linear-algebra

[stoic python]

Pencil

[lavish jewel]

the n subscript specifies the maximum degree

[wintry steppe]

Oh so F[x] means degree till infinity? Meaning all the polynomials possible with any degree?

[lavish jewel]

i think in any case, the usual definition requires the degree to be finite

one just mentions a specific degree explicitly

[wintry steppe]

Ooh okay 👍

[wintry steppe]

A matrix A ∈ M3×3(F3) is diagonalisable if and only if it has n distinct eigenvalues. I conclude this is false bcs the definition of diagonalisable says that it must have n distince eigenVECTORS right?

[native rampart]

Well,The if part is correct

If there are n distinct eigenvalues,The matrix has to be diagonalizable (because there will be n distinct eigenvectors)

Only if is false

Take Identity

[novel needle]

can someone explain the AB-BA part

i mean can someone provide a prove to this

[spare widget]

if (AB)^H = B^HA^H and we have A^H=A, B^H=B then (AB-BA)^H = B^HA^H - A^HB^H = -AB + BA = -(AB-BA)

[novel needle]

😮

i love you @spare widget

[slate hatch]

Our professor uses |v| to signify the length of a vector. Most sources I have come across use | |v| |. Is there any difference between the two?

[fair siren]

Does anyone know what Xi means in terms of matrices?

[lavish jewel]

should be the same thing. sometimes they are called different names, like magnitude vs norm, but they signify the same thing

[slate hatch]

okay, thanks

[spare widget]

could mean column vector or row vector

[lavish jewel]

we need a little more context

[fair siren]

This is everything that was given lol

[slate hatch]

What's this called?
ŵ=1/(| |w| |)w

[spare widget]

normalization

it produces a unit length version of w

[slate hatch]

Thanks mate

[spare widget]

I already addressed those here: #linear-algebra message

[wintry steppe]

I was reading definitions of eigenvectors and eigenvalues. According to my understanding, eigenvectors are those vectors that change their length by a scalar after a linear transformation.

I'm a bit confused about eigenvalues though. (Correct me if I'm wrong) Eigenvalues are the scalars that change the length of the eigenvector? And when the linear transformation takes the form of n×n matrix, it's the matrix product Av, v in Vector Space V?

[dusky epoch]

"over any field F" is sus

[wintry steppe]

Ah wait let me edit my message again

[dusky epoch]

we have a vector space V and a linear map T: V -> V

if, for a nonzero x in V, it happens that Tx is a scalar multiple of x, i.e. that there exists lambda (then necessarily unique) s.t. T(x) = λx, then x is called an eigenvector of T, and the scalar λ is called an eigenvalue of T

[wintry steppe]

Okay so, I tried to go to basics again of Linear transformation. By definition, it's a transformation that satisfies the addition property and scaling of vectors.

Can we always think of Linear transformations as matrices? That would make sense that why T(v+w) = T(v) + T(w) holds true then, and why T(cv) = cT(v)

[zinc timber]

??

what do u mean by your last sentence?

[native rampart]

All Linear transformations have a matrix representation

So yes

[lavish jewel]

in fin dim

[native rampart]

mb

[zinc timber]

you can probably extend to inf ones given a Schoder basis

[native rampart]

What if there is no basis

[wintry steppe]

Ooo I'll look into that

[grand hare]

given a,b,....l ∈ R if det(X) = 4, how can i find the determinant of Y by properties of determinants and without substituting the values of variables

[zinc timber]

ZF worshipper

[native rampart]

Clearly,You can't cut a sphere and get 2 spheres

[zinc timber]

or can you

[halcyon spindle]

I feel like if you cut a sphere, say in two pieces. You can squish each part into a sphere.

Wait, a sphere is not squishy.

[zinc timber]

that's not what they meant@halcyon spindle

[halcyon spindle]

ah, my bad then.

[zinc timber]

google Banach Tarski paradox

[halcyon spindle]

" set-theoretic geometry"

[zinc timber]

look for row/col manipulation

[short canopy]

so, i'm looking for an explaination (graphical if possible) of why the distance between point P and Q in R^n if P = (x_1, ... , x_n) and Q = (y_1, ... , y_n) is sqrt((x_1 - y-1)^2 + ... + (x_n - y_n)^2)

i understand this in graphically on R^2 if for example two points have the same distance on one of their coordinates

basically, the difference between hipotenuses

but what if two of those points don't have the same coordinate, we use the sine rule to find that distance?

i understand that here, the distance between the two points is basically the difference between both hypotenuses

but what if

do we use the sine rule to find the distance of the two points? (the red)

[grand hare]

afaik row operations take an effect to the determinant, do column manipulation also behave similarly?

[zinc timber]

they do

[grand hare]

if im not mistaken, i would only look out for multiplying rows/cols (det = kdet) and interchanging cols (det=-det) right?

[grand hare]

so i got det = -2/3 since i had to halve the 1st row and multiply the first column by 1/3

and i had to interchange two columns

[grand hare]

could you help me on a problem that's also under matrices? it's in #help-0

[spare widget]

Do you know why the length of a vector is sqrt(x1^2 + ... + xn^2)?

[short canopy]

yes

[spare widget]

If you do then form the vector P-Q or Q-P and compute its length

[short canopy]

from the pytagoras theorem

sure, this makes sense in my head if the vecto has the same direction but one is bigger than the other

but if they don't have the same direction, is not that intuitive to me

it doesn't help that i don't have a proof of this in the book i'm reading

[spare widget]

you should look into the geometry of vector addition

[fringe fjord]

P-Q is the vector you can add to Q in order to get P -- in other words the vector that will end at P if you put its beginning at Q.

[spare widget]

Try drawing a parallelogram with a vertex at 0 and the other two vertices being P and Q

You'll note that one diagonal is the sum P+Q and the other is P-Q or Q-P

And you can see that the P-Q/Q-P is the one between P and Q

Try drawing it

Thus its length is the distance between the two

[grand hare]

how do i find invertible matrix X such that XA=B

also it is given that these two are row equivalent
as well as the question expected that the ff elementary matrices are to be used to answer this problem

[short canopy]

right, now it makes more sense

one of the legs is x_1 - x_2 and the other is y_1 - y_2

then you use pythagoras to get the blue distance

bruh

[lavish jewel]

please try #prealg-and-algebra

[exotic lantern]

K

[grand hare]

i need help here pls

[zinc timber]

as a hint, note that $XA = B$ means $Xa_1 = b_1$ and $Xa_2=b_2$ where $a_i$ is the i'th column

[zinc timber]

i.e. $X\m{a_1 & a_2} = \m{b_1 & b_2}$

[zinc timber]

find a matrix that sends a1->b1 and a2->b2

[grand hare]

so something like making a a system of linear equations

[zinc timber]

kinda no

think about a LT with that prop

[grand hare]

do the elementary matrices help in finding M?

[zinc timber]

you can choose your own basis

[zinc timber]

for approach, no

there's also another eat to cheat

[grand hare]

it's just that the question implied that it expects for the elementary matrices to be used in finding the invertible matrix x

[zinc timber]

I can write it out but you aren't probably allowed to do that

[grand hare]

to do what

[zinc timber]

.

will give you an easy way to find X

by extending to a basis

[grand hare]

we havent gone into basis yet so

[zinc timber]

yeah guessed that much

[grand hare]

i just dont know how the elementary matrices are gonna be of use

[grand hare]

i dont exactly know where to begin

or more generally, what steps are to be taken

[spare widget]

Do you know how to solve Dy = f using elementary row ops?

[grand hare]

dy=f?

[spare widget]

D being a matrix

You can write XA = B as A^TX^T = B^T, then you can take out columns one at a time

which results in the usual setting

They seem to suggest something else in your exercise though, so maybe theirs is easier

[elder cliff]

Hey can anyone help me with this? Or have a problem like this worked out

Bro is this even English let’s be real here

I’ve legit been staring at this problem for 15 minutes

[spare widget]

the definition for matrix multiplication is

$(AB){ij} = \sum_k A{ik}B_{kj}$

[stoic python]

criver

[spare widget]

Then do a similar thing for BC but with indices j,l

Finally write out the same thing for ABC

[elder cliff]

Hmm okay

[pulsar wedge]

If the cardinality of a span== cardinality of a vector space is this equivalent to saying a basis

[subtle walrus]

saying a basis what?

[elder cliff]

Can I get help wit this

[icy totem]

??

[nocturne jewel]

help with what?

proving asso of composition? or the explanation?

[elder cliff]

Proving

Not sure what I’m supposed to do

[spare widget]

expand the definition of (f o g)(x)

after that you just have to apply it twice

[elder cliff]

Meaning?

I’m mega stuck

[nocturne jewel]

$(g\circ f)(x):=g(f(x))$

[stoic python]

Mosh

[spare widget]

you may want to read some of this also:

https://en.m.wikipedia.org/wiki/Function_composition

[elder cliff]

Thankyou

Still not sure but I appreciate the help

Gotta go see my professor

[spare widget]

To understand your exercises you need to know what the notation means, i.e. look up the definitions.

[elder cliff]

Any good youtubers who cover them?

[spare widget]

Do you not have a textbook?

though wikipedia is good enough for simple stuff like matrix multiplication and function composition

[elder cliff]

I do, but I’d rather have a video I learn better that way

[open locust]

In the following four pages (https://a.uguu.se/xtoGDuQ.png , https://a.uguu.se/jOUqqFxV.png , https://a.uguu.se/wZqCLKtg.png , https://a.uguu.se/rCDaQHB.png) I am rather baffled by the import of the writing on pages 109-110 starting at the definition of $(u^\ast, u) = \langle Ku^\ast, u\rangle$ and onward. What exactly is the point of the derivations on page 110, and are they related to the bidual mentioned towards the end there or do they have some kind of meaning unto themselves?

[stoic python]

joesmith1042

[spare widget]

This K looks like the metric tensor

[open locust]

@spare widget I'm not familiar with that

[spare widget]

I am not certain it is just fromreading this though, but you canessentuallytransform between vectors and covectors using it

[open locust]

I see

[spare widget]

e.g. say you have so,e basis g1, ..., gn

then G_{ij} = <gi, gj>

And one defines <u,v> = u^TGv

And you can treat Gu as the components of a covector

Since (Gu)^Tv

I am thrown off by the notation they use in what you linked though

What bookis this from?

[open locust]

It seems like (maybe I'm wrong here) that the author showed that if we "started" with the f^i's and set K^{_1} f_i = e^i for all i, we get that the u_i's are equal to the alpha_i's, so that kind of shows that the representation of f that we get is the same regardless of whether e_i or e^i are the basis for U.

@spare widget Functional analysis by E. Suhubi

[spare widget]

No idea

[open locust]

That last page and a half seems like it isn't making a really coherent point

Should I just ignore it? I worked through all the equations and confirmed them, I'm just not sure what I'm supposed to take a way from that whole passage.

[spare widget]

I am not sure either, but I can give you the metric tensor interpretation: you can basically change between the components of a vector and functional using the metric tensor

Typically one uses upper indices for a contravariant vector

And lower for a covector

[open locust]

@spare widget That seems to be what he is discussing there.

[spare widget]

e.g. see 2.10.4 for instance

[open locust]

The components of the vector seem to be equal to the components of the functionals when representing the functional in the dual basis

[spare widget]

alpha are the components of the functional

[open locust]

And you get the components of the vector by using K, or the metric tensor

[spare widget]

if you wanted to get a contravariant vector from saud functional you would do K^{-1} alpha

[open locust]

It seems a bit weird though, since how do we know that K gives those coordinates no matter what?

[spare widget]

K in the setting I know is just the grammian

[open locust]

I think I need to read through this again but hearing your explanation helped me think a bit better about this. Thank you!

[spare widget]

above 2.10.5

they claim it exists because the two are isomorphic

They don't specify what it is though

[open locust]

Yeah, I do agree with its existence

[spare widget]

the fi(ej) = delta_ij is just a requirement for biorthogonal bases

e.g. if you had e1,...,en (not being the canonical basis) expressed inthe coords (1,0,0...,0), ..., (0,0,...,1)

Then if you put those as columns of a matrix

E = [e1 | e2 | ... | en]

Then if you set now F = [f1; f2; ...; fn] as rows

[open locust]

@spare widget What confuses me is, the alpha_i come from a particular basis, e_i. In this proof, he starts with f^i and then maps it via K^{-1} to a new basis, e^i. But then he says, look, the alpha_i and the u_i (components of vector representation with respect to e^i) must be equal.

[spare widget]

You get F * E = I

But then F = E^{-1}

[open locust]

So I guess, this proves that K is in fact an isomorphism, or something?

@spare widget I see what you're saying in your explanation sort of

[spare widget]

They just take it to be

[open locust]

@spare widget Then I don't understand why he is showing what I just said, that alpha_i = u_i.

[spare widget]

What page

[open locust]

110, right before the paragraph about the bidual.

With the original basis e_i, f(e_i) = alpha_i.

So he then shows that if you start with f^i and use K to get e^i = K^{-1}(f_i), you end up with f = sum u_i f^i = sum alpha_i f^i

So alpha_i = f^i

[spare widget]

no alpha_i != f^i

[open locust]

My fault, alpha_i = u_i. typo.

So the components of a vector in the original basis equal the components of our function in the dual

regardless of the basis e^i

But, I guess, what's the importance of that?

Like, why does it matter that we have this equality?

[spare widget]

The components of a vector inarbitrary bases are definitely not equal

I am reading to figure out what the bases are here

[open locust]

Okay. Thank you!

[spare widget]

So the second half of the first page and the first half of the second establish that U^* is isomorphic to F^n

Then they take the canonical basis of F^n and through the isomorphism to U^* they find corresponding elements from U^* that are f^1,...,f^n

suchthat f^i(e_j) = delta_{ij}

[open locust]

Right, I do understand the U^* and U isomorphism part

And also the finding of those elements you mentioned

[spare widget]

Note that e_i is from U and f^j is from U^*

[open locust]

Yes 🙂

I understand up to that point.

[spare widget]

Ok here's the detail

Now they pick e^i from U such that e^i = K^{-1} f^i

[open locust]

Yes.

[spare widget]

Note that f^i is from U^* but e^iis its corresponding vector in U

[open locust]

Yes.

[spare widget]

Then they just show that the components wrt e^i are the same as wrt f^i

[open locust]

Right.

[spare widget]

you can think of this as having the equivalent basis in U instead of U^*

[open locust]

The components of f represented in its basis f^i are either the alpha_i 's (f(e_i) = alpha_i) or u_i's (components of u in e^i)

But what is the meaning of such an equivalency?

[spare widget]

sure, the detail is that one is a functional and the other is a vector

so you have a 1-1 relation between functional and vector

as far as I can tell it's mostly formally relevant

As one is from U^* while the other from U^n

[open locust]

Hm, so showing alpha_i = u_i means that f and u are isomorphic?

I thought he assumed that in the first place

[spare widget]

The thing is that a vector can be consumed by a functional

And previously you did not have a way for a vector to "consume" a vector

So by using this correspondence you dohave such a way

Say you are given u^i, v^i wrt e_i, vectors in U

[open locust]

(I'm wrong, he didn't assume it in the first place)

[spare widget]

using u_i = sum_j K_i^j u^j you can find u_i in the basis e^j

[open locust]

Right

[spare widget]

But you know that this matches the components of the functionalwith respect to f^j in U^*

[open locust]

Yes

[spare widget]

Thus f(v) = <Ku,v> = u^TKv

[open locust]

Okay I see

[spare widget]

note that u with lower indices is u in the basis e^i, while with upper indicesit is wrt e_i

[open locust]

Yes 🙂

[spare widget]

when it has lower indices it is equivalent to the components of a functionalin U^* wrt f^i

[open locust]

Right

[spare widget]

So fromthis you can turn vectors into functional and vice versa

[open locust]

Interesting.

[spare widget]

Allowing you to pair those

[open locust]

Thank you for the explanation! This helped me a lot

Studying this stuff on my own can be hard sometimes. This book is pretty good overall but sometimes I get stuck like this

[spare widget]

And at some point you will probabky have K explicitly defined as the metric tensor

[spare widget]

Honestly the book sounds all over the place

[open locust]

Really why?

[spare widget]

If I hadn't seen this before I would be totally lost

[open locust]

Oh yeah... well... I found it because I was trying to deal with another messy book that was much worse

[spare widget]

It just introduces concepts with no motivation whatsoever, ina by the way manner

[open locust]

Yeah. It's mostly written pretty well but occasionally it's confusing like this

Thank you again for helping me out with this, I really appreciate it 🙂

[spare widget]

Regarding this specific part I can suggest looking into some physics intro on tensors

should clarify things better

[open locust]

Oh okay!

[spare widget]

They have covectors, contravariant vectors, metric tensor etc there

[open locust]

I can look into it. Thanks for the recomendation 🙂

[spare widget]

this is with some illustrations, but it is rather slow: https://www.youtube.com/watch?v=8ptMTLzV4-I&list=PLJHszsWbB6hrkmmq57lX8BV-o-YIOFsiG

you may have to watch it at 2 speed if you have the time to at all

it's pretty basic

but gives some geometric intuition

[open locust]

Oh cool, thank you! I also have a book on manifolds that has some of this but it looks complicated

[spare widget]

this is also a nice "cheat sheet":
https://www.mathematik.hu-berlin.de/~wendl/pub/connections_appendixA_2.pdf

much more condensed

[open locust]

Thanks 🙂

[compact tartan]

hello hello

so my idea here was, S transitions from w to v

[spare widget]

S * w1 = v1, S * w2 = v2

[compact tartan]

that S*w1 =v1

[spare widget]

find w1, w2

[compact tartan]

but that isnt right i dont think

[spare widget]

yes those do not work

you don't need matlab for a 2x2 system

[compact tartan]

this is the answer key

this is what the method we said gives

[spare widget]

I am guessing the answer key is wrong

ok

I figured out

how they got those nubmers

[compact tartan]

i hope so, because i am very confused

oh?

[spare widget]

$\begin{bmatrix} \vec{v}_1^T \ \vec{v}_2^T\end{bmatrix} S = \begin{bmatrix} \vec{w}_1 & \vec{w}_2 \end{bmatrix}$

[stoic python]

criver

[spare widget]

idk why

[compact tartan]

thats really odd

[spare widget]

but that's what their answer corresponds to

[compact tartan]

ill keep looking at it, thank you very much 🙂

[spare widget]

Maybe they are treating {v1,v2} and {w1,w2} as the two bases

Then I think it may make sense

$\vec{w}1 = S{11} \vec{v}1 + S{21} \vec{v}_2, \quad \vec{w}2 = S{12} \vec{v}1 + S{22} \vec{v}_2$

[stoic python]

criver

[spare widget]

Thisis what they mean

[compact tartan]

ohhhhh that would make a lot more sense

thank you for looking into it for me

i appreciate it!

[spare widget]

I didn't realize they were treating those as basis vectors

[wintry steppe]

I'm learning affine spaces and subspaces now, and in this diagram, I'm unable to really understand that how exactly a-b gives the displacement vector for showing how displaced plane P2 is
[Source: Wikipedia Affine Spaces]

[wanton night]

I’m using Cholesky factorization to generate random data, but I noticed that some correlation matrices don’t result in data with the same correlations.

https://mlisi.xyz/post/simulating-correlated-variables-with-the-cholesky-factorization/

Is there a test I could do to determine if a correlation matrix is feasible?
For example, intuitively I know this doesn’t make sense:
c = [ 1, 1, 1,
1, 1, 0
1, 0, 1 ]

[wanton night]

I believe I answered my question. The corr matrix must be positive definite

[nimble sierra]

Hi, looking at matrix homework question and struggling to explain why changing x in Ax=b equation when there are infinite solutions. I understand it's just moving down the line that the matrix represents just don't know how to express that mathematically other than just pointing at a line on a graph

would appreciate any help

[dusky epoch]

can you show 3a?

assuming you still need help with this

[nimble sierra]

here is 3a myb

@dusky epoch

[dusky epoch]

consider $A \bmqty{8\-2}$

[stoic python]

Ann

[nimble sierra]

I was having trouble wording why changing the variables in 3b) doesn’t alter the result. I don’t know how you would explain that, other than calculating the answer by substituting A into the equation. Sorry if my question was unclear.

[dusky epoch]

have you learned about linear maps?

[nimble sierra]

Not for this unit

[dusky epoch]

have you at least learned that $A(x+y) = Ax + Ay$ for any matrix $A$ and vectors $x, y$ (whenever they are sized so as to make everything make sense)?

[stoic python]

Ann

[nimble sierra]

Yeah I understand that bit

[dusky epoch]

great

so that brings me back to this:

[dusky epoch]

.

also consider $A\bmqty{y\z}$

[stoic python]

Ann

[wintry steppe]

I think computing the determinant of matrix A could also support the answer

[nimble sierra]

but the question asks to not use calculation, so I'm confused on what it wants the working out to look like.

sorry if that sounds dismissive

[dusky epoch]

A*[y;z] is equal to b by construction

A*[8;-2] is easily seen to be [0; 0]

[nimble sierra]

I understand that point, but doesn't that result (A* [8;-2]) rely on substituting the matrix that A represents into the equation to solve A*x? If that makes sense.

[dusky epoch]

no

well

...i mean like ok

a particualrly anal teacher might yell at you for that since it is technically a calculation

[nimble sierra]

teacher explained it as: leaving A as the letter 'A' and changing only the left hand side to equal b

Feels like question is trying to be obtuse

[lavish jewel]

i guess they mean to generalize the result?

something like Ax = b and Ax' = 0, so A(x+x')=b?

[dusky epoch]

but how do you know Ax'=0 without calculation?

[lavish jewel]

yes, i'm also stumped there

[nimble sierra]

I was googling for the answer before and one solution was https://math.stackexchange.com/questions/2340563/how-to-prove-that-multiplying-the-equation-ax-b-from-the-left-with-invertibl

[lavish jewel]

i can't think of any argument for that that doesn't rely on calculating something

[nimble sierra]

but we haven't covered anything like the question in the content so far

[dusky epoch]

maybe applying linearity is in and of itself an act of calculation

this question feels really obtuse

[lavish jewel]

maybe we're all calculations

yeah the wording is a little too vague

[nimble sierra]

mini dose of existentialism with my homework

yeah

[dusky epoch]

maybe it's on purpose

[nimble sierra]

The question wants you to show that A*[y+8;z-2]=b without substituting in the values for A and performing the matrix multiplication. So, leave A as the letter 'A' and manipulate the left hand side until you can say that it equals b.

Is verbatim from the teacher also

so some calculations are allowed (?) just not involving A maybe

[dusky epoch]

well how else do we get that A*[8;-2] = 0??

[nimble sierra]

yeah seems hard to answer with equations, feel like screenshotting a desmos screen and visually proving it is easier than trying to rearrange 2x2 matrix to equal a 2x1 one

[wintry steppe]

In that sense without substituting A in the matrix equation Ax=b, you can find determinant of A, which is 0, and say equation Ax=b has infinitely many solutions for x

[weary flare]

already happy that lin alg exists

no more solving systems of equations

[icy totem]

??

[dusky epoch]

have you made any attempt?

[rough spoke]

Hey does anyone know how to do this question??

[astral sedge]

The basis for the row space of a matrix A is the set of non-zero rows in rref(A).
What if we take the rows in the original matrix A that corresponds to these non-zero rows in rref(A), do they also form a basis for the row space of A?

[spare widget]

not necessarily

an elementary operation allows you to swap rows

[rough spoke]

I am super confused can plz show how to do

[zinc timber]

rref can help

[rough spoke]

what is rref?

[zinc timber]

,w rref

[stoic python]

Results provided by WolframAlpha

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

Requested by Ryuzaki#5323

[astral sedge]

Right. But won't we have different non-zero rows in rref then?

[astral sedge]

Row reduced echelon form

[wintry steppe]

Have you learned what happens to the determinant if you multiply a row or column by something? This could be useful in solving this efficiently.

[astral sedge]

How come?

[rough spoke]

what would the span be though?

[wintry steppe]

||I'm thinking it may be possible to factor something out of that last column and then multiply the determinant by that at the end||

[icy totem]

yes , I know it but I couldn't apply it for this question

[spare widget]

say I have [v1; v1; v2] and I move [v1; v2; v1] then [v1; v2; 0] etc

then I will get first and second row form a basis in RR (provided v1 and v2 are lin indep)

but in the original matrix first and second row are v1,v1

which do not

[astral sedge]

Oh

Thanks!

So as long as I dont relace the rows, my claim is valid?

[spare widget]

I don't think it is

[rough spoke]

@spare widget could you use microsoft paint by change to solve this question because its hard for me to understand sorry this is my first time taking linear algebra

[spare widget]

because even without replacement it should depend on how you sum things

which row you add to each row

[spare widget]

after all the "replacement" is just row addition and subtraction

if I have r linearly dependent rows in the beginning, RREF would require me to zero some out and have them at the end

so I don't think it works as nicely

[spare widget]

do you know the definition of span?

[rough spoke]

kind of

its between the vectors i think

[spare widget]

look up the definition of span in your textbook

[astral sedge]

Cool thanks again

[spare widget]

you can't expect to solve a problem where you don't know what it says

for the case I would suggest trying (0,0), (1,0), (0,1), (1,1) and figuring out what is the convex set between those points

[rough spoke]

The span of a set of vectors is the set of all linear combinations of the vectors

[spare widget]

ok, write that out explicitly through mathematical notation

i.e. span(v1,v2) = {v = ? : ...}

[rough spoke]

yes thats the part i dont know

[spare widget]

https://en.wikipedia.org/wiki/Linear_span#Definition

[rough spoke]

this is the first question of the chapter

[spare widget]

do you know what a linear combination is

[rough spoke]

can you maybe do the question so I could work backward plz

Well, a linear combination of these vectors would be any combination of them using addition and scalar multiplication

[spare widget]

write it out mathematically

[rough spoke]

→v1, →v2, →v3

[spare widget]

???

$span(\vec{v}_1, \vec{v}_2) = { \lambda_1 \vec{v}_1 + \lambda_2 \vec{v}_2 ,: , \lambda_1, \lambda_2 \in \mathbb{R}}$

[stoic python]

criver

[rough spoke]

im soo sorry sir, i just need to see it in normal writing so i can understand easir

[spare widget]

now the question is - are v1 and v2 linearly independent?

if they are, then the set above spans the whole R^2

[rough spoke]

kk

but how do i use thatb in my question

Im am dying rn

Math can be soo stressful

[spare widget]

maybe try to work it in reverse

set v1 = (3, -2)^T

v2 = (0,1)^T

now graph some points

e.g.

[rough spoke]

what is T?

[spare widget]

transposition - I mean that the vectors are column vectors

it's irrelevant

(0,0) -> 0 * v1 + 0 * v2 = (0,0)
(1,0) -> 1 * v1 + 0 * v2 = v1
(0,1) -> 0 * v1 + 1 * v2 = v2
(1,1) -> 1 * v1 + 1 * v2 = v1+ v2

try drawing those point on a piece of paper

[rough spoke]

do i just graph random points?

[spare widget]

and then connect them with edges

[rough spoke]

ohh

[spare widget]

yes, in order to understand how linear combinations work

[rough spoke]

so 3 is x and _2 is y on graph?

[spare widget]

yes

[rough spoke]

okok

now what

[spare widget]

v1 = (3,-2) first component is along x second is along y

then draw also v2

and finally construct those 4 points

[rough spoke]

ok bet

[spare widget]

(0,0) -> 0 * v1 + 0 * v2 = (0,0)
(1,0) -> 1 * v1 + 0 * v2 = v1
(0,1) -> 0 * v1 + 1 * v2 = v2
(1,1) -> 1 * v1 + 1 * v2 = v1+ v2

and connect them

then try to figure out what happens when you vary (s,t) in [0,1]^2

what kind of set do you get

[rough spoke]

yeah thats the part idont know

i havnet learneit yet'

[spare widget]

just draw what I mentioned

and you'll figure it out

one builds the linear combination

s * v1 + t * v2

the question is what kind of set does this span when one varies s and t

to figure it out try plotting several points (s,t) = (0,0), (s,t) = (1,0), (s,t) = (0,1), (s,t) = (1,1)

[rough spoke]

ok

[spare widget]

the point of the exercise is to familiarize yourself with the meaning of span

[rough spoke]

so it made like a line

[spare widget]

what coordinates did you get for the four points?

[rough spoke]

so the other points are (0,1) (-1,2)

i mean (1,0) and (2,-1(

[spare widget]

how did you get that

s * v1 + t * v2

(s,t) = (0,0), (s,t) = (1,0), (s,t) = (0,1), (s,t) = (1,1)

just plug in and compute

[rough spoke]

what is s,t?

[spare widget]

real numbers

[rough spoke]

is and anyway I can get on a call sir?

[spare widget]

you know how to compute s* v1 + t * v2?

[rough spoke]

no

[spare widget]

you don't know how to sum vectors?

[rough spoke]

i do

[spare widget]

do you know how to multiply a scalar by a vector

[rough spoke]

but i dont know wt s and t are

ohh

no

i only know vector to vector

[spare widget]

do you have a textbook?

$s \vec{u} = s\begin{bmatrix} u_1 \ \ldots \ u_n \end{bmatrix} = \begin{bmatrix} s u_1 \ \ldots \ s u_n\end{bmatrix}$

[rough spoke]

no

[stoic python]

criver

[spare widget]

find one

[rough spoke]

ok

but is there anyway you can give me the answer

so i can work backwards and correlate with my textbook

[spare widget]

Hoffman and Kunze, Strang, Friedberg, Shifrin, Axler, any would do

[rough spoke]

k

[spare widget]

this would also do: https://tutorial.math.lamar.edu/classes/calcii/VectorArithmetic.aspx

[rough spoke]

i have the david poole textbook

[spare widget]

then look up vector arithmetic in it

[rough spoke]

\

dude what formula do i use to find the answer

can u put the numbers in for me

becuase my textbook is bot

[spare widget]

no, that would be didactically improper and then you'll be back tomorrow with a variant of the same question

learn how to add vectors and how to multiply vectors by a scalar

https://tutorial.math.lamar.edu/classes/calcii/VectorArithmetic.aspx

[rough spoke]

sdflkdsfjilodfsjgpoik'fdjg'poil;dfkg;hlb\

[spare widget]

once you do that, you will be able to better appreciate what linear combination and linear independence means

[rough spoke]

ahhhh its cuz of people like you Math is not fun

[spare widget]

have it your way, have a nice day

[rough spoke]

bro plzzz just help ,me

im sooo sorry for wht i said and done Im bad at math sir

@spare widget im actually sorry tho

[spare widget]

idk what you're sorry for, my answer remains the same regardless - practice some vector arithmetic

once you understand how that works, compute some linear combinations as described

I gave you a resource: https://tutorial.math.lamar.edu/classes/calcii/VectorArithmetic.aspx

use it

do some problems

then go back to your problem and find out whether v1 and v2 are linearly independent

if they are then they span the whole R^2

then for the example part try plotting the 4 points corresponding to (s,t) = (0,0), (s,t) = (1,0), (s,t) = (0,1), (s,t) = (1,1)

finally quantify what the set (s,t) in [0,1]^2 represents

I can't help you beyond that

[still lodge]

not entirely LA but

anyone got neat ideas for storing matrices in a text file

doing this rn which is id rows columns a1 a2 a3... ....

[wintry steppe]

Probably see how numpy does it

[quartz compass]

why would you want to do this in the first place

[still lodge]

cs project

csv files are definitely easier but it's a group thing and im on the half of the group that was assigned to try txt

[still lodge]

any resources/pointers? im not very well versed in anything python

[wintry steppe]

...neither am I

Anyway, in real life, you can compress the text file by implementing your favourite compression algorithm by hand. If your text is guaranteed to be plain text with the numbers 0 to 9 and spaces + line breaks, you can compress it by using more letters of the alphabet

Also for sparse matrices, instead of storing every entry, you can store the position and the value of each entry

[still lodge]

i'll implement this eventually

[wintry steppe]

If you're fine with lossy compression, consider looking up on various matrix decomposition methods, in which SVD is one such algorithm

Many considerations, depending on the nature of your project

[still lodge]

well it's still early so im just getting basic stuff implemented like matrix addition, multiplication, random generation etc.

but i'll have to think about stuff like that eventually

[compact tartan]

@spare widget just wanted to thank you again for helping me last night, i understood the whole section i was struggling in cuz of ur help

[drowsy gulch]

Let A be a square matrix $n \geq 2$ with all minors of order $n-1$ equal to eachother. Show that $det(A) = 0$

[stoic python]

Alphara

[drowsy gulch]

can someone provide some guidance for this proof?

[native rampart]

Consider Minors obtained by erasing a_11 and a_12

You get that first column of minor obtained in first case= first column of minor obtained in second case

Which is to say Column 1 and column 2 of matrix are equal except for a_11 and a_12

@drowsy gulch

mb I misunderstood minor

[drowsy gulch]

oh hmm let s see

[native rampart]

I think the approach will be similar

Consider Minors formed by erasing a_11 and a_12

[wintry steppe]

I was studying cross products and dot products, but I can't really get the notion of them. Why exactly are they useful?
Things I noted while learning about them:

1. 2-d cross products show how close two vectors are to each other (assuming both vector lengths to be equal) as the formed parallelogram's area would be less on the graph.
2. Cross products also give a notion on the position of respective vectors. If the product is positive (of two vectors x and y), then x will always lie on the right side of y, and if the product is negative, thwn x will always lie on the left side of y, on the graph
3. I really can't seem to get the intuition how the projection and dot product are equivalent. Also, are there more notions of these two concepts?

[nocturne jewel]

cross product gives a mutually orthogonal vector

dot product generalizes a lot to other inner products / how we define geometry of vectors

[drowsy gulch]

alright

i found C1 = C2 except a_11 and a_12 ; and now @native rampart i would do minors of a_21 and a_31 to find L2=L1 which would mean 2 columns/lines are equal which would mean det(A) = 0

am i getting this right?

yea i was afraid to point that

[native rampart]

I thought minors were the submatrix

mb

[drowsy gulch]

hmm

i do not see why this d be wrong tho

[native rampart]

Because minors are the determinant of the submatrix This argument doesn't even make sense

[drowsy gulch]

ok, then what would work in this case?

i thought that maybe showing 2 colums are identical would instantly mean det(A) = 0

[drowsy gulch]

but i m clueless, do u have another idea?

[spare widget]

Maybe swapping a row and showing that dets are equal

swapping a row negates the determinant

[native rampart]

Ok does that work

[spare widget]

But if you can show that those are nevertheless equal after the swap then it should work

I am not sure whether that follows from all minors being equal though

I was also thinking about using the laplace expansion, but idk how that would work out

Let det of any minor be equal to d

Then the determinant expansion rt first row would be

$d\sum_j (-1)^{1+j} a_{1j} = d\sum_j (-1)^j a_{2j}$

[stoic python]

criver

[spare widget]

does this help, I guess not

[drowsy gulch]

trying it

[spare widget]

You can write asimilar thing for all rows

I am gessing that if you somehow group the systems it turns out that the vectors are linearly depedent

e.g.

$(-1)^{i+1}\begin{bmatrix} 1 & -1 & 1 & -1 \ldots & (-1)^{n+1} \end{bmatrix} \cdot \begin{bmatrix} a_{i1} & \ldots & a_{in} \end{bmatrix}= (-1)^{k+1}\begin{bmatrix} 1 & -1 & 1 & -1 \ldots & (-1)^{n+1} \end{bmatrix} \cdot \begin{bmatrix} a_{k1} & \ldots & a_{kn} \end{bmatrix}$

[stoic python]

criver

[viral olive]

Hey guys i have a problem with defining an correct answer for a i guess almost solved problem.

I hope people can read this my biggest problem is how to show that there is an map which is an additive inverse for any element of another random map of the same Set.

I would have done by saying the additive inverse of the same element would g´(x)= sgn(-1)g(x) the problem is the field T is not defined at all. So all elements could be inverse to themselves as much as i know. So i feel hard to define an actual answer. (English isnt my main language so if i was unclear with my question. Just point it out)

[royal wave]

Hey everyone, I'm a bit clueless here. I have an intuition this is not true but can't think of a counter example. If anyone can help I would appreciate it! :

[spare widget]

Scaling columns and rows is not the same

you can find plenty of counterexamples

Pick any S that is not diagonal

Then $(T S){ij } = t{ii}s_{ij}$ while $(ST){ij} = s{ij} t_{jj}$

[stoic python]

criver

[spare widget]

Because t_{ii} != t_{jj} for ageneral diagonal matrix

[zinc timber]

[spare widget]

he proceeded to try to dm me too, unfortunate

[zinc timber]

[zinc timber]

they think helping them is our liability

[viral olive]

thats kinda messed up

[zinc timber]

ban worthy

@eddd

[spare widget]

I think it's just kids being forced to study something they don't want to, so I can understand their frustration, but yeah I can unfortunately not put in the effort required from their side

[zinc timber]

yeah but if someone is genuinely trying to help, you mustn't be rude

oh well

[spare widget]

I don't really mind that

[zinc timber]

[spare widget]

It's like when your kid doesn't get a toy they wanted and they throw a tantrum, they eventually learn not to

[zinc timber]

[spare widget]

there's no point in getting offended at them

[viral olive]

At least you hope they learn.

[spare widget]

anyways enough derai;ing of the thread from my side

[spare widget]

Look up linear functionals

dual and so on

[viral olive]

thank you

[karmic hazel]

Is the solution to Q30(b) correct?

[zinc timber]

looks ok

but check the 2nd life from bottom

[karmic hazel]

Ya right it shouldn't be there.

Thank you so much!

[lavish jewel]

i find \ni used as s.t. a bit weird 😛

[karmic hazel]

Isn't that symbol universally used to write 'such that'?

[lavish jewel]

it's also used in the same way as \in

i wouldn't know which one is more prevalent tbh

i have seen more s.t. outside of sets, and | or : within

(it's just a tangential comment btw, the proof is fine, as ryu said)

[karmic hazel]

Oh okay. I haven't seen \ni being used the same way as /in. Anyway I am not pursuing a degree in mathematics so I don't have much experience with symbols.

[lavish jewel]

what do you think?

[solid jetty]

positive

strong

idk the last one

[lavish jewel]

try picturing where those data points would lie on the graph

maybe put them in using paint

[solid jetty]

first one I near the line

[lavish jewel]

and ask yourself, what would need to happen for the association to be better, worse, or about the same

[solid jetty]

and 2nd one is not near the line

[lavish jewel]

not near, but is it more or less close?

[solid jetty]

less close

[lavish jewel]

i would still claim that it's about the same distance from the line as the other points

if you placed more points exactly on the line, the association would be better. if you placed points off the line and/or far away from it, it would get worse

here you put one point almost on the line and another still kinda close to it

i wouldn't expect that to change the result much

the two new points look kind of similar to the other points when you compare their distances to the line

[solid jetty]

is it the same

[lavish jewel]

that's my impression, yes

[solid jetty]

ok

[plush umbra]

what does it mean to actually multiply or divide vectors?

[spare widget]

what definition of multiplcation/division

[plush umbra]

To change the magnitude of something

[spare widget]

what is the definition that you use

mathematically

[plush umbra]

No idea not at that level yet to understand what it means to define something mathematically

[spare widget]

how do you compute this multiplication/division that you want the meaning of

you can have elementwise multiplication (hadamard product), dot product, cross product, exterior product, etc

[hardy inlet]

is this a safe claim to make in my notes i'm typing?

or is something weird with the conjugates

[gray dust]

yes

[hardy inlet]

"In the definition above, the order of the vectors does not matter, because <u,v> = 0 iff <v,u> = 0" the book never explicitly explained that iff in the previous parts, I guess its just assumed by the conjugate symmetry property since the conjugate of 0 = 0

[nocturne jewel]

In definitions, if is always iff

[hardy inlet]

yeah orthogonal iff <u,v> = 0. i now understand what the non-definition paragraph was saying

[grave kettle]

why are linear transformations ‘tranaformations’

not ‘functions’

[nocturne jewel]

cause they arent functions?

take any non-injective T, it's many to one

since $\operatorname{Ker}(T)\neq{0}$

[stoic python]

Mosh

[nocturne jewel]

@grave kettle

[hard drum]

Functions don't need to be injective lol

[nocturne jewel]

you can find many linear transformations that aren't functions, hence why we don't call them all functions

[hardy inlet]

what does the box in the bottom mean when it says "the converse is true in Rⁿ"

does that mean if \norm{u + v}^2 = \norm{u}^2 + \norm{v}^2 then they're orthogonal?

[hard drum]

for example?

[hardy inlet]

how is <v,u> = conjugate <u,v> = Re<u,v> (the real part of the equality)

[nocturne jewel]

0 mapping

many-to-one

[gray dust]

thats a function

[hard drum]

Yeah so that implies you think functions can't be non-injective or smth

[nocturne jewel]

Oh wait I have it backwards right

[solid leaf]

Wait so are all linear transformations functions? I’m a first year Lin alg student sligjtly confused now

This is still true right ^^

[hard drum]

Linear transformations are generally defined as functions between two vector spaces which are also linear

[gray dust]

@grave kettle @solid leaf transformation/function/map are synonyms

[solid leaf]

Ok I thought so

[gray dust]

not even a linalg thing, if i said f(x)=0 isnt a function in #prealg-and-algebra ppl would call me insane

[hard drum]

It's not quite that, the point is that z + its conjugate = 2Re(z) for any complex number

[gray dust]

or f(x)=x^2 isnt a function

[nocturne jewel]

K, and I acknowledged I had it backwards.

[hard drum]

Just because (a+ib) + (a-ib) = 2a

But then if we are working over real inner product spaces, everything is real anyway, so the conclusion holds iff <u,v> = 0, that is, iff u and v are orthogonal

The reverse Pythagorean theorem is basically that if a triangle obeys the Pythagorean theorem then it must be right angled

[gray dust]

not even that, transformation/function/map are synonyms

[nocturne jewel]

Holy I get it

[hard drum]

This shows that if we are working over a real inner product space we have |u+v|^2 = |u|^2 + |v|^2 then u and v must be orthogonal

@hardy inlet

If you need any further explanation/detail feel free to ask

[hardy inlet]

Im blanking on this part

oops i wrote that backwards

Re<u,v>

[hard drum]

Yeah that isn't true in general

[hard drum]

I mean all they say there is that 2 Re<u,v> = <u,v> + <v,u> and ye that's explained here and below

It's just that if we work over a real inner product space then, well, <u,v> = <v,u> = Re<u,v> automatically cause everything is real and symmetry of the inner product etc

[hardy inlet]

yeah if u treat it as real to begin with that makes sense

and i get what u mean with z + *z = 2(Re z)
where *z is the conjugate

[hard drum]

sure

[hardy inlet]

but what does that have to do with conj<u,v>

[hard drum]

<v,u> is the complex conjugate of <u,v> (part of the definition for complex inner products)

[hardy inlet]

This isnt a proper proof cause I started with what we want but theres no was of simplifying the conjugate of uv?

this true?

[hard drum]

Not quite, <u,v> = Re<u,v> + i Im<u,v>

But yes we have <u,v>* = Re<u,v> - i Im<u,v>

[hardy inlet]

ah right the i

ok so why doesn't reverse Pythag hold for complex vector spaces then?

oh okay so the complex converse says the only the real component is guaranteed to be zero, we know nothing about its complex part?

therefore they're not always orthogonal since if the complex isn't zero, the whole thing isn't zero, and hence it isn't orthogonal

Is the only thing in parenthesis the orth_u(v)?

cause in my stats class we were doing linear algebra to derive the linear regression curve and somewhere was an "orth" and the professors like "u guys know what an orth is right?" (nobody knew)

and when I google orth linear algebra it doesn't have anything describing that. Does this exist?

is from his own notes:

[analog nacelle]

yes

$u - \frac{\langle u,v\rangle}{|v|^2}u$ is the orthogonal component
What you're saying is that the orthogonal projection of a vector $u$ onto $v^\perp$ is the projection of $u$ onto $v$ subtracted from $u$.

[stoic python]

Kirby

[hardy inlet]

is "orthogonal component" the "orth"

[analog nacelle]

I should say orthogonal projection, but yes

[hardy inlet]

and is the $\operatorname{orth}_vu$ a real notation?

[stoic python]

MattDog_222

[analog nacelle]

I've never seen it personally, but it might just be something your professor does
In most cases, it's best to double check with your prof, but they seemed to define it in their notes as \vec{x} - proj(\vec{x})

which is what is expected

[hardy inlet]

so is the orthogonal projection facing this way? (or the other)

[analog nacelle]

yes

the way you've written it is correct

[hardy inlet]

and vector cv = proj_v(u) correct?

[analog nacelle]

yeah I believe so

[hardy inlet]

(i always did those wrong)

cause its like read backwards

proj_v u is "projection of u onto v"

[wintry steppe]

Given any two 3D-vectors, how do you calculate the pitch, roll, and yaw to rotate one vector about another?

[grim leaf]

H+K uses $N_f$ to denote the "null space of $f$". Does $\ker(f)$ mean the same thing?

[stoic python]

totally not anamono

[grim leaf]

he doesn't mention anything about kernel but wikipedia says kernel and null space are the same thing, i just wanna make sure im not mixing up two different meanings

[hardy inlet]

null space is the kernel of a Transformation. but I dont think all kernels can be called nullspaces. I'll let someone else confidently answer; and or ask in Abstract algebra

[native rampart]

Well if T is a linear transformation, nullSpace(T)=Kernel(T)

Kernel is a more general term

I guess H+K calls it NullSpace to emphasise it's a vector space

[gray dust]

@grim leaf same thing. nullspace is used primarily in linalg, kernel is a more general term in abstract algebra

[grim leaf]

gotcha, thanks guys

[hardy inlet]

At the bottom of the proof, how does v=u1-u2??

[gray dust]

wording is important

the eqn holds 'for every v in V'

so in particular it holds for v=u1-u2

[hardy inlet]

ah, and then <a,a> = 0 implies a = 0

thanks

[gray dust]

no prob

[hardy inlet]

and to clarify, we're letting v = u_1 - u_2 yes? basically a counterexample sorta?

[gray dust]

not a counterexample, its plugging a specific thing into a 'for all' statement

for example if all dogs are bad and ben is a dog, then i can conclude ben is bad

[hardy inlet]

isn't it sorta the inverse of that
we think all dogs are good, but ben (u1-u2) is bad, and ben is a dog, so not all dogs are good?

[gray dust]

its not, see the format of the proof

it shows a statement holds 'for all v in V'

so it holds for any specific v we pick

the dog thing works the same way

[hardy inlet]

but like I didnt think we were "picking" any v. I thought we were picking a u1 and u2

like theres only one v that equals u1-u2

[gray dust]

i can really only just reiterate whats said

https://i.gyazo.com/6c6c7f29116f745198112aa02bd5a42b.png

this is shown 'for every v in V'

then we 'take v=u1-u2'

ie plug that specific quantity for v into the eqn

[hardy inlet]

and thus $\langle u_1 - u_2,, u_1 - u_2\rangle = 0 \implies u_1 - u_2 = 0 \implies u_1 = u_2$

[stoic python]

MattDog_222

[gray dust]

yes

[hardy inlet]

but to me that feels like a counterexample. Its like "Oh yeah, you're saying this holds for every v? Well what about v = u1-u2? gottem!"

[gray dust]

it doesnt lead to any contradiction

its a common logic thing that i tried to make feel familiar with the dog thing

going from general to specific

if the logic is still too unfamiliar, we can set v=u1-u2 at the start instead of using v as a stand in

[hardy inlet]

like i get think i understand the logic i just dont understand what its called, "proof by __"

i guess its a direct proof idk

[gray dust]

idt theres a name for a proof containing this step

but idt it needs one either, the step of going general to specific should feel intuitive after enough use

[hardy inlet]

bc to me it feels like a 'general counterexample' like 'you give me any u1 and u2 that u think this holds for and I'll give you this v'

[gray dust]

heres a simpler example that follows what u feel but doesnt contradict anything

x^2>=0 for any real x

[hardy inlet]

but thats true

[gray dust]

what if i set x=3? that should be a counterexample right?

[hardy inlet]

9 >= 0

[gray dust]

just by virtue of me setting a specific x value

[hardy inlet]

no you're giving 1 example of it being true

you need to give me infinite examples of it being true

[gray dust]

exactly i didnt contradict anything

[hardy inlet]

but it felt like we're giving 1 example thats false (thus it doesn't hold for every v)

[gray dust]

but u somehow feel differently about setting a specific v for a statement which was already shown true for all v

[hardy inlet]

my brain: If its true for every v then it must also be true for v=u1-u2, but when we do this u1 = u2 hmm wait yeah i guess

we know $0 = \langle u_1 - u_2, u_1 - u_2\rangle$ since we proved it in the first part

[stoic python]

MattDog_222

[hardy inlet]

which clearly says u1=u2

wait we're not proving for every v in V

we're using every v in V

[gray dust]

our goal is u1=u2 (uniqueness of u)

[hardy inlet]

yeah and we're saying the only possible way that <v u1> = <v u2> is if u1 = u2, since we know it equals 0 by the first proof

brain: At least one case we have u1=u2 as the only solution therefore for all we need u2 to also equal u1

well wait we aren't saying u1 = u2 = the only solution

we're saying u1 = u2

then the first part, the existance theorem shows that u does exist

so then u1 = u2 implies uniqueness; idk it still feels a little fuzzy

[gray dust]

technically the u1=u2 bit means 'at most one vector has the property of u'

together with existence of u it gives uniqueness

[hardy inlet]

yeah like, if theres a solution then they're the same; so we first showed theres a solution

[gray dust]

'at most one' involves supposing 2 things have a certain property then showing theyre equal

[hardy inlet]

Alright thanks for helpin im gonna move on from the page

[gray dust]

ur welcome

[hardy inlet]

kinda whack tho that not only does it prove its unique but what it equals

[grim leaf]

in the second to last line, why is it evaluated at 0 to b

and not a to b

[hardy inlet]

looks like a typo

[grim leaf]

kk yeah that's what i thought too

[hardy inlet]

theres also another typo there with no x

[grim leaf]

oh right didnt catch that

[hardy inlet]

$V = \mathcal{P}(\mathbb{R})$

[stoic python]

MattDog_222

[hardy inlet]

is a shorter way of saying "Let V be a vector space of all polynomial function over the field of real numbers"

[grim leaf]

P denotes the power set?

[hardy inlet]

no the set of polynomial functions

[grim leaf]

ah kk

[hardy inlet]

$\mathcal{P}_n(\mathbb{R})$ is polynomials of degree $n$ or less

[stoic python]

MattDog_222

[grim leaf]

oh interesting

[hardy inlet]

so like $p(x) = x^2 + 5 \in \mathcal{P}_2(\mathbb{R})$ but $p(x) = x^3 \not\in \mathcal{P}_2(\mathbb{R})$

[stoic python]

MattDog_222

[grim leaf]

wait ok bit tangent but then is basis denoted with $\mathscr B$ or $\mathcal B$

[stoic python]

totally not anamono

[hardy inlet]

I dont think there's a set symbol to denote a basis

usually "Let B be a basis of V such that B = (v1, v2, ... , vn)"

[grim leaf]

H+K uses this fancy B

lmao just not sure if it's scr or cal

im tempted to say scr

[hardy inlet]

not sure, depends on their font but I dont think it matters

[grim leaf]

kk

[hardy inlet]

You just need to say its a basis

[grim leaf]

yeah

[hardy inlet]

WAIT

WAIT

[grim leaf]

waiting

[hardy inlet]

THE basis

its a basis

[grim leaf]

yeah

[hardy inlet]

[grim leaf]

oh

lmfao

¯_(ツ)_/¯

[hardy inlet]

let me just let B = (1,1), (1,0) be THE basis of R²

xD

but it should say "a" basis

[grim leaf]

yeah

[hardy inlet]

what is C^3

cause its certainly not complexes

(to my unextensive knowledge)

[grim leaf]

i think it is just complexes

R is a subset of C

[gray dust]

bases generally arent unique but idt the wording deserves changing too much

THE refers to how the vectors are defined

[hardy inlet]

yeah R is a proper subset of C, so R doesn't span C

but a1 a2 and a3 are all real coordinates

i don't deal with complex stuff much tho so i could be wrong

Robake is that a basis of complex^3? I wanna know

[gray dust]

yes

[hardy inlet]

but theres no imaginary component

[gray dust]

so?

[hardy inlet]

how are you gonna form (i, 1+i, 5) from that

or anything with an i

oh

u multiply by i

thats cheating

xD

[gray dust]

the algebra is a bit hard but heres an easier example

i,j,k is a basis

[hardy inlet]

perhaps use e1 e2 e3?

since i is conflicting

[gray dust]

eh u get the idea

theyre all real yet a basis

[hardy inlet]

so like would it be e1i + e2 + e2i + 5e3?

[gray dust]

yes but just say (1+i)e2

[hardy inlet]

ah that would make sense

that's a cheesy scalar

but I guess it makes sense cause dim C3 != 6

they do say computation is for insight

[gray dust]

C^3 is usually taken as a space over C

[hardy inlet]

as in scalars pulled from C?

[gray dust]

yes

so dont fear writing complex coefficients here

[hardy inlet]

I took abstract algebra and remember approximately 1% of it but is there ever a use to use non complex non real fields as scalars

or maybe other things cant even be scalars

i guess rationals could

but maybe a vector space over general linear matrices or something

[gray dust]

yes theres quite a bit to be done with finite fields

[hardy inlet]

is modulo ever used in the context of linear algebra vector spaces

like $\mathcal L (Z_5)$

[stoic python]

MattDog_222

[zinc timber]

what's L here?

[hardy inlet]

i guess the set of linear maps from Z/5Z to Z/5Z

[gray dust]

a simple example w/ modulo is encryption matrices

[hardy inlet]

ah cause u cant undo modulo fully

[zinc timber]

F

lmfao

[hardy inlet]

isn't it a field?

[zinc timber]

it is

[gray dust]

no, it can be undone

[zinc timber]

🤣

[hardy inlet]

it cant fully be undone tho right

[gray dust]

u just need to know the matrix used to encode the msg

[hardy inlet]

like if u have 4, was it 4, 9, 14, 19,...

[zinc timber]

they are considered same

same eqv class

[hardy inlet]

yeah so like u cant undo it right

since its ambiguous

[hardy inlet]

would that we like enigma?

[zinc timber]

so for encryption scheme like m^e mod n, we have an restriction that m<n

[gray dust]

as long as the encryption matrix is invertible, u can decode

if it wasnt then communicating w/ ur friends would suck

[hardy inlet]

i wonder if Z/2Z is useful since in CS XOR is useful for encryption

[zinc timber]

you can look at GF(2⁸)

[hardy inlet]

whats a GF

[zinc timber]

field of order 2⁸?

idk what GF stands for tho,

it just means an extension of Z/2Z of order 2⁸

[gray dust]

galois field

[zinc timber]

ig

[hardy inlet]

ah a finite field

idk of a field with 256 elements

[zinc timber]

given any prime p and n>1, you can always find a field of order pⁿ unique upto iso

[hardy inlet]

oh right

and 2^8 is its prime factoring

[zinc timber]

here p=2, n=8

[hardy inlet]

i wrote some program once but it doesnt' have a p value

or is it the sum of the n's

[zinc timber]

what?

[hardy inlet]

well like the p doesnt matter right

[zinc timber]

why are you partitioning 8?

p is any prime, not necessarily 2

[hardy inlet]

i thought it was the fully prime factorization with the product of the powers

[zinc timber]

[hardy inlet]

so like p1^a * p2^b * ... is func(a)* func(b) * ...

[zinc timber]

[hardy inlet]

72 = 2^3 * 3^2 so the number of subgroups up to isomorphism is this function evaluated at 3 times it evaluated at 2

System.out.println(numSubgroups(3) * numSubgroups(2)); prints 6

if u have Z_81 its 3^4 so numSubgroups(4) = 5 subgroups up to iso

n = 1 --> 1
n = 2 --> 2
n = 3 --> 3
n = 4 --> 5
n = 5 --> 7
n = 6 --> 11
n = 7 --> 15
n = 8 --> 22
n = 9 --> 30
n = 10 --> 42
n = 11 --> 56
n = 12 --> 77
n = 13 --> 101
n = 14 --> 135
n = 15 --> 176
n = 16 --> 231
n = 17 --> 297
n = 18 --> 385
n = 19 --> 490``` this pattern

e.g. something with order 2^{19} has 490 subgroups up to isomorphism

i might be rambling on nonsense

but something has to do with those numbers in group theory

[zinc timber]

oh you are finding non isomorphic subgroup of given order

[hardy inlet]

welp it was fun toying around with groups but i need to do some linear algebra now :/
I'll be back

and yeah I guess my function name is misleading

I should call it partitions(int n)

but when i wrote it i didn't know thats what they were

so the intersection is empty? It doesn't say subseteq

and if U is a subset (not a subspace) it doesn't need to contain 0

but like if V is R3 and U is the XY-plane, then U^ is Z-axis, and the intersection contains zero, no?

[using U^ as U^{\perp}]

[zinc timber]

lol, some authors use $\subset$ to denote a subset

[zinc timber]

and $\subsetneq$ for strict subset

[zinc timber]

proper subset

[hardy inlet]

oh thats lame

so it can contain 0

[zinc timber]

indeed

well

[hardy inlet]

wack

[zinc timber]

that's exactly zero

oh

no

wait

[hardy inlet]

subset

not subspace

[zinc timber]

U is subset not subspace

[hardy inlet]

its empty OR zero right?

[zinc timber]

so yeah, it can be empty

[hardy inlet]

but U can also be a subspace and thus has 0

[zinc timber]

yes

when U is a subspace, the int is {0}

[hardy inlet]

would e be a subseteq too

[zinc timber]

yes

\subset mean subset here, proper subset or equal

misleading symbol if you are used to \subseteq

[hardy inlet]

ok and I have this. is this the proj_uv kinda deal or the orthogonal component?

[hardy inlet]

like earlier when i had this in stats, is this P_U?

are we using P_U as the proj_uv from calculus?

Like it doesnt have an arrow showing if it came from 0 or v