linear-algebra

[rose umbra]

I believe its the first one but i dont know why, can anyone tell?

[teal grotto]

all symmetric matrices over the reals are diagonalizable but just because a matrix is not symmetric doesn’t mean it’s not diagonalizable.

[lavish jewel]

i agree it seems to be the first that is false

[teal grotto]

the last three options are true, so the first must be false.

[lavish jewel]

ninja'd

[teal grotto]

lol

that just doesn’t seem satisfying tho

i think it’s because C!=E that the first one is wrong. you would need a change of basis matrix to find out what [T(v_1)]_E is

[lavish jewel]

right, you get coords in the C basis

you'd need to transform them once more still to get to E, e.g. with a matrix whose columns are the u_i

[wintry steppe]

How to solve these equations for x y z w

[nocturne jewel]

is theta known?

[wintry steppe]

No actually i am proving that inverse of rotation matrix exist by finding unknowns in the inverse matrix

By comparing it to R(theta)R(theta) inverse=I

[lavish jewel]

an easy way by inspection would be to try stuff like letting y and w be cos theta and sin theta, for example

that'll give you cos^2 theta + sin^2theta = 1 for the last eq

idk if that works for the others tbh

[teal grotto]

it works for the second equation

[wintry steppe]

Is there any other way to prove inverse of rotation matrix exist

[lavish jewel]

rotate it back in the opposite direction? finding the determinant (reasonable in 2D)

[teal grotto]

take x=cos(t) and z=-sin(t) for the other equations

[wintry steppe]

Okay

[teal grotto]

i was about to suggest case work 🤮 thank god you said something before i did.

[lavish jewel]

[wintry steppe]

Actually problem is . We know the inverse of rotation matrix is the transpose of itself thats why i k we can take x=cos(theta) z=-sin(theta) and this will easily solve the problem . But can we prove it without this eqn stuff n equating it with I

[lavish jewel]

why not do the usual inverse of a 2x2 matrix?

[teal grotto]

i believe this is a special case of a more general property of orthonormal matrices

an orthonormal matrix is one whose rows and columns are orthonormal vectors. if you have an inner product space and an orthonormal matrix A, then AA^t = Id = A^tA.

[wintry steppe]

Actually i haven't come across the concept of orthonormal matrices 😅. Just started the 2 chap of s lang intro to linear algebra

[teal grotto]

well, i mean, you secretly have if you’re working with rotation matrices

if you can show that the dot product of any column with itself is one and the dot product of any column with a different column is 0, then you’re done.

[nocturne jewel]

The inverse of a rotation matrix is just rotating the opposite way

also it's the 2x2 matrix, so the formula for the inverse is known explicitly

$\begin{bmatrix}a&b\c&d\end{bmatrix}^{-1}=\frac{1}{ad-bc}\begin{bmatrix}d&-b\-c&a\end{bmatrix}$

[stoic python]

Mosh

[wintry steppe]

@teal grotto okay i will try it anyways thank you all of u

@nocturne jewel okay

[quiet heron]

Hey guys, I have a question about the rank of a matrix. Is the rank of a row space simply equal to the number of independent rows? Also, are the column rank and row rank always the same? Is the rank of the matrix just equal to the rank of the row or column space?

[wintry sphinx]

the inverse of a rotation matrix is its transpose

[nocturne jewel]

Yeah ik

[wintry sphinx]

and you prove this by noting that the columns are orthonormal

[nocturne jewel]

Im aware....

I was answering the person that actually had the question

[teal grotto]

@quiet heron

1. the dimension of the row space is the number of linearly independent rows.

2. yes. one way to see this is because of gauss jordan elimination and row equivalence.

3. the rank of a matrix is the number of linearly independent columns, so yes it would be the dimension of the column space.

[quiet heron]

Gotcha. Thanks for the answer!

What's the difference between dimension and rank?

[teal grotto]

the rank of a matrix is the number of linearly independent columns of a matrix, or equivalently, the dimension of the image. you don’t talk about the rank of a subspace, you talk about the dimension of a subspace.

[nocturne jewel]

dimension is the size of a basis
rank is dimension of the column/row space

[quiet heron]

Gotcha

[gloomy nebula]

What's the difference between part a & b? For a, I determined that B is in the span.

Then, for the next part, can't I just set B=[3,2,1] and call it a day?

[native rampart]

B could also not be [3,2,1]

[hollow finch]

i think they might want you to check if no matter what the values of a,b,c are, it will still be in the span

another way to phrase it is "does there always exist a solution to the system regardless of the value of B?"

[gloomy nebula]

How did you figure that out so quickly? I just rrefed the matrix

[dire thunder]

figured out what?

b) actually asks you whether (X_1, X_2, X_3) is basis for R^3

[gloomy nebula]

That [a,b,c] is in the span of those vectors.

Before starting part b, I assumed there would have to be restrictions for a,b,c. Turns out after rrefing the matrix, It works for all values

[dire thunder]

if there is no stated restrictions on [a,b,c] u assume that a,b,c are any from you field

and it happens so that indeed any vector of the form (a,b,c) is in the span

[gloomy nebula]

Where did you get LI from?

Bc you said basis and I only wanted to show that B is in span

[dire thunder]

LI?

oh

[dire thunder]

you have list of three vectors spanning R^3

it is forced to be linearly independent

since R^3 has dim 3

dimension is basically the length of smallest spanning list

[gloomy nebula]

ahhh rank-nullity I see

[dire thunder]

wym rank nullity

no

just definition of basis

[gloomy nebula]

the matrix has 3 pivots

[storm vessel]

Help me plis 😦

[nocturne jewel]

Find how to represent $\begin{bmatrix}4&3\2&1\end{bmatrix}$ as a coordinate vector of B

[stoic python]

Mosh

[tall oasis]

how would i prove or disprove that A and B are similar ?
they have the same Eigenvalues, Eigenvectors, Dimension , Rank and Determinant.
so i am thinking they are similar but not sure how to prove it.

[lavish jewel]

two matrices are similar if there exists a P such that A = PBP^-1

if they have the same eigenvalues and eigenvectors, you already have that AP = BP = PD, with a diagonal matrix D that has the eigenvalues along its diagonal

maybe you can use that in some way

[tall oasis]

i knew this part had no idea a about the digonal matrix thing, but i am not sure how to find P, for 2x2 matrix it is usually easy to figure it out but for a 4x4 matrix it seems really hard.
is there a method or an algorithm i can use ?

[lavish jewel]

you can do it in software

also, are you sure they have both the same eigenvalues and eigenvectors?

that just kinda sounds like they're the same matrix

[tall oasis]

yes i calculated it by hand and then used an online calculator

the issue is this question has only 2 points on it, and it just seems too hard for 2 points so it feels like i am missing something obvious

[lavish jewel]

i guess it could be issues with numerical precision, but

[zealous junco]

cant you find their span?

[tall oasis]

A and B are defined for rational numbers does it make sense for P to be non rational ?

[lavish jewel]

that makes a HUGE difference

i forgot it was over Q

ignore the image i shared

[tall oasis]

ya it is in the picture 😅

[lavish jewel]

i was negligent

[zealous junco]

wait ignore i am dumb af

[tall oasis]

row reductions would change the eigenvalues

[lavish jewel]

seems kinda weird to me, if BP = AP and P is a full rank matrix, then you can multiply by P^-1 from the right and get that A = B

but i never with with Q, so i defer to someone who knows what they're doing

[tall oasis]

ok ,thanks for trying

[zealous junco]

could u use jordan decomposition? I dont know much about it though

[tall oasis]

I just googled it and if my prof. expects me to do this for 2 points on an exercise sheet then he is an asshole 😂

[ebon veldt]

The number of eigenvectors aren’t the same right?

[teal grotto]

so they have the same eigen values. are they both diagonalizable?

[tall oasis]

That is true i just double checked their geometric multiplicity is not equal

[tall oasis]

neither is diagonalizable but i think i have an answer now they are not similar because B has a geometric multiplicity = 1 and A = 2

[wintry steppe]

I am not able to proceed any further can someone help ?

Chapter - Vectors
Topic - Triple Product

[lost dragon]

@wintry steppe
From step a) to b), the “plus” changed to a “cross”.

[wintry steppe]

Can you use doodle or any marking so I can know which step are you talking about ?@lost dragon

[primal fable]

What's the reasoning behind the cross product existing only in the R^3 and R^7?
I think the best explanation I've seen is that when we add say some 4th perpendicular axis, say w, we need (?) 3 more dimensions as w forms cross products with the other axes.

But why this is necessary isn't clear to me.

Like why can't the following be true instead:
$$\hat{x} \cross \hat{w} = \hat{y}$$
$$\hat{y} \cross \hat{w} = \hat{z}$$
$$\hat{z} \cross \hat{w} = \hat{x}$$

[stoic python]

Thomas

[primal fable]

Or something similar

[glacial mango]

How do you show that a linear map with a trivial kernel is surjective (assuming domain and codomain are of equal dimension)?

[teal grotto]

i wanna say division algebras right? somebody more qualified can correct me if i’m wrong tho

like, the cross product is based off of some properties of quaternions and octernions or something and it only plays nicely in some dimensions.

[nocturne jewel]

dimension formula

[glacial mango]

So that gives you that the dimension of the codomain and the dimension of the range of the linear map are equal.

How does that prove surjectiveness?

[teal grotto]

because the image is a subspace of the vector space with the same dimension as the vector space

[nocturne jewel]

$dim(V)=dim(W)=dim(Im(T))$ if $T:V\to W$ is injective

[stoic python]

Mosh

[nocturne jewel]

so W and Im(T) are isomorphic

[glacial mango]

So I'm trying to do it from definition, if I take an arbitrary vector in the codomain how do I know something maps onto it given the assumptions

[nocturne jewel]

oh duh, if $dim(V)=dim(W)$, then V is isomorphic to W, which implies there is an invertible map between them. So the map is injective and surjective

[stoic python]

Mosh

[nocturne jewel]

basically it's proving that characterization of invertible operators holds so long as domain and codomain are isomorphisms

[glacial mango]

I'm trying to prove that first line. You're assuming it.

well kind of

I'm trying to prove that if I have a fixed linear map and the kernel is trivial, then it's an isomorphism.

Without the theorem that there exists an isomorphism if the domain and codomain are of equal dimension

[teal grotto]

$\text{im}(T)\subseteq W$ and $\dim\text{im}(T)=\dim W$ so $\text{im}(T)=W$

[stoic python]

coycoy

[glacial mango]

OK, I think that's what I'm looking for.

[nocturne jewel]

yeah there we go

dimension formula then the subset stuff

[teal grotto]

or you could go overkill and apply the first isomorphism theorem

[glacial mango]

Is there no way to prove it from the definition of surjectiveness?

[teal grotto]

i mean, that is the definition…the image is the whole space W

[glacial mango]

well I guess that gives it to you

[nocturne jewel]

surjective maps are characterized by im(T)=codomain

[glacial mango]

Yea I just realized that Axler defines surjectiveness like that

I thought it was like the set-theoretic definition.

Which is corollary now

I guess

[nocturne jewel]

ker(T)={0} means injective
im(T)=W means surjective
both means bijective

[teal grotto]

only when dealing with linear transformations from two isomorphic vector spaces

[glacial mango]

I was trying to construct, I guess explicitly, the vector that maps onto the arbitrary vector I chose, but I got stuck because I don't know what maps onto the basis vectors in the codomain for an arbitrary linear map.

So I guess, how do I prove that ^ ?

Now that it's shown that the map is surjective as defined with imt(T)=W

[teal grotto]

choose a basis for v, call this v’=(v1,…,vn)

since T is, in particular, injective, then w’=(T(v1),…,T(vn)) is a basis of W

[glacial mango]

Wait, how do you know that it maps onto a linearly independent set?

[teal grotto]

injective maps always map linearly independent families to linearly independent families

[glacial mango]

OK, I'll review that theorem after. I'll take it for granted for now.

[teal grotto]

it’s not too hard to show.
anyway, take your w in W. write it in terms of the basis vectors of w’

then use linearity of T and apply T inverse to w and the expansion of w in terms of w’

[glacial mango]

dope

thank you

[teal grotto]

yessir

[idle tide]

can i swap columns in a matrix?

[nocturne oracle]

no its illegal

[teal grotto]

you can but you will most likely change the kernel of the matrix

[idle tide]

whats the kernel?

[teal grotto]

the set of vectors x such that Ax=0

[nocturne jewel]

kernel / nullspace

it's a LinAl term, further down the line

[idle tide]

yeah i have no clue what that means looool

[teal grotto]

the reason you can swap rows in for instance, row reduction, is because row operations don’t change that set.

[nocturne jewel]

they just started LinAl coy btw

[wintry steppe]

[teal grotto]

then disregard

lo siento

[idle tide]

ok, so just stay away from column swapping for now, got it

[glacial mango]

Does column swapping preserve anything?

[wintry steppe]

[glacial mango]

What about row swapping?

[idle tide]

row swapping is fine i think

[teal grotto]

i believe it preserves the image

[wintry steppe]

it preserves the absolute value of the determinant

[glacial mango]

row swapping changes the sign, right?

[wintry steppe]

column swapping preserves the image, yes

[teal grotto]

swapping anything changes the sign of the det

[idle tide]

wait im confused, does row swapping change anything? like if i row swap do i have to do anything

what's the det?

[glacial mango]

It changes certain properties about the matrix that you will learn about later.

[teal grotto]

oh no

[glacial mango]

I guess no need to worry about it now.

[idle tide]

oh so i shouldnt column OR row swap for now

ok

got it

[teal grotto]

no no no. row swap definitely for row reduction

[glacial mango]

If you're just doing row-reduction, it's fine.

[idle tide]

oh

ok

yeah im doing gauss-jordan elimination

[glacial mango]

But like if you were going to calculate the determinant (whatever it is) you want to be careful with that

[warm yarrow]

Row swapping would just be writing the linear equations in a different order

[wintry steppe]

the determinant is a function of square matrices that tells you various things such as invertibility and, at least in R^n, how the linear mapping changes the volume of the unit cube

[idle tide]

okkkkkkkkkk that makes sense why when i column swapped the answer made no fucking sense

[teal grotto]

let’s just dump all of linear algebra on him rn

[warm yarrow]

And swapping columns would be something like x y z to x z y

[glacial mango]

the rank-nullity theorem in essence is....

[wintry steppe]

the spectral theorem in C states...

[teal grotto]

a special case of the first isomorphism theorem

[glacial mango]

stfu

[wintry steppe]

the determinant is very geometric and you should not let anyone convince you otherwise

[idle tide]

@teal grotto that's basically what that last 9 hours of my life has been...my book threw this shit expecting that I already knew how to do it. IDK if we were supposed to learn this in alg 1 or not because we didnt, and im sure i didnt miss an intro lesson to this unit, last lesson i was working with complex numbers and imaginary units which the book throroughly explained but i literally have been stuck on matrices and gausshittian elimination for the last day and a half

[wintry steppe]

why are you sullying me coycoy

[idle tide]

or im prob just stupid

[glacial mango]

*linear algebra

[wintry steppe]

agreed

[teal grotto]

idk. what even is a volume

[wintry steppe]

[mild crypt]

This is a chat

[glacial mango]

some differential form, right?

[wintry steppe]

integral of 1 over the unit cube

taken against the standard volume form in R^n

[teal grotto]

damn higher math strikes again

[wintry steppe]

(so more correctly, the integral of the volume form dx^1 \wedge \cdots \wedge dx^n)

this is basically second year calculus

[glacial mango]

I thought it was pre-calculus?

I got this in a Leap Pad in pre-school.

[arctic basalt]

precalc has no integrals

[glacial mango]

real shit though, wtf is a form?

[teal grotto]

idk. i just haven’t seen volume other than area under a curve made precise yet

[wintry steppe]

it's a thing to be integrated

woops wrong message

meant to reply to the previous

a differential k-form, morally, assigns to each point a tool that measures the volume of infinitesimal k-dimensional parallelepipeds at that point. when you sum up all of these infinitesimal volumes (integration) you get the area of a surface

(take "area" and "volume" to be synonymous in this explanation)

[teal grotto]

could u give a small example

say, a one form

[wintry steppe]

in a few minutes i just got out of the shower

i might need to handwave a little on the "infinitesimal" part since that's formalized using the tangent space

meh tangent space to R^n is easy to get

[teal grotto]

i’ve heard it described as something like equivalence classes of functions or something but i didn’t get too familiar with them

[wintry steppe]

ok so

i can probably get away with not even mentioning tangent space, but for the sake of generalization i will

[stoic python]

SSussy

[wintry steppe]

so it's just like a copy of R^n at p

[stoic python]

SSussy

[wintry steppe]

(star means dual space)

[teal grotto]

ok gotcha

[wintry steppe]

example time

so you know how some people say "dx is a one-form," right?

time to unwrap that

[glacial mango]

OK so we have vector spaces parametrized by vectors with linear maps for each one

[glacial mango]

this is going to be a good day

[wintry steppe]

i'll do things in R^2 to keep the notation simple

the expression $dx$ stands for the one-form on $\bR^2$ defined by $$(dx)_p(p, (v, w)) = v$$

[stoic python]

SSussy

[wintry steppe]

that might take a second to decipher

remember first that dx takes in a point p, and then it takes in an element of the tangent space to R^2 to at p

the idea is that, once eating a point p, we want to measure things in the tangent space to p

dx, for example, measures the "horizontal component" of a vector in the tangent space at p

[teal grotto]

oh shit

[wintry steppe]

wanna guess what dy does?

[glacial mango]

= w?

[wintry steppe]

yes

[glacial mango]

wet

[teal grotto]

lol

[wintry steppe]

we have notions of continuity and smoothness for differential forms

[glacial mango]

I'm not sure if I got an intuition about dx, but I mostly follow. lol

Like imagining a tangent space to a point in R^2 is a bit tricky

[wintry steppe]

just imagine a copy of R^2 centered at p

[glacial mango]

Yea OK, in some orientation

[stoic python]

SSussy

[teal grotto]

this seems like it could get really complicated really quick

maybe not idk

[wintry steppe]

for higher degree forms (not yet defined) it just gets complicated in notation

[teal grotto]

yea that’s what i meant to say. a lot of notation going on

[wintry steppe]

right

at the start it seems like just a lot of notation and formalism

[glacial mango]

yea wtf can I do with this shit?

[wintry steppe]

integrate

[glacial mango]

And who was high enough to think about this?

I'm going to take every element of a group and assign a group to it. I'll call it the 'tangent group'.

[wintry steppe]

just wait until you hear of algebraic topology

[glacial mango]

Next, I'll assign isomorphisms to each group....

totally high

I hate algebraic topology. It's just algebra.

[teal grotto]

elie cartan was apparently

[glacial mango]

It makes sense though

AT, makes sense, not forms

You want to count holes on a smooth space? Rigidify it with something like a simplicial complex and you can define a hole.

forms are fake tho

[teal grotto]

facts lol

[low pond]

Hi guys, I have a question that I can't solve it, can I ask for some help here?

[glacial mango]

go for it, we've left the solar system already

[low pond]

The question is to find the cartesian coordinates of this parameterization

[sharp lodge]

My brain is checked out, can i post a problem here and come back tomorrow to see if anyone solves it?

I hope i wrote it right

[wintry steppe]

hold on my carbon monoxide detector is going bonkers i can't keep explaining the forms stuff

low battery

[idle tide]

FINALLY

I KNOW HOW TO DO IT

I CAN USE GAUSS-JORDAN ELIMINATION

IT HAS TAKEN EXACTLY 11 HOURS AND 58 MINUTES

BUT I GOT IT

[teal grotto]

[glacial mango]

these are the moments to grind for

[nocturne jewel]

matrix algebra time

[wintry steppe]

[halcyon pollen]

i am learning linear algebra btw

is there any way to pratice those ?

[half ice]

Those?

[halcyon pollen]

like the concepts i learn

i'm learning from youtube and other resources for now

just i want to solve some problems and do stuff around

[half ice]

Get a textbook. Maybe some engineering book that includes linear algebra?

It will have problems to do

[halcyon pollen]

you know brilliant.org?

there are some problems in it

but is that payable?

[half ice]

Payable?

[halcyon pollen]

like is it a paid service or free one?

[half ice]

Some of it is free

There's free textbooks, which is why I mention it kek

[halcyon pollen]

[halcyon pollen]

can you link me any of those ?

[half ice]

I just Google. This is how I find textbooks.
https://www.google.com/search?q=engineering+linear+algebra+textbook+pdf&client=ms-android-samsung-ss&sxsrf=ALeKk01nw1EwMBzB3OVSur6eBRCPuE_ZBw%3A1624593204685&ei=NFPVYLypKYa-tAajoZrwDw&oq=engineering+linear+algebra+textbook+pdf&gs_lcp=ChNtb2JpbGUtZ3dzLXdpei1zZXJwEAMyBggAEAgQHjIGCAAQCBAeOgQIABBHOggIABAIEAcQHjoECCEQClD4_gFY4YECYP2DAmgAcAF4AYABwQGIAfcDkgEDMi4ymAEAoAEByAEIwAEB&sclient=mobile-gws-wiz-serp

Sorry for the awful url

[halcyon pollen]

[halcyon pollen]

nah lol i don't mind it

thanks for that i found a book idk how it's enough for me

i'll learn it anyway

[hard portal]

can someone help me in finding basic applications of matrix methods except circuit diagrams,crypto,traffic flow,balance equations... someone pls?

[nocturne jewel]

solving any system of linear equations, multivariable curve fitting can be done w/ matrices

[delicate oxide]

Computer Graphics

[late canyon]

I have a problem, I am given this matrix with K=Z_2

And I have to find a matrix B

with

And I have no clue how the image of A looks here

nvm, I found the image of A

But now I'm stuck again, how do I construct a function whose kernel is that?

[whole peak]

@late canyon You mean something like this?
https://en.wikipedia.org/wiki/Kernel_(linear_algebra)#Computation_by_Gaussian_elimination

[late canyon]

Yes

<@&286206848099549185>

[wintry sphinx]

well, if it helps, Ker(A^T) = Im(A) perp (i.e. the orthogonal complement of Im(A)) for all matrices A

or wait did I mess that up

idk

nope

[whole peak]

https://www.youtube.com/watch?v=ZvAkIf7oBHw

[late canyon]

I have the inverse problem, but I think I've solved it

[hollow finch]

i believe this implies that BA=0

but since ker(B)=Im(A), dim(ker(B))=4 so you do need rank(B)=3 by rank-nullity

[craggy gulch]

Can pivots in REF be ANY number? Including negatives?

Example:
[ 1 2 0 ]
l 0 -1 1 l
[ 0 0 -1 ]

[nocturne jewel]

no

just 1

[craggy gulch]

SO i'd have to make r2 column 2 & r3 column 3 into 1s?

[craggy gulch]

You sure you aren't talking about RREF?

I know RREF all pivots must be 1s and on top and below the 1s has to be 0s

So then basically if they do have to be 1s. Then for both r2 c2 and r3 c3 all I have to do is do is -1R2 -> R2 & -1R3 -> R3

[nocturne jewel]

yes pivot columns are when there's 1 1 in it and the rest is 0

[dusky epoch]

pivots can't be 0 but other than that yes they can be anything

[craggy gulch]

ty all

[dusky epoch]

the subspace theorem?

that's

the definition

is it not?

you don't prove a definition

[craggy gulch]

This matrix in the 3rd row doesn't have a pivot in column 3. So would this mean I won't be able to solve my system?

[ 1 -1 2 4 ]
l 0 5 -5 -7 l
[ 0 0 0 -7 ]

[dusky epoch]

is this the augmented matrix of your system?

[craggy gulch]

Original was:

[ 1 -1 2 4 ]
l 2 3 -1 1 l
[ 7 3 4 7 ]

[dusky epoch]

is this the augmented matrix of your system?

[craggy gulch]

Where column 4 is (4, 1, 7) xyz

yes

[dusky epoch]

okay

then the system is inconsistent.

you will only be "unable to solve it" if a mysterious force prevents you from writing down the word "inconsistent"

[craggy gulch]

lol i see

So look at this..

I got to this point in my matrix right:

[ 1 -1 2 4 ]
l 0 5 -5 -7 l
[ 0 10 -10 -21 ]

Would it had been correct to go about and saying.. 10R1 + R3 -> R3 ... or not? I first went with -2R2 + R3 -> R3

Doing it as 10R1 + R3 -> R3 would had gotten me a Z = answer?

[lucid glacier]

Yo who is the manager of quadratic and bilinear forms

I just wanna talk

[halcyon viper]

What's that about?

[lucid glacier]

Just an immensely annoying subject someone decided to squeeze into my linalg 2 curriculum

[whole peak]

@craggy gulch Why not: R3 := R3 - 2*R2 ?

Oh I see, the 0 row

[craggy gulch]

Yep

So this must be inconsistent like ann said

[old dirge]

Sorry didn’t mean to take these sideways. I am not sure how to get those values in part E. Or just not sure how to read it. Everything else makes sense. Can someone break down the values found for E? This is a study guide btw

[limber sierra]

whats theorem 6.5.3 say?

[short lodge]

How would you start this? New to linalg and proofs in general

[limber sierra]

$(v_1, \dots, v_n)$ spans $V$, which means any vector $u \in V$ can be written as [u = a_1v_1 + a_2v_2 + \dots + a_nv_n] for some scalars $a_1, a_2, \dots, a_n$. what you want to show is that there exist scalars $b_1, b_2, \dots, b_n$ such that:[
u = b_1(v_1 - v_2) + b_2(v_2 - v_3) + \dots + b_{n-1}(v_{n-1} - v_n) + b_nv_n]

[stoic python]

Namington

[limber sierra]

this is just unpacking the definitions

in other words, we want to show we can choose $b_1, b_2, \dots, b_n$ so that[a_1v_1 + a_2v_2 + \dots + a_nv_n = b_1(v_1 - v_2) + b_2(v_2 - v_3) + \dots + b_{n-1}(v_{n-1} - v_n) + b_nv_n]

[stoic python]

Namington

[limber sierra]

(since the LHS is equal to u)

oh that second image got cut off a bit

hopefully you get the idea

\begin{align*}
&a_1v_1 + a_2v_2 + \dots + a_nv_n \=& b_1(v_1 - v_2) + b_2(v_2 - v_3) + \dots + b_{n-1}(v_{n-1} - v_n) + b_nv_n\end{align*}

[stoic python]

Namington

[limber sierra]

to start with this, its probably best to try rewriting the RHS in a simpler form; say, by distributing each b_k into the brackets. then figure out what each b_k must be to "match" a_k based on that.

does that all make sense?

and do you see why, once you do this, it proves that the list spans V?

[short lodge]

I'm about to go to sleep but yeah it makes sense, thanks

[short lodge]

Since you can write an arbitrary vector u (that's in V) with a linear combination of the new list, the list spans V

[arctic basalt]

,iam studying

[stoic python]

You already have the selfroles studying!, do you want to remove them? (y(es)/n(o))
(Tip: use ,iamnot to remove roles without this prompt.)

Session timed out waiting for user response.

Jun 26, 2021, 3:24 AM

[frosty vapor]

use #bots

[dreamy iron]

Hi #linear-algebra

I need some clarification on some notation…..

Why are they writing it this way?

T(e_i) is a sum wrt to the basis vectors of the codomain space, so there’s like \plus signs tucked inbetween the entries…..

and a matrix has no \plus signs hidden in it

I’ve seen so many sites do this and lecturers also…..

AS A POINT OF CONTRAST,
this is how Axler writes it, which is more circumspect (and less clear), but also more accurate

[odd pelican]

T(e_i) is indeed a sum over the codomain basis vectors, which would mean it represents a column vector. The collection of these column vectors is what makes T as a whole. That is, the columns of T are where each basis vector in the domain gets mapped to

[dreamy iron]

$$\begin{bmatrix} a_{1,1} w_1 \ + \ a_{2,1} w_2 \ + \ \vdots\ + \ a_{m,1} w_m \end{bmatrix}$$

[stoic python]

a rainbow powered ninny

[dreamy iron]

But no one writes a column matrix that way

$$\begin{bmatrix} a_{1,1} w_1 \\ a_{2,1} w_2 \\ \vdots\\ a_{m,1} w_m \end{bmatrix}$$

$$\begin{bmatrix} a_{1,1} w_1 \ a_{2,1} w_2 \ \vdots\ a_{m,1} w_m \end{bmatrix}$$

[stoic python]

a rainbow powered ninny

[dreamy iron]

My thinking is that the notation in the green box is convenient, but ultimately incorrect……is my thinking wrong? is my question.

[limber sierra]

i dont follow

whered the + come from

[odd pelican]

Remember each of the $w_i$ represent a column vector with a single 1 in the $i$th row (I’m tacitly assuming an orthonormal basis). So if
$T(e_i) = a_{1,i}w_1 + \dots + a_{n,i}w_n$, then we can write this as
$T(e_i) = \begin{pmatrix}
a_{1,i} \ \vdots \ a_{n,i}
\end{pmatrix}$

[stoic python]

MemeMachine420

[limber sierra]

whered the + come from

[dreamy iron]

T(e_1) = a_1,1 w_1 + ,,\cdots, , + a_m,1 w_m

[limber sierra]

no?

[dreamy iron]

Its a linear combo of the basis vectors of w_i

[limber sierra]

each column is the image of a single basis vector

[dreamy iron]

Lmao. seriously. The plus signs just vanish and you say this is a column vector?

Okay. Ive not seen that convention before

[limber sierra]

where did the plus signs come from?

do you not know what a standard basis is?

[dreamy iron]

Yes, lol.

[limber sierra]

compute [\begin{pmatrix}a&b\c&d\end{pmatrix}\begin{pmatrix}1\0\end{pmatrix}] and [\begin{pmatrix}a&b\c&d\end{pmatrix}\begin{pmatrix}0\1\end{pmatrix}] for me

[stoic python]

Namington

[limber sierra]

what do you get?

[odd pelican]

It’s not that the plus signs vanish, it’s that each $w_i$ is already a column vector. For example,
$aw_1 + bw_2 = a\begin{pmatrix} 1\ 0 \end{pmatrix} + b\begin{pmatrix} 0\ 1 \end{pmatrix} = \begin{pmatrix} a\ b \end{pmatrix} $

[dreamy iron]

Axler, page 70.

[limber sierra]

missing a \

before third begin

[stoic python]

MemeMachine420

[odd pelican]

Was fixing it

[limber sierra]

thats the exact same definition

except it takes an arbitrary basis rather than fixing the standard basis

[dreamy iron]

Okay. Yeah. This is making sense.

i see it

Yeah. Imma LaTex some examples to be perfectly sure, but I think that’s what I was missing

[odd pelican]

Ye, so the original notation will all the $T(e_i)$ is just saying that each of those is a column vector and together they make up the entire transformation $T$

[stoic python]

MemeMachine420

[dreamy iron]

\begin{bmatrix} a \ c \end{bmatrix} and \begin{bmatrix} b \ d \end{bmatrix}

[stoic python]

a rainbow powered ninny
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[dreamy iron]

lol. I didn’t even put the dollar signs

[hollow finch]

$\begin{bmatrix}a&-b\b&a\end{bmatrix}$ and $\begin{bmatrix}a&b\-b&a\end{bmatrix}$ can both be used to represent the complex number $a+bi$. Is one clearly better or more standard than the other?

[stoic python]

nix (@ me for the love of euler)

[hollow finch]

basically which isomorphism is better i guess

[native rampart]

idt one is better than the other

I would personally prefer the first one

[wintry steppe]

first one feels more natural

[lavish jewel]

[wintry steppe]

$e^{i\theta}=\cos\theta+i\sin\theta$ under the second correspondence is the matrix $$\begin{pmatrix}\cos \theta & \sin \theta \ -\sin \theta & \cos \theta \end{pmatrix},$$ which represents a rotation \textit{clockwise} by $\theta$ degrees. but we always work with counterclockwise rotations.

[stoic python]

TTerra

[wintry steppe]

or something

,ti

[stoic python]

The current time for TTerra is 03:40 AM (EDT) on Sat, 26/06/2021.

[wintry steppe]

^this is my excuse if i just said something dumb

[lavish jewel]

that seems like a weird argument 😛

i woulda just said the second looks more like a - ib

i have no time related excuse other than i am dumb 24/7

[wintry steppe]

weird arguments are how i do mathematics

[lavish jewel]

proof by confusion

[wintry steppe]

idk

there really is not much of a difference between the two identifications

just transpose one of them xd

i guess you could say the first one is more natural because a complex number a + ib is also an element of R^2, so a column vector (a, b), and the first column of the matrix rep should be that?

who knows

[lavish jewel]

so that would be part of my argument tho

[wintry steppe]

i can say for certain that the first one is more standard

[lavish jewel]

transpose is "but now it's in dat dual space tho" moment

[halcyon pollen]

guys what's the difference between row echelon form and reduced row echelon form?

[half ice]

The word "reduced"

[halcyon pollen]

https://tenor.com/blwSX.gif

[halcyon pollen]

gaussian elimination and gauss jordan elimination

i learned both is like the same i think

i guess i'll head to Crout's method

[half ice]

From Wiki, you only need that pivots exist, and zero rows are at the bottom

REF isn't a very strong condition haha. It's just enough to read off some matrix properties

[halcyon pollen]

like an identity matrix right?

[half ice]

It becomes "reduced" when

• Pivots are 1
• Any column with a pivot has only 0s otherwise

[halcyon pollen]

like that right?

[half ice]

So for example:
4 1 2 3
0 3 3 3
0 0 0 2
Is REF

This matrix will RREF as
1 0 0 0
0 1 0 0
0 0 0 1

Your textbook may disagree slightly on these definitions, I'm going off wikipedia here

[halcyon pollen]

yeah i see wait a minute

[halcyon pollen]

it's similiar to that right?

[half ice]

That picture is an RREF.

[halcyon pollen]

this ?

REF?

both RREF and REF does the same thing i can't classify what's the difference between them

[half ice]

See my examples. The REF is definitely not an RREF.

4 1 2 3
0 3 3 3
0 0 0 2
Is REF
This matrix will RREF as
1 0 0 0
0 1 0 0
0 0 0 1

I see a distinct difference between those two

[halcyon pollen]

so REF has the lower diagonal variables as 0s but RREF is an identity matrix right?

[half ice]

1 0 0 0
0 1 0 0
0 0 0 1
Is an RREF matrix, but is not an identity matrix

[halcyon pollen]

uhh

in the last row

there must be an one in the 3rd column

but it's in zero

here's an simple explanation

[half ice]

Uwat

[halcyon pollen]

umm

[teal grotto]

this is only for square matrices

[halcyon pollen]

i'll learn those things again and come back

now i get row echelon from

i were confused with the namings btw

[halcyon pollen]

thanks for your help 😄

[half ice]

Haha np. Feel free to ask if you have anything else!

[halcyon pollen]

don't worry i have tonnes of problem coming

[dreamy iron]

I saw this in a lecture and I just wanted to know if I typed it correctly.

I think the w_k are all out of place….. none of them should be there.

[native rampart]

Yea,They shouldn't exist

[dreamy iron]

I actually write out a gory example in section 4. https://www.overleaf.com/read/wkktvyvtcntr

Any guesses why he would write them in: start at 6:55

[dreamy iron]

https://youtu.be/9zPeRXnzEdE

[native rampart]

Yea T(v_i) is that

That part's correct

[dreamy iron]

Actually…..I think I figured it out. He was doing this.

But he was writing the sums vertically….

Im going to rewrite the matrix but ill throw pluses in between the entries vertically

[lavish jewel]

ok, it was taken care of

[wintry steppe]

If f: E->F, h:F->G are linear and h is injective. How do i prove that rg(f)=rg(hof)?

i started by saying let v_i=f(w_i) be a base of ImF, then h(v_i)=hof(w_i) => Imh(v_i)=Im(hof), but i fail to prove that Imh(v_i)=Im f

what is rg? rank?

[lavish jewel]

prolly range?

[wintry steppe]

they already used im

also range of f is in F but range of h o f is in G

[lavish jewel]

[wintry steppe]

an easier way to do this is to use the rank nullity formula on f and on h o f, and this works under the assumption that E is finite dimensional

rank

sorry i come from a french system so half the terms im just guessing how theyre written in english lol

usually rank or rk

sometimes dim im if you're feeling groovy

[wintry steppe]

so is this a good start

(maybe base=basis in english)

[lavish jewel]

im(h o f) need not equal im(f), btw

only the dimension of the image is equal

h: F -> G injective means dim(G) >= dim(F)

[wintry steppe]

both terms are common

okay yea, i feel like i should use somewhere that Ker h=0

if you want to do it by choosing a basis you could use the fact that h will take linearly independent sets of vectors to linearly independent sets of vectors

in particular

since your v_i's are linearly independent, your h(v_i)'s are, and that tells you something about the dimensions

i saw that but how does it help me

so dim h(vi)>=dimE?

wait no

no i think its that actually

[lavish jewel]

you don't know anything about the relationship between E and h

[wintry steppe]

v_1, ..., v_d is a basis of im f, and h(v_1), ..., h(v_d) are linearly independent vectors in im(h o f), so what does that tell you about the dimension of im(h o f)?

wouldnt i know something since vi's are the basis of E, and dimh(vi)<=dimG

what are the assumptions in the first place on the dimensions of the spaces?

[lavish jewel]

h cannot be applied to the vi's unless E is related to F somehow

[wintry steppe]

its <=rkhof

[wintry steppe]

they said the v_i's were in im f

[lavish jewel]

they also said they are a basis of E

so idk which one htey mean

[wintry steppe]

vi's are supposed to be the basis of Imf

[lavish jewel]

so NOT of E

[wintry steppe]

also vi=f(wi)

oh yea that was a mistake on my part

[lavish jewel]

aight, then yea

[wintry steppe]

i confused them

[wintry steppe]

i dont follow lmao, yea refered to what lol

[lavish jewel]

v_i = f(w_i), with w_i as basis vecs of E and v_i as basis vecs of F

hmm

[wintry steppe]

yea thats how i defined them

[lavish jewel]

that's not true either tho

[wintry steppe]

why

[lavish jewel]

cuz we dunno if f is injective

[wintry steppe]

oh yea, damn

the argument you were going for doesn't need the w_i's to form a basis anyways

at least how i see it

unless im high

yea i dont think it needs to

at least i didnt envision it that way but idk

[wintry steppe]

what i was going for with this is that rank(f) = d <= rank(h o f)

okay yea i follow

to finish it off you wanna show the other inequality, that rank(f) >= rank(h o f)

that's one way to continue the argument you started

[lavish jewel]

less obscure than what i was thinking

[wintry steppe]

the point of the exercise is that injective maps preserve the dimensions of finite-dimensional subspaces

thats since ker h if h injective is equal to 0 right?

i guess it could follow from that

if you used rank nullity

yea

[wintry steppe]

i dont see thus

this

same thing, choose a basis of im(h o f) and do some stuff

it's the exact same thing i did earlier

okay ill try, thanks

on a side note, do you see how i could show that rk h(vi)=rk f?

to finish my inital proof

im not really sure what rk h(vi) means

vi=f(wi) => h(vi)=hof(wi)

i don't either but it should somehow turn out to be the other thing lol

[lavish jewel]

if v_i are a basis of F, you can express the f(w_i) as linear combinations of v_i

then h is injective, so h(v_i) yields a set of linearly independent vectors, as many as "went in"

wave your jazzy hands

buncha sums and linearity

[wintry steppe]

so the dimension is the same?

understood it finally

[wintry steppe]

nice

[wintry steppe]

actually what i did isnt good, i didnt manage to properly make an application for a base of Im(hof) so that at the end i get rk(hof)<=rkf

could (h^-1 o f)(a_1) be a base of Im(hof)

[wintry steppe]

.

[gloomy nebula]

Why is this false?

rref of 3 unknowns is a diagonal matrix since each row is nonzero

which means x,y,z, the unknowns each are assigned a value

[limber sierra]

for a counterexample, consider the following system

x = 1
y = 1
z = 1
z = 0

[gloomy nebula]

why are there two z's

[limber sierra]

unless its supposed to be EXACTLY three nonzero rows

in which case hmm

[limber sierra]

anyway this is a false statement

nothing in the original question said that the system had to have 3 equaitons

although in fact, if the system does have 3 equations, R will be 3x4

so it cant be a diagonal matrix

(since its an augmented matrix, so theres 3 rows - 3 equations - but 4 columns since theres 3 unknowns + a solution column)

as an example, consider the system x = 1, y = 1, z = 1.

the augmented matrix of this - which is already in RREF - is [\begin{pmatrix}1&0&0&1\0&1&0&1\0&0&1&1\end{pmatrix}]

[stoic python]

Namington

[gloomy nebula]

That's one solution

can we think of another solution that's not (1,1,1)?

[limber sierra]

that isnt meant to be a counterexample to that image

just the specific message i replied to

[gloomy nebula]

I mean

It can't have 2 solutions

It's either 1 or infinite

so it can't be at least 1

[limber sierra]

"at least 1" means either 1 or infinite, yes

(assuming your underlying field is infinite)

[glacial mango]

Is it false because you can have two identical rows?

[gloomy nebula]

they can't be identical because the matrix has to be rref

[glacial mango]

oh, right right

[limber sierra]

they can be identical in the system, just not in R.

anyway, heres a useful theorem

(rouche-capelli)

[gloomy nebula]

is a scalar is 1/2, then this fails.

I just want to make sure. Scalars can be rational right?

[limber sierra]

yes

to be more explicit, if youre considering this as a subspace of ℝ³, then that means your "base field" or "scalar field" is ℝ

so your scalars can be ANY real number

including rationals like 1/2

[gloomy nebula]

i have not taken abstract yet

[glacial mango]

so that fails closure under scalar multiplication

[wooden pike]

ℝ

i didn't know that was a unicode char

[teal grotto]

what is the character

[nocturne oracle]

naruto from naruto is my favorite character i really relate to him hes really well written and hes so strong and cool

[limber sierra]

of all the things to praise naruto for, its prose is pretty low on the list

[limber sierra]

symbol for the set of real numbers

a ton of mathematical symbols have a unicode character btw

ℝ is fairly mundane, theres some pretty damn specific shit

theres a triple contour integral character

∰

also stuff ive never seen like "Dot Plus" ∔

thats its official name, its classified as a mathematiocal operator

so its not some weird language that has an i with a line through it

⊌

apparently this \cup with a ← denotes a "multiset"

there is no \cup with a →, from what i can tell

unicode is weird.

⋇

"Division Times"

its like \pm but for * and the division sign

also ⋻ and ⋼

two separate "contains with vertical bar at end of horizontal stroke"s

anyway thats enough ranting about unicode

but it has weird stuff

[teal grotto]

haha ik that, was asking for the unicode char. thanks tho

[limber sierra]

ah

U+211D "Double-Struck Capital R"

but i use wincompose to type it, i dont memorize the code

[teal grotto]

thanks, will check that out

[dire thunder]

unicode is exhaustive

set of all chars is less than set possible to be represented by unicode

[old swift]

that’s why ascii sucks

[lethal inlet]

if I want to get length 1 of my vector

v1 = (1,1, 0,0) I've to do this right ?

v1 / | v1 |

where |v1|= sqrt ( < v1, v1> )

what I'm I doing wrong here ?

[dire thunder]

what do you mean

v/|v| has norm of 1

@lethal inlet what is \tilde v_1 denoting?

[lethal inlet]

v1 * 1 / |v1|

/ divide

[dire thunder]

what

it is meaningless then

since as it is written \tilde v_1 is scalar

and notation is bad also

[lethal inlet]

oke good

but what im I doing wrong

calculation wise

[dire thunder]

you did not done anything wrong

v/|v| produces vector of length 1

assuming |v| != 0

you should have v/|v| as (1/sqrt(2), 1/sqrt(2), 0,0 )

[topaz wadi]

Can someone explain this statement

Ping me if ur willing to help

Nvm I think I got it

Actually I don’t get the second sentence (“Therefore....”)

[nocturne jewel]

what are you confused on exactly?

[topaz wadi]

Like the mapping V to V

Let’s say I chose two different sets of Basis

[nocturne jewel]

yep

[topaz wadi]

How would I have the same determinant

[nocturne jewel]

matrix representations of a transformation have to preserve eigen, so they have to have the same det is my guess

so phi with respect to basis B and basis C, have to give matrices with the same determinant, regardless of the fact B and C are different, since they're both phi (loose terminology)

[topaz wadi]

Haven’t gotten to Eigen vectors yet

[lavish jewel]

the idea can be followed through with change of basis matrices

[nocturne jewel]

edd faster than I can google

[topaz wadi]

Something like this?

[nocturne jewel]

A is similar to B iff $B=P^{-1}AP$

[topaz wadi]

What would the mapping in the second line be with respect to this image

[stoic python]

Mosh

[nocturne jewel]

if A and B are the matrix representations, P is the change of basis matrix b/w them

[topaz wadi]

Yeh I get that

So B and A are in the same vector space?

[lavish jewel]

A, B, and P are all V -> V

[topaz wadi]

Ok

[topaz wadi]

So in this image B and B’ are in the same vector space

If I’m talking about this scenario that we are discussing

[lavish jewel]

idk what the image is trying to show tbh

[topaz wadi]

Like change of Basis

[lavish jewel]

aight

so yeah

[topaz wadi]

They are just representing linear transformations with arrow marks

[lavish jewel]

if you start at the upper left

going down is A

going right, down, left is P B P^-1

[topaz wadi]

Yeh

So ur setting B’ is B right

In that image

[lavish jewel]

they don't necessarily have to be, to be fair

[topaz wadi]

Ok

[lavish jewel]

as long as P is some isomorphism between them

which here is the change of basis

[topaz wadi]

But in the end the mapping is still V to V tho?

[lavish jewel]

yes

[topaz wadi]

I see

[lavish jewel]

in R^n, for example, B and B' would be the same

[topaz wadi]

Yeh

[lavish jewel]

but it doesn't necessarily have to be, as you pointed out

just gotta be nicely, invertibly mapable

[topaz wadi]

So A and Lamda with respect to this image are in the same vector space

[lavish jewel]

hmmm

in R^n, for example, yes

[topaz wadi]

When correlating our discussion with this diagram

[lavish jewel]

well

A isn't in a vector space, yeah?

it's a linear transformation

A and Lambda are endomorphisms or something

[topaz wadi]

Kk

But they would have the same determinant?

[lavish jewel]

that's the idea

[topaz wadi]

If we consider them all to be in same vector space

[lavish jewel]

this is what google gave me

[topaz wadi]

Haven’t gotten to eigen values yet

But thanks

[lavish jewel]

there's something simpler

assume there are A and B that are similar

[topaz wadi]

I think I got it

[lavish jewel]

this means A = P B P^-1

take the dets

[topaz wadi]

Det(S inverse) would cancel with det(S)

[lavish jewel]

yeah, exactly

the det of the product is the product of dets

so det A = det B

[topaz wadi]

Yeh I’m actually trying to find a geometric meaning to all this

U know changing space and stuff

I was trying to compare the areas

Or volume in that matter

Dealing with R^2

So a mapping V to V

Using Lambda and A

Would still result in the same determinant

Or the areas would remain unchanged

Correct?

[lavish jewel]

mhm

[topaz wadi]

Ok thanks

[gloomy nebula]

Do I start this by rrefing [X Y]?

[teal grotto]

hint: the coefficients of X and Y give you the two equations. the answer is the spoiler'd message
||x_1 + 2x_3 = 0 and x_1 - x_2 + 2x_4 = 0||

[wintry steppe]

rref

[glacial mango]

I'm trying to understand why that's the case.

Can we bump this down a few dimensions? Maybe 2D vectors?

I don't know if the analogue should still use a system of two equations.

two 2D vectors, 2 equations, 2 unknowns?

OK, so X and Y span the solution space.

So I'm trying to understand how you can just take the solution space and form a system that has it as a solution space

I'm trying a trivial example: x-y=0

The solution space is spanned by [1,1]

But I can't just say that x+y=0 is a system of equations with [1,1] spanning the solution

what breaks down in this trivial examples?

[glacial mango]

Also, I tried this and didn't get X and Y back, but my linear is rough and I might be stupid (not sure).

wait, I got that for the solution of the system coycoy gave.

something fishy is going on here

OK so how did you derive that system of equations?

[lavish jewel]

my guess is they want some null space action

so rrefing wouldve worked

mhm

you could do it in a different basis, maybe

[teal grotto]

u mean 4 x 1 matrices?

yea... X is a 4 by 1 matrix

np

[lavish jewel]

anyway

2 eqs with 4 variables is a 2 x 4 mat, call it A. we wanna have something like Ax = b, and they're asking for x

so x is indeed spanned by 4x1 vectors

as someone said above, this is done by just taking those vectors straight up

you can use the null space to give infinitely many sols tho

they didn't say it's homogeneous

[teal grotto]

oof mb

[lavish jewel]

you can take Ax = b, and just let b be any linear comb. of the columns of A

then b is for sure in the span of the cols of A

and x is the corresponding linear comb, of which there are infinitely many. BUT, they WANT it to be in the span of the rows

so set the null space components to 0

since x = x_row + x_null, yeah?

for x to be in the span of X and Y, x_null has to be 0

i'm just waking up, so do correct me if i messed up

[glacial mango]

Aren't they asking for A? The system of equations?

[lavish jewel]

A alone does not describe a system of equations

the system of equations requires A and b

without b, you don't have equations

and notice they ask for "a" system

meaning there's more than one

[glacial mango]

Hmm... I think I understand. And the solution given takes b = 0?

to keep it simple

[lavish jewel]

no

if you put b and you follow the instructions

the only solution is the 0 vector

i doubt they want that... thought it is technically correct

[dusky epoch]

what's the question again?

is it this?

[lavish jewel]

this one here

[dusky epoch]

and y'all are arguing about whether or not the system has to be homogeneous?

it MUST be

[lavish jewel]

i mean, homogeneous works, but then the solution has to be x = 0

you could make up any system

[dusky epoch]

the solution set of our system is known, and it is {c1X + c2Y | c1, c2 in R}

[lavish jewel]

as long as x is in the row space, with no component in the null space

[dusky epoch]

the solution set of a non-homogeneous system doesnt contain the zero vector and so isnt a subspace of its domain and so cannot be the span of anything

...........

[lavish jewel]

[dusky epoch]

what's your point

[lavish jewel]

the wording tho, they don't say the solution has to be the span of the 2 vectors, but be in the span

[dusky epoch]

same thing

[lavish jewel]

i see

i'm pretty sure they want the opposite of that

going by ann's insight, you either let b = 0 or let b be some generic vector in the column space of A

and then the sol is 0 in the first case, cuz there has to be no component in the null space, or some linear combination (which includes 0) of X and Y

[dusky epoch]

https://tenor.com/view/von-karma-ace-attorney-manfred-von-karma-head-bang-ace-attorney-von-karma-gif-19478022

[lavish jewel]

do correct me :x

[dusky epoch]

our system must be Ax = 0 because x=0 must be one of its solutions

because the solution set is {aX + bY | a, b in R}

which contains 0

i fail to see how that could be in any way hard to grasp

who said it wasn't correct

[lavish jewel]

ann, in what i wrote above, i let x be any aX + bY to consider the whole solution space. could you point out for me what was wrong? i fail to see it myself 3:

[dusky epoch]

it's x that is aX + bY

not b

[lavish jewel]

yes, sorry

that was a typo

or well, to be precise, i put b in the column space, but they require the solution to be in the row space

in this case, A is invertible, so kinda the same thing, no?

[dusky epoch]

who said A was invertible

its not even square

[lavish jewel]

yes, but if x is spanned by X and Y, it's in a dim 2 subspace

and so is the output

the pseudo inverse is the inverse if x_null = 0

x is, for example, in R^4, but we only care about the dim 2 subspace of R^4 spanned by X and Y

the null space is its orthogonal complement

the pseudo inverse can map im(A) to the row space, which is aX + cY

[halcyon pollen]

what's the difference between regula falsi method and bisection method?

[halcyon pollen]

is that calculus?

[lavish jewel]

after finally waking up and thinking about it for a while longer, yeah

big oof from my side

you want b = 0 and the null space to be spanned by X and Y

[glacial mango]

I want that Kermit gif back.

[lavish jewel]

due to the same reasoning i gave above, but done correctly :x

[halcyon pollen]

[snow jetty]

two similar real matrices are congruent? how can it follow from the spectral theorem that just states that all eigenvalues are real, that eigenspaces are orthogonal, and that there is an orthogonal basis?

[coarse sandal]

How does one find a vector that is in the span{cosx,sinx} and what makes a vector not in that span?

[nocturne jewel]

somethings in a span if it can be written as a linear combination of the spanning set vectors

and vice versa for not in span

those are examples of vectors not in the span..

yeah..

and I was waiting for the person to respond

[coarse sandal]

sorry i was away

[nocturne jewel]

Ok

[coarse sandal]

so would it be k1cos(x) + k2(sinx) ?

something like that?

[nocturne jewel]

k_1 k_2, but yes, assuming the k's are from your scalar field

[coarse sandal]

how would you determine what k1 and k2 are if you dont know what x is?

[nocturne jewel]

You're just asked for 2 functions in the span of them

so pick scalars..

[coarse sandal]

so would any scalars work?

[nocturne jewel]

any from the scalar field, yes

[coarse sandal]

so would (1,0) and (0,1) work?

[nocturne jewel]

yes, since the spanning vectors are in the span themselves

[coarse sandal]

cause (1,0) = 1cos(0)+0sin(0)?

[nocturne jewel]

why did you change x to 0?

$\cos(x)+\sin(x)\in span{\cos(x),\sin(x)}$ for example

[stoic python]

Mosh

[coarse sandal]

im not sure myself

the concept of spanning set kinda confuses me

[nocturne jewel]

it's just the set that contains all linear combinations of the vectors in the set

[coarse sandal]

so a vector b is in the span{cos(x),sin(x)} if and only if k1cos(x) + k2sin(x) = b

[nocturne jewel]

no

[coarse sandal]

k1 and k2 being scalars

[nocturne jewel]

yes, assuming the 1 and 2 are subscripts

[coarse sandal]

does it matter what x is? or just any x?

[nocturne jewel]

it's just x

[snow jetty]

the result should be a function

so "x" does not matter, it is just a variable.

if two functions are equal they are equal for any "x" in their domain of definition

[coarse sandal]

oh okay

so 2cos(x) + sin(x) would be in that spanning

then x^2 or x^3 would not be in the spanning

thank you

[snow jetty]

f(x) = 0, cos(x) and sin(x) also work

[snow jetty]

then there are scalar fields where 2 does not really exist (for example F2 = {0,1} s.t. 1+1=0)

[nocturne jewel]

$\mathbb{Z}_2$

[stoic python]

Mosh

[snow jetty]

ye

[coarse sandal]

This would not be a vector space because u + (v+w) != (u+v)+w

where u,v,w are vectors of V

is that correct or did i do something wrong?

[stoic python]

Commander Vimes

[sudden narwhal]

V is a vector space if $0_{\mathbb{R}^2}\in V$ firstly

[dire thunder]

while first coordinate of x+(y+z) would be 2x_1+4y_1+4z_1

so yes

[stoic python]

Herels

[coarse sandal]

thank you

[old terrace]

\begin{equation*}
\begin{bmatrix} 1 & 0 \end{bmatrix} \equiv \begin{bmatrix} 1 & 1 \end{bmatrix}
\begin{bmatrix}
1 & 1 \
7 & 1
\end{bmatrix}^ n \mod x
\end{equation*}

[stoic python]

LordPoseidon

[old terrace]

find n for a given x

a sublinear algorithm should do

[it is given that gcd(x,6) = 1, therefore such an n exists for any x]

[quartz compass]

looks like n+1 will be the smallest divisor of phi(x) just thinking about what happens when you diagonalize it

it looks like it's asking what's the smallest power that makes $1+\sqrt{7}$ and $1-\sqrt{7}$ both equal 1 mod x. Basically a discrete logarithm type of problem

[stoic python]

Merosity

[quartz compass]

since that's exponential time, you're probably not going to find a sublinear algo lol

[merry imp]

when i searched this type of problem up, the technique shown was to determine the rank of the matrix with these vectors as its rows. however, in this text row reduction hasn't shown up yet, so that's not the way to go for this i think. the relevant stuff i have access to is a) if dimV=n, then each basis of V consists of exactly n vectors, and b) if v_1, v_2, ..., v_n in the n-dimensional vector space V are linearly independent then they constitute a (unique) basis for V.

i've determined the vectors of V have the form [a,b,a,b] but i don't know how to proceed? does this mean it's of dimension 2?

[nocturne jewel]

yes

since span(v1,v2,v3)=span(v1,v2) cause the 3rd vector is 1st + 2nd

[merry imp]

i feel kind of stupid, how do we know that span(v1,v2) has dimension 2?

[nocturne jewel]

cause the basis of a spanning set is just the set of vectors

[merry imp]

it's the fact that the vectors themselves are in R^4 that's putting me off

[nocturne jewel]

which has 2 cause we removed the 3rd vector to make the set independent

$V=\text{span}([0,1,0,1],[1,0,1,0]) \ \text{dim}(V)=2$

[stoic python]

Mosh

[woven haven]

(I presume that the conversation between sm and Mosh is at least on hold since it's been 20+ minutes since the last message. Forgive me if I'm interrupting here.)

I'm struggling a bit with this problem, but I have a proposed solution that I'd like to run by you all.

I'm going to draft it into latex to make it easier to read actually ill brb

[teal grotto]

@woven haven it would be pretty immediate using the dual basis for V from beta’

[woven haven]

@teal grotto I dont believe we've ever mentioned dual bases in class so I'm hesitant to use that idea.

[teal grotto]

mmm. that makes it really easy. i could also see it being done by induction

but anyways, curious to see you approach

[woven haven]

A bit wordy, I know, but here's my idea:

If the change-of-coordinate matrix is the identity matrix, then the bases are the same. To show this, I merely use the fundamental property of the change-of-coordinate matrix (i.e. changing bases) to show that the resulting phi representations are always the same. If that's the case, then the bases of the phi transformations must be the same.