linear-algebra

[swift plaza]

I think you're forgetting that $\frac{1}{\det(A)}$ is not a matrix but a scalar

[stoic python]

killer_memestar:

[wintry steppe]

yes

That gets the answer

But why we took it as C

@wintry steppe Thanks

And can you remention what is this thing called?

which thing?

The n thing

And why we took it as C

I don't think this theorem has a name

let's just say it's a property of matrices

and determinants

Ahh

But why did we do that

And is it valid everywhere

[stoic python]

Blitzkrieg:

[wintry steppe]

^ this is true

Ahh right

[thorn robin]

we call it "the determinant is multilinear as a function of the column vectors"

[wintry steppe]

Gotcha

[thorn robin]

you scale one column by c and you get a factor of c out

[wintry steppe]

@wintry steppe thanks!!

That's a very nice interpretation @thorn robin

@thorn robin Thanks!!

I finally reached somewhere

Thanks you guys

Although I am still not clear about stuff but atleast I know

[wintry steppe]

I need Linear Algebra textbooks

Any ideas?

[sonic osprey]

The one by Strang is pretty popular

[wintry steppe]

With a free pdf

[sonic osprey]

libgen.is

[wintry steppe]

Thank you

Idk how good it is, but there's Gelfand

[eternal widget]

yo

[indigo cradle]

hello could i have some insight to extra credit part of question 10

i understand for I+I its 16

but idk how the answer get 4 and 0

those numbers seem real precise and im wondering if theres something im missing

i tried working out some of them using the big formula

they work but i cant put it into words why it does

jus feels like theres something im suppose to like realise but im just blind atm

[pallid swallow]

If there is a 2-cycle then 0.

Otherwise, the permutation must be a 3-cycle.

Notice that swapping rows would only change the sign of the result.

Those observations should help @indigo cradle

[indigo cradle]

is cycles a term, i have not learnt of it. may i know what it means?

[pallid swallow]

http://mathworld.wolfram.com/PermutationCycle.html

@indigo cradle

Yeah, I think they explain it okay here

[quaint heart]

@indigo cradle think of the map which sends 2 cycles to 1 in Z/2Z

Notice that that's a group homomorphism

And use the first isomorphism theorem

[indigo cradle]

@quaint heart sorry im not familiar with those terms

@pallid swallow is there any relation between adding an identity to a 3-cycle permutation such that they all give 4?

[pallid swallow]

Well, if your matrix has lots of 0s, then your determinant has only a few terms

I think you can simply cofactor expansion on the 1-cycle

[pallid swallow]

Hmm I think if you write out the entire determinant in terms of all the permuations of elements...

it might help with this

You are literally selecting a subset of the cycles in the permutation

@indigo cradle

[indigo cradle]

hmm actl i think i have formed some sort of understanding after thinking towards permutation cycles

3-cycle permutations are formed by fixing one and mixing the others

so when they add to the identity

theres one row with 2

and the other rows have 1s put in place such that

theres only 2 ways to pick them

[wintry steppe]

Hello, how can I show that, given a linear maps matrix in standardbasis B=..., the eigenvalues are 0 and -3 without calculating the cartesian polynomial.

?

[sonic osprey]

by showing that B has a null space and by showing that B - 3I has a nullspace

[wintry steppe]

Okey, thank you @sonic osprey

[grand island]

I am trying to get my head round the maths in Modern Robotics and Control, and its a bit beyond my linear algebra abilities. For equations like this one, with the quadratic form for kinetic energy, what are the underpinning maths pre-reqs to understand it? Is it multivariable calculus meets matrices of some kind? https://cdn.discordapp.com/attachments/480050247029227522/614366743930077204/unknown.png

[wintry steppe]

I guess

Both are quadratic forms, one just happens to have some derivatives in it

[glossy edge]

Hey so I have this question where I have to define this linear transformation

[dusky epoch]

,rotate 90

[glossy edge]

Excuse the spots, they are because my camera is broken

But the L is the transformation

[dusky epoch]

ok

[glossy edge]

Here's what I found

Is it correct?

[dusky epoch]

uh

show work?

[glossy edge]

Alright

I just did the second part by deriving a polynomial of degree 3 and then multiplying by X+1

Actually this was an exam question in January

[undone garnet]

is there another way to prove that ${e^x, e^{2x}, e^{3x}}$ is linear independent

[stoic python]

Nguyễn Thành Trung:

[undone garnet]

my way is using determinant

[pallid swallow]

linearly independent as functions under addition?

[undone garnet]

equivalent prove that $a.e^x + b.e^{2x} + c.e^{3x} = 0$ if and only if $a=b=c=0$

[stoic python]

Nguyễn Thành Trung:

[pallid swallow]

Yeah, functions under addition.

[lone cove]

take derivatives of that
but determinant of the wronskian is always the best way

[undone garnet]

yeah

that was what i did

but I want another way

[pallid swallow]

I think you can multiply by e^-kx and integrate

[lone cove]

dont even need to integrate

just lim x to - inf

[pallid swallow]

yeah that would do

[undone garnet]

don't get what you say

[pallid swallow]

So multiply by e^-x

Then let x go to -inf

therefore a=0

Then multiply the original expression by e^(-2x)

[undone garnet]

😮

and then

yeah yeha

I got it

wow

that's a way

well

how about integral

can you show me?

[pallid swallow]

Well, probably that would include the complex numbers and stuff like that

(for each bit to cancel out to 0)

was thinking about fourier transforms and how they calculate each coefficient

[sonic osprey]

I mean, you can plug in x = 0, 1, 2 and just show that the corresponding system of equations for a,b,c is linearly independent

[undone garnet]

@sonic osprey I did that

that's the way I did

[sonic osprey]

Okay sure

You just said you used determinant so

[pallid swallow]

would be better to plug in 0, ln(2), ln(3) since that gives nice small integers

[undone garnet]

well

using exp(-kx)

we can solve for e^x, e^(2x), e^(3x), ..., e^(nx)

$(b_1, b_2, ..., b_n)$ is linear independent\
$a_1 = b_1; a_2 = b_1 + b_2;...;a_n = b_1+b_2+...+b_n$\
Is $(a_1, a_2, ..., a_n)$ linear independent?

[stoic python]

Nguyễn Thành Trung:

[pallid swallow]

yeah it is.

Proof sketch: Suppose we have a nonzero solution to $c_1a_1+...+c_na_n=0$, then we can convert it to a nonzero solution of $d_1b_1+...+d_nb_n=0$.

[stoic python]

Element118:

[undone garnet]

wow

how stupid I am

[pallid swallow]

Well, not very, lots of us miss out easy things from time to time

[glossy edge]

Is the correct representation of the linear transformation

[pallid swallow]

for?

[glossy edge]

I will send it

For this?

[pallid swallow]

$V=\mathbb{R}[X]_{\leq 3}$ (polynomials of degree at most 3), $L:V\rightarrow V$ \
$L(P(x))=\alpha P(x)+(x+1)P'(x)$

[stoic python]

Element118:

[pallid swallow]

Looks right

[glossy edge]

Alriiight thank you

[wintry steppe]

Hello guys, I have problem with a question regarding reflection in a plane: x+2y=0. First should I find the ON-bases containing eigenvectors to the linear map F, then should I find F:s matrix [F]_s (wrt. standardbasis). I find the ON-bases but I am not sure how to get the matrix [F]_s

I'm using the identity matrix to get the answer but I'm not sure how to find the matrix (-1,0,0);(0,1,0);(0,0,1) which is supposed to be The matrix with respect to the ON-bases

[noble swallow]

Are you finding the reflection with respect to the plane?

@wintry steppe

[wintry steppe]

@noble swallow Yes, "Let F be the linear map which is given by the reflection in the plane x+2y=0 (coordinates in standard basis S).

I don't know how to find the Matrix [F]B of F with respect to the ON-bases given in the first exercise

[noble swallow]

@wintry steppe there are some ways, depending on what you have done:

1. Have you written the reflection in terms of the projection on the plane?
2. Have you found the matrix of F with respect to the standard basis?

[wintry steppe]

I have taken the normal vector + a basis of the plane of the reflection ( 2 vectors) that also are orthogonal, (-2,1,0) and (0,0,1). I get these 3 vectors and normalising the 3 vectors. After normalising them I want to get the matrix matrix for F with respect to the ON-basis I just wrote to you, the normalised 3 vectors

@noble swallow

[noble swallow]

@wintry steppe ah ok

Let's say your basis is
{w1,w2,v}

Where w1,w2 are an orthonormal basis of the plane

And v is a normalized vector orthogonal to the plane

Now, what F(w1), F(w2), F(v) would be?

[wintry steppe]

I'm not sure

[noble swallow]

@wintry steppe think about reflecting them with respect to the plane

[wintry steppe]

@noble swallow I think I found a video about this

[noble swallow]

Ah nice, I hope you clarified it

[wintry steppe]

https://www.youtube.com/watch?v=fvXG-DVx6z0

It only worked for my (0,0,1) vector though

[noble swallow]

Mm, ok, let's see the others then

Since w1 and w2 are on the plane, reflecting them leaves them unchanged

So F(w1)=w1, F(w2)=w2

Do you agree?

[wintry steppe]

Yes I understand that

[noble swallow]

While since v is orthogonal to the plane, F(v)=-v

Now it should be easier to write the matrix

[wintry steppe]

I see

And in what case does F(q) =0?

Not in this exercise but in others maybe?

[noble swallow]

Well it depends on what F is

Vectors which are sent to 0 vector by F are elements of the so called kernel of F

However then in your problem the matrix of F w.r.t the basis {w1,w2,v}
is {(1,0,0),(0,1,0),(0,0,-1)}

[wintry steppe]

I would also write like that but they have written it like this

(-1,0,0);(0,1,0);(0,0,1)

I meant;

Doesn't we have two orthogonal vectors and one normal vector?

[noble swallow]

That matrix is just explained with a permutation of the basis

It's the matrix of F w.r.t the basis {v,w1,w2}

[wintry steppe]

So the normal vector => F(v) = -v

[noble swallow]

v is always the orthogonal one

Yeah, it's the eigenvector associated to -1

[wintry steppe]

Okey, I think I got it now. Thank you very much for the help! 🙂

[noble swallow]

No problem!

[undone garnet]

how can we prove ${\sin(x), \cos(x), \sin(2x), \cos(2x), ..., \sin(nx), \cos(nx)}$ is linear independent

[stoic python]

Nguyễn Thành Trung:

[pallid swallow]

fourier transforms

well, unless you are proving that for fourier transforms

[undone garnet]

fourier transform?

how?

[pallid swallow]

Okay, write out your linear independent equation

then multiply by f(kx) for your choice of function and scaling factor

and integrate from 0 to 2 pi

[undone garnet]

don't get what you say

[pallid swallow]

write out $a_{\sin x}\sin(x)+a_{\cos x}\cos(x)+...+a_{\sin kx}\sin(kx)+a_{\cos kx}\cos(kx)+...=0$

[stoic python]

Element118:

[dusky epoch]

uhhh

those are some

weird subscripts you got there

also wow TWO ellipses

sheesh

[pallid swallow]

2 ellipses?

oh yeah

[dusky epoch]

yes, two ellipses

[pallid swallow]

it's the plural form of ellipsis

[undone garnet]

let $a_{\sin(kx)} = a_k, a_{\cos(kx)} = b_k$

[pallid swallow]

english is weird

[stoic python]

Nguyễn Thành Trung:

[dusky epoch]

why not do something more sensible like... yeah that

[pallid swallow]

okay sure, that might be more sensible

[dusky epoch]

$\sum_{k=1}^n (a_k \sin(kx) + b_k \cos(kx)) = 0$

[pallid swallow]

But ultimately, we want to show all the $a_k, b_k$ are 0

[stoic python]

Ann:

[dusky epoch]

that's your eq

[stoic python]

Element118:

[dusky epoch]

yes

[undone garnet]

yeah

[pallid swallow]

What we do for fourier transforms work here

multiply the entire thing by sin(kx) (or cos(kx))

integrate

win

[dusky epoch]

well

...ok yes

integrate over [0, 2pi] or something

[undone garnet]

sin(kx)?

[dusky epoch]

yes

[undone garnet]

k = 1 to n?

[pallid swallow]

for each k

[undone garnet]

which k we will choose?

[dusky epoch]

let k ∈ {1, ..., n} yes

[pallid swallow]

well, it generalises

if you do it properly

[dusky epoch]

let it be arbitrary

[undone garnet]

$a_1.\sin(x)\sin(x) + b_1\sin(x)\cos(x) + a_2\sin(x)\sin(2x) + b_2.\sin(x)\cos(2x) +... $

[stoic python]

Nguyễn Thành Trung:

[pallid swallow]

yay

Okay, maybe it might be a better idea to get some intuition about specific cases first

let's integrate both sides

[undone garnet]

so integrate both side from 0 to 2pi

$a_1.\pi + b_1.0 + a_2.0 + b_2.0 + ... $

[stoic python]

Nguyễn Thành Trung:

[undone garnet]

oh wait

how $\int_0^{2\pi}\sin(x)sin(kx)dx = \int_0^{2\pi}\sin(x)cos(kx)dx = 0$ for k = 2 to n

[stoic python]

Nguyễn Thành Trung:

[pallid swallow]

yeah, that you need to prove

[undone garnet]

sorry but

how $\int_0^{2\pi}\sin(x)\sin(kx)dx = 0$?

[stoic python]

Nguyễn Thành Trung:

[undone garnet]

for k >= 2

I just used my calculator

oh

I just solve it

[pallid swallow]

Do a substitution? pi-x?
Hence equal to
integral from 0 to 2pi sin(x)sin(k(pi-x))dx
If k is even, then
integral from 0 to 2pi sin(x)sin(k(pi-x))dx=integral from 0 to 2pi sin(x)sin(-kx)dx

[undone garnet]

😮 another solution

oh I just found another solution for this problem

[pallid swallow]

yeah exactly the same

[solid stratus]

It says that in 2 dimensions, the only possible way for there to be no solutions to a set of equations is if the lines are parallel, later on it gives 3 equations where the lines aren't parallel but there aren't any solutions, I don't get it

The first paragraph that comes after "The singular case"

The lines are

$x+2y=2$

$x-y=2$

$y=1$

[stoic python]

spammy:

[sonic osprey]

The only possible way for there to be no solutions to a set of two equations in two dimensions is if the lines are parallel

[cedar solar]

Can someone give an intuitive explanation of what dual spaces are?

I’m afraid of misunderstanding it

[wintry steppe]

so. You have a space of vectors. And you have linear bounded functionals on that space. You can attach to them a new structure of a vector space. This is a dual space

Idk what dual relates to, but I think of it as someone in the background pulling the strings

[cedar solar]

So let’s say my vector space is R^2

[quaint heart]

If you fix a basis of R^2, that gives you a dual basis of (R^2)*

[cedar solar]

Then the functional is all combinations ax+by=c for some fixed c? If the vectors are (x,y)

[quaint heart]

No

The dual basis functionals are the functions that map one basis vector to 1 and the rest of the basis vectors to 0

Do you know what a linear function is @cedar solar

Because from what you just said it doesn't seem like you do

You should read up on them

[cedar solar]

Derp i dont know why i didnt click that wiki link

Ok yeah so functionals are linear functions f that satisfy f(av+w) = a f(v) + f(w)?

[quaint heart]

Functionals are just linear functions to R

[wintry steppe]

functionals are linear functions from our vector space to the field of scalars

for example to R when we work in Euclidean space

[quaint heart]

Smh @wintry steppe

You don't have to be general

[wintry steppe]

I distinguish between real and complex spaces

I think that's important even if you are not being general

[quaint heart]

I guess that's fair

[cedar solar]

So the dual space says these functionals can be treated like vectors, and the dual space is the resulting “vector space”?

[wintry steppe]

pretty much

[cedar solar]

Ok thanks i think i understand it now

[wintry steppe]

they have nice properties so they are an object of study

[quaint heart]

If you fix a basis of your vector space

You get a basis of your dual space

As I mentioned earlier

[cedar solar]

Right, since without a basis we cant write the functionals into nice equations?

[quaint heart]

(assuming finite dimensional)

[cedar solar]

Oh while you two are here i have another question if thats ok

So i learned about the adjoint, but i dont understand what it represents or why it is useful

Why do we care about adjoints?

[wintry steppe]

adjoint in the sense of 'inverse' of a linear transformation?

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

this or

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

this

[cedar solar]

I think the adjoint was defined as the conjugate transpose

Hold on lemme read those links

[wintry steppe]

so

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

[cedar solar]

I think so

Like ok, <Ax,y> = <x,A* y>. That’s nice but how to interpret it and why is it useful

Like i went through the motions w the proofs but i never understood the intuition and i’d like to fill that gap

[wintry steppe]

useful for math?

[thorn robin]

that's kinda like integration by parts

[wintry steppe]

you mean applications ?

in physics for example?

[cedar solar]

Hmmm yeah any application is fine. But rather than application i’m concerned more with the intuition

[slow scroll]

there are relationships between orthogonal fundamental subspaces of matrices that the adjoint helps describe. e.g. Col(A) is orthogonal to Ker(A*) err well thats not really intuition i guess

[cedar solar]

@thorn robin what do you mean by it’s like integration by parts?

[thorn robin]

I would have to think about it a bit to write down something explicit
but you can imagine that A is a differential operator and the inner product is an L^2 inner product on functions
then that equation for the adjoint moves the derivative from one function to the other a la IBP

[cedar solar]

I guess what i want is like, how determinant is area of parallelogram for R^2. That kind of intuition but for adjoints

[wintry steppe]

intuition about definition

[cedar solar]

Brb reading differential operator

I mean like how 3B1B showed how linear transformations distort squares into parallelograms. It took that for it to click for me

[wintry steppe]

but that's just geometric interpretation

[thorn robin]

you can just think about the case of dimension 1
the conjugate lets you find something real from something complex by considering z\bar z

so in general you get self-adjoint operators by considering $AA^*$ or $A^*A$

[stoic python]

Seoin:

[thorn robin]

self-adjoint basically means real

[wintry steppe]

I don't think you'll find any great intuition behind adjoint (especially geometric), but you can interpret it as a certain operation on matrices/operators which we can exploit to a certain degree. And that operation is described by the property that <Ax, y> = <x, A^* y>

[cedar solar]

Oh, so after googling around, someone gave intuition for adjacency matrices for graphs

It’s equivalent to reversing all the edges

But of course only for the R case and not C

Also another explanation is, if A: X->Y, then A* : Y*->X*

But then what does the conjugate transpose of a vector space signify, intuitively

[wintry steppe]

adjoint is defined for linear transformations from the vector space to itself

now you mean transpose which is technically the same for real spaces

[cedar solar]

Does it make sense to talk about the transpose of a vector space?

[wintry steppe]

[cedar solar]

Ok since nothing shows up when i search for transpose of vector spaces i assume it is nonsensical

So transpose is sometimes called the dual. Does this have anything to do with dual spaces?

[thorn robin]

https://math.stackexchange.com/questions/2658473/explicit-relation-between-dual-and-adjoint-of-a-linear-map

I didn't read it in depth and I can't really tell you off the top of my head

but if A : X -> Y then the dual map A' : Y' -> X' has as its matrix in the dual bases the transpose of the matrix of A

so yes as you said you might as well call the dual space the transpose

but for real vector spaces the adjoint Y -> X also has the transpose as its matrix

[cedar solar]

Hm, ok

So i was previously unfamiliar with the dual basis, and after reading up on it, it seems odd to me that the dual basis is defined by the kronecker delta

How do i reconcile that the “transpose” of a vector space is the basis for functionals?

[sonic osprey]

It's not really that weird, if you know what the kronecker delta is saying

[cedar solar]

I dont understand how that relates to basis for fuctionals

My thought process is, if functionals are linear functions that map to R, then why is the dual basis uniquely defined by the bases of the vector space

Why cant i choose any random bases that span R^n?

[sonic osprey]

Also transpose of a vector space makes no sense

You can choose any basis of R^n and that will give you a basis of your dual space

That's the point

[cedar solar]

Since each column is a linear function

I mean why does it depend on the basis we pick for R^n

[sonic osprey]

It doesn't

[cedar solar]

And what issues arise if i think of the dual space as the “transpose” of a vector space? Like yeah maybe not entirely correct but it seems to be the intuition

[thorn robin]

I think it makes sense. In a basis, elements of V are column vectors and elements of the dual are row vectors

[cedar solar]

So the basis for V* is defined by v_i v*_j = delta_{ij} right?

So once we pick the basis vectors for V

[sonic osprey]

escape those asterisks

[cedar solar]

Then we have chosen the bases for V*

[sonic osprey]

No, V* is just an n dimensional vector space

There's a function from the set of bases for V and the set of bases for V* that's very nice and makes a lot of sense

Also, thinking of it as the transpose isn't too bad because of what Seoin said

And furthermore, taking the double dual of the vector space gets you back to what you started with, just like transpose does

[cedar solar]

Oh wait

So the dual basis is not the basis vectors for the dual space?

Im confusing myself

Wiki says it is the basis for the dual space. But then why? Since any arbitrary n-dimensional vector is a functional on R^n

Or am i misunderstanding functionals

Sorry if these questions seem dumb. My textbook did not cover dual spaces. I really appreciate everyone’s help

[sonic osprey]

@cedar solar The dual basis is A basis for the dual space

Vector spaces have many different bases

Given a basis for V, the dual basis is one basis for V*

[cedar solar]

Can V be a basis for V*?

[sonic osprey]

what

[cedar solar]

Er i mean

Can the basis vectors of V be the basis vectors of V*?

[dusky epoch]

[sonic osprey]

As I said

There is a function from bases of V to bases of V*

This function is exactly the dual basis

So if you have a basis of V, applying this function will give you a basis of V*

[cedar solar]

Ohhh ok

And bear with me

So i figure my misunderstanding was thinking that the kronecker delta definition is THE only basis vectors allowed, but the emphasis on A means other basis vectors exist

But the kronecker delta definition is there to get us the transformation to get us from the bases of V to the bases of V*?

[sonic osprey]

Yep, that's how you get the dual basis

[cedar solar]

So if $f_i v_j = \delta_{ij}$ then if we string together the vectors into matrices F and V, how do we get the basis vectors for V*?

[stoic python]

BamBam:

[cedar solar]

Actually rather than annoying all of you with these silly questions, can you recommend a book that covers dual spaces?

[sonic osprey]

Most of them? I'm not sure

The questions are fine, I was just a bit busy

[wintry steppe]

just grab any functional analysis book

https://www.mimuw.edu.pl/~aswiercz/AnalizaF/lecture.pdf

[sonic osprey]

I'm not sure what you mean by string together the vectors

[wintry steppe]

or maybe some lecture

[sonic osprey]

And F and V are what

[wintry steppe]

also cool remark that dual space is always Banach in that link
in general space of continuous linear functionals from V to W is complete if and only if W is complete (here W is a field of real/complex numbers)

[cedar solar]

F is the matrix where f1, f2, ... are the columns. So if V is v1,v2,... as columns we can give a coordinate vector in the standard bases and converts it to the new basis right?

I just looked up my previous textbook and i completely missed the chapter on dual spaces

Fuck im dumb

It was marked as optional

And i skipped it lmao

[sonic osprey]

Yeah usually it's not really covered in first courses

[cedar solar]

Im gonna revisit that chapter

[sonic osprey]

the elements in F are your basis?

the f_i are your basis for V*

[cedar solar]

So what i dont get is why cant we pick arbitrary u1,u2,... as the bases for V* since they’re all linear transformations and hence make up functionals

Im probably misunderstanding quite a lot

Ok actually i should shut up and open the textbook

Thanks for the help, everyone!

[sonic osprey]

@cedar solar The same reason you can't just pick random vectors in V and have them be a basis

[cedar solar]

I mean, arbitrary but linearly independent

But yeah imma read this chapter for now

[sonic osprey]

@cedar solar I mean you can

Yeah go ahead

[quaint heart]

@sonic osprey if you can pick n random vectors, they will with probability 1 be a basis

[sonic osprey]

not if your field is finite

[quaint heart]

True

[cedar solar]

Ok i think i have a better grasp of what transposes mean, intuitively

Is there an intuition behind something like $A^TA$?

[stoic python]

BamBam:

[gilded junco]

"The sum of any two irrational numbers must always be irrational"

I'm puzzled as to how to write this in terms of logic, like If A then B ?

A ^ B

I decided that it was synonymous with
"IF you sum any two irrational numbers THEN the result will be irrational"

but that isn't particularly helpful, so i've deduced that must be wrong

[thorn robin]

@cedar solar well for a vector you have v^T v = |v|^2 yeah? so you can think of A^T A and A A^T as bearing some kind of relation to a norm squared of A

[cedar solar]

@gilded junco i think the logic form is correct, but the statement is not. Consider pi + (-pi)

[gilded junco]

@cedar solar , it's important you understand the stance I am writing this from. I want to know the logic form, so I can find the negation of it, so I can try proof by contradiction

[cedar solar]

Hmmm thats an interesting take, analogous to squared norms

[gilded junco]

I'm looking for a how-to on how to deduce logic forms for that kind of statement

[cedar solar]

Try #proofs-and-logic since it has no bearing on linear algebra

[gilded junco]

oh sorry!

[iron oar]

If $A^TA = \begin{bmatrix} a&0&0\0&b&0\0&0&c \end{bmatrix}$ where a, b and c != 1, A is still orthogonal, right?

[stoic python]

UglyPuppet:

[pallid swallow]

no...

[iron oar]

Damn.

[pallid swallow]

orthogonal matrices have orthoNORMAL columns

[iron oar]

Oh, okay. I didn't realise they needed to be orthonormal. Thank you!

[pallid swallow]

misnomer, but oh well

[iron oar]

Actually, I'm not trying to determine if A is orthogonal, just if the columns make an orthogonal set.

[wintry steppe]

just multiply A^T A

if it's a diagonal matrix, then the columns are orthogonal

[pallid swallow]

Not sufficient

It needs to be an invertible diagonal matrix

well

hmm

that just requires all columns to be nonzero

argh 0 is so annoying

[iron oar]

Okay, that makes sense, because if I normalised the column vectors first, I would be getting the identity when A^T A.

[wintry steppe]

yeah just diagonal with all nonzero diagonal elements

[iron oar]

Thank you! I appreciate your help!

[glossy edge]

hello, how do i define a linear map that transposes a 3x3 matrix?

[dusky epoch]

wdym

do you want the transpose considered as an operator on $\bbR^{3 \times 3}$?

[stoic python]

Ann:

[dusky epoch]

or what

[glossy edge]

define the linear transformation L : R^(3x3) --> R^(3x3) : A --> A^T

it takes a 3x3 matrix input and outputs its transpose

[dusky epoch]

well

that alone already defines your map just fine

it's linear

what else are you after

[glossy edge]

can't i represent it with a matrix?

i thought that's what they meant in the question

[dusky epoch]

you'll need to fix a basis of R^3x3

ok uh

why don't you

post the question exactly as stated

so that we don't have to play broken telephone

[glossy edge]

it's in dutch

i will translate it

define the linear transformation L : R^(3x3) --> R^(3x3) : A --> A^T
define the subspace of R3x3 U:= {U := {A ∈ R3×3|A is symmetrical}
W := {A ∈ R3×3 |A is skewsymmetrical}.

it could be that the question is not correctly formulated because it was students who wrote it down from memory

so if you think that defining that linear mapping doesn't make sense then maybe you are right

[dusky epoch]

what's the question?

[glossy edge]

a) prove that 1 and -1 are eigen values of L and that U is equal to the eigenspace
E1 and W is equal to the eigenspace E−1.

b)calculate dim(U) en dim(W).

c)prove that L is diagonalizable. give the characteristic polynomial φL(X) of
L.

how do i find all of those things without a matrix?

[dusky epoch]

you don't need a matrix for any of these

verifying that 1 and -1 are eigenvalues can be done by showing ker(L - I) and ker(L + I) are nonzero

dim(U) and dim(W) don't even have anything to do with your transformation

for diagonalizability just present an eigenbasis for L

[pallid swallow]

Eigenvalue?

[glossy edge]

but where do i go from ker(L-1)?

[pallid swallow]

Just construct eigenvectors

a) 1 is eigenvalue because L(I)=I, -1 is eigenvalue because L([0,1,0;-1,0,0;0,0,0])=[0,-1,0;1,0,0;0,0,0]

[dusky epoch]

do you know what ker() is

just yknow

write out

[glossy edge]

oh I is the identity matrix

[dusky epoch]

the definition

I is the identity transformation

it has the same matrix in all bases anyway

[glossy edge]

alright

how do i find the characteristic polynomial of L?

also idk how to prove that the kern is not empty

[pallid swallow]

I guess we shouldn't do that

Suppose L(A)=A

Then, well, A is symmetric

And if A is symmetric, then L(A)=A

Likewise, L(A)=-A and skew-symmetric

b) just find a basis for each of these

c) just show that dim(U)+dim(W) is big enough, then φL(X) only have roots 1 and -1, then you can find the algebraic multiplicity

[glossy edge]

alright thanks

[nimble galleon]

I can't solve question. 5 and I am not sure how to translate it . It's in Hebrew

[pallid swallow]

Try geometry-trigonometry

#geometry-and-trigonometry

I thought we translated it already?

Find the points on y=x+2 that are distance 5 from (7, 8)

[livid falcon]

First question says to find "a" so that f is a linear function (mapping, transformation(?)), why is ( f(0) = 0 ) enough proof ? why is there no f(λ1u1 + · · · + λnun) = λ1f(u1) + · · · + λnf(un)?

[pallid swallow]

f(0)=0 is a requirement for f to be linear.

Afterwards, you can check that a=0 gives f a linear function.

Wait, that seems to be what that is saying

[livid falcon]

oh wait, I thought I didn't have to do the verification part, my bad, and thanks

[dusky epoch]

toutes les applications linéaires vérifient f(0) = 0 ; si la tienne ne le vérifie pas, elle n'est pas linéaire

[solid stratus]

Do I need to know b?

I'm not given b

That's a bit blurry

For question 2

b has been given multiple definitions in the chapter, I could try with the last definition?

Ok so I was able to find a solution using the last definition of b, so I might have understood the question correctly

[nimble galleon]

Well we translated it but haven't solved it and then tons of people jumped in and spammed their questions

[plain fjord]

I am not sure what is meant with 'b' in that question... since there is b1, b2, b3... not b

[livid falcon]

Is writing: Ker(f) = {(x,x,x) with x a real number } correct here?

[noble swallow]

Yep

You can also write it as span{(1,1,1)}

[wintry steppe]

Hello, I need help with showing that a function is linear. I know the definition: including 0, closed for addition & multiplication. But how do I do it with functions like this?

[thorn robin]

you need to show that f(au + v) = af(u) + f(v) for all scalars a and all vectors u, v

in that definition, an arbitrary vector is of the form u = lambda_1 b_1 + lambda_2 b_2 + lambda_3 b_3

so write down another arbitrary vector in this form and try to verify the equation I posted

[wintry steppe]

@thorn robin Should I start with the left or right side? I've seen examples of where they write it as a matrix but I don't know where to start.

[thorn robin]

the left

[wintry steppe]

f( (lambda1+alfa1)(b1u1)+...))?

[thorn robin]

you start by writing f(au+v) where u = lambda_1 b_1 + lambda_2 b_2 + lambda_3 b_3 and v = mu_1 b_1 + mu_2 b_2 + mu_3 b_3 and a is any scalar

and you work with it until you show that it equals af(u) + f(v)

[wintry steppe]

And does this change the right hand side?

[thorn robin]

I'm not sure what you mean

[wintry steppe]

Ok, I'll try it!

[spice storm]

Please for the love of god. Why is elementary row operations so hard for me.

[dusky epoch]

what are you finding hard about them

the execution or knowing what to do when

or is it something else

[spice storm]

I don't know when to add or multiply. Like when I do 3R_1+ R_2---> R2

[dusky epoch]

so knowing what to do when

[spice storm]

yes

[dusky epoch]

yeah well that depends on what you want to accomplish really

which is most likely Gaussian elimination

[spice storm]

That's probably it. But prof didn't name it that method

[dusky epoch]

well ok you're using it to solve systems of linear eqs right

[spice storm]

Right now I have 2 X 2

[dusky epoch]

by first writing your system as a matrix

[spice storm]

Yes

[dusky epoch]

and then doing the row operation thing on it

okay alright

so the general Gaussian elimination algorithm is actually better illustrated on higher matrix sizes, paradoxical as that might sound

like 3x3 at least

[dusky epoch]

okay so here's the idea basically

this is our ultimate goal: $\left[\begin{array} {cc|c} 1 & 0 & ? \ 0 & 1 & ? \end{array}\right]$

[stoic python]

Ann:

[spice storm]

Yes

[dusky epoch]

once we make this happen, we can read off the answer in the entries i've marked with question marks

okay

so to do this

we want to use row operations to get each column to a state where it has one entry equal to 1 and the rest equal to 0

there are of course many different ways you could go about this for any given system

since you're doing this by hand, there will sometimes be shortcuts that specifically keep all the coefficients as integers, which can help reduce fuckups

[spice storm]

I think my problem is trying to decide which row to have the first leading zero term

[dusky epoch]

it doesn't matter. you can make it work no matter which row you choose for that

[spice storm]

Oh okay

[dusky epoch]

so anyway, one natural first step is to multiply the first row by 1/5.

this will give you a 1 in the first column, which is a good sign

because then you can add -3*row 1 to row 2, and make row 2's first entry zero

[spice storm]

So when you do that. Do you just multipy 1/5 to 5 and 10?

[dusky epoch]

you multiply the first row by 1/5.

and that means exactly what it says.

[spice storm]

so it will 1 on top and 5/3?

[dusky epoch]

the resulting row will be [ 1, 2 | 4 ].

the ROW. not the COLUMN.

[spice storm]

That's where i get confused

am I doing the row or column when doing these calulations

[dusky epoch]

always the row. you're doing ROW OPERATIONS, after all.

[spice storm]

Okay. so I will never touch the column

[dusky epoch]

you see that vertical line separating the last column?

[spice storm]

Yes

[dusky epoch]

that's a decoration to remind you that you can't mess with columns

but you can mess with rows

[spice storm]

Okay

[dusky epoch]

and by mess i mean apply elementary transformations of course

[spice storm]

so the first row is 1 2 and 4

Okay. I will try this. Thanks @dusky epoch. Will message back if I have any trouble

[carmine terrace]

Hey guys, I have a little trouble with this Matrix/Matrice?

I just learned it today so I'm not exactly sure how to proceed

x-3y = -3, x-3y = 12

| 1 - 3| -3
| 1 - 3| 12

I multiply by negative 1 top row, add to row 2

1 -3 -3
0 0 -15

[dusky epoch]

yeah

[carmine terrace]

that's where I'm confused

[dusky epoch]

that's where you see that second row and realize you're done

[carmine terrace]

I didn't see an example where both my variables in a row goes to 0

[dusky epoch]

the system is inconsistent

because you have a zero row w/o a zero constant term

the corresponding equation is 0=15

which is obviously bullshit

[carmine terrace]

ah ok thanks, we still haven't received our textbooks so I was trying to jump ahead.

[timber blade]

not sure why [-2,3] isn't the domain, can anyone help me?

[sonic osprey]

count

[timber blade]

?

-2 is an x value that works, so does 0, so does 1, so does 2, so does 3

the graph shows those, so that's the domain i put in

and it still does not work

[gray dust]

what's the maximum x value of that curve?

[timber blade]

3 is the maximum x value of the curve

[gray dust]

look at the graph again

[timber blade]

nevermind

it's 4

that worked

[graceful osprey]

👏

[timber blade]

i didn't know it was like that, thought it was just the x values that produce a whole number or smth for some reason

[graceful osprey]

Just a smol remark, look where the axis arrows are

[timber blade]

it's also asking me to estimate the value of f(-1) but -1 isn't on the graph unless they want me to mathematically determine the function of the graph and then solve for x=-1

[gray dust]

f(-1) means the y value at x = -1

they just want you to eyeball it

[timber blade]

ye i did and it's still wrong

looks like .2 or .25 or something but

[gray dust]

err... try again

[timber blade]

LMAO

ye -.2 worked

[gray dust]

👍🏽

[soft ingot]

Can someone help me with the problem I'm not sure if the answer is "no solution" or "all real solutions"

[feral mountain]

Is there an L that makes 6L<6L?

[soft ingot]

Wdym?

[feral mountain]

Is there a real value L can take that makes 6L<6L?

[soft ingot]

don't get what you mean by real value

[gray dust]

he's making a distinction from real values of L and possibly complex values of L

[feral mountain]

I just assumed squiiishi was working in the real numbers

[gray dust]

if you haven't learned complex numbers yet, don't worry about the "real" part just yet

[feral mountain]

"all real solutions"

[sage mauve]

there is no order to complex numbers

[soft ingot]

it's "inequalities"

[sage mauve]

which question is it?

[soft ingot]

23

[sage mauve]

what do you not get?

[soft ingot]

I can't decide if the answer is "all real solutions" because there isn't an equal to sign there

But the two sides equal each other

[gray dust]

so is there any way the left side can actually be less than the right side?

[soft ingot]

There isn't

[sage mauve]

there you go

[gray dust]

so how many solutions do you have?

[sage mauve]

how can it be every real if nothing at all works

[soft ingot]

One

Thanks guys!

[sage mauve]

no there are no solutions

is 5<5?

no

5=5

5 is not less than 5

so NO L satisfies 6L<6L

NONE

[soft ingot]

But the sign is <

[sage mauve]

yes

so?

[soft ingot]

Oh

does the sign matter at all if the two side says otherwise

[sage mauve]

what do you mean?

[soft ingot]

The equaltion is equal but the sign is <

Do I ignore the sign?

[wintry steppe]

< is trichotomus (I believe it was called)

It means that for a, b only one from those 3 things can happen

a<b, a=b, b<a

6L = 6L, so you can't have 6L < 6L

[soft ingot]

O shi

[gray glen]

mous

but ye

[feral mountain]

sets are trichotomous over an order thingy

[wintry steppe]

Fair

[gray glen]

I'm pretty sure relations can be trichotomous

[feral mountain]

yeah it's a relation that's trichotomous oop idk

[surreal grove]

When you feel like you don't know shit trying to reduce a matrix down to ref, and basically stumble on the rref

I confirmed that the rref was just a diagonal of 1s also

[undone garnet]

prove that if $\operatorname{rank}(M)=1$ then $M^2=\operatorname{trace}(M).M$

[stoic python]

Nguyễn Thành Trung:

[undone garnet]

M in M_n

I thought that this only true when M in M_2 but this one is M in M_n

[brittle juniper]

See if M is diagonalizable

[undone garnet]

how?

I mean

If M is diagonalizable

How can we prove that equation

[brittle juniper]

There exists P in GL_n such that P¯¹MP is diagonal
and it appears that trace(M)=trace(P¯¹MP)

trace(M) is the unknown eigen value

(P¯¹MP)² has zeros everywhere except for 1 diagonal entry, which is the unknown eigenvalue squared

(P¯¹MP)² = λP¯¹MP, where λ is the unknown eigenvalue, which is also trace(M)

and also (P¯¹MP)² = P¯¹M²P

P¯¹M²P = trace(M) P¯¹MP

M²=trace(M) M

that's a bit messy

[noble swallow]

Mm, I think that's also true in general

If M has rank 1 then dim(Ker(M))=n-1

Hence, we can choose a basis of Ker(M) which has n-1 elements

Then we extend it to a basis B of all the space adding one more properly chosen vector

Then, with a change of basis to this basis B we have that
M = S^-1 A S where A is the matrix of L_M wrt to the basis B

So, A has the first n-1 columns filled with zeros and only the last column equal to some (a_1,...,a_n)

tr(M)=tr(A)=a_n

And M^2 = S^-1 A^2 S

Now, A^2 has the first n-1 columns filled with zeros and the last column equal to a_n(a_1,...,a_n), hence A^2 = a_n A

So, M^2 = a_n (S^-1 A S) = a_n M = tr(M) M

@undone garnet

[pallid swallow]

If $M$ is rank 1 square matrix, then $M=uv^T$ for column vectors $u, v$. \
$M^2=uv^Tuv^T=u(v^Tu)v^T=(v^Tu)uv^T=(v^Tu)M$, \
so all you need to show is $v^Tu = trace(M)$.

[stoic python]

Element118:

[pallid swallow]

@undone garnet

which isn't too hard

[noble swallow]

Wow great solution!

[slow scroll]

@undone garnet where do your problems come from?

[gaunt mulch]

A matrix has eigenvalues –1 and –2. The corresponding eigenvectors are [1−1] and [1−2] respectively. The matrix is

help!!!!!!!!!!!!!!!!!!!!!!!!!!!

[noble swallow]

@gaunt mulch let's call this matrix M

[gaunt mulch]

ok

is formula is MX=lambdaX

lambda for eigen value

[noble swallow]

The matrix of L_M with respect to the basis (1,-1),(1,-2) is
[(-1,0),(0,-2)]

[native lodge]

You have the answer already from MX=ΛX

Just apply X^{-1} on the left on each side and you will have isolated M

[gaunt mulch]

thank you both of you

[noble swallow]

Ah he was looking for M lol my bad

[native lodge]

The result you reach is pretty important. It’s a very special change of basis

[gaunt mulch]

ok

[native lodge]

Quickly though, M=?

[noble swallow]

He needs X for this, will he be equipped?

[native lodge]

He has the eigenvectors and values, so he’s ready to go

Though I sense you want to see me get the formula in detail again lol

[noble swallow]

(I have a feeling he's in a test)

[gaunt mulch]

nope bro

[noble swallow]

Ah I hooked ya!

[gaunt mulch]

but test going to happen today only

😋

[native lodge]

But M = what? I just want to see you got it

[noble swallow]

Yes @native lodge!

[gaunt mulch]

i assumed m is the original matrix

[native lodge]

In terms of X and lambda I meant

I wanted to see you complete the formula, by isolating M

[noble swallow]

Reminder: in the text you'll need to give the answer

[gaunt mulch]

ok

something just like this

the middle one

[native lodge]

We have MX=ΛX, so isolate M, assume X is invertible (when we have distinct eigenvalues and a corresponding number of eigenvectors, X is always invertible)

[gaunt mulch]

ans is [0 1 /n -2 -3]

[native lodge]

Your final answer won’t have variables in it, only just numbers

Unless I misunderstand what /n means

[gaunt mulch]

/n means next line

[native lodge]

Going over how to get M=X^{-1}ΛX takes a while, Mat lol

[noble swallow]

Amphy isn't it MX=XΛ ?

[native lodge]

Oh, maybe

I would have to think

But I don’t want to right now because late night now

I think it’s that then

X lambda sounds correct

I haven’t memorized the formula yet, since I don’t have to use it often

[noble swallow]

Ok yeah I mean otherwise we can't make the Λ sandwich

[native lodge]

With practice I would be able to dedicate it to memory

I mean you can always do X inverse lambda X

The way to remember which order it really is, is to see how the lambda X multiplication works out

Mx=λx so then make a matrix of x’s

Nah, don’t want to end up doing a formula demonstration when I said I wouldn’t today

[noble swallow]

Lol we'll catch up when you'll be rested

[native lodge]

Checking the notes in my personal server, it is indeed XΛ

[noble swallow]

Where's @gaunt mulch though?

[native lodge]

https://media.discordapp.net/attachments/569771149849002024/598343816147238913/401544836907270155.png

[gaunt mulch]

here

i was solving next problem so went offline for a bit

[noble swallow]

I see, so this is all cleared up?

[neat olive]

sorry for the super beginner question, but how do you get the points on the xyz space from a vector equation?

[native lodge]

Equation such as?

[neat olive]

for example, cv + dw where v=(1,1,0) and w = (0,1,1)

the vectors themselves are arbitrary

[native lodge]

Are you given c and d or are you given the result of the combo?

[neat olive]

for all c and d

it's just a combo vector

[native lodge]

Like you just have to find the final result of this combo? Yes

It’s going to be symbolic then

[neat olive]

let's just suppose c =d=1

plugging in 1 into the combo vector wouldn't work, right?

[native lodge]

You example would give:

[c, c+d, d]

[neat olive]

yep

but does that describe the points on the xyz space?

[native lodge]

It will not, to describe all points in 3D, you need three independent vectors

[neat olive]

because [c, c+d, d] is a vector..

right, but it would create at least a plane, right?

[native lodge]

At most

[neat olive]

but how to find the points on that plane

[native lodge]

The symbolic vector gives all points on that plane

[neat olive]

wait seriously?

does that mean all planes and lines with 3 vector components pass through (0,0,0)?

[native lodge]

Not all planes will go through the origin

[neat olive]

how would we describe x+y+z=1 then with vectors?

but plugging in zero for c and d would give 0 in every circumstances

no matter the vector...

[native lodge]

Planes and lines in R^3 that go through the origin fulfill one condition for being a subspace of R^3 though, you will see that as you go through your linear algebra course.

We would need to find a basis for that plane

Then take all linear combos of the basis vectors to get that plane

But the origin isn’t included in that plane, correct?

[neat olive]

any linear combination will pass through the zero vector, but the zero vector doesn't necessarily mean 0,0,0 on the xyz space, right?

which would mean we can't technically convert x+y+z=1 for instance without losing data when converting that form into a vector

[native lodge]

Too much big thonk required at 1 am
Time to

[neat olive]

LOL

alright thanks for the lead

good night floof!

I saw this

From what I'm understanding from this, you can find the general shape of whatever you have, but it doesn't necessarily correspond 1 to 1 on the xyz plane. Correct me if I'm wrong

[noble swallow]

@neat olive What do you mean? Are you still talking about a plane like x+y+z=1 ?

[neat olive]

I'm talking about linear combination vectors in general

If we have any c and d and have the linear combination [c, c+d, d] for instance, how would that combination vector correspond to a plane on the xyz space

if we have any c and d, we would get [0,0,0] which is the zero vector

but does the zero vector necessarily mean the origin on the xyz space?

a counterexample would be x+y+z=1 where because of the right side (1), it would not pass through the origin

so I was wondering what combination vector would describe the x+y+z=1 plane

[noble swallow]

The plane x+y+z=1 can't be only described by linear combinations of a set of vectors.
It is rather obtained by a translation of a set of linear combinations

As you can see, x+y+z=1 is a parallel plane to x+y+z=0

[neat olive]

yes

[noble swallow]

It can be described as a translation of the latter, which in turn can be expressed as a set of linear combinations (as it includes the vector (0,0,0))

[neat olive]

how would you indicate the translation in vector form?

[noble swallow]

Ok so we are looking for a vector wise formulation/description of the subset of $\bR^3$ given by $\pi = {(x,y,z) \in \bR^3: x+y+z=1}$

[stoic python]

Mat:

[noble swallow]

We initially simplify the problem attempting to describe the plane $d(\pi) = {(x,y,z) \in \bR^3: x+y+z=0}$

[stoic python]

Mat:

[neat olive]

Right

[noble swallow]

Which is called the director space of pi

d(pi) can be described as a set of linear combinations of two vectors, we need to identify them

[neat olive]

Ooh

[noble swallow]

We look for two linearly independent vectors which lie on the plane

Would you be able to find them?

[neat olive]

I'm stuck.... x+y+z=0 passes through the origin, and doesn't the vector start from the origin too?

It seems like everything is dependent

@noble swallow

[noble swallow]

Here the term "vector" is somewhat abstract and is simply the name for an element of a vector space. But not to digress, in our specific case the vector space is
R^3={(x,y,z): x,y,z in R} and any of its elements (x,y,z) is called a vector of R^3

[neat olive]

Can I just make up a random vector that seem to work? like (1,0,0) and (0,1,1)

[noble swallow]

Exactly

But we need x+y+z=0

We can just substitute two arbitrary values for y and z and retrieve x

[neat olive]

y=z=1, x=-2

what about this

[noble swallow]

Nice, so we have (-2,1,1)

Another one?

[neat olive]

um 1,1,-2? or is that too cheeky

[noble swallow]

No that's fine, the important requrement is that the two vectors are not a multiple of each other

[neat olive]

Right!

[noble swallow]

with (-2,1,1) and (1,1,-2) it doesn't seem to be the case

We found a so called basis of d(pi)

[neat olive]

ooh

[noble swallow]

We can write \
$d(\pi)={(x,y,z) \in \bR^3: x+y+z=0} \= {(x,y,z) \in \bR^3: (x,y,z)=c(-2,1,1)+d(1,1,-2),~ \text{for some} ~ c,d \in \bR}\
={c(-2,1,1)+d(1,1,-2): c,d \in \bR }$

[stoic python]

Mat:

[neat olive]

right!

that makes sense

[noble swallow]

A notation for the latter is span({(-2,1,1),(1,1,-2)})

Now, we take a vector on x+y+z=1

Like before

Any suggestion?

[neat olive]

1,0,0 0,0,1

[noble swallow]

Let's take (1,0,0)

We can describe $\pi$ as \
$\pi = d(\pi) + (1,0,0) = \operatorname{Span}{(-2,1,1),(1,1,-2)} + (1,0,0)$

[neat olive]

oh so that's how we describe the translation

[stoic python]

Mat:

[noble swallow]

Yes, if you take a vector v on pi you can express any vector on p as
(some vector on d(pi) ) + v

To visualize this you can even take the lower dimensional case of two parallel lines, of which only one through the origin, on the plane

[neat olive]

if we were to convert d(pi) into a linear combination vector however, we would get something like (-2c+d, c+d, c-2d)

[noble swallow]

The situation is the same

Yes probably

[neat olive]

OH so we just have to take the linear combination and indicate that we translate +(1,0,0)?

[noble swallow]

Yes, translate is more of a geometric term, in the environment of a vector space we are taking the so called "lateral class" of d(pi) represented by (1,0,0)

[neat olive]

so other than taking the lateral class, there's no way to describe a plane that does not pass through the origin, right?

at least in linear algebra

[noble swallow]

Yes, there are other ways, this is just one of the possible formulations

Do you know about matrices?

[neat olive]

yep!

[noble swallow]

Ok so any plane for the origin is a set of solutions of a homogeneous linear system

Like, d(pi) is the set of solutions to x+y+z=0

Such a set can also be described as the set of vectors which are sent, through matrix multiplication, by some matrix A to the vector (0,0,0)

In this case, the matrix would be $A=\begin{pmatrix} 1&1&1\0&0&0\0&0&0\end{pmatrix}$

[stoic python]

Mat:

Mat:

[noble swallow]

The set on the right is called Kernel of A

Is it ok for now?

[neat olive]

I'm sorry for using up all of your time D: but could I bring the entire thing to a close with a question? With any combination vector in R^3 in its simplest form span(u, v), it would pass through (0,0,0) in the xyz space, right?

Simplest form as in without taking the lateral class, et cetera

[noble swallow]

Yes, a set of all linear combinations always includes the 0 vector as it is the trivial linear combination where all coefficients are 0:
0 = 0u+0v

The bold 0 is the 0 vector

0=(0,0,0) in R^3

[neat olive]

and that would mean any plane described using the combination vector would pass through the origin, which would mean x+y+z=1 cannot be expressed through just a combination vector

at least in the form cu + dv+ ew etc..

[noble swallow]

Yes, the difference between the two planes is that the first is a "linear subspace" of R^3, while the second one is not

That's just a term you can look up

[neat olive]

Alright, thank you so so much for all your time and effort~!

You're a lifesaver!

[noble swallow]

You're welcome, it's been a pleasure

[undone garnet]

@noble swallow sorry, how M = S^(-1)AS?

I don't understand

[gray dust]

@undone garnet look up "matrix diagonalization"

[undone garnet]

but

rank(M) = 1

how can we be sure that

M is diagonalizable?

for any M

[noble swallow]

@undone garnet
M=S^(-1)AS is not diagonalization

In the argument

That says that M is similar to the matrix A

Where A is the matrix of L_M wrt the basis B described previously

It's just a change of basis

And yes we only assume that rank(M)=1

[solid stratus]

Are the elements of F^n called vectors or lists?

[dusky epoch]

depends on context but probably vectors

[solid stratus]

Thanks!

"I gave every possible way of constructing a tetronimo a unique representation as a list/vector in R^4"

[undone garnet]

if all eigenvalues of a matrix are equal zero, can we imply that its characteristic polynomial is X^n?

[wintry steppe]

real eigenvalues ?

[undone garnet]

A is real matrix

all eigenvalues are equal 0

[wintry steppe]

all real eigenvalues ?

[undone garnet]

equal 0

I mean

lambda_1 = lambda 2 = ... = lambda_n = 0

0 is real and complex too

I don't understand a lot your question

[wintry steppe]

all complex eigenvalues are equal to 0, this is what you mean

[undone garnet]

no

I don't know

just

[wintry steppe]

yes

[undone garnet]

all eigenvalues are equal 0

nothing more

[wintry steppe]

what is the characteristic polynomial?

[undone garnet]

no

[sonic osprey]

@undone garnet The answer to your original question is yes

[undone garnet]

is it possible that characteristic polynomial x^n

how?

@sonic osprey

[wintry steppe]

I wanted to travel you through but of course 'no'

[sonic osprey]

What do you mean how?

[undone garnet]

how to prove it?

[sonic osprey]

What's the definition of the characteristic polynomial

[undone garnet]

the polynomial

eigenvalues are roots

[dusky epoch]

What's the definition of the characteristic polynomial
the polynomial

[grave plank]

would studying a theoretical focused linear algebra book be beneficial for applied mathematics?

[wintry steppe]

with arbitrary fields? Probably not but who knows

[native lodge]

I would think maybe

depends on if you enjoy some more theory and are ok with less numbers in the examples

[wintry steppe]

Help!! I've been stuck for days..

answer is this but i can't solve it. 😦

[half ice]

@wintry steppe
Which part? Are you able to graph the feasible region?

[wintry steppe]

Nope i cant...

i dont get how to draw constraint 1..

i drew this..

[half ice]

You could always draw it as a function.
-4A + 3B ≤ 3
B ≤ 1 + 4/3 A

Looks like yours is more accurate

[wintry steppe]

hmmmm... so to test my feasible points i will usee these coordinates??

(0,0),

(3,0)

(0, 1)

[paper egret]

given a set of columns of matrices with actual values, is there like a certain way to show the set of columns of matrices is linear independent?

or like a way that is very convincing

like i know this is definitely linearly independent

but i cant just write its linearly independent

do i have to prove that the solution is unique?

cuz thats ugly

[torn hornet]

well, you would want linear combinations of the vectors to not give 0 unless the coeffs are 0 right

[jagged saffron]

is question 2 even right? say we have (1 2 3) ( 1 2 3 )( 0 1 4) ( 0 0 1) then they arent linearly independent

[torn hornet]

so write that statement as a matrix, and see if you can figure soemthing out from that

hint:||no null space||

[jagged saffron]

i feel like i kinda get what he's trying to ask

[paper egret]

ok gimme a sec

how do i formally answer this

like we did it in class, but it didnt seem formal at all

like shit i know for sure x_1 and x_2 are 0, 0

[torn hornet]

so linearly independent means:

[paper egret]

but that ain't cnovincing enugh

[jagged saffron]

well row reduce is a way

of doing it

gaussian elimination

its not a lot of work in this case

[paper egret]

do i have to show that (0,0) is unique?

[torn hornet]

$a \begin{bmatrix} 1 \2\end{bmatrix} +b\begin{bmatrix} 4 \4\end{bmatrix} = 0 \implies \begin{bmatrix} a \b\end{bmatrix} =\begin{bmatrix} 0 \0\end{bmatrix} $