linear-algebra

[nocturne jewel]

Assuming you checked the input vectors form a basis, then yes

[rose umbra]

not quite sure, do i need to find Te,e?

so it will give me T

[nocturne jewel]

I mean, if the input vectors form a basis, then you're done, as you can represent any other vector as a linear combination of the basis vectors (since basis spans V)

so finding the representation of the vector under the input basis will tell you how it transforms

[wicked lion]

where?

from 1:30 onwards

it just shows 3d graph

[nocturne jewel]

literally right before the timestamp you gave me

[wicked lion]

no numbers lol

[nocturne jewel]

yeah.. he animated it, cause Grant animates everything

[wicked lion]

u mean this?

[nocturne jewel]

No literally right before the timestamp you gave

he animated the transformation

[wicked lion]

theres nothing

at 1:58

or 1:57

[nocturne jewel]

1:50

[wicked lion]

nothing lol

nothing

lol

are we watching the same video?

[nocturne jewel]

I watched the one you linked

[wicked lion]

there's nothing at 1:50

its just the graph

[nocturne jewel]

1:50 is the animation of $T[v]=\begin{bmatrix} 1&0&-1\1&1&0\1&0&1\end{bmatrix}^Tv$

[wicked lion]

oh

[stoic python]

moshill1

[nocturne jewel]

too lazy to retype the matrix, so just transposing it

[wicked lion]

i feel like

im stupid man

whats wrong with my brain

god

why do i just, lack math comprehension

lol

[rose umbra]

i still cant see it

[wicked lion]

it takes me so damn long to understand stuff

much longer than the average human i think

[rose umbra]

what do you mean by vector under the input basis

[nocturne jewel]

so if I have a basis for $\mathbb{R}^3, B={v_1,v_2,v_3}$, which need not be the canonical basis, then I can write any vector $w\in\mathbb{R}^3$ as $w\in span(B)\implies w=c_1v_1+c_2v_2+c_3v_3, c_i\in\mathbb{R}$

[stoic python]

moshill1

[nocturne jewel]

so the co-ordinate vector of w under B is $(w)_B=[c_1,c_2,c_3]$

[stoic python]

moshill1

[rose umbra]

right

[nocturne jewel]

so then applying the matrix of the transformation to (w)B, you get the transformed vector of w in the other basis, which we assumed to be canonical

also how to swap between bases

(I will be honest, I suck ass at change of basis so I can't properly explain it lol)

[wintry steppe]

change of basis is very important

[rose umbra]

whats the formula for it

[nocturne jewel]

It's the theorem I posted from my prof's notes

[wicked lion]

How can I calculate the initial velocity needed to fire a projectile such that it will fly in a straight line towards a given target?

[wintry steppe]

with #calculus

[stoic python]

squirtlespoof

[wintry steppe]

sounds right

just carefully write it out using the definition of the transpose of a map

how would you carefully prove that $f\circ i = f|_{W}$

[stoic python]

Terra

[wintry steppe]

both as functions defined on W

you'd show that they agree on all elements of W, right?

which is basically what you've done

so the transpose of i takes any function on V to its restriction to W

[stoic python]

squirtlespoof

[wintry steppe]

.

right

your explanation feels like you're not confident in the result

but you're not wrong anywhere

in fact you're completely correct lol

it could just be shortened quite a bit, you could probably get away with just writing something like

given $f\in V^*$, $i^t(f)=f\circ i = f|_W$, so $i^t$ takes functionals on $V$ to their restrictions to $W$

[stoic python]

Terra

[wintry steppe]

always be confident in your writing, especially if it's being graded

the former

in the latter, "restriction to W*" makes it sound like you're talking about functionals defined on the dual spaces

but they're really elements of those dual spaces

[runic frost]

Hi! i have a question. How would i form the diagnoal matrix of eigenvalues if the eigenvalue has a multiplicity greater than one?

So lets say i have a 3x3 matrix A with eigenvectors 2 and 3 and the characteristic polynomial is P(L) = (L - 2)^2(L-3) and the eigenspace for L1 has dimension 2 and eigenspace for L2 has dimension 1. How would i construct the diagonal matrix in this case?

[faint lintel]

well the diagonal would contain 2 2's

[hexed mural]

Question regarding taking a Cross product. If I have a vector V1 that is in a space defined by an orthonormal set of basis vectors in R^3. If I take a V1 (cross) with some other Vector V2 I get some orthogonal vector V3. Now with V3 is there a way calculate where this vector lands for any orientation? Meaning if I want to get the projection of V3 on any of the basis vectors I don't want to have to take a dot product with every combination after the fact.

[faint lintel]

so D_1,1 and D_2,2 would both be 2 @runic frost

and then D_3,3 would be 3

[runic frost]

What about the other vector that's constructed with the bases of the eigenspaces though?

[faint lintel]

wdym

[runic frost]

*matrix i mean

[faint lintel]

do you mean finding Q such that Q D Q^-1 = the original matrix?

[runic frost]

yea!

[faint lintel]

pick an order for your basis

and just write the columns

it's just like how you would write the matrix representation for any linear transformation wrt any basis

[faint lintel]

yea

[runic frost]

so like that?

[faint lintel]

yessir

exactly like that

and use a matrix calculator to verify that's right

[runic frost]

great

so then when i go to calculate A

it would just be PDP-1

and then if i wanted A^K , i'd just do PD^KP^-1

[faint lintel]

exactly

and that last result is one of the upsides of diagonalization

cause powers of diagonal matrices are super super easy to calculate

[runic frost]

yea, this isnt too bad

i like linear algebra, its a cool class now that i kind of understand it but the calculations make my head hurt sometimes

[faint lintel]

the concepts

The calculations

[runic frost]

exactly!

like i read up on eigenstuff a few days ago cause i was still confused

and now they make sense

[faint lintel]

I will say it can get much worse

[runic frost]

😭

[faint lintel]

if your class does Jordan Canonical Form

💀

[runic frost]

this is only an introductory lin alg course

[faint lintel]

ok you probably won't touch it then

[runic frost]

thankfully i get a break from linear algebra for a semester

then im back to it in the winter

[faint lintel]

it's basically the generalization of diagonalization

for when the polynomial splits

but the algebraic multiplicity isn't nessessarily equal to the geometric

and the calculations

[runic frost]

yea that scares me

[faint lintel]

I do not recommend

I did not do well on that quiz

Ok now I got a question of my own

for 2b

[runic frost]

i havent been doing bad in this class at all, im doing pretty well its just the professor has ruined it for me 😭, shitty textbook and incomprehensible lectures

[faint lintel]

shouldn't the answer be (1, -2)

[runic frost]

[faint lintel]

not (1, 2) like my book suggests

[midnight forge]

oof, i finished a sloppy proof for a linear algebra problem

[runic frost]

alr im gonna do some homework, my exam is tommrow 😭

i just finished writing up basically every i have to know

thanks so much @faint lintel

[nocturne oracle]

yw

[nocturne jewel]

If a normal operator P has eigenvalues of either 0 or 1, show that P^2=P. Do you need P to be normal for this..?

I dont see how P being normal is relevant

cause P[v]=0 -> P[P[v]]=P[0]=0 so P^2=P
P[v]=v -> P[P[v]]=P[v] P^2=P

[misty storm]

fellers, what does this mean?
The written portion reads "why the usual operations of..."

[faint lintel]

wait yoooooooooooooooooooooooooooooo
is matrix multiplication defined in terms of inner products?

like matrix multiplication of complex matricies

is each entry in the resulting matrix

an inner product of a row and column

[zealous junco]

no

[faint lintel]

damn

[zealous junco]

but

[faint lintel]

how do I do matrix multiplication for complex numbers

[zealous junco]

symmetric positive definite are

all inner product

[faint lintel]

what?

[zealous junco]

Every inner product, if you choose a basis right

then you can write that inner product as a SPD matrix multiplying the coordinate w.r.t that basis

and similarly every SPD matrix induces an inner product, and u can easily check that

but be carefule i dont think they are 1-1 correspondence, since say B = {u1,...un} is orthonormal w.r.t some inner product, then u could also choose for instance B = {sigma(u1),...,sigma(un)} the permutation of the original ordered basis to again get the same matrix, which is the identity

[faint lintel]

Hm ok

I was just thinking because like

Dot product is the inner product for real vectors

But you can also do matrix multiplication as dot products

So I was wondering if that generalized

[zealous junco]

oh shoot

i think I misunderstood what you meant

XD

yes then each entry is dot product of rows and columns

of the 1st and 2nd matrix, and yea dot product is a specific inner product

[faint lintel]

So if I have a matrix of complex numbers

I would do inner product of complex vectors?

[zealous junco]

then no unfortunately

only for reals

[faint lintel]

Shit

Ok neat

I thought I had found a cool generalization 😔

[zealous junco]

yea matrix multiplication just happened to have this somewhat neat way of representing

there is just so many ways you can think about it

[faint lintel]

Yea

[zealous junco]

i mean the simplest nontrivial example right would be [i i] [i i]^T and u get -2 but <x,x> > 0

[nocturne jewel]

W is the set of even functions from R to R, with typical operations of + and *

[misty storm]

thx

[winter harbor]

Ok so

This is clearly a linear transformation of complex vector spaces right?

Oh

Lmao

ok so

Can I tell my idea anyway?

Lmao

Just prove that if φ(v) = 0, that would mean that for every w in V

Then the inner product of v and w

Would be 0

You would have to prove

That if this was the case

Then v=0

And since φ is clearly a linear transformation of complex vector spaces

Then it would be an injection

Was this your idea?

Haha nice

:)

Remember the following thing

If you have an inner product space

This induces a metric

So like

In particular

<v,v>^2 = 0

But this implies that v=0

Because of the metric induced by the inner product

I mean

I just used that

Given an inner product space

Over the complex numbers

You can put a norm on it

In particular a metric

But whatever

Given by |v|=√<v,v>

And by the axioms of a norm

If this is 0

Then v=0

You could use this

[zealous junco]

hello

[winter harbor]

Anyway

[zealous junco]

yea all NVS are metric spaces

[winter harbor]

Good luck with the other problems

I mean

Knowing that inner product => induced norm => induced metric

Is really nice

You should keep this in mind

Because you can use this whenever you wnt

[nocturne oracle]

am i being dumb or is there a way to show this without plugging in the lin comb of basis vectors for x and y $\langle {x|y} \rangle$ = $\sum_{i = 1}^{n} \langle {x|x_i} \rangle\overline{\langle {y|x_i} \rangle}$

[stoic python]

gokugamer69

[nocturne oracle]

id rather not spend like half a page just working it out

[winter harbor]

Does make sense tho? With x_i I suppose you mean the i-th component of x in a basis B.

But you can't take the inner product of a scalar like x_i with x

But yeah

There's something similar to this

[nocturne oracle]

oh yes sorry

[winter harbor]

That is called the hermitian form or hermitian inner product

Suppose that we are in C^n

A C-vector space that is has complex dimension n

And I choose a random

Orthonormal basis

Then

Given an inner product $\langle \cdot , \cdot \rangle : \mathbb{C}^{c} \times \mathbb{C}^{n} \rightarrow \mathbb{C}$ in the following way.
\
\
For any given $u,v \in \mathbb{C}^{n}$, we have that $\langle u, v \rangle = \sum\limits_{i=1}^{n} u_{i} \overline{v}_{i}$

Try to prove this

Btw I am sorry

[stoic python]

MisterSystem

[winter harbor]

Ok

Notice how you are just summing the inner product

Of each component

If your basis is orthonormal

You can always do this

And represent any given inner product this way

You could also do this

For any general C-vector space that has finite dimension

And you choose an orthonormal basis

Same thing

But yeah

[nocturne oracle]

oh wait, because they are orthonorm, the norm of \langle {x_i|x_i} \rangle is just 1, and the rest of the parts go to 0. i forgot

i forgot i was working w an orthonorm basis

[winter harbor]

Yeah

Exactly

This is the reason why this works for even general finite dimensional C-vector spaces

If you fix an orthonormal basis

Same argument

Any inner product can be expressed in this way

And btw

(This is pretty much the reason why working with orthonormal basis is so useful when we deal with inner product spaces)

[nocturne oracle]

yeah ic ic

ty

[winter harbor]

Ah

Thanks

[nocturne oracle]

what

why r u saying thanks lol

[winter harbor]

Idk idk 😳

I guess I am used to saying thanks lmao

I am so used to not knowing how to properly respond to stuff

Being shy is a bitch

Ok anyway

Lmao

[nocturne oracle]

🥺👉👈

[crisp cloud]

is f induced by a matrix?

I want to say no, but someone correct me if I'm wrong

[wintry steppe]

are you asking if f is linear?

the answer is no

[crisp cloud]

cool, just checking

thank youuu

[crisp cloud]

so when I RREF this, this is, T would be onto, correct?

[dire thunder]

@crisp cloud what do you mean

[solid flower]

[viral sonnet]

hey everyone, I just came up with this result while trying an ex: kerf = imf

is that possible?

[wintry steppe]

sure, consider {0} -> {0}

[viral sonnet]

the function was not 0

😦

[native rampart]

Consider T:$R^2->R^2$ whose matrix is given by:
$\begin{pmatrix}
0 & 1\
0 & 0\end{pmatrix}$

[stoic python]

Buncho Drunk

[native rampart]

T(1,0)=(0,0) and range(T)=span{(1,0)}

[viral sonnet]

ah I see

thx a lot

[crisp cloud]

nevermind, I figured it out, thank you though!

[zealous junco]

for this question it is sus

what does $(S_i)^V = [(S_V)i]{S_V}$ even mean, isnt that just the ith standard basis vector

[stoic python]

Anticipation

[crisp cloud]

can someone show me how to do this

[lavish jewel]

find an orthonormal basis for the plane first

[crisp cloud]

could I say something like {( 1,2,3) (4, 5, 6) (3,-2,1)}?

[lavish jewel]

3,-2,0 is definitely not on the plane

[crisp cloud]

woops i meant 1 not 0

[lavish jewel]

are you just making it up?

[zealous junco]

ok they are just std basis vectors

[crisp cloud]

yes

because literally my teacher said "it can be anything" smh

[lavish jewel]

that would explain why it's all wrong 😛

use the definition of a plane

[crisp cloud]

how should I go about doing that?

[lavish jewel]

they're not wrong in that the plane has infinitely many bases, and infinitely many orthonormal bases

[zealous junco]

is polar decomposition useful for anything

[lavish jewel]

i've never seen it used directly

everything is SVD 😛 but idk about any nice case where it would be faster and more efficient to use the polar decomp directly

[zealous junco]

ye true just corollary

was wondering if it showed up anywhere in analysis of convergence or whatever

cuz it seems pretty general

[crisp cloud]

wait @lavish jewel so when I make a basis, would {( 2,3,0) (0, 4, 8) (3,-2,1)} be right?

[lavish jewel]

your basis will only have 2 vectors

[crisp cloud]

ooo.... wait that changes everything lol

[lavish jewel]

i suggest you first read the definition of a plane

[zealous junco]

I know by definition of matrix multiplication u can prove that

(AB)* = B* A*

but i was wondering is it also ok to prove it like this:

Assuming I proved that for orthonormal basis B,C if $A = [T]{B,C}$ then $A^\star = [T^\star]{C,B}$. This means that $[(TU)^\star]_{B} = [TU]B^\star$ and also $[(TU)^\star]{B} = [U]_B^\star[T]_B^\star$

[stoic python]

Anticipation

[zealous junco]

where this is only true if B is orthonormal, but thats enough because for any basis u can represent any matrix as a linear transformation

because the thing is (TU)* = U* T* is easy to prove using inner products

[native rampart]

That works

[zealous junco]

nice, thanks!

[wintry steppe]

I want to prove that the set of differentiable real-valued functions f on the interval (0,3) such that f'(2) = b is a subspace of R^(0,3) if and only if b = 0.

Here is my attempt: Let U = { f | f : (0,3) -> R, f' exists, f'(2) = b}

First assume that b = 0. Then define f(x) = 0, so f'(x) = 0 and f'(2) = 0, so 0 in U.

Now let f, g in U. We f(x) + g(x) = (f + g)(x) by definition, and that (f(x) + g(x))' = f'(x) + g'(x) = (f' + g')(x), and also f'(2) + g'(2) = 0. So f' + g' exists and (f + g) in U.

Then let a in R. So (af)(x) = af(x) by definition, and (af(x))' = af'(x). Then af'(2) = 0, and so U is a subspace if b = 0.

Now we go the other way and assume that U is a subspace. This means that 0 in U. So let f(x) = 0, and f'(x) = 0. This means that f'(2) = 0 and f'(2) = b, which implies that b = 0. And we're done both ways.

[dusky epoch]

feels wordy

[wintry steppe]

Okay, first can you tell me if it works?

[dusky epoch]

seems like it does

[wintry steppe]

Thanks, now how would you make it less wordy?

[dusky epoch]

i probably wouldn't bother giving an explicit name to the zero functino

function*

idk

im a little too tired to come up with a less wordy prof

proof*

[wintry steppe]

But you need the zero function?

[native rampart]

Just do it in one step instead of doing f+g and cf as 2 separate steps

Let f and g be in U,then f'(2)=0,g'(2)=0 implying (cf+g)'(2)=0 for all c in F,f in U,g in U

Put c=-1 and g=f to get that 0 is in in U

The other direction doesn't need any changes

[wintry steppe]

I have a question

How do I prove that f(x) = 0 is the "0 in U"?

That it couldn't be something else

Should I proceed by contradiction or something?

[native rampart]

Pick a function g in U

Note that (g+f)(x)=g(x) for all x if you take f(x)=0

Implying f is the identity wrt that operation

[wintry steppe]

That operation?

You mean addition in the set?

[native rampart]

Yes

[wintry steppe]

I want to show that if U = {(x,0,0) in F^3 : x in F} and W = {(0,y,0) in F^3 : y in F} then U + W = {(x,y,0) : x,y in F}

Here's my attempt: First we show that U + W is a subset of {(x,y,0) : x,y in F}

Let u in U, w in W and then u = (u_1,0,0), w = (0,w_1,0). Then by definition, U + W = {u + w} = {(u_1,w_1,0) : u_1,w_1 in F}

This is clearly a subset of {(x,y,0) : x,y in F}

Next we will show that {(x,y,0) : x,y in F} is a subset of U + W, and that will give us equality

We observe that if we pick x = u_1, y = w_1, then we are done.

[native rampart]

Works

[wintry steppe]

Thanks

[wicked lion]

Does anyone know how I am supposed to predict the velocity of a bullet moving from one starting vector to an ending vector, given a static speed value and a gravity value?

[native rampart]

Kinematics?

[wicked lion]

If I'm starting at v(0,0) and I want to end at another vector, but I can only move at a certain rate of time, with gravity integrated into it

Well I'm not takinginto account fluid or air resistance lol

This is just a simple parabola, but with vectors and gravity integration

[quartz compass]

well the horizontal velocity remains constant and all that changes is vertical velocity which you can model as just how much it gains from falling, which is just gt

[nocturne jewel]

$\vec{v}(t)=\begin{bmatrix}v_{ix}\v_{iy}+gt\end{bmatrix}$

[stoic python]

moshill1

[verbal vessel]

Can anyone help me with proving this equation?

$sqrt(1-norm(u)^2)sqrt(1-norm(v)^2)<=1-inner_product(u,v)$

[stoic python]

Alireza

[limber sierra]

$\sqrt{1 - \norm{u}^2}\sqrt{1 - \norm{v}^2} \leq 1 - \langle u, v \rangle$

[stoic python]

Namington

$\sqrt{1 - \norm{u}^2}\sqrt{1 - \norm{v}^2} \leq 1 - \langle u, v \rangle$

[limber sierra]

is this what you wanted?

[verbal vessel]

Yes exactly

I'm having difficulty proving it

[limber sierra]

maybe im missing something

the left hand side will sometimes be imaginary

how are you defining ≤?

[verbal vessel]

Ah I missed some information, norm(u) and norm(v) are <= 1

[limber sierra]

ah

[errant minnow]

That looks similar to $\sqrt{\norm{u}^2\norm{v}^2} \leq \langle u, v \rangle$

[stoic python]

Meow0thing

[limber sierra]

yeah my immediate temptation is to rewrite the RHS using one-argument-linearity

[errant minnow]

I think something like that is one of the more standard proofs

[limber sierra]

and try something like that

[verbal vessel]

That looks like the opposite of Cauchy-Schwarz

[errant minnow]

ah oops but yeah that shape was pretty familiar to me

Maybe you can somehow formulate it into that?

If yes then you'd have solved it

[limber sierra]

eh

[errant minnow]

That'd be my instinct

What is one-argument-linearity?

[limber sierra]

that might actually not be the best approach looking back

cauchy schwarz should work here with some massaging

via inner product properties

but i dont think its very clean

[errant minnow]

Hmm... is it ok for me to post my question now, or is the previous question here still somewhat active

Alireza is not replying, he's probably trying out that way as we write

[verbal vessel]

Yeah, but I haven't gotten to the answer yet, it seems difficult

You can ask if you want, you might get the answer quickly :)

[errant minnow]

Ok I'll formulate the question then

Might take a while too

Topic: intrinsic rotations and extrinsic rotations (euler angles)
Question: Why is it that intrinsic rotations are basically extrinsic rotations, just in reversed order?
Details: In one of my courses, we were tasked with showing that rotations are not commutative, by performing intrinsic rotation. This task is solved, but has only sprung forth questions, like "how does intrinsic / extrinsic rotation work" "why doesn't it work like what we thought it should work like" and more. We thought that since the y axis got rotated away, you would have to determine the new y axis, and then change the matrix to the changed bases, such that the rotation now rotates around the y axis. Instead, they just changed the order of rotations, and somehow that works. We tried googling and wikipediaing it but none of the answers really explained it.

Like, the difference between intrinsic and extrinsic rotations is that intrinsic rotations basically rotate the coordinate axis with them and all further rotations are done on the rotated axises

While with extrinsic rotations you keep rotating on that "global" axis, like, the one that would be if they were "unaffected" by the rotations

Like, why does simply changing the order of rotations already achieve that

I'm not too sure how to get across my confusion or my question. I can draw some pictures to illustrate what is happening in each case I just don't know why or where to look

I'm not even sure if it's linear algebra, but it certainly is matrices

Ima do some pictures

Intrinsic rotation in order of black red blue

extrinsic rotation in order of black red blue

Note how the red turn thingy that signifies the second rotation is at different places

Why does simply changing the order of matrices change whether it's intrinsic or extrinsic? That is the question

And how are they even correlated

Ah wait in the last pic I think y' also needs to change

Yee

... Ping me when you reply, I'll take a break for now and listen to some music

(Thanks in advance)

Mm doesn't look like it's working. Whatever, I'll just try asking the prof that then or whoever else I can contact. Or maybe the solution will explain it.

[wintry steppe]

Anyone knows the proof or where I could find a proof for the theorem there

[quartz compass]

the first thing comes from looking at the diagonal of the matrix having all the lambda terms on it, the second thing comes from plugging in lambda=0

[wintry steppe]

@quartz compass Could you explain aa little more please im really dumb

what equation do you use to plug lambda=0?

[prisma pier]

does anyone have any ideas for how to prove that involutions are diagonalizable without using stuff involving minimal polynomials or cayley-hamilton

I managed to do it for 2x2 using stuff with adjugates, but I think it'd get really messy and undoable if I tried the same method for nxn

[prisma pier]

nvm I got it

[snow jetty]

Hello. I am trying to understand why is the multiplicity of an eigenvalue in a minimal polynom corresponds to the size of the largest Jordan block. Particularly I did not get this argument.

Don't they mean that if I multiply a matrix by an inversible matrix the minimal polynomial does not change? but that's not true...

[gray dust]

why y'all deleting msgs @last mist @snow jetty

[snow jetty]

not to block "open" questions 😄

[last mist]

^^^

[gray dust]

it's not necessarily blocking, and feel free to ping helpers if you've not received a response

[snow jetty]

ok 🙂 in summary: R^n can be a vector subspace of R^m if n ≤ m as it is not empty (zero vector) and stable to linear combinations; very formally it would rather be a subset of R^m isomorph to R^n)

[snow jetty]

Can you explain me why?

I mean

Ok, it is due to the Dunford decomposition, but...

[stoic python]

mors

[snow jetty]

Ok I will try this

[lavish jewel]

looks kinda like this?

you can check out neumann series

[snow jetty]

I will check that out 🙂

[next vapor]

Say V is a finite dimensional vector space over R, let T : V → V be a linear operator,
W a non-trivial T-invariant subspace of V such that W itself has no proper non-trivial T-invariant subspaces. Show that dim(W) is 1 or 2

been stuck on this for a while now could i get a hint

[snow jetty]

let me check that out

[next vapor]

I know the only subspaces of 1 dimensional vector spaces is the trivial one

so thats a possibility out

but how do u tackle the dimension 2 case

[snow jetty]

... and the subspace itself?

[next vapor]

yea

[dusky epoch]

delirium i think itd be better to show instead that dim(W) > 3 is impossible

[next vapor]

but thats also trivial

[dusky epoch]

consider a T-invariant subspace W such that dim(W) > 3, find a smaller T-invariant subspace W' strictly contained in W

showing dim(W) > 3 is impossible directly gives you dim(W) ≤ 2

you can make examples for which dim(W) = 2 such as a rotation in R^2

[next vapor]

okay, lemme give it a go

[snow jetty]

I think I have it, but there is one nuance I am not sure about

Could I apply the geometric series (with all the limit things) to the matrices? Like I did here

[sacred crescent]

If N is nilpotent why do you need an infinite sum?

[snow jetty]

I want to understand this assumption

But if I want to say that this polynomial is actually an inverse, I have to make sure that it's the same thing as (I + tN)^-1

That's why I use the geometric series thing

[dusky epoch]

expand $(I+tN)(I - tN + t^2N^2 - \dots + (-1)^{n-1} (tN)^{n-1})$

[stoic python]

Ann

[dusky epoch]

i assume n is the nilpotence index of N

[sacred crescent]

^ there is no series here, and you plug in t = 1/lambda

[snow jetty]

indeed

[dusky epoch]

that second matrix is just (1/λ)I

[snow jetty]

Yes, that's stupid

It's wrong in fact

This makes more sense

[next vapor]

okay so if i have a W with dim>3 or >=4 then it has a basis {v1,...v_n} with n at least 4, so i can just take that basis with 1 fewer vector and get another T non invariant vector space of dimen at least 3 which contradicts the assumption

this seems too simple tho

[dusky epoch]

you cant just take a random basis no

you need to consider the characteristic polynomial of T restricted to W

[next vapor]

hmm alright

[snow jetty]

So the proof would be just this?

[dusky epoch]

guess so? idk

i'm a little busy with class shit at the moment sorry

[snow jetty]

hm...

There should be only one generalised eigenvalue for the restriction of T to the subspace W

If I have dimension 1 then the algebraic multiplicity in the characteristic polynomial will be 1, so 1 eigenvalue and 1 linearly independent eigenvector for this eigenvalue

If I have dimension 2 then I can take just one vector and it still stays invariant

Strange, no idea (i'm no way an expert, just studying 😅 )

[snow jetty]

so this question is actual

[snow jetty]

You must show that the vectors are linearly independent and span R^2

That means, that you can express any vector (x, y) as a(1,1) + b(2, 3) where a and b are scalars

Which means to solve a system of equations a + 2b = x; a + 3b = y

Then you will have a solution like a = "a formula with x and y's" and same for b

If you manage to do it you prove it's the base; to prove they are linearly independent take x, y = 0, plug into the obtained formulas and if it implies a = 0 and b = 0 you show they are linearly independent.

In fact in R^2 it works if and only if the vectors are not colinear, so for {(1,1), (2,3)} it's ok but for {(1,1), (2,2)} it won't work (you can't express for example (2,3) as their linear combination)

check out some youtube for this

if you want to do a matrix thing solving the system is same as finding the solution for a matrix equation ([1,2] [1, 3])*(a,b) = (x,y)

[wintry steppe]

Hello guys, Ive to prove that this is a subspace from R3

Thats what Ive. Should I just replace the Uc with R3?

[next vapor]

are you sure you're asked to prove it's a subspace or work out the values of c for which it's a subspace?

[earnest vessel]

You wrote that $x_1 + x_2 + x_3 = c$ and $y_1 + y_2 + y_3 = c$ implies $x_1 + x_2 + x_3 + y_1 + y_2 + y_3 = c$, which does not make a lot of sense, right?

[stoic python]

edelopo

[wintry steppe]

Well it says I have to look if it is a subspace from R3

[wintry steppe]

I looked at a youtube video. It did not had the c, thats why I just added the c

so is it maybe c^2?

[earnest vessel]

It says "Find out for which values of c the following set is a subspace of R^3"

Why do you think it is c^2?

You are adding both equations, right?

[wintry steppe]

so it is c+c right?

[earnest vessel]

Yep

[wintry steppe]

oh but do i have to replace the Uc with R3 because it wants the subspace of R3

[earnest vessel]

No, what they want you to find out is whether U_c, which is a subset of R^3, is actually a linear subspace

Do you know the definition of linear subspace?

[wintry steppe]

no unfortunately not

[earnest vessel]

Well, then you should definitely look that up hahaha

[wintry steppe]

But are my answers right?

[earnest vessel]

No, they are not

[wintry steppe]

D:

not even one?

[earnest vessel]

You didn't give one answer, though

You gave three arguments which form a proof that U_c is a subspace of R^3 for all c

But the arguments are wrong

And so is the conclusion

[snow jetty]

You are asked for which c does it work

[wintry steppe]

with arguments you mean the calculation i used right

[earnest vessel]

Yeah

[snow jetty]

It is obvious, you add two of such (x1, x2, x3) and (y1, y2, y3) and the result should satisfy as well the relation

The addition is defined term by term

[wintry steppe]

so x,yec?

Im a bit confused

[earnest vessel]

Go look up the definition of linear subspace

Otherwise it won't make sense

[wintry steppe]

ok

[snow jetty]

Suppose (x1, x2, x3) and (y1, y2, y3) satisfy the relation x1 + x2 + x3 = c and y1 + y2 + y3 = c. Then (x1 + y1, x2 + y2, x3 + y3) must satisfy (x1 + y1) + (x2 + y2) + (x3 + y3) = c, otherwise it's not a subspace. That means "the law of internal composition a.k.a sum is well defined

you also have to prove that the subset is not empty, and the 0-vector must be in the subspace

I think now the value of "c" is more than obvious

[wintry steppe]

but thats what i also did in 2.

[wintry steppe]

.

[nocturne jewel]

you said c+c=c, which is not true

[snow jetty]

c + c = c if and only if c = 0

the question is "for which "c" is it a subspace"

"für welche" = "for which" si jamais

[nocturne jewel]

yes that matrix will help determine independence and span, however independence should be obvious

since (2,3) != k(1,1)

[snow jetty]

if you manage to get it in the REF it's done, the range is 2 so finished for today

It depends on how it was presented in your course

In fact the assertion is way too obvious

Another argument is that "two linearly independent vectors in a 2-d space are for sure a base, by definition of dimension"

[wintry steppe]

I think the resolution was bad and i made it better

there is no c+c=c

[nocturne jewel]

Ok, why do you have 0,0,0 are in U_c?

when you dont know if the zero vector is in U_c

in all that working, you have not said what c needs to be for it to be a subspace

[snow jetty]

in any subspace the null vector is present

[nocturne jewel]

Im aware.

[snow jetty]

otherwise it's not an abelian group so not a vector space by definition

[wintry steppe]

but i thought if x is 0 then c is zero then Uc has a zero vetore

[nocturne jewel]

right, so write out that c=0

[wintry steppe]

ok

[nocturne jewel]

if you keep c then you are asserting that U_c is always a subspace

which it clearly isnt since 0 isnt in U_1

since 0+0+0 != 1

[snow jetty]

suppose c = 0 it works otherwise it doesn't

[nocturne jewel]

Yes, that;s what I just said

[wintry steppe]

so if c=0 i dont have to write it down anymoreright?

[nocturne jewel]

yes, or if the question is determine which c makes U_c a subspace, at the end explain why U_0 is the only subspace

[wintry steppe]

So i have to remove the c's from 2. and 3. because c=0

[nocturne jewel]

yes, assuming the question is to also then show it's a subspace

[wintry steppe]

the question is to show that c element of R is a subspace from R^3

[nocturne jewel]

c is a scalar, not a vector space, let alone a vector

[crude falcon]

If I want to find a subspace W with dimension2 such that dim(W intersection V) = 1, and dim(V) = 2

if I make the vectors of span of W the same of V, then the intersection would be 1 right?

[snow jetty]

If you take have W = span{v1, v2} and V = span{v1} then the intersection will be of 1 dimension

[next vapor]

okay so if i consider the characteristic polynomial of T_w here its gonna have degree of at least 4, and if a lambda is a root of that poly i can get out an eigenvector in W and i know the span of that eigenvector will a T invariant space, but this might not work for even degree right cuz it might not have any roots in R

[slim gyro]

could anyone explain why we care about knowing the row/column spaces of a matrix? i don't really understand what they tell us about the matrix

[lavish jewel]

they tell you which types of matrix products can be inverted exactly

[slim gyro]

wait i dont see the relation between those two

[lavish jewel]

if , in Ax = b, b is in the column space of A, then you can find at least 1 exact solution

the exact solution would be a vector x in the row space of A, plus any arbitrary linear comb. of vectors in the nullspace of A

similarly with A^T, but now using the column, row, and left null spaces

this means you can actually invert matrices that are not square in some cases

[slim gyro]

ohhh alright

are there any connections between the column/row spaces of a matrix and the fact that every matrix can be interpreted as a linear transformation? im not sure how those two concepts are related and it's what im most curious about

[wintry steppe]

column space of a matrix is the image of its associated linear transformation

[slim gyro]

ohh nice

[lavish jewel]

and the row space is its domain, so to speak

what is not in the row space, is in the null space

and gets mapped to 0

since the row space and the null space are orthogonal complements of each other

[wintry steppe]

is there a name for that part of the domain (that is not mapped to null space)

coimage or something

[lavish jewel]

codomain?

or coimage

i dont remember

[wintry steppe]

coimage is a quotient tho

[lavish jewel]

let me look it up

[slim gyro]

also, just to clarify, the column space is the image of the entire domain of the linear transformation?

[wintry steppe]

yea

[slim gyro]

cool

[lavish jewel]

is it preimage?

[wintry steppe]

preimage of codomain except {0} lol

[lavish jewel]

mhm

[slim gyro]

is there a way to figure out the image of any subset of the domain using the fact that the column space is the image of the whole domain?

since it's easy to determine the column space of a matrix

[wintry steppe]

the image of a subset could be almost anything

[stable kindle]

i'm not sure there is

idk much tho

[lavish jewel]

if you can express that subset of the domain in terms of the right singular values of the matrix, yes

the right singular values are a basis for the row space and the null space

or in other words, the SVD of a matrix reveals all this stuff directly. what gets mapped where and how

[slim gyro]

looked up right singular values and they seem to be related to eigenvectors? im a bit far from that but maybe in a few weeks i can really learn what all this means

[lavish jewel]

not directly

[slim gyro]

since ill have learned eigen stuff by then

[lavish jewel]

the singular vectors are related to the eigenvectors of A^T A and A A^T

[slim gyro]

yea, that's what came up when i googled, are singular values different?

[lavish jewel]

since a generic matrix A may not have eigenvectors

but all matrices have an SVD

[slim gyro]

ohhh ok

[next vapor]

could someone double check this if possible? idk if im on the right track here <@&286206848099549185>

[next vapor]

this is the problem

[wintry steppe]

Over R every irreducible polynomial is of degree 1 or 2

[next vapor]

yea but that doesnt guarantee me real roots for degree=4 for instance so the argument fails there

like i can have 2 complex conjugate eigenvalues right

so it works for the odd degree case but not sure how to tackle the even degree

[wintry steppe]

in even degree it might just have subspaces of degree 2, i don't remember how to prove tho

[next vapor]

what has subspaces of degree 2?

u mean dimension?

[wintry steppe]

yea dimension

[next vapor]

so if the characterstic polynomial is of even degree then all non trivial subspaces are of degree 2?

[wintry steppe]

not necesarily, there could be some of degree 1

You know that every real polynomial can be factored into factors of degrees either 1 or 2 right?

[next vapor]

yea

[wintry steppe]

i think you can do this

Let v be a non zero vector of V

then if V has dimension n, (v, Tv, ..., T^n v) is linearly dependent

so there are coefficients not all 0 such that a0v + a1Tv + ... + anT^n v = 0

since not all coefficients are 0, there is an ai that is non-zero such that no higher term coefficient is 0 (for example an if an is not 0)

So aiT^i v + ... + a0v = 0

and this i is greater than 0 cuz since v is non-zero then if i=0 then would have a0v=0 which would mean a0 =0 which would contradict linear dependence

So we can factor this

You get 0 = c(T-y1I)...(T-ykI)(T²+j1T+h1I)...(T²+jmT+hmI) v

since v is nonzero that means one of the factors is not injective..

[coral geyser]

this was calculating determinate of A

can someone tell me where i went wrong

answer is detA = 20

i got det A = -20

WAI

WATI

IM STUPID

WAIT I THINK I SEE IT

[next vapor]

okay you kinda lost me carla lol

i dont see how to use this to find the dimensions of the subspaces of the invariant subspace

[next vapor]

also could someone explain where this falls apart? Ann said I cant just take a random bases but i dont understand why

[next vapor]

i gotta go but im still kinda stuck, if anyone can offer a hint please ping me

[dusky epoch]

if you have W = span{v_k | k = 1,2,...,n} a T invariant subspace, where the v_k are just whatever that makes a basis, then in no way can you claim W' = span{v_k | k = 1,2,...,n-1} is also T-invariant

i mean, you can't say that $Tv_1$ necessarily has a zero coefficient on $v_n$ in its decomposition!

[stoic python]

Ann

[dusky epoch]

concrete example: R^6, standard basis, Te_2 = e_2 + 8e_4, Te_k = e_k for k = 1,3,4,5,6
W = span{e1, e2, e3, e4} is T invariant
but span{e1, e2, e3} is not

@next vapor

[next vapor]

i see

[next vapor]

am i on the right track here?

[dusky epoch]

kinda

every real polynomial factors into linears and irreducible quadratics

[next vapor]

yea

but im unsure how to tackle a charc polynomial of the form, say (x^2+1)(x^2+3)

how do i go about finding the dimensions of the subspaces of something like that

[dusky epoch]

you can pass to the complexification of your space

and show that if v is an eigenvector of a real matrix A with eigenvalue lambda then conj(v) is also an eigenvector but with eigenvalue conj(lambda)

and so span{real(v), imag(v)} is A-invariant

or T invariant whatever

[last egret]

Hi everyone.. could anyone answer me why linear algebra is useful in group theory?

[next vapor]

hm im probably being dumb but i cant figure out why that implies span{real(v), imag(v)} is T invariant

[wintry steppe]

Let V vector space. Let G_0 be empty list then its lin indep. For every ordinal w, if G_w is linearly indep. and not basis, then can choose a v_{w+1} not in span of G_w and define G_{w+1} as G_w union {v_{w+1}} which is linearly independent then. For a limit ordinal w, we define G_w as union of all previous G's. If all G's before it are linearly independent so must be G_w clearly i think. Let {G_a} be the well ordered set of all of these (linearly independent) G's up to at most the cardinality of 2^V. Then the union B of all of them must still be in this set. So this B should be a maximal linearly independent subset of V, because a subset with higher cardinality is linearly dependent, so B is a basis.

Is this correct?

[sleek spruce]

this is a determinant

i can not

get my brain to understand

why it's ad-bc

it's slanted, so if i unslant it, it would just be a * d

as the area of the parallelogram or rectangle

the length of the base is A and the height is d

so why do we subtract bc?

[stable kindle]

the long side might be d but you want the perpendicular height

it probably works out, but i don't think tilting it is the nicest way

[sleek spruce]

but even if you unslant it

isn't it the same area

[stable kindle]

well

hmm

well first you have to set the short side down on the y-axis

[wintry steppe]

draw some triangles

[stable kindle]

and that changes the height

[sleek spruce]

i have and

i can't get my mind

ive tried multiplying unit vectors

with linear transformatons

and it doesn't add up

i had the unit square as 1 and then a linear transformaton of 1,2 and 3,1

and i got -3

[stable kindle]

after you set the short side down and compensate for that, you can unslant it or slant it all you want and you're right the area won't change, but you have to compensate for the tilt first

[sleek spruce]

which i know is cause of orientation but i dont understand it

i can't get my head to understand but can someone point out what bc is

[stable kindle]

it just is

[sleek spruce]

my brain wants to keep subtracting d by b

oh

i think i get it

lol

sorry

but to be clear

is it because the origin height to (c,d) is different than the height to (a,b)

[stable kindle]

i don't understand, sorry

[sleek spruce]

maybe im just dumb

lol

nvm i think it's trivial

[stable kindle]

no, they're good questions

[sleek spruce]

they're the same height i just looked

because it's b+d

[stable kindle]

i don't understand what you're doing but ok

[sleek spruce]

like

im unslanting it

if i unslant it

b would be 0

so if i do that, then the area would just be a*d

but then i don't understand why if i have some vaue at b

it becomes ad-bc

because the height would still be the same on both sides

[stable kindle]

well you seem to be using height to mean one of the sides

but you don't want that, you want the perpendicular height

[sleek spruce]

yes

why

[stable kindle]

you want the total height, really

so if you increase b

[sleek spruce]

i have a bad foundation on geometry lol sorry

[stable kindle]

the bottom left corner stays at the origin

[sleek spruce]

but why is it that we cant use the value of y

and instead need the perpendicular height

[stable kindle]

well it's like

[sleek spruce]

i view area as like

the (c,d) vector going a length

[stable kindle]

if you have a triangle with two sides that are both 1, it can have any area from 0 to 0.5, but if you have a triangle with one side that's length 1 and then a height perpendicular to that side that's length 1, you know the area has to be 0.5

and increasing b increases the perpendicular height here? i think?

[sleek spruce]

is it a viable idea to find the distance

of (c,d)

and multiply that by a

or the distance of (a,b)

[stable kindle]

there's no particular reason why that should do the thing you want it to do, i think

that i can see

[sleek spruce]

cause i can do area like

3*2

like length of 3

and length of 2

[stable kindle]

but that's perpendicular

[sleek spruce]

covers an area of 6

[stable kindle]

that's for perpendicular things

and you're not using anything that's definitely perpendicular that i can see

[sleek spruce]

oh

i think ill just think of bc

as like

accountability for the slantness

i guess

my ape brain can't comprehend it

[stable kindle]

i don't know what you're doing at all in any way

so i can't say whether that's a good idea or not

[sleek spruce]

i was trying to understand why i couldn't just do

a*d

[stable kindle]

yeah

[sleek spruce]

and not subtract bc

[stable kindle]

but like

[sleek spruce]

but you said it's cause of slantness

and i was wondering why the slanted height isn't hte same as a normal height cause

[stable kindle]

i don't think that's what i said exactly

[sleek spruce]

if you go like x length, isn't the area the same regardless

[stable kindle]

look

firstly

why should it be ad

[sleek spruce]

because you have base a

and height d

[stable kindle]

no you don't

[sleek spruce]

wdym

[stable kindle]

if you set the side that's just a so it's flat on the x-axis, the perpendicular height won't necessarily be d

why should it be

[sleek spruce]

because the lengt from the origin is base a

and height d

[stable kindle]

why should the height be d

[sleek spruce]

because that's the length from the origin

[stable kindle]

no it's not

wait, even if you untilt it so the short side is on the x-axis the side won't even be a

the length of the short side is sqrt(a^2 + b^2), by pythagoras

and the length of the long side is sqrt(c^2 + d^2)

[sleek spruce]

can you multipoly those two

and get the area

[stable kindle]

no

because they're not perpendicular

if they were perpendicular i could

[sleek spruce]

wtf

bruh

[stable kindle]

if it was a rectangle

ok let's try this i think i have a simple but long proof

hear me out?

[sleek spruce]

yea

[stable kindle]

ok so imagine you go from the top-right corner and you drop a vertical line down to the x-axis, and a horizontal line left to the y-axis

you have a rectangle, right?

[sleek spruce]

yea

but you'd have some area missing

[stable kindle]

so what if we just cut out the rest of the bits

so the big rectangle has area (a+c)(b+d)

because the vertical line hits the x-axis at (a+c, 0)

and the left side of the rectangle is at (0, 0)

[sleek spruce]

why isn't it

b+d,0

wouldn't the tew new formed vertices be

(a+c,0)

and

oops

(0,b+d)

[stable kindle]

yes

so (0, b+d) is the top left corner

so (a+c) is the width and (b+d) is the height, right?

[sleek spruce]

yes

[stable kindle]

ok

so now we want to take pieces away so we have the parallelogram

so let's drop a vertical line from the bottom right corner, now

[sleek spruce]

so just

subtract the triangles?

[stable kindle]

more or less, but there's a bit more to it

ok just follow pls

so dropping a vertical line from bottom right corner

you have a triangle underneath the parallelogram, with a right angle at (a, 0)

yes?

[sleek spruce]

yea

[stable kindle]

ok so the area of that will be 0.5ab

half base times hieght

[sleek spruce]

yea

wait wouldn't it be

ab/2

oh

[stable kindle]

same thing

[sleek spruce]

ok

didn't see b

sorry

[stable kindle]

and then if we take a horizontal line rightwards from the bottom right corner also, then we get a triangle to the right of the parallelogram that hits the right side of the big rectangle

with a right angle at (a+c, b)

[sleek spruce]

yea

d

well

yea

[stable kindle]

so the height will be (b+d) - b = d

and the width will be (a+c) - a = c

so the area of this right triangle will be 0.5cd

[sleek spruce]

wait

yes

yea

[stable kindle]

ok

so we've cut out two triangles like that, but if you visualise clearly there's still a bit below the right triangle and to the right of the bottom triangle that we still need to cut out

it will be a small rectangle in the bottom right corner of the big rectangle

[sleek spruce]

yea

[stable kindle]

ok so the area of this small rectangle will be base * height: ((a+c) - a)((b+d) - b) = (c)(d) = cd

[sleek spruce]

yes

[stable kindle]

yes?

ok

so now we've done all the stuff that's to the bottom right of the parallelogram

so we just need to do the same thing on the top left

so the triangle to the left of the parallelogram will have area 0.5cd

the triangle above the parallelogram will have base (a+c) - c = a and height (b+d) - d = b, so area 0.5ab

and the small rectangle in the top left corner of the big rectangle will have area ((a+c) - c)((b+d) - d) = (a)(b) = ab

so now we've cut out six pieces from the big rectangle, and that just leaves the parallelogram

so the area of the parallelogram = big rectangle - bottom triangle - bottom right rectangle - right triangle - left triangle - top triangle - top left rectangle

= (a+c)(b+d) - 0.5ab - 0.5cd - cd - 0.5cd - 0.5 ab - ab = (ab + ad + bc + cd) - 2ab - 2cd = ad - ab - cd + bc which is completely wrong, what have i done

[sleek spruce]

lol

i think i get your steps tho

i want to do this on my notebook because

i kind of lost you when you were doing the

coordinates lol

[stable kindle]

yeah it'll be easier with pictures and actually drawing the lines and things

[sleek spruce]

tyty

[stable kindle]

(a+c)(b+d) - 0.5ab - 0.5cd - bc - 0.5cd - 0.5ab - bc = (ab + ad + bc + cd) - ab - 2bc - cd = ad - bc

[sleek spruce]

ii got cb+cd+ab

as the triangles

[stable kindle]

ok so that's the right answer now

yeah that's definitely not quite right

ok so

[sleek spruce]

the bottom right triangle is .5ab

[stable kindle]

basically

[sleek spruce]

and the top triangle

is .5ab

[stable kindle]

i got the rectangles wrong

[sleek spruce]

and the left triangle is cb+cd

[stable kindle]

so both the small rectangles are bc

which i just got completely wrong for some reason

i am very tired

the area of the parallelogram:
= big rectangle - bottom triangle - bottom right rectangle - right triangle - left triangle - top triangle - top left rectangle
= (a+c)(b+d) - 0.5ab - 0.5cd - bc - 0.5cd - 0.5ab - bc
= (ab + ad + bc + cd) - ab - 2bc - cd
= ad - bc

[sleek spruce]

ty

gonna

work it out and see if it matches lol

[stable kindle]

so

[sleek spruce]

can i ask how it's (a+c)(b+d)

for the parallelogram

[stable kindle]

the area of the big rectangle that you're cutting bits away from is (a+c)(b+d)

[sleek spruce]

but why

the top right edge is (a+c,b+d)

[stable kindle]

well drop a vertical line from the top right corner and you have (a+c, 0), right?

[sleek spruce]

oooh

[stable kindle]

as the bottom right corner of the rectangle?

[sleek spruce]

yea

i thought it was parallelogram

but it's the big rectangle

got it

tyty

[stable kindle]

yeah

i mean if i could just get the parallelogram that way there'd be no point lol

but anyway i just realised that this was all entirely pointless and you probably didn't need to be convinced that this worked, you wanted to see where it all came from

which this doesn't really help with

because it's not exactly transparently insightful

it is in fact opaque as cardboard

[sleek spruce]

i need to see the proof yea

then i'll be good

also thanks whoever changed my name

cross star

[stable kindle]

oh well this is definitely a proof

so if that's all you want

[sleek spruce]

yea

just needed some sort of explaination

then im good

for now

[stable kindle]

but anyway yeah they change your name if it's too hard to type, because then ppl can't ping you

but yeah if you find that this doesn't actually explain anything, because it doesn't, come back later and ask someone else and they'll probably give you a better answer

[sleek spruce]

bruh

i got it

thank you

lol

[wintry steppe]

Let V a vector space, P the set of linearly independent subsets of V partially ordered by inclusion. Let C be a chain of P. Then its union must also be linearly independent i believe. That means every chain of P has an upper bound in P, so by Zorns lemma there is a maximal element B. If B was not basis then let v not in span of B then B u {v} linearly independent contradicts that B is maximal. So B is a basis.

Is this correct?

[prime drum]

That seems correct to me, but i'm not exactly an expert with this subject

[stable kindle]

i think it's good??

[wintry steppe]

im sus cuz my teacher said proving this for infinite dimensional is hard but this is such a simple proof lol

[prime drum]

well that doesn't exactly say it works for infinite dimensions

[wintry steppe]

why not?

i said V any vector space

[prime drum]

oh cause i skipped a word while reading, ignore me lmao

[winter harbor]

https://youtu.be/zRcQnXvJGGY

Maybe this should be helpful

But yeah

You are pretty much correct

[wintry steppe]

yay

[winter harbor]

Just notice the following

You are implicitly using the fact

That the set of linearly independent subsets of V

Is non empty

This is true ofc

[wintry steppe]

yea it clearly is

[winter harbor]

I mean

[wintry steppe]

empty set

is alwats independent

[winter harbor]

It uses the fact that V is defined over a field

[wintry steppe]

oh

i did say vector space

[winter harbor]

There's a fun thing

When you study modules

That there are some modules that don't have linearly independent subsets

For a general ring

[wintry steppe]

free modules do tho right

[winter harbor]

So you have to think a little bit more

[winter harbor]

Yeah

[winter harbor]

Yeah so, try to think why it should be non empty

Using the fact that you are working with a vector space

So it is a module over a field

[wintry steppe]

module over field is same as vector space tho?

[winter harbor]

Yeah

[wintry steppe]

empty set is always a subset of V

and its linearly independent

[winter harbor]

But yeah

You are right

The empty set is in it

I guess it should work

[wintry steppe]

i thought about it just didnt write