oh, that thing

yeah, just a weirder norm

[rare spade]

Yeah the context is in fourier series in [-pi, pi] -> R

1, cos(𝑥),sin(𝑥), cos(2𝑥),sin(2𝑥), . . . , cos( (𝑘 − 1)𝑥),sin( (𝑘 − 1)𝑥)

[wintry steppe]

oh, that

ok

[lavish jewel]

oh lol

[wintry steppe]

so what's the problem? write out the norm

[rare spade]

is orthogonal with this

[wintry steppe]

what's the integral of 1 on [-pi, pi]

[wintry steppe]

then you get this.

[lavish jewel]

$\Vert 1 \Vert_2^2 = \int_{-\pi}^{\pi} 1^2 dx$

[stoic python]

Edd

[rare spade]

Hmm

[rare spade]

Why is it two different results then

I don't get it

[stable kindle]

??

[rare spade]

$2\pi \neq 1$

[stoic python]

Timur

[stable kindle]

the same thing in different norms can have different things

[wintry steppe]

whatever you're reading is supposedly using the notation || ||_2 for the vector 2-norm, and || || for the function 2-norm

[lavish jewel]

because the two spaces are completely different

one of them is of square integrable functions

[rare spade]

Ohh

[lavish jewel]

the other was some euclidean R^n space

[rare spade]

Nice it is only written as a side note in the notes that were given out -.-

[lavish jewel]

and this is still not what you had above

but in general, that'll depend on the choice of normalization for the period

[rare spade]

Yeah the intervals differ (the last ss was from a different example)

Yep I get it now

[lavish jewel]

aight

[wicked lion]

Could someone show me some references (written or video format) on dot product and how they work with the different left/right handed systems?

I know I could google one, but I'm worried about the information being incorrect (since I don't know the material, I won't be able to know whether or not it's valid info).

[limber sierra]

what is meant by "the different left/right handed systems"?

dot products dont care about orientation

maybe youre thinking of cross products?

[wicked lion]

Oh yeah

Left and Right handed coordinate systems

https://www.youtube.com/watch?v=TGbMzoJqV7c

More useful in 3d stuff

[rare spade]

If I have a space with differentiable functions $f: [0; \pi] \rightarrow \mathbb{R}$ with the $L^2$-inner product $\langle f,g \rangle=\int_0^\pi f(x)g(x)dx$.
Then for $f(x)=1$ the projection of f(x) along sin(x) and sin(3x) is

Does it look correct I just wanna confirm

[stoic python]

Timur

[rare spade]

forgot sin(x) and sin(3x) -.-

[stiff marlin]

This is probably me being silly but how does the calculation for a1 work?

Apologies if this is the wrong channel haha

[pallid rampart]

Ok so

If a cubic polynomial has roots $\alpha_1,\alpha_2,\alpha_3$, then the polynomial is $(x-\alpha_1)(x-\alpha_2)(x-\alpha_3)$ right?

[stoic python]

Whoever

[stiff marlin]

Yes

[pallid rampart]

Now expand this

[stiff marlin]

I understand that bit, it is just the calculation for a1, I don't really understand the summation thingies

[pallid rampart]

Do you not know what that means or why a_1 is equal to the sum

[stiff marlin]

Kind of?

I just don't really understand the language

[pallid rampart]

Ok

[stiff marlin]

I understand it when it is written like xy + xz + yz

But obviously I want to use this kind of language because it is better shorthand

[pallid rampart]

So let's start out in general I guess

A sum $\sum_{i=a}^bf(i)$ is just defined to be $f(a)+f(a+1)+\cdots+f(b-1)+f(b)$

[stoic python]

Whoever

[pallid rampart]

$\sum_{k>j}^3$ is the same thing as $\sum_{k=j+1}^3$

[stoic python]

Whoever

[stiff marlin]

That makes sense

[pallid rampart]

Alright

[stiff marlin]

Yeah because it isn't bigger than or equal, alright yes that makes sense

[pallid rampart]

Yeah

Exactly

So if the bottom part was k≥j then the sum will be from j to 3 instead of j+1 to 3

[stiff marlin]

It would just be j to 3 twice

In the above thingy that I posted

Yeah

[pallid rampart]

Ok so if you expand the sum out then you will get $\alpha_1\alpha_2+\alpha_1\alpha_3+\alpha_2\alpha_3$

[stoic python]

Whoever

[pallid rampart]

Oh one thing to mention

If j=3 then you might get $\sum_{k=4}^3$

[stoic python]

Whoever

[pallid rampart]

In which case the sum is defined to be 0

It is defined to be 0 whenever the bottom limit is strictly greater than the top limit

[stiff marlin]

Oh alright, I will be sure to note that

So how would you do a double summation?

[pallid rampart]

Well

The same way

[stiff marlin]

It would just be that thingy you put above

And then sub in the already known values of a3 and a2

[pallid rampart]

$\sum_{j=1}^3\sum_{k=j+1}^3\alpha_j\alpha_k=\sum_{k=2}^3\alpha_1\alpha_k+\sum_{k=3}^3\alpha_2\alpha_k+\sum_{k=4}^3\alpha_3\alpha_k$

[stiff marlin]

Oh alright, and then the k = 4 one is 0

[pallid rampart]

Yeah

[stiff marlin]

So you are just left with the first two

I see my mistake now

Thank you for explaining this!

[stoic python]

Whoever

[pallid rampart]

Sorry I made a typo

You're welcome

[stiff marlin]

Oh that is fine, I get the idea

I'm trying to teach myself maths

[pallid rampart]

Nice

[stiff marlin]

Currently at A level and they aren't very rigorous, we were taught this but I just didn't understand it when using summation notation

[pallid rampart]

Ah yeah it's really frustrating when people just use notations they haven't defined

[stiff marlin]

Haha yeah it is

Do you have any advice for learning all of this notation stuff?

[pallid rampart]

Google, ask people like on this discord, or read more books

[stiff marlin]

Yeah that is true

I always seem to pick books that are either too difficult to follow or ones that are a cake walk

Video lectures are a good source

[pallid rampart]

True

What do you know right now

I'm not really familiar with the A level thing

[stiff marlin]

Are you in the US?

[pallid rampart]

Yes

[stiff marlin]

It is kind of similar to there, we do trig approximations and binomials and basic differentiation and integration

I could send you a picture of the contents page of my maths syllabus if you want?

The A level course goes more into harder concepts but the language used it really vague

[pallid rampart]

Huh I see

I mean maybe you should continue reading about calculus

Maybe also read number theory, linear algebra

Just realized your problem isn't linear algebra

[stiff marlin]

Yeah I messed up with the channels and then realised after and kind of just tried to own it haha

For some reason my brain was like preliminary algebra

Also, now that you mention channels, we probably shouldn't be talking in here

My fault haha

[pallid rampart]

no big deal

[stiff marlin]

Thank you once again, now back to this book

[wintry steppe]

can someone help me with this

why wouldnt the rank of A be 0? why is it 1?

nvm now that i look at it the null space is linearly dependent

[wicked lion]

Finally did it

[wintry steppe]

Cool

[lapis cosmos]

Since the rank A is at most 3 therefore there exists the 3*3 “sub” matrix B of A such that the rest 4 rows of A can be linear generated by rows of B. Then rankA =rankB now dim N(A)=dim N(B)=2 therefore rankB =3-dim N(B)=1

I made it unnecessarily complicated... just that rankA is n=3 -dim N(A)=3-2=1😂😂

[lavish jewel]

you need 9 degrees of freedom for all 3x3 matrices

what you found is for symmetric matrices

[wintry steppe]

Not the. Basis are not unique except for dim 0 (and dim 1 over Z2)

[lavish jewel]

looks ok then

[sonic osprey]

It's true for all vector spaces

[boreal crane]

Jup its true for all K-Vectorspaces the Dimension theorem is Just your equation: dim(S+U)=dimS+dimU-dim(S∩U)

[broken sun]

Hello. Is there an example of a vector space $V$ and subspaces $U_1$ and $U_2$ of $V$ such that $U_1 \times U_2$ is isomorphic to $U_1+U_2$ but $U_1+U_2$ is not a direct sum?

[stoic python]

MathPhysics

[lilac stratus]

no, that's not possible

since $dim(U_1 \times U_2) = dim(U_1) + dim(U_2)$ and $dim(U_1 + U_2) = dim(U_1) + dim(U_2) - dim(U_1 \cap U_2)$

[stoic python]

Shika-Blyat

[lilac stratus]

and being isomorphic <=> same dimension

so if they're isomorphic, then $dim(U_1 \cap U_2) = 0$

[stoic python]

Shika-Blyat

[lilac stratus]

and well, then that's a direct sum

@broken sun

[broken sun]

Thanks. But, I did not assume that the vector space is finite-dimensional. Do these identities hold for infinite- dimensional case?

[lilac stratus]

the identities does hold, but then i can't anymore deduce that $dim(U_1 \cap U_2) = 0$.

[stoic python]

Shika-Blyat

[lilac stratus]

In the infinite dimensional-case, U_1 x U_2 ~ U_1 + U_2 but U_1 cap U_2 != 0 is possible yeah

[sonic osprey]

In that case you can and its not too hard, just take R^infinity and U_1, U_2 to be infinite dimensional overlapping subspaces

[lilac stratus]

or like U_1 = R[X] and U_2 = R_n[X] for any n

[broken sun]

Any two overlapping subspaces work?

[lilac stratus]

mmh, yeah

as long as atleast one of them is infinite dimensional

[broken sun]

I cannot see. Can you please elaborate?

[lilac stratus]

So if you take $E = \bR[X]$, and like $U_1$ any infinite dimensional subspace of $E$

[stoic python]

Shika-Blyat

[lilac stratus]

And let $U_2$ be any subspace of $E$

[stoic python]

Shika-Blyat

[lilac stratus]

Then, by taking a basis $(e_i)$ of $U_1$ and a basis $(f_j)$ of $U_2$, then $(e_i) \times (f_j)$ is a basis of $U_1 \times U_2$. But since $card((e_i)) = dim(E)$, and $card((f_j)) \leq dim(E)$

[stoic python]

Shika-Blyat

[lilac stratus]

so $card((e_i) \times (f_j)) = dim(E)$

[stoic python]

Shika-Blyat

[lilac stratus]

and $card((e_i) \times (f_j)) = dim(U_1 \times U_2) = dim(E) = dim(U_1) = dim(U_1 + U_2)$

[stoic python]

Shika-Blyat

[lilac stratus]

but that assumes thatt $U_1$ and $U_2$ have basis, so like if you want to avoid this, just let $U_1$ be a subspace with an explicit basis, just like $E$ entirely with the basis $(X^n)_{n \in \bN}$, or any subspace generated by an infinite subfamily of this basis

[stoic python]

Shika-Blyat

[lilac stratus]

and take $U_2$ to be the subspace generated by the family of polynomials you want

[stoic python]

Shika-Blyat

[lilac stratus]

is this clearer ? @broken sun

[broken sun]

Thank you very much for your help.

[lilac stratus]

what the values of the first row of AB are going to be ?

ok, how do you computed it ? @hollow grove

I mean, what are the calculations you did to get these values

[zealous junco]

just take first row of A

you did $6R_1+9R_2+0R_3-7R_4$ where $R_i$ is rows of B to get the first row of AB

[stoic python]

Anticipation

[zealous junco]

i.e. lin combo of the rows of B by coefficient of first row of A

[lavish jewel]

a nice and compact way is to start from wAB and force it to be a row from AB, which is trivially done by setting w equal to a canonical basis vector and associating (AB)

then just associate (wA) = v and you're done

if you choose the vector w = e1 = [1,0,0], you get the solution anticipation gave

[quasi vale]

wdym?

[sonic osprey]

you need all three lines to intersect for it to be a solution

like the bottom intersection is only a solution to the red line and the purple line

so if your system is all three lines, then its not a solution

You're basically looking at two different system of equations here

solution 1 is a solution to the system that has the purple line and green line

but it isn't a solution to the system that has all three lines, since it doesn't satisfy the red line

similarly, solution 2 is a solution to the system that has the purple line and green line

but it isn't a solution to the system that has all three lines

[zealous junco]

this lemma and proof r both valid right? and could i simplify

[sonic osprey]

this is vaguely how projective geometry works

you say that parallel lines intersect at "infinity"

[sick tartan]

is it okay to use linear algebra approach if your subject is calc 3?

[dire thunder]

wym

[shut sigil]

By any chance would anyone know what the answer to this would be?

or how to solve it?

[gritty swift]

wrong channel, but you just take the slope then solve for k

[shut sigil]

that's what I've been doing but for some reason it still doesn't add up

wait nvm

I just realized my issue

[gritty swift]

$\frac{2k-3k}{8 - (k-1)} = 2/5$

ok lol

[stoic python]

uli

[quartz compass]

really depends on what they're solutions to

[dusky epoch]

(x,y,z) and (X,Y,Z) are solutions to what?

a system of linear equations?

great

your answer for b is incorrect

[wintry steppe]

it is actually possible. Consider the system of linear equations 0=0 over Z2

[dusky epoch]

we're working over the real numbers.

no finite field stuff.

if 25 planes have two points all in common then they also all pass through the line that goes through those two points @wintry steppe

[zealous junco]

ah i guess i need to show B_2\B_1 is lin indep

and $B_2\backslash B_1 \cap W = \emptyset$

[stoic python]

Anticipation

[wintry steppe]

Anyone good at vectors, dm me - I can pay

[zealous junco]

nvm this doesnt work

[uneven shore]

<@&286206848099549185>

can anyone help me simply this expression

please ping me if anyone can help

[wintry steppe]

yea, do you know the axioms of a vector space?

[uneven shore]

nooo 😭😭

[plain saffron]

From

#❓how-to-get-help

Rule 4

If your question has not been answered for a minimum of 15 minutes, you may use the Helpers tag once. Please do not try to bump your question using this ping unnecessarily. Do not abuse this ping. Do not individually ping users with the Helpers tag without their express permission.

[limber sierra]

that said, do you know how to simplify if x, y, z are regular ol' variables instead of specifically vectors?

since the process is the same

[uneven shore]

oooohhh

yeahhhhhh

[uneven shore]

is this rightttt

[limber sierra]

yes.

[uneven shore]

okay thank you!!

[dusky epoch]

one small typo

the arrow isn't supposed to jump from the y to the 5

[uneven shore]

Oooh okay thank you!!!

[zealous junco]

i think this proof is correct, anyone can check?

this is question

[next vapor]

how do i show that u1 tensor v1 is equal to u2 tensor v2 if and only if there exists a c in the field such that u1=c u2 and v1=c^-1 v2, where u1,u2 and v1,v2 are in finite dimensional vector spaces U and V?

im guessing this is from the definition of a tensor product but really not sure how to link it togethor

[wintry steppe]

I need help with my math

[next vapor]

channel is in use, please try another one

[wintry steppe]

it could mean scalar multiplication by 0 but that's trivial

[native rampart]

You want u_1(x) v_1(y)=u_2(x) v_2(y) for all x,y in domain for u_1 tensor v_1 to be equal to u_2 tensor v_2.
Now pick a y such that atleast one out of v_1(y) and v_2(y) is nonzero . Let v_2(y) be nonzero. Then v_1(y)=cv_2(y) for some c in F,i.e.,c u_1(x) = u_2(x) for all x

You end up with u_2=cu_1 and cv_2=v_1

[nocturne jewel]

Im confused on how singular value decomp works algorithmically, I think I got the sigma and Q* matrices, but how do I get P? $A=P\Sigma Q^*$

[stoic python]

moshill1

[nocturne jewel]

My notes say compute $w_i=\frac{1}{\sigma_i}Aq_i$ for the columns of Q then extend to an orthonormal basis

[stoic python]

moshill1

[next vapor]

wait wdym by the first line, You want u_1(x) v_1(y)=u_2(x) v_2(y) for all x,y in domain?

[native rampart]

Yes

iirc, f tensor g (x,y)=f(x) g(y)

[next vapor]

uh we didnt really define it like that hm

the way it was defined was a quotient space of the free vector space F(VxU) quotiened by the subspace U with elements of certain properties

honestly idek how to treat elements of the tensor i just know their properties and how theyre defined

[native rampart]

Yea,in context of finite dimensional spaces,free space just means a normal ordinary vector space

There are no non free finite dimensional vector spaces

[next vapor]

wait this isnt their product right

its the cartesian product?

[native rampart]

f(x) and g(y) are elements in F. f(x) g(y) is just a product in F

[next vapor]

i figured out the reverse direction its just the forward one

[native rampart]

(x,y) is a member of a Cartesian product

[next vapor]

hmm

okay so their tensor being equal implies that for every bi linear map in their dual space F we have F(u1,v1)=F(u2,v2) right

[native rampart]

I think so

[next vapor]

but then what do i do from here lol

[native rampart]

I think you take a particular bilinear map which simplies your situation

Let's say e_1,e_2,...e_k is basis for V

Consider the bilinear form f such that
f(u_1,e_1)=1,f(u_1,e_2)=1... f(u_1,e_k)=1
and f(x,v)=0 for all x not in span{u_1} and for all v

This will force u_2 to be cu_1 since otherwise one side of this bilinear product will be zero,while the other is nonzero

[next vapor]

so that means f(u1,v1)=a1+...an, where v=a1e1+...an e_k right

and since we must have F(u1,v1)=F(u2,v2) this means that u2 has to be in the span of u1

okay i see it

but then how do we show the same for v2=c^-1 v1

like what input of F do i consider

[native rampart]

Repeat the same thing

Consider f such that f(e'_1,v_1)=1,f(e'_2,v_1)=1... f(e'_m,v_1)=1 where that's the basis and f(x,b)=0 b not in span {v_1}

This would imply v_2=c'v_1

[next vapor]

okay yea that makes sense

idk how on earth i was meant to come up with that

[native rampart]

now f(u_2,v_2)=cc' f(u_1,v_1) for any bilinear form

=f(u_1,v_1)

Now just take a nonzero bilinear form and you end up with cc'=1

[native rampart]

Yea, That's a very common FDVS technique

Choose a function that makes your life easier

[next vapor]

like, that specific function selection is common?

or the technique

[native rampart]

Both

[next vapor]

fair

thanks alot

[nocturne jewel]

Is there a way to find an orthonormal basis consisting of eigenvectors of a matrix without finding an eigenbasis and orthonormalizing it?

(Question for ref)

[spiral star]

not in general. but you wont have to use gram schmidt in this case

[nocturne jewel]

Yeah nudge on how I dont need gram-schmidt please?

Cause Ik C*=C so ik that there is an orthonormal basis by Real Spectral

[spiral star]

C is self-adjoint so its eigenspaces are orthogonal

you have 4 distinct eigenvalues

so one eigenvector from each space

any choice of an eigenbasis will automatically be orthogonal you just have to make them unit length

in general, for self adjoint maps, you only have to orthogonalize the vectors from the same eigenspace

so if i have v1,v2 from eigenspace E1 and v3 from eigenspace E2 then i only have to orthogonalize {v1,v2}

(as an example)

[nocturne jewel]

Ok might've just messed up finding the eigenvalues then

[spiral star]

let's be lazy for a moment

uuuh

,w eigenvalues {{a,0,b,0},{0,a,0,2b},{b,0,a,0},{0,2b,0,a}}

[stoic python]

Results provided by WolframAlpha

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

Requested by Flow#8160

[nocturne jewel]

yeah I messed up the chacteristic poly lol

[spiral star]

that looks distinct

[nocturne jewel]

I got complex lambda

[spiral star]

anyway, every eigenbasis will be orthogonal since C is symmetric

and you have distinct eigenvalues

so you wont have to worry

[nocturne jewel]

[acoustic zodiac]

how can i show that a projector is orthogonal over a certain subspace?

[nocturne jewel]

P is orthogonal over V iff $\langle u, P[v]\rangle=\langle P[u],v\rangle , u,v\in V$

[stoic python]

moshill1

[nocturne jewel]

(iirc that that's the characterization)

[north steeple]

Shouldnt mine be right as well?

[acoustic zodiac]

no, because from a) you can't get (0,1,0)

[north steeple]

ok thanks

what about this one?

im just checking to see if they got it wrong

[acoustic zodiac]

in this case you have that every vector follows (a, a, 2a, d)

therefore you get: a(1,1,2,0) + d(0,0,0,1)

so the dimension is 2

[wintry steppe]

Ok

Wow

Great 👍

C

[north steeple]

wow ok

but this one im sure that I cannot be wrong

how am i wrong

[wintry steppe]

A+C=D

Ur right

[north steeple]

i just plugged in 0 for t and the deteminant is 0

[acoustic zodiac]

plug it for -1

[north steeple]

why

[acoustic zodiac]

to check if answer d) is correct

[north steeple]

but what is 0,-1

how do u even use that

[acoustic zodiac]

it's 0 and -1

not a decimal

[wintry steppe]

Yep

[north steeple]

oh i thought they ment like a pair

wth

ok

last one

[wintry steppe]

Where is it

[north steeple]

waitr

[wintry steppe]

K

[north steeple]

im not sending it

because i feel dumb

i got it wrong but it was so easy

lol

[wintry steppe]

Just send it

[north steeple]

i dont like this class

im happy i have no more math

[wintry steppe]

K

[acoustic zodiac]

math is fun tho

[north steeple]

some math

but once we got to vector spaces

i was like i hate this

[acoustic zodiac]

vector spaces make a lot of sense once you get them

you just have to get them

it's not like, idk, calculus, even if you know how to integrate, lots of integrals are going to be hard

[lilac stratus]

you can ? [1, 1, 1] - [1,0,1] ?

[acoustic zodiac]

oh shit you are right, guess the teacher was wrong

[lilac stratus]

Yeah I think that's also a basis

[acoustic zodiac]

it definitely is, it's just not the usual one

[lilac stratus]

(pong @north steeple)

[north steeple]

i dont wanna learn it tho

like i dont think ill ever really use it for my career

[acoustic zodiac]

anyways yeah, a) is indeed a basis

[north steeple]

wait

so mine is also correct

i jus saw that now

[acoustic zodiac]

yep

[north steeple]

ok

[lilac stratus]

yeah, pinged you for that reason lol

[north steeple]

i was gonna say i swear its correct

oh lol thanks

[acoustic zodiac]

well let me think for a sec

[north steeple]

i remeber doing it

[acoustic zodiac]

i mean, to get something like (3,1,3) you'd have to first do (1,1,1)-(1,0,1) and then add 3(1,1,1)

which means you can't get all vectors in the subspace with just those two vectors

[north steeple]

oh

so then its wrong

[acoustic zodiac]

well, not necessarily, this is a bit subtle, in fact

[north steeple]

im gonna email anyways to try and get a mark for it

[acoustic zodiac]

a basis of E is "a set of vectors that belong to E, that span E, and that are linearly independent"

a) does span E using a finite number of vectors, (1,1,1) and (1,0,1) obviously belong to the space, and they are linearly independent

therefore it does follow the definition of basis

[north steeple]

yes that does make kinda make sense

[rough stag]

Anyone can help me with 10-12?

[north steeple]

thanks

[acoustic zodiac]

#multivariable-calculus

[rough stag]

Thanks. Wasn't sure which section this would belong.

[wintry steppe]

I've been stuck on this for a while

how do i manipulate the right side to get x'?

[wintry steppe]

Anyone good at vectors, dm me - I can pay

[nocturne jewel]

No

[zealous junco]

Try showing $(A-uv^\top)(x+\alpha A^{-1}u) = b$

[stoic python]

Anticipation

[wintry steppe]

let me look

[zealous junco]

thats because $x' = (A-uv^\top)^{-1}b$ and you want to show $x' = x+\alpha A^{-1}u$ also, so you equate the 2 and try to show they are actually equal

[stoic python]

Anticipation

[wintry steppe]

hey can i dm you if i get stuck @zealous junco ?

[zealous junco]

yea

[safe locust]

anyone know?

[wintry steppe]

find eigenvalue and eigenvectors first

[safe locust]

ok

now what

[wintry steppe]

D = diag(eigenvalues) and P is the eigenvectors

[safe locust]

so D is the eigenvalues diagonalized ?

[wintry steppe]

yes the eigenvalues will be along the diagonal of D

[safe locust]

ohhh

nvm nvm

i was confused what u were saying

okok

[safe locust]

@wintry steppe you know how to solve this one?

<@&286206848099549185>

[safe locust]

anyone lol?

[wintry steppe]

Remember the definition of a linear transformation

[safe locust]

so how would i do this prob?