linear-algebra

[dire bough]

It can be either positive semidefinite or negative semidefinite

[knotty raven]

im confused about v and v-perp dimensions, dont want spoilers though but like where should I start to figure out their relation to eachother?

[red prawn]

think about the 3-dimensional case

[knotty raven]

like if v is the plane then v-perp is a line perpendicular to the plane, I feel like if v is a subset of R^n with dimension k then dim(v-perp)=n-k but dont know how to prove that.

[gray dust]

i'd first prove a more general result given subspaces M & N, dim(M+N)=dim(M)+dim(N)-dim(M cap N), which reads as dim(sum)=sum of dims-dim(overlap)

[knotty raven]

alright

I feel like I can do that directly with the definition of span

then the rest would come easy because I can easily show that there is no overlap between v and vperp

thanks!

I feel like theres an easier way just talking about sets tbh

[red prawn]

I think it will work out as well

[knotty raven]

|a|+|b|=|a U b| + |aintersectionb| right

then rearrange

and its the same thing as the general result he talked about

[red prawn]

With a lot of linear, there's freedom to prove things by speaking of elements, or alternatively talking about sets as the spans of elements.

Or dimensions

[knotty raven]

can I talk about dimension like I would cardinality?

[red prawn]

A lot of the same stuff works because of what @gray dust said.

[gray dust]

I can easily show that there is no overlap between v and vperp
not really, v cap vperp={0} by definition

[knotty raven]

oof

[red prawn]

but that's okay

[knotty raven]

dimension still 0

so dimension is only similar to cardinality then I gues

[gray dust]

dim=card(basis)

[knotty raven]

omg

[red prawn]

Here's more than you would want to know about it, by H. Whitney: https://graal.ens-lyon.fr/~abenoit/algo09/matroids.pdf

[west spade]

Any intuition for why a complex linear transformation might not be diagonalizable if it has a duplicate eigenvalue? I don't really see how it differs from the case where eigenvalues are distinct

[lean lynx]

so eigenvectors with distinct eigenvalues are linearly independent

if all the eigenvalues are distinct, that gives you a basis you can diagonalize by

namely the eigenvectors corresponding to those eigenvalues

but when the eigenvalues duplicate, this argument no longer works

if an eigenvalue had degree k, you need k linearly independent eigenvectors corresponding to that eigenvalue

this doesn’t always happen, but it obviously has to happen if k = 1 always

and here’s a simple matrix that isn’t diagonalizable

$A = \begin{pmatrix} 1 & 1 \ 0 & 1 \end{pmatrix}$

by the characteristic polynomial, the only eigenvalue is 1

[stoic python]

doubledual:

[lean lynx]

but you can quickly check the only eigenvectors with eigenvalue 1 are in the span of (1,0)

so it’s not diagonalizable

diagonalizable is equivalent to saying there’s a basis of eigenvectors, and that’s usually how I think of it

@west spade hope that helps!

[west spade]

So two eigenvalues of 1 does not mean a 2-dimensional subspace of eigenvalue 1

[lean lynx]

yes exactly!

degree of eigenvalue in characteristic polynomial =/= degree of eigenspace

the degree in the characteristic polynomial is an upper bound though

if you have k linearly independent eigenvectors with eigenvalue lambda, use them to make a basis. then compute the characteristic polynomial. you get a factor of (x-lambda)^k, in the characteristic polynomial

so the degree of lambda is >= k

[west spade]

I know the matrix you mention is invertible, but is there any way in which "both" eigenspaces of eigenvalue one are dependent? It feels like there's some way in which there is kinda an eigenbasis it's just deficient

[lean lynx]

there’s only one eigenspace, and it has dimension 1

that space in the above example is span(1,0)

[west spade]

Yeah, I got that. Was just spitballing

[lean lynx]

have you learned about jordan normal form yet?

[west spade]

I ask because I'm wondering whether transformations of finite order must be diagonalizable so that's kinda the flavor I was getting at

[lean lynx]

I don’t think I could explain that well, but it’s the moral answer to this question

[west spade]

I'll look into it

[lean lynx]

basically it says any transformation has a matrix in this form

where the lambda_i are the eigenvalues, and the amount of times they show up is their degree in the char-polynomial

those square submatricies on the diagonal are called “jordan blocks”

[west spade]

There doesn't need to be ones in those slots, right? Could be zeros?

[lean lynx]

yeah right, it depends on the dimension of the eigenspace for the eigenvalue

and other stuff too

if your characteristic polynomial is (x-1)(x-2)^2, there’s a few possible jordan normal form matrices corresponding to that

one is the diagonal matrix

the other is the one where you have a 1, then a 2x2 jordan block with eigenvalue 2

you get a jordan block for every linearly independent eigenvector basically

[west spade]

That actually is enough to prove what I want, so I'll just get to this first

Because the blocks will never have finite orders with 1s above the diagonal

[lean lynx]

cool, glad it helped 🙂

if i remember right, there’s always are jordan normal form for complex matrices, and for real ones you need some other conditions

so you should be good for your particular problem

[west spade]

Yeah, trying to classify representations for finite commutative groups and this does it

[lean lynx]

ok i just checked: there’s a JNF if the characteristic polynomial splits over the field you’re working in

so obviously good in C, and not necessarily good in other fields

[west spade]

only if too, right? take your completion to be the base field and take the roots you just added in the diagonal

[lean lynx]

yeah that’s right

there’s actually a nice proof of JNF using fairly simple ring theory stuff, lemme see if I can find a reference...

that proof is better than the ones found in linear algebra books

[west spade]

yeah, I've definitely put the cart before the horse in terms of my LA education so that would be perfect

[knotty raven]

@red prawn Thanks for sending me that paper!

[lean lynx]

dummit and foote covers it in chapter 12

you need to also know how modules work, which you should be able to handle if you haven’t run into them yet

[knotty raven]

Ill probably look at it and see if I should review it byself or if I should take it into office hours some day

[red prawn]

@knotty raven Woah he was in to graph theory up until this point. According to Wiki, at this time he got interested in smooth functions, and the following year he did the Whitney Embedding Theorem

It's so surprising to me, that he could go from primarily interested in dots and lines and counting, to investigating extremes of just how bent and twisted shapes can get

[knotty raven]

I really want to learn spectral graph theory

it feels like it will be a cool way to use all the linear algebra im learning

plus matrix theory seems cool I like playing with different types of matrices

[red prawn]

There's some crazy hard math related to that...

[uncut fjord]

if a matrix is diagonalizable and its eigenvalues are all the same value, does the matrix have to be diagonal?

I want to say it does but I'm not 100% sure if it's true, because I've tried a few matrices and any that satisfy the multiple eigenvalue condition are never diagonalizable because they don't have enough independent eigenvectors

does anyone have any intuition in what direction I should follow to create a proper proof?

[odd kite]

if a matrix is diagonalizable it can be written $P\Lambda P^{-1}$ where $\Lambda$ is diagonal right? And the eigenvalues are on the diagonal of $\Lambda$ right? And if the eigenvalues are all the same then $\Lambda = \lambda I$, right? So then what is $P\Lambda P^{-1}$ ?

@uncut fjord

[stoic python]

Timon:

[uncut fjord]

oh wow that gets you lambda I right? because $P\lambda I P^{-1} = $P\lambda P^{-1} = \lambda PP^{-1} = \lambda I$

[odd kite]

yes

[uncut fjord]

wow I never thought to break it down the diagonal as lambda I

thanks you

[odd kite]

yw

[tame cave]

Hi I’m currently taking 3rd years university course which is Linear algebra course(even tho I’m doing duo, high school and Uni(UCLA) so technically I’m 1st years Uni and G12 student) but that’s not the problem, so in the next chapter after the Gaussian Elimination, we will learn to through a new chapter, and that chapter include Rienman Sum theory which i don’t remembered much , so can someone here explaining or give me a briefly explanation about the Rienman sum for me thanks
P/s I did remembered Intergral

[native rampart]

#calculus

Not exactly linear algebra

[tame cave]

#calculus
@native rampart yes but I also mean to ask the related of the Rienman sum and Linear Algebra because I’m going to study about that soon, so I want a briefly understanding before studied

[wintry steppe]

Finding the general solution for this:
$$y_1' = -y_1 + 3y_2$$
$$y_2' = 2y_1 - 2y_2$$

I find eigenvalues for matrix:
$$\begin{pmatrix}-1&3\ 2&-2\end{pmatrix}$$

Which are 1 and -4.

So general solution is:

$$y = c_1e^t\begin{pmatrix}3\ 2\end{pmatrix}+c_2e^{-4t}\begin{pmatrix}-1\ 1\end{pmatrix}$$

[stoic python]

ryаn:

[wintry steppe]

Is this correct?

(btw the vectors in the solution are the eigen vectors)

[native rampart]

Do you understand the derivation for this?(Yes,This is correct)

@wintry steppe

[wintry steppe]

@native rampart I know the derivation of the eigen values and eigen vectors.

The final part is from some weird equation iirc.

For finding y(0) = (5, 0), is it just plugging c1 and c2 with 5 and 0?

Or does it just get worse?

[native rampart]

Just plug

[wintry steppe]

Oh ok, thats good.

[flat sedge]

@uncut fjord supposing that $\lambda$ has multiplicity greater than 1, just calculate the eigenvectors corresponding to that eigenvalue. If the dimension of its eigenspace is equal to the multiplicity, then it’s definitely diagonalizable.

[stoic python]

Majez_tic:

[hollow cobalt]

Hey I need help with this question... I am trying to solve it for a while but I am unable to. I tried to learn it through videos but all is going in vain
This is the question 👇🏻

Investigate for consistency of the following equations and if possible then find the solutions:
2x-3y+7z=5
3x+y-3z=13
2x+19y-47z=3

Can anyone help me? Also if you will explain it a little I would be very grateful to you because I have similar questions in my plate that I need to solve. xx

[flat sedge]

its consistent if there exists a solution

have you tried putting into an augmented matrix and row reducing

[pure wasp]

Can someone teach me how to do vectors?

[rough olive]

It's been a while since I've linear algebra and I was wondering about this particular part in my book:

[pure wasp]

I understand the whole concept and like addition and stuff but I just dont understand it past that

I need it for my computer science at uni

yeah ik

im asking just for someone to help me with them

[rough olive]

I don't get the need for transposing the vector w. Wouldn't it give the same result if you just take the dot product of the given column vectors w and x_n as they appear above? Why is w transposed times x_n equal to the equation above whilst the dot product of w (not transposed) and x_n isn't?

[pure wasp]

I was just asking if someone could go through my material with me

and just explain the stuff I dont understand

[rough olive]

In comment 1.7 as the book suggests I read, it says the following, but I could use someone to ELI5 what the author means:

Yes but my question is why does the author transpose the column vector w?

why is that? as long as the amount of columns and rows is the same surely the matrix multiplication is defined, no?

[pure wasp]

damn this guys annoying

[rough olive]

So since there are two rows in the column vectors

but only one column

the matrix multiplication is undefined unless one of them is transposed?

have I understood this correctly?

that's what it says above, right?

in the transposed w there are two columns and 1 row.

and in x_n there is 1 column and 2 rows

Ok cool, it makes sense now. It's coming back to me. It's been a while since I've had the linear algebra course, but we're going to be using matrix notation and rules in this course, so it might be a good idea to read my notes from my LinAlg course. Anyway, thanks for the help! :)

[cerulean cloud]

https://cdn.discordapp.com/attachments/486841106298961930/778687723758288937/image0.jpg

how do i go from a graph to a system of inequalities

[slow scroll]

read pinned post

[nocturne jewel]

How do you go from a matrix to the basis of row,col,null ?

[round coral]

@nocturne jewel do gaussian elimination

[nocturne jewel]

Yeah i know it's something todo with the rref, but i cant remember exactly

row is take pivot rows of the original matrix?

[round coral]

yeah when you get the rref form you can easily see the linearly independent rows and columns

[nocturne jewel]

null(A) is augment rref w/ 0, then take the vectors which make the "plane"

not sure what to call it past 3D lol

[round coral]

if you want the dim of null space then you can use rank nullity theorem.

[nocturne jewel]

no i need bases of row, col, null space

[round coral]

as I said the linearly independent rows form a basis of row space, lin indep col for the col space

[acoustic zodiac]

The dual space is essentially just the set of all linear forms, right?

[round coral]

the dual space of say V , call it V* is the vector space of all linear functionals on V. that is the set of all linear maps from V to F

where F is the field

[acoustic zodiac]

so it essentially is that

the set of all linear forms

[rough olive]

Hey, so I'm trying to understand what has been done here:

[stoic python]

Resident Tracksuit Advisor:

[main moth]

Could somebody explain this to me?

Apparently the answer is ƛ=0 but I don't really understand why

[limber sierra]

a matrix with linearly dependent rows always has a 0 eigenvalue

there are a variety of proofs/justifications of this fact, its not particularly difficult to see

[main moth]

i understand that

but i don't see the intuition 😅

[limber sierra]

if a matrix has a linearly dependent row, then its determinant is 0

you're familiar with that fact, right?

[red prawn]

(1,0,0) - (0,1,0) is a vector that is sent to zero

i.e., an eigenvector with eigenvalue 0

[limber sierra]

and ab = 0 implies either a = 0 or b = 0

so take determinants

[main moth]

if a matrix has a linearly dependent row, then its determinant is 0
@limber sierra what's the reasoning behind this?

sorry if i'm a bit slow on this lol

[limber sierra]

are you familiar with the determinant?

[main moth]

yes

oh wait

i see why

[limber sierra]

recall that "determinant 0" means "not invertible"

they're the same thing

and certainly if a matrix has a linearly dependent row

[main moth]

if you reduce to rref you'll end up with a zero along the diagonal

[limber sierra]

it can be row reduced into something with a 0 row

[main moth]

so det = 0

[limber sierra]

right

[main moth]

yep

[limber sierra]

which cant be invertible, hence has determinant 0

[main moth]

yeah

ok

that makes sense

thanks 😄

[red prawn]

So any time you see linear dependence among rows or columns, you don't have to calculate the full determinant formula, you already know the det is 0

[gloomy nebula]

I know that U is isomorphic to W. That's it lol

[gloomy nebula]

an isometry is an isomorphism between two vector spaces such that B(v1,v2)=B(L(v1),L(v2)) for all v1,v2 in V

[crystal oracle]

@simple hornet Sorry for a supre old reply I am on chapter 8.B

[acoustic zodiac]

Given a set of vectors, how can I determine the subspace they form?

[marble lance]

Do you mean their span?

[acoustic zodiac]

well the problem asks for the subspace. but the span is a subspace, so i guess

[marble lance]

Well, the smallest subspace that contains those vectors will be the span.

Idk what else you could mean

[acoustic zodiac]

let me translate it, wait a sec

Find the linear subspace of $R^4$ generated by the following vectors: $((1, 2, 0, 0), (0, 3, −1, 0), (0, 0, 5, 4).$

Solution: $E = {(x^1, x^2, x^3, x^4) ∈ R^4 / 8x^1 - 4x^2 - 12x^3 + 15x^4 = 0}$

[stoic python]

rcatalang:

[marble lance]

Yes, they mean the span

So $E = { a(1,2,0,0) + b(0,3,-1,0) + c(0,0,5,4) | a, b, c \in \bR } $

[stoic python]

Lunasong:

[torpid horizon]

can someone check my work for a math problem

[acoustic zodiac]

alright, yeah, according to $\sum_{i=1}^{i=n}a^i\cdotv_i$ right?

where that equals the span ofc

[marble lance]

Yes

[acoustic zodiac]

how do you get from that to the set conditions, though?

[torpid horizon]

does anyone know how to do these math problems?

[marble lance]

See #❓how-to-get-help

And that is not linear algebra

[torpid horizon]

i cant find the correct subject ... :/

[acoustic zodiac]

seems like abstract algebra to me

[wintry steppe]

read pinned message

[marble lance]

Let (a, 2a+3b,-b+5c, 4c) = (x1, x2, x3, x4). Then a = x1, c = x4 / 4. So x2 = 2a + 3b = 2x1 + 3b, so b = x2 / 3 - 2x1 /3. Then x3 = -b + 5c = 2x1 / 3 - x2 / 3 + 5x4 / 4.

Then multiply through by 12

[acoustic zodiac]

oh

damn

it's pretty simple lol

thanks

[marble lance]

Np

[wintry steppe]

Hey can I get help with this, not sure how to do a) i

Since there two different size of parameters

Is it just multiplying the two matrices?

[main moth]

sorry if this is obvious but why is this true/false?

theoretically couldn't have an eigenvalue that corresponded to more than one eigenvector?

or am i misunderstanding lol

[wintry steppe]

Can i get some help understanding 3.b from the question I posted above?

[hollow finch]

@wintry steppe what does it mean for W to be a subspace of R4?

[wintry steppe]

I'm not entirely sure, I imagine it could be turned into a matrix

[hollow finch]

nonono dont overcomplicate it

what is the def of a subspace

no matrices are necessary for this at all

[wintry steppe]

Its like zero vector, additoin and scalar

[hollow finch]

what about them

be specific

[wintry steppe]

Like i gotta like prove it for each

[hollow finch]

prove what

[wintry steppe]

That zero vector exists in that subspace
Addition works in that subspace
And scalar multiplication works in that subspace.

But idk, how that links with a?

[hollow finch]

not quite "works"

theres a specific thing we're looking for

[wintry steppe]

Oh, that it maintains consistency kinda

[hollow finch]

go back to your textbook if you need to

[wintry steppe]

Yeah I'll look into it again.

[thorn lichen]

not really sure what to do here, but I have A and C

and I guess y is B no?

[round coral]

y is the value that c attains at the corresponding values of x

that is c(-2) =5, c(-1) =2 , c(1)=1.... Anyone correct me if I am wrong

[warm harbor]

hello my lovely people! can anyone explain me why E is a subspace?

Because when I write the matrice as span, the condition c+d is not always satisfied no?

[round coral]

you have to say it is a subspace, not just a vector space, the basis of vector space may not be the basis of its subspace

[warm harbor]

sorry i meant subspace yeah mb not vector space

[round coral]

for that look if it is contained in the vector space, then see if all the operations are inherited as well from the vector space

[warm harbor]

sorry can you explain what you meant?

[round coral]

look over the definition of subspace again

[warm harbor]

yeah i know what the definition is. do you mind explaining to me why what I did was wrong then?

[half ice]

You created a basis for M2×2(R)
I don't see how it relates in any way to E.

Worth knowing is the subspace test, which is very general. You only need to use a few facts to prove something is a subspace

[warm harbor]

it wasn't my intention to write E as a basis. Since any spans is a subspace, that's why I tried to write E as spans and I did that because that's also how one of a similar question from my textbook was answered:

[half ice]

Yeah, that would work. However, what you have above is not a span of E

[warm harbor]

hmmmm

[half ice]

Here's a hint. Use d = -c

[warm harbor]

kk

oh ok

yeah i get it now

still though, could you explain to me why what I wrote previously is not a span of E?

[half ice]

Because I can set a,b,c,d such that the result is not a member of E

I'll use E to represent the set of matricies I guess, lol

For example
a = b = c = d = 1

[warm harbor]

yes

[half ice]

Gives the matrix
[1 1]
[1 1]
Which is not a member of E because c + d ≠ 0

[warm harbor]

kk, i get it now, thanks for your help @half ice !

[half ice]

Np! Feel free to ask if you need anything else!

[pearl elm]

is there some trick to expressing a vector in a span as a linear combination of all the other vectors in the span?

[red prawn]

You wouldn't want to use... all of the elements in the span. That would be a rather large linear combination

[pearl elm]

what if you have to use all the elements

seems like trial and error

[half ice]

@pearl elm
Yes. It is equivalent to solving a matrix equation and so can be done with row reduction

[pearl elm]

wdym

just do row reduction?

[half ice]

Like expressing (2,3) using (1,2) and (5,6) is solving the augmented system:
[1 5 | 2]
[2 6 | 3]

Which probably has some ugly solutions but that's what I get when I come up with the numbers randomly

[pearl elm]

How would you go about this problem

hold on

i messed up a sign

[half ice]

What's it asking?

If the system is linearly independent?

[pearl elm]

write some vector in the span S as a linear combination of the others

[round coral]

is S the set of these three vectors?

[pearl elm]

yes

[round coral]

so just write a linear combination of them, what is the field? Is it R

the vector you would get will be in span S

1. (1,0,3) + 2 .(2,-1,1) -1. ( 5,-4,-5) = (0,2, 10)

[pearl elm]

i think i got it

derped

[crude oasis]

is $\lambda v + (1- \lambda)w$ a theorem for an affine set?

[stoic python]

bakes:

[crude oasis]

going through a textbook solution and v and w were plugged into that equation and resulted in 1 and then it says by the theorem its an affine subset

[wintry steppe]

what does it mean to call a vector a theorem

[dusky epoch]

@crude oasis wat

[crude oasis]

sorry if i explained it horribly but the question was to show A is an affline subset and near the bottom where it says to plug into the theorem idk what theorem its talking about

[wintry steppe]

is there context for this question? is this from a textbook or other source you can link us to?

like i don't see any "theorems" being used

what are the other steps? what comes before what you showed us?

[crude oasis]

sorry for late response but this was the question, maybe that solution is just incorrect i was just confused why it said at the bottom plug into that theorem

[wintry steppe]

is this one of those shitty online solutions manuals

hey kanishk

[round coral]

what is the specialty of affine subset? It can't contain origin.

Hi

[crude oasis]

haha looks like it might be , becuase i have no clue what it was doing

[wintry steppe]

me neither

[round coral]

the fact that you are given sum of \lambda _i =1 is a hint, a big giveaway. Don't go on solution.

[wintry steppe]

the poor formatting of the solution leads me to believe it's one of those crappy solution manuals online that someone wrote on a $ per solution basis

[crude oasis]

gotcha, thank you i didnt know if i was missing something from the chapter

[wintry steppe]

hi guys

i have an exercise that i have no idea how to solve

maybe some one of you guys knows hoy

how*

It's problem 2

[fading plaza]

Okay, I just grab my paper and pen.
Use coordinates vertor? What does that mean to you?

[wintry steppe]

@fading plaza can I call you and I tell you details?

[fading plaza]

Nah, like you can show your work here and we can go on together

Like you can use the vectors and compare them if they are linear independent to each other

[wintry steppe]

The thins es that the aren't vectors

They are the components x y z of the vectors in subspace P

X=p1 and so on

From that you are supposed to find a basis

It would be easy if there were two independent parameters instead o t and t^2

[fading plaza]

I'm kinda working on it with this
p1= (1 0 1)(1 t t²)^T
p2= (0 1 -3)(1 t t²)^T
p3= (1 1 -3)(1 t t²)^T

So that you can compare it with the basis
e1=1
e2=t
e3=t²
Of IP_2

[wintry steppe]

Don't understand

[fading plaza]

You can set the x,y and z components as 1,t and t²

I'm not sure if this helps.

[rugged scarab]

I am currently practicing some electrostatics problems and got stuck because of math.
Mostly vectors.

I found this site for multiplication https://nrich.maths.org/2393

but I am not sure if a.b is the same as a * b

<@&286206848099549185>

and is there anything else must-know about vectors and trig?

[raw sand]

scalar product is just the dot product

the two main operations youll probably see are dot and cross products

[round coral]

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

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

[raw sand]

a dot product makes a scalar from two vectors and the cos of the angle formed by the vectors

a cross product makes a vector from two vectors and the sin of the angle formed by the vectors

[round coral]

@rugged scarab Would love to have a pancake in this cold. There's load of material on wikipedia check out the links. You will understand it better there than asking here

[rugged scarab]

thanks, I think I have a clearer picture on what to do next

@rugged scarab Would love to have a pancake in this cold.
@round coral same. Started studying almost 5h ago and still haven't eaten anything today. Math definitely isn't helping my apetite, though

[true vessel]

I need assistance understanding what the heck my book is talking about. It uses a notation that, to quote the book, “isn’t what most linear algebra books do”.

The problem lies on this page, and the notation it sets up (poorly imo) is very confusing

It’s a big block of text that claims to give “rules of covariant indices” when in reality it summarizes their concept without giving any rules.

Later, it makes references to something that it doesn’t appear to go over

“L83” and the footnote are where my problems lie

[red prawn]

Check the last paragraph in the first pic. "...covariant notation is an important aid in discriminating between objects that are fundamentally vectors and others that are not." Upper index things are for vectors, which you will think of as tangent vector fields to your given manifold. Lower index things are for covectors - things that you input a point, vector or vectors, and it gives you a number, vectors, etc. The vector product behaves less like a tangent vector; it gives you a vector perpendicular to the inputs, and he's pointing out that you can observe this distinction in terms of violation of notational rules (what things are allowed to live upstairs vs downstairs)

If you're wondering where L.83 comes from: L.82 is a Definition. Applying the dot*e_k to both sides yields L.83.

[stray granite]

anyone

?

[wintry steppe]

I have a problem in understanding this definition. By ""There exists j", we can choose an arbitrary j?

[nocturne jewel]

j in {2,3...m} just means that one of the vectors except v_1 is dependent on the others

[wintry steppe]

@nocturne jewel "one"?

[nocturne jewel]

yeah

like single

if j = 2 then v_2 is linearly dependent

[wintry steppe]

for example in the list {(1,2), {2,4}} that is linearly dependent, we can verify that (1,2) is in span(2,4) and (2,4) is in span (1,2) though @nocturne jewel

[nocturne jewel]

yeah

[wintry steppe]

then?

there are more than one?

[nocturne jewel]

yeah, so that set has 1 linearly independent vector

[trail mulch]

at least 1

[nocturne jewel]

cause the 2nd one is already accounted for in the span of the 1st

[wintry steppe]

at least 1
@trail mulch hmm that was what I was wondering

now I'm aware, thanks

[red prawn]

@wintry steppe If, for example, they had instead wanted to assert the existence of more than one object, the statement would be more like "there exists a set of v such that the following conditions hold and this set has more than one element" or some other such quantification.

[half ice]

@stray granite
It's interesting and all, but have you got any thoughts on the question, or anything you've tried? What's your sense of what a subspace is?

[opaque plover]

Is this the correct way to find all X for the matrix?

[gray dust]

@opaque plover yes but line 4 equates 3x2 matrices, should give a system in 6 eqns

[opaque plover]

ye if I solve the first column i get the first 3 x and the second column for the next 3 x

[crude oasis]

What can you conclude from the Fundamental Theorem of Algebra about factorization of real polynomials

[wintry sphinx]

that if the degree of a polynomial is d, then there are at most d real roots

[pearl elm]

Ok so I’m derping real hard on how to do row reduction in this case

Err number 47

Do I just solve for r since there seems to be no clear way to do row manipulation all the way?

I also skipped 42 for now for the same reason

yea im still stuck. Trying to solve for r doesn't seem to do anything

[plain ibex]

You could try taking the determinant of the matrixes and finding the values of r for which the determinant is 0

[pearl elm]

have not covered that section yet

is there another way?

so im forced to calculate a determinant here?

[plain ibex]

No, probably just Row Reduce. Sorry 😔

[slow scroll]

find a linear combination of (1,2) and (2,1) giving (-1,7). Then use those coefficients to compute the value of r that makes it linearly dependent

[pearl elm]

so we are ignoring the last row?

why that specifically?

anyway this problem doesn't row reduce to an obvious form where we know what r is. I have tried it like 3 times already.

maybe just take determinant without doing row reductions? would that work?

[quartz compass]

doing what kxrider says and ignoring the last row just ensures the first 2 components work out

then whatever the bottom entries are will just go along for the ride to give you the value of r

[pearl elm]

so does this always work?

[quartz compass]

I don't think you understand what's happening here, cause your question doesn't make sense

[pearl elm]

ok so for number 47

[quartz compass]

I'm not psychic, try to describe what you're thinking this does

[pearl elm]

so basically, lets say I am trying to solve 42, would i exclude the row that includes r and do row reduction and solve?

[quartz compass]

like let's say 3 of the first and 2 of the second gave you that vector

what would r be?

yeah @pearl elm that would work for 42

[pearl elm]

so what do i do after the row reduction with the 2x3

it doesn't seem like it helps

basically like... all i got was "here is a linearly dependent matrix and here are your solutions to it"

[slow scroll]

you should have c_1, c_2 such that c_1(1,2) + c_2(2,1) = (-1,7), right?

[pearl elm]

oh yea

well

not quite

oh wait

you meant an augmented matrix

i fucked up then

lol

give me a sec

[slow scroll]

c_1, c_2 are the solutions to
c_1 + 2c_2 = -1
2c_1 + c_2 = 7

[pearl elm]

ok so how does that help me find r

[slow scroll]

add the third components back in and compute c_1 (1,2,-1) + c_2 (2,1,-3) = (-1, 7, r). There is only one thing that r can be here

[pearl elm]

Hold on

Does this look right

Hold on I made a fix

uh i feel like i should of done the augmented row reduction just with the 3x3

what was the point of this

ok

i just used the original matrix and did augmented row reduce

got the answers i was looking for doing that

[half ice]

Another option is just to put all 3 in a matrix and row reduce to identity. At some point you'll be forced to divide something with r in it, just remember that you can't divide by 0

As such, row reduction to identity is impossible for some r

[slow scroll]

@pearl elm
It looks like ur working too hard here. You have
5(1,2,-1) - 3(2,1,-3) = (-1,7,4) = (-1,7,r)
so what does r have to be?

[pearl elm]

I figured that one out already

Answer in book is r = 11 but I get -9

[elfin schooner]

helo

[pearl elm]

maybe they did more row reduction and got diff answer for r

[elfin schooner]

you can have multiple answers for r

aye

you can maybe verify your solution if that makes you feel better

[elfin schooner]

this is a different question right?

[pearl elm]

ye

[elfin schooner]

do you wanna verify it?

a computer can do that easily

[pearl elm]

how would i verify that answer

also

not sure if my intuition is right for this answer

number 49

[elfin schooner]

use the definition of linear dependence

and solve them system of equations

are you solving the matrix by substitution method?

erm >.<

[pearl elm]

i just go right to solving it if i don't see an intuitive row reduction

[elfin schooner]

isnt it boring 😆

[pearl elm]

im not sure if that intuition for 49 is correct tho

[elfin schooner]

okay yeh, its not possible

[pearl elm]

so it is neither linear dependent or independent in this case

or just not linear dep

[elfin schooner]

by not possible, i mean the solution of (1) and (2) is not a solution of (3)

[pearl elm]

yea

[elfin schooner]

aye, yep

its lin indep

wait HANG ON

[pearl elm]

how would i know its lin indep

doesn't all the scalars have to be 0

[elfin schooner]

hey, i didnt notice this earlier

um

why is it an augmented matrix

[pearl elm]

cuz solving for r

[elfin schooner]

OH OKOK

the fourth row shouldnt exist then, but i get u

[elfin schooner]

yes

no other solution is possible

[pearl elm]

so it is neither lin dep or ind

[elfin schooner]

no its lin ind

[pearl elm]

how so

[elfin schooner]

im gonna say the first vector is v1, second v2, and third v3 okay?

you solved x1v1 + x2v2= v3

and you saw that there are no possible x1s and x2s such that this holds true

im gonna multiply the entire equation by x3

[pearl elm]

wdym by x3, there is x1 and x2

[elfin schooner]

multiply it, what do u get

[pearl elm]

im still a little confused by what you mean

[elfin schooner]

sorry my keyboard is kinda broken i cant type x

let x3 be an arbitrary real number, okay?

now i multiply it on both sides, that still preserves the equality

can you type the new equation?

[pearl elm]

oh wait

the third column vector doesn't belong in the existing set of vectors?

[elfin schooner]

earlier you were solving the third column vector as a linear combinatiion of the first two

now, we are considering the third column vector as a vector in its own right

[pearl elm]

yea it is an indepedent vector

[elfin schooner]

and lin ind is a set of all independent vectors

uhh- can we join on vc? i cant type well

[pearl elm]

i think i get it

the third vector is its own independent vector, which is why we can't get a solution in the augmented matrix, rendering the set S as linearly independent

[elfin schooner]

YES!

there's one tiny part remaining- that you show that the first two vectors are independent as well

[pearl elm]

Does this look right?

So far anyway

The answer in book says solution is all reals

But I don’t see how it’s getting that

[slow scroll]

because (0,0,0,0) ruins any chance for linear independence on its own

[pearl elm]

but can't you just throw in a zero vector into any matrix

if I remove it from this I still get a 4x3

if I had a 3x4 then it is lin dep for sure

[slow scroll]

you can't just remove vectors from your list, even if its 0

[pearl elm]

ahhh ok

Also does this look correct

It’s a diff solution than book but I am using x3 and x4 as free variables

[slow scroll]

if you have any collection of vectors v1, v2, ..., 0,..., vn containing the zero vector, then there are always nontrivial linear combinations of these vectors giving the zero vector:
c_1 v_1 + ... + c0 + ... + c_n v_n = 0
where you take all of the c_i = 0, and c != 0. If you were to remove 0 from the list, then v1, v2, ..., vn very well could be linearly independent.
that's what i meant by (0,0,0,0) "ruining any chance for linear independence"

[pearl elm]

thanks for clarification

so for that problem i chose different free variables

than book did

i think its correct. I just end up with fractions cause I just kinda brute forced it

lol

[slow scroll]

,w rref [[1,-2,-1,-4],[2,-4,3,7],[-2,4,1,5]]

[stoic python]

Results provided by WolframAlpha

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

Requested by kxrider#0326

[slow scroll]

x2 and x4 are free here

[pearl elm]

i probably just should do row reduction

i was lazy

[wintry steppe]

hiii i'm not sure if this is allowed

but i have a final in 2 weeks

and i was just wondering if someone has time to tutor me tmmrw

paid of course!

please dm! thanks!

[wicked birch]

Umm

I’m not sure

I could

But

I would rather not earn

I would do it voluntarily since I don’t wanna break tos @wintry steppe

So yeah

[slow scroll]

i don't think rules are broken unless you do their homework for them

[wintry sphinx]

honestly you could probably just join vc when there are people in it and start asking away

[pearl elm]

lmao row reduction makes these problems much easier. I'll finish the last 6 problems in this section later. Thanks

[wintry steppe]

linear algebra is so nice with my prof this semester

[rain echo]

I'm trying to prove the Cayley Hamilton theorem a certain way

by showing that if p is a polynomial, and c(x) = det(xI-A), and if p is divided by c to get a quotient q(x) and remainder r(x) then p(A) = r(A)

I'm stuck though

[wintry steppe]

elaborate

why should this help prove CH?

[rain echo]

this implies Cayley Hamilton cause if p(x)=c(x) then r is 0 so c(A)=0

[wintry steppe]

how is this different from the usual false proof of just plugging in A?

[wintry steppe]

The cardinality of the set Map(X, X) of maps from X to itself is n! if X is a finite set with cardinality n, right?

[rain echo]

the false proof is just saying det(AI-A) = det(0) = 0. This proof is showing that for any polynomial p which divided by c(x) (the characteristic equation) gives a remainder r, p(A) = r(A). It isn't cheating to let p(x)=c(x) as a special case of this, once it has already been proved

[wintry steppe]

<@&286206848099549185>

[dusky epoch]

@wintry steppe no, it's n^n

n! is the count of bijective maps

[wintry steppe]

But don't they have to be bijective? nani

[dusky epoch]

do you think non-bijective maps don't deserve to be called maps?

[wintry steppe]

But it's the set of maps from X to itself

How can a map from a set to itself not be bijective

[dusky epoch]

very easily

take X = {1, 2, 3, 4, 5, 6} for the sake of example

you can have a map f: X -> X which sends everything to 1

[wintry steppe]

oh ok

Thanks

Can you tell me the definition of a natural map?

[dusky epoch]

uhhhh

i think that might be a bit context-dependent, it's not an actual rigorous term unless you get into category theory or something

[wintry steppe]

I can give context if you want

[dusky epoch]

sure

[wintry steppe]

https://i.imgur.com/LXXbvD8.png

[dusky epoch]

yeah "natural" is kinda... informal here ig?

it's a bit hard to pin down exactly what it means since its meaning is a bit subjective

[wintry steppe]

Do you have a best guess maybe

[fringe matrix]

hey

[dusky epoch]

uh oh stinky

we got a cheater here

[wintry steppe]

Me?

This is from notes

I can PM you the whole notes if you want..

[dusky epoch]

no it was someone who's b& just now

anyway yeah like

idk

dont take the word "natural" too seriously i guess?

its more of a "doesnt require much additional legwork to construct" thing

[wintry steppe]

Ok thanks

What does it mean for a map to be factorized?

[half ice]

@wintry steppe
"Natural" is informal to mean "It is an easy thing to recieve from our construction". There's no special property of this map.

I don't know what factorization of maps is in general, but I'm guessing in linear algebra it can be thought of as writing a matrix as a product of two matricies instead. Have a picture of that one too? Lol

[wintry steppe]

I do yes

Do you want it?

[half ice]

Yeah just to get some context haha. Sorry to make you work so hard

[snow yacht]

yes i totally agree with you.

[wintry steppe]

Hey I can't complain here thanks for helping

[half ice]

Ah okay. Much like one can factorize
6 = 3×2
We can take any function f and break it into two parts, such that their composition is that function again:
f = i∘π
Special thing to note, if π is surjective and i is injective we can still always do this, via that picture's construction

These functions don't have to be linear, but if they are we can represent this whole thing with matrix multiplication instead.

[wintry steppe]

Oh ok thanks

Can we always take any function f and do that?

[half ice]

Trivially, yes haha.
x² = x² ∘ x

It's boring but it works

The injective/surjective thing makes it a little more interesting

[thorn lichen]

?

[half ice]

@thorn lichen
I've actually never understood that notation. Does that mean:

"expressed in the basis B, [1,2] is the same as, expressed in the standard basis, [1,2,1]"?

[thorn lichen]

i think

im trying to find B

[half ice]

If that were the case, then there's two "vectors" such that:
2a + 3b = i + 2j + k

[stoic python]

Blu3_bear:

[thorn lichen]

how do i find them?

@half ice

[half ice]

There's no two unique such vectors haha. Just any will do

Why not set a few and see if you have to decide the last entry?

[thorn lichen]

but like

ok nvm

i see

is the basis jsut two vectors

or do i combine them?

i used [ 1 1] and [ 3 1 0]

[half ice]

How do you add those? Haha

You care about:
2a + 3b = i + 2j + k

The right side has vectors in R³ so naturally the left must as well

[thorn lichen]

dont understand

is a b not the variables for the vector?

[half ice]

Yes a and b will be vectors in R³

[thorn lichen]

ohhhh

gotcha

so i can use [0 1 0] and [1/3 0 1/3]

[half ice]

Clever solution haha

Yeah that works

[thorn lichen]

thanks !

[hollow finch]

if i know two matrices A and B are similar, is there a way to find the "change of basis matrix" P such that PAP^-1=B?

[crystal oracle]

Maybe calculating SVD of both matrices can help. Then make a matrix that takes the first list of right singular vectors to the second list of right singular vectors as the change of basis matrix.

I am not actually sure, but this is an idea

Ups: actually, nevermind

[hollow finch]

also @gritty kelp sigma is the same dimension as A so not always square. you also must have made a mistake because the eigenvalues of A^TA and AA^T are always nonnegative

[hollow finch]

hm well if theyre both square then SVD may be a bit heavy handed. maybe investigating the jordan forms

$$TAT^{-1}=B$$
$$T(PJP^{-1})T^{-1}=P'JP'^{-1}$$
$$(P'^{-1}TP)J(P^{-1}T^{-1}P')=J$$
$$(P'^{-1}TP)J(P'^{-1}TP)^{-1}=J$$

[stoic python]

nix:

[hollow finch]

i decided to use T instead. does it follow that P'^{-1}TP must be the identity matrix?

[round coral]

do you want to prove that to change the basis of a square matrix/ operator , you need to multiply with the identity matrix both right and left

the identity matrix which maps the old basis vectors to the new basis and its inverse which maps the new basis vectors to the old basis vectors

that's right

@hollow finch I would suggest looking at this through linear maps/transformations. It will be easier to understand this fact

[native rampart]

That matrix could just be in the centraliser of J

[round coral]

what is centraliser?

[native rampart]

basically matrices I,such that IJ=JI

The transformation could have the exact same matrix in another basis

[round coral]

Yeah! That's quite obvious. We prove that the first we study transformations

[hollow finch]

hm wait so could i do this

$$(P'^{-1}TP)J(P'^{-1}TP)^{-1}=J$$
$$TP=P'$$
$$T=P'P^{-1}$$

[stoic python]

nix:

[hollow finch]

i guess that would make sense lmao

[wintry steppe]

https://i.imgur.com/fLmPvkH.jpg

Is the yellow area shaded in correctly?

[limber sierra]

i dont know what youre trying to do

what is the yellow area supposed to represent? why is this in #linear-algebra ?

[wintry steppe]

I don't know, linear algebra class

The yellow area is supposed to represent R x R as in the top right

But I don't think it's correct, it should be all 4 quadrants highlighted?

read the pinned message

Where should it go?

RxR is the entire plane

[wintry steppe]

idk, probably not here

Yeah where though?

pinned message

[wintry sphinx]

I disagree with the first

the first one is obviously linear algebra

solving a system of linear equations

[limber sierra]

i would not consider "solving a system of linear equations" to be linear algebra ceteris paribus

certainly if youre using linear algebraic techniques

like row reduction or w/e

[wintry sphinx]

perhaps, but I can write the first question as "find the x that minimizes ||[4 - (3x + 1)]||^2

[limber sierra]

alright, tell that to the typical 14 year old who comes into #linear-algebra thinking it'll help them with their algebra 1 homework

[serene tulip]

I am struggling with this question

I am suppose to be able to find dim(L(V))

in terms of dim(V)

Then I assume k>dim(L(V)) so k can generate the Z function

[rain echo]

I want to show that A and A*A have the same null space

it's obvious that NS(A) is a subset of NS(A*A)

and if AAx = 0 taking the conjugate transpose on both sides gives xA*A=0

oops discord is weird

[wintry steppe]

backslash the *s

[rain echo]

that was supposed to be A*A=0

ok

so if A*Ax = 0 taking conjugate transpose gives x*A*A=0

which means also (Ax)*A=0

so x COULD be in the null space of A as well

but x*A could also be zero

which is A*x is zero

[wintry steppe]

hint: ||look at <x, A*Ax>||

[rain echo]

k I'm gonna try that thabks

but based on my manipulation, doesn't that mean that A* and A also have the same null space?

cause that doesn't make sense

ah thanks for the hint I think I have it

can you check

[wintry steppe]

sure

[rain echo]

so if x is in the nullspace of A*A then <x,A*Ax> is just <x,0> which is 0, but it is also |Ax|² therefore Ax=0 and x is also in the nullspace of A

[wintry steppe]

correct

[rain echo]

yay thanks

k this led to a curiosity of mine

if A is square do A* and A have the same nullspace?

here is my reason why I think this should be the case:

we already see that A*A has the same nullspace as A, and if x is in NS(A*) then A*Ax = ((A*Ax)*)*

which is

(x*A*A)*

which is

((Ax)*A)*

[wintry steppe]

@rain echo in general, the kernel of A* is the orthogonal complement of the image of A (try proving this if it's new to you), which should help you find a simple counterexample

[rain echo]

nice I think I will try to prove that

does that imply that their null spaces are disjoint?

actually no it doesn't

[wintry steppe]

all vector spaces have zero 😉

[rain echo]

well I meant besides 0

[wintry steppe]

lol

[rain echo]

and even so it's wrong

[wintry steppe]

i guess you meant direct sum?

[rain echo]

what about their direct sum?

that it is the entire space?

[wintry steppe]

ah sorry i see what you mean

[rain echo]

yeah

[wintry steppe]

1 sec

it's been a while since i've dealt with this kinda thing

[rain echo]

I still find it extremely satisfying that the euclidian inner product just accidentally also is similar to matrix multiplication which gives stuff like NS and RS are orthogonal complements and like unless the vectors were lists of numbers idk if that would be the case

well let me think

I guess rowspace has no meaning unless each vector is a list of elemenrs from a field

[wintry steppe]

just take A to be the zero matrix lol

took me longer than it should have

[rain echo]

omg hahahahaahga

well also

just take rank(A)<n/2

[wintry steppe]

i think in some specific situations (e.g. A unitary, maybe even normal or self-adjoint too - play around with it) you can say something nice though

[rain echo]

well I know

that

nvm

k I have an inner products question

I have this idea that I used as intuition when learning quotient soaces

spaces

I want to see if you think it's truw

if L is a subspace of K

hold on its hard to word this

so like

if you have an inner product defined

then the orthogonal complement of L has a basis which is linearly independent over L

I'm wondering

[wintry steppe]

wdym linearly independent over L?

[rain echo]

oh that's the terminology my book ussz

it means that no linear combination of the set of vectors is in L unless all coefficients are zero

[wintry steppe]

i see

[rain echo]

my book uses that to define what the quotient space iz

is

well not to define it but to understand it better

[wintry steppe]

so a set S is linearly independent over L if its image in K / L is linearly independent

[rain echo]

yes exactly

[wintry steppe]

ok i get it

[rain echo]

in fact the way I understand it is that you have a basis for K which contains a basis for L, and the basis for K/L is the classes of vectors which contain each basis vector for K that wasn't one of the vectors in L

so I was wondering

the orthogonal complement is a very nice example of this

so like if x and y have the same image in K/L then their projection to the orthogonal complement of L is the same

so I was wondering, if L and M are subspaces of K whose direct sum is K

and a basis for M is linearly dependent over L

does there always exist an inner product on K such that M and L are orthogonal complements?

[wintry steppe]

@frosty valley channel occupied

@rain echo gimme a second to take this in

[rain echo]

alright

[wintry steppe]

if x and y have the same image in K/L then their projection to the orthogonal complement of L is the same
i'm not so sure about this

[rain echo]

why not?

if B is a basis for the orthogonal complement, then x and y having the same image in K/L is the same thing as their B coordinates being identical and their L coordinates can differ by whatever

similarly, the projection of x and y onto the orthogonal space is determined by their B-coordinates

I'm using the expression "B-coordinates" loosely here, I just mean, in a basis for K which contains the basis B, the coefficients of the vectors in B for the representation of x will be the same as for y

[wintry steppe]

oh i was having a brain fart oops

forgot how quotient spaces work for a sec

yeah i see it

now lemme think about the inner product question

[rain echo]

ok

[wintry steppe]

it's a good question

[rain echo]

:)

[wintry steppe]

just throwing an idea out there

let $\ell = \dim L$ and $m = \dim M$, and choose an isomorphism $\Phi : L \oplus M \to \bR^\ell \oplus \bR^m = \bR^{\ell + m}$ with $\Phi(L) = \bR^\ell \times {0}$ and $\Phi(M) = {0} \times \bR^m$. this defines an inner product on $L \oplus M$, and maybe you get $L^\perp = M$ with it

[stoic python]

TTerra:

[rain echo]

after this I have an interesting fact about linear algebra that I already proved but it's a very sexy theorem

ok let me read for a sec

[wintry steppe]

i'm leaving a few details out that i don't wanna type in lol

[rain echo]

the little L didn't show up on line 1 I assume it's supposed to be there

[wintry steppe]

where

[rain echo]

it starts off "= dimL"

oh

[wintry steppe]

it looks fine for me

[rain echo]

wait

I was viewing limited

[wintry steppe]

lol

i feel like this will work

ofc you should write out what such an isomorphism is and what not

[rain echo]

yes I get the point you are making

let me think on it for a sec

it seems a little too magic

[wintry steppe]

a lot of linear algebra does 😌

[rain echo]

hagag

ok

so

I like how you put us in a very easy vector space

now I feel like

let's just let l+m=n to make things easy

[wintry steppe]

all of finite-dimensional linear algebra is just reduction to euclidean space

[rain echo]

you're right

i just didn't take advantage of thay

but is this true in Rⁿ?

[wintry steppe]

which is why you study infinite-dimensional spaces where fucking everything goes wrong

[rain echo]

well fourier is very sexy though

[wintry steppe]

alright so the isomorphism construction i just did kinda ignored the condition "a basis for M is linearly dependent over L"

[rain echo]

well that is kind of assumed

just unsaid

[wintry steppe]

ye i'm just pointing it out

[rain echo]

also

I'm pretty sure just by geometry this should hold on Rⁿ

right?

would that prove it?

I just assume yes

[wintry steppe]

you mean whether or not there's an inner product on R^n that makes L and M orthogonal complements?

[rain echo]

yeag

I can't prove it

but it would be so shitty if that was false

[wintry steppe]

let me write down some more detail for the isomorphism thing i proposed

just to make sure nothing goes wrong

[rain echo]

it looked straight forward

especially considering L and M have bases which are linearly dependent over each other

which must be the case in order for you to say that phi(L) is R^l x {0}

I think so at least

alright here is a sexy theorem

and if you row reduce with the euclidian division algorithm it becomes really easy to see why

if you have n, n digit numbers represented in base b

and you put their digits as entries in an nxn matrix

if all n of those integers were divisible by k

then the determinant of the matrix will be divisible by k

[wintry steppe]

Let ${v_1, \dots, v_\ell}$ be a basis of $L$ and let ${w_1, \dots, w_m}$ be a basis of $M$, and consider $L \oplus M = K$. Define an isomorphism $\Phi : L \oplus M \to \bR^\ell \times \bR^m$ by $$ \Phi\left( \sum_{i=1}^\ell a^i v_i, \sum_{j=1}^m b^jw_j \right) = (a^1, \dots, a^\ell, b^1, \dots, b^m). $$ Consider the unique inner product structure on $L \oplus M$ turning $\Phi$ into an isometric isomorphism, namely $$ \langle v, w \rangle = \langle \Phi(v), \Phi(w) \rangle. $$ Then, since $\Phi(L) = \bR^\ell \times {0}$ and $\Phi(M) = {0} \times \bR^m$, the subspaces $\Phi(L)$ and $\Phi(M)$ of $\bR^\ell \oplus \bR^m$ are orthogonal complements. Since $\Phi$ is an isometry of inner product spaces, $L$ and $M$ must be orthogonal complements. (Note that we didn't need to impose any conditions about bases of one space being linearly independent over another.)

sorry im just gonna drop that

outta nowhere lol

bot slow?

[rain echo]

is the bot gonna pop up?

I can't read that

[wintry steppe]

idk it's been laggy recently

give it a moment

[rain echo]

oh ok

I will wait

[wintry steppe]

👻

[stoic python]

TTerra:

[wintry steppe]

good god that is long

note that im considering the internal direct sum (L, M are subspaces of K, adding to K with trivial intersection) - this doesn't really matter but i just wanna point it out

[rain echo]

omfg

[wintry steppe]

so the answer to your question is yes

[rain echo]

beautiful

I love it

[gentle yacht]

hello

[rain echo]

and it's so simple

[wintry steppe]

my intuition was basically to pretend that everything is subspaces of euclidean space

[rain echo]

it seems obvious now

[wintry steppe]

now how can this go wrong in infinite dimensions

[rain echo]

hahaha

I'm bad at thos3

this is where you lose me

although

[wintry steppe]

alright here is a sexy theorem
and if you row reduce with the euclidian division algorithm it becomes really easy to see why
if you have n, n digit numbers represented in base b
and you put their digits as entries in an nxn matrix
if all n of those integers were divisible by k
then the determinant of the matrix will be divisible by k
this sounds neat, although i have no clue how to approach it lmao

[rain echo]

might be useful to try some special cases

like R^infinity

and maybe functions from R to R roo

then try to abstract

do you want me to take you through the proof with you?

[wintry steppe]

\ell^2 would be good to start with, i think

[rain echo]

of the determinant thing

[wintry steppe]

it's okay, thanks though

[rain echo]

what's \ell^2

[wintry steppe]

set of sequences converging in the 2-norm

more explicitly

well i guess it's basically R^\infty

but slap a hilbert space structure on it so it becomes just a teeny bit nicer

$$ \ell^2 = \left{ (a_1,a_2,a_3,\dots) \in \bR^\bN : \sum_{n=1}^\infty a_n^2 < \infty \right} $$

[stoic python]

TTerra:

[rain echo]

oh that's cool

I've never been that vector space before

[wintry steppe]

i'd look at something like that if i want to think about infinite dimensional examples to finite dimensional things

ell^2 isn't very... unwieldy

unlike certain spaces

[rain echo]

yeah

it's pretty cool though

like

[wintry steppe]

if you take functional analysis you will probably see this every single day

[rain echo]

such a wildly defined vector space

hahaha

not taking that for a while

am I wrong or is functional analysis all about infinite dimensional space?

[wintry steppe]

that's not that far off

finite dimensional inner product space stuff is generally well-understood

so one prefers to study infinite-dimensional inner product spaces

that's where the topology and analysis comes in

thus placing functional analysis somewhere in the intersection of linear algebra, topology, and analysis

[rain echo]

that sounds scaryyyy

but also fun

cause finite dimensional inner product spaces are like really happy and nice things

they are a comfort zone

[wintry steppe]

interesting things happen when you get out of your comfort zone

[rain echo]

I mean that's what math is all about

[wintry steppe]

func anal is more analysis-y than linear algebra-y in my experience

a lot of its examples come from and are motivated by analysis

all of finite dimensional linear algebra is just fucking with bases, so when you get rid of a finite basis....

[rain echo]

hahaha I guess so

but if you are concerning yourself with inner products I bet most of the structure comes from linear algebra

so by definition wouldn't it always be linear algebra-y

[wintry steppe]

inner product gives you a norm, which gives you a metric, letting you do topology/analysis

if you've ever heard of a banach space or a hilbert space

it's when the metric induced by the norm is complete (banach), and in this case, if the norm comes from an inner product, you call it a hilbert space

[frosty valley]

can I post my question now?

[wintry steppe]

if ethan has no mroe questions, go ahead

we were basically just chatting about math lol

[frosty valley]

Does that mean that the vector wasn't turned into probability vector? which means it doesn't have to be equal to 1

[odd kite]

it normalizes eigenvectors by default

prob. whats happening

result of scaling an eigenvector is still an eigenvector

for reasons that should be fairly obvious

so it scales them so that the norm is 1

[frosty valley]

so it means it's the scaler value which makes the difference otherwise both are the same

[odd kite]

yes, they both correspond to the same subspace and have the same eigenvalue

the scale factor is arbitrary basically

[rain echo]

@wintry steppe I guess I never saw the connection between linear algebra and metric spaces

but it was there all along

very interesting

[wintry steppe]

why is v1, v2, ..., v(k-1) LI?

i think it might be due to linear dependence lemma, but i'm kinda unsure

[rain echo]

cause of how k is defined

worded differently

vk is the first vector that is in the span of the previous ones

that means before that point

no vector was in the span of the previous ones

to say it loosely it "added to the span"

and k is the smallest integer where the vector added is "useless"

[wintry steppe]

i don't get it either way

[rain echo]

so let's build up the set of all vectors v

starting with v1 and add v2 then add v3 and so on

if the entire set (up to m) is linearly dependent

then there is some point

when as you are building up the set, the vector you add is in the span of the vectors you have already added

so maybe v1 and v2 are LI, so v2 is not in the span of v1

and then you add v3, and v3 isn't in the span of v1, v2, v3 so everything is STILL LI

but then maybe v4 is v1+v2+v3

and if that is the first vector that you add that has this property

then you know the vectors that came before have to be independent

cause if they were dependent, then one of them would be a combination of the others

but that contradicts the fact that k is the SMALLEST integer for which they are independent

make sense?

[wintry steppe]

totally

thank you, I was stuck on this

[rain echo]

nice

there's another proof that the eigenvectors are independent too if you want that

I find it a bit more straightforward

[wintry steppe]

I saw it by using induction, but I was trying to get it in this way

[rain echo]

induction gives you an argument that's veryyy similar to this one

[wintry steppe]

could you disclosure?

[rain echo]

what do you mean?

want the induction proof?

[cunning arch]

I'm a bit lost on what I'm supposed to do with the A^5 part, I know what to do if it's just A, but what do I do with the 5

[wintry steppe]

yes!!

[slow scroll]

A^5 means you apply A five times to a vector

A^5v = A(A(A(A(Av)))))

[cunning arch]

Ooooh ok, thank you, let me go give that a try

[slow scroll]

np

[rain echo]

but use the fact that Av=-v

to make it easier

[rain echo]

alright here it goes. base case n=1 is pretty obvious, any set of 1 vector that isn't zero is linearly independent. Even for 2 you can see that if v1 and v2 are linearly dependent, then v2 is a scalar multiple of v2, but that implies that it corresponds to the same eigenvalue as v1 which is not the case. Inducvitve hypothesis is that this holds up to n distinct eigenvalues where n>=2. Then suppose you have a combination a1v1+a2v2+...+anvn+a(n+1)v(n+1)=0. Make 2 mental copies of this equation and manipulate them both in different ways. if you multiply this equation by A you get each term multiplied by a different eigenvalue. If you multiply the original equation by the n+1th eigenvalue and subtract this equation from the first one, you get that the term with v(n+1) cancels since you subtract lambda(n+1) of it from lambda(n+1) of it. For all of the other terms you get ai(lambda(n+1) - lambda(i))vi. but now you have a combination of the first n vectors which gives zero. since the inductive hypothesis is that these are independent, that gives you that each coefficient must be zero. since the difference of distinct eigenvalues is not ever zero (I mean they are distinct real numbers) that leaves that each a_i must be zero. therefore the original equation gives you simply a(n+1)v(n+1)=0 which gives you a(n+1)=0 and that proves the linear independence

sorry the notation is so weird

idk how to use the bot

also sorry, just a picky note, I said one time in the proof that the eigenvalues are real numbers

that obviously doesn't have to be the case

they are elements of whatever field you are working with

[wintry steppe]

put single dollar signs around things you want inline, double dollar signs around things you want presented in the middle

like

here is inline: $G = \bZ / p\bZ$, and here is display mode (i think it's called that): $$ \frac{1}{2}E''(0) = \int_0^a \left\langle V, \frac{D}{dt}\frac{dc}{dt} \right\rangle , dt $$

[stoic python]

TTerra:

[wintry steppe]

after that it just boils down to how much latex you know

messed up a sign on the displayed formula but whatever

[rain echo]

0

I guess I should learn that

it's worth it to be able to visualize the formulas in a discussion

[wintry steppe]

#linear-algebra message
here's an example of using it

learn by osmosis

@wintry steppe where are u from

why

your name reminds me a portuguese name

im canadian

[rain echo]

canada > US (does that break the offensive rules of the server?)

[wintry steppe]

it's the truth though