linear-algebra

[wintry steppe]

i meant isn't not tre

true

because it'll change by a scalar

so it can't be the same line

[tame mural]

huh -_-a

[gray dust]

@tame mural no. take rotation

[tame mural]

doh have to change my thinking

[gray dust]

@wintry steppe it's true but you just need 1 eigenvector that has a nonzero eigenvalue, then a line that gets mapped to itself is the span of that eigenvector. you can show it; let a linear map A have an eigenvector x with an eigenvalue L!=0. show that A(span{x})=span{x}

[blissful pagoda]

I asked this like 12 hours ago but I'm very unsure about my answer so I just wanted to see if I could confirm it by any chance

Axioms 4, 5, and 8 fail.
4 fails because u + 0 ≠ u in this vector space. u + 0 = 0
5 fails because 4 fails. It uses the zero vector definition in 4.
8 fails because (c+d)u ≠ cu + du in this vector space. (c+d)u = 0

[tame mural]

my intuition is that it's a vector space

is that the cop of beware of dangerous opinion 😄

but I'm used to a more fuller incantation like "A vector space over a field"

[gray dust]

wrong statement not opinion

[tame mural]

I interpret the two cases as definition for a transformation, and they're asking if after the transformation we still have a vector space

R^2 over R is of course a vector space

and when people don't specify the underlying field, I assume it's just from the same set

[gray dust]

@blissful pagoda 8 fails but not for that reason. it's bc c(x,y)+d(x,y)=0 for any c,d,x,y and so isn't equivalent to (c+d)(x,y)

[little cliff]

I don't know what you mean by a "transformation." They're giving you non-usual rules for the addition

So, for example, you should be able to derive from the axioms of a vector space that
(1+1)(x,y) = 2(x,y) = (2x,2y)
but on the other hand you also have
(1+1)(x,y) = 1(x,y)+1(x,y) = (x,y)+(x,y)=0

[blissful pagoda]

I'll correct that, that's what I wanted to say but I didn't know how to thanks

[little cliff]

since (2x,2y) doesn't equal (0,0) in general, this doesn't satisfy the axioms

[tame mural]

Well

What I thought was that this is a a trivial vector space

with only [0, 0]

So it satisfies all the axioms

[gray dust]

meow, that's nowhere near what the question is asking. R^2 IS an R-vector space with usual rules of adding/scaling. read the above operations on R^2. they're not the usual ones. you check the vector space axioms and you see R^2 with these rules fails to be a vector space

also for future reference one doesn't need to check every single axiom. just showing a set fails to satisfy one of them is enough for the set to fail to be a vector space

[blissful pagoda]

I said
"4 fails because u + 0 ≠ u in this vector space. u + 0 = 0
5 fails because 4 fails. It uses the zero vector definition in 4."
Is this correct because even though the axiom 4 fails, I know I still have (0, 0) in the vector space (since that's what the addition rule stated in the question says) and (0, 0) is normally the zero vector

Because axiom 5 actually would hold as adding two vectors does equal (0, 0) as axiom 5 stated, it's just that axiom 4 fails, the definition of the 0 vector that axiom 5 uses the definition of

[gray dust]

no. (0,0) is the 0 vector in R^2 with the usual rules of adding/scaling

this is R^2 with different rules. you showed there doesn't exist a 0 vector so you can't even hope to satisfy axiom 5

[blissful pagoda]

Okay makes sense, thank you

[gray dust]

you're welcome

[thorn lichen]

can someone explain what I did wrong on b

so im solving for the coefficients?

hmmm

[silk dragon]

Can someone point me in the right direction with this problem? https://prnt.sc/vw1thl

https://i.imgur.com/8i9pZ8M.png

hmm my images are not showing up

[thorn lichen]

so got 5/2, -3/2, and 2

is the solution a vector of those numbers?

[opaque plover]

Can someone help me with this please, I cant find an approch except x1 + ... + xm = 0 and y1 + ... + yn = 0 and 0+0=0, but I don't think thats a valid proof.

[tame mural]

@silk dragon Try to figure out the 2x2 matrix without thinking about the right vectors that would satisfy it. Because of Bw = 0 we know there's a non-trivial nullspace, which means our matrix only has 1 row to think about.

[silk dragon]

That makes a lot of sense thanks @tame mural. Is it then possible that Bv = -v cannot hold true?

[tame mural]

Depends on if it's a trick question, but I found an answer that works

So I don't think it's a trick question

[silk dragon]

stumped

the u and w vectors are easily doable with the information you've given me but I'm not too sure about the v vector

So maybe I am wrong about my matrix B

[silk dragon]

Still haven't found it, can you enlighten me @tame mural

[silk dragon]

Are you sure the answer is a nonzero vector @tame mural

[errant vapor]

how do you use the system on equations again

[worldly edge]

Anyone know where to start with linear algebra, a book recommendation would be nice to practice shit fast and easy (got an exam coming up in 2 weeks)

really tried wrapping my head around it with youtube and some books but they're too complex for a beginner with all the symbols 😂

[tame mural]

But if you're studying for a test, what could be better than your own textbook

Esp. if test is in 2 weeks

[worldly edge]

there's no textbook

[tame mural]

-_-a

[worldly edge]

only lecture notes

[tame mural]

weird @_@

what sort of teacher wouldn't assign a book

[worldly edge]

ikr it's so dumb, he recommended like 5 books lol

[tame mural]

3B1B on YouTube is recommended for digesting what books say

Gilbert Strang will walk you through all the major forms and how to hand-compute

[worldly edge]

omg i didnt know 3b1b had a linear algebra playlist

thank you

[shrewd mortar]

im not sure if someone just deleted their question about if they know one eigenvalue of a 2x2 matrix, whether they can determine the other by computing det(matrix), and yes

[hollow finch]

Also the trace

"My" prof really emphasized techniques that require the least amount of work possible. his tests and examples usually had matrices where you could gleam the eigenvalues just from the trace and determinant.

or just inspection in general

[shrewd mortar]

lol

yeah

determine the eigenvalues by just visualizing the transformation bro

smh

[hollow finch]

true. idk why everyone doesnt just do that

[silk dragon]

@shrewd mortar would you mind giving me the answer to the question I posted above? I still have not found it :c

[shrewd mortar]

@silk dragon right so what kind of vectors are u, v

for M

hint: they're [something]vectors of M

or uh of B

lol

[silk dragon]

well, nonzero...?

[shrewd mortar]

actually how much LA have you covered yet

do you know what eigenvectors are yet

[silk dragon]

yes

[shrewd mortar]

okay

well then, u and v are eigenvectors of M right

of B*

[silk dragon]

sorry, i don't see how you drew that conclusion

[shrewd mortar]

tell me the definition of an eigenvector of a matrix M

[silk dragon]

alright.

I get it now.

[shrewd mortar]

okay

now do you know that det(M) = (some function of its eigenvalues)

[silk dragon]

yes

[shrewd mortar]

yep, their product

[silk dragon]

yup

[shrewd mortar]

so if you let B = (a, b ; c, d) then det B = ad - bc = 1*-1

as these are the eigenvalues of B right

[silk dragon]

Yes, right.

[shrewd mortar]

and do you know about trace B in terms of eigenvalues

[silk dragon]

have not learned trace, no

[shrewd mortar]

uh well i mean you can bash it already

[wintry sphinx]

wait

I'm confused

how do you have a 2x2 matrix with Bu = u, Bv = -v, and Bw = 0?

[shrewd mortar]

might also help to just think about the char poly of B and get another equality for the coefficients from that

[wintry sphinx]

how are you going to give a matrix like that

[silk dragon]

One of my questions was if one of them is impossible

[wintry sphinx]

or are you trying to prove that's impossible

oh okay

[silk dragon]

is one of them impossible?

[shrewd mortar]

o lol

[silk dragon]

that's probably my main question

[shrewd mortar]

wait what was the orriginal question precisely again

[silk dragon]

well

the question is as shown

but they never said all of them were possible

https://cdn.discordapp.com/attachments/540211747613704221/784586171841511484/8i9pZ8M.png

[shrewd mortar]

well yeah actually just think about, we have a condition on the det already, and now think about what the last equality gives for the det

@silk dragon are you familiar with the last equation being equivalent to another property

[silk dragon]

Which property?

[shrewd mortar]

something to do with invertibility

you have Bw = 0 for nonzero w

if there did exist a B^-1, what could you do with that

[wintry sphinx]

another thing you could observe is that a 2x2 matrix can't have 3 distinct eigenvalues...

[shrewd mortar]

lol

[silk dragon]

🤦

[shrewd mortar]

this is pretty simple too though

so, you should just know this equivalence btw

if the kernel is nontrivial then it's noninvertible

(kernel = nullspace, different terminology is all)

so if you have indeed Tv = 0 with v nonzero

if there was a T^-1 you would get T^-1Tv = T^-1 0

[silk dragon]

right

[shrewd mortar]

then v = 0

contradiction

[silk dragon]

yeah that makes sense

[shrewd mortar]

so we showed nontrivial kernel => noninvertible, can you show the other direction

[silk dragon]

well if determinant != 0

[shrewd mortar]

(hint: noninvertible means det 0)

[silk dragon]

yes

[shrewd mortar]

(now think about det in terms of eigenvalues)

[silk dragon]

both the eigenvalues should equal the determinant

[shrewd mortar]

i.e. you are now showing noninvertible => nontrivial kernel

[silk dragon]

multipluied

[shrewd mortar]

which is the same as det = 0 => nontrivial kernel

[silk dragon]

👍

[shrewd mortar]

so what is det in terms of the eigenvalues again

[silk dragon]

thank you

[shrewd mortar]

it's just their prroduct right

[silk dragon]

yep

[shrewd mortar]

right so if det = 0 what does that mean

wrt eigenvalues

[silk dragon]

if det = 0 its non invertible

[shrewd mortar]

well yes

but we're trying to show noninvertible => nontrivial kernel

[silk dragon]

and then you have

[shrewd mortar]

if det M = 0 then what does that say about the eigenvalues of M

[silk dragon]

det(a-xi) = 0

x being the eigenvalue

[shrewd mortar]

well just think about it like this

det(M) = l_1l_2...l_n with l_i the eigenvalues

= 0

hence there's at least one eigenvalue equal to...

[silk dragon]

0

[shrewd mortar]

indeed skip

and yeah that's all you wanted to show

[acoustic path]

Nicely explained

[shrewd mortar]

0 is an eigenvalue

hence nontrivial kernel

as there's vectors such that Mv = 0*v = 0

[silk dragon]

hence the contradiction for a nonzero vector

i understand

[shrewd mortar]

there's no contradiction in this direction

we just showed it directlyy

[silk dragon]

no i mean

all 3 can't be found

[shrewd mortar]

ah indeed

since we have that det(M) = 1*-1 (the eigenvalues) but also det(M) = 0

which is impossible

[hollow finch]

its not too difficult to come up with an example of a 3x3. if you dont want to trivially use a diagonal matrix and you have some specific u,v,w in mind you can use diagonalization (assuming u,v,w are linearly independent)

i doubt you would actually want to do that but yeah lol

[blissful pagoda]

when finding row and column space do i perform row operations until REDUCED ROW echelon form or just ROW echelon form

[gray dust]

doesn't matter

[hollow finch]

i suppose it doesnt really matter

[round coral]

it depends, if you can see in RE form, then there's no need for RRE

[hollow finch]

as long as you can see which columns are linearly independent and get the rows to be linearly independent then i guess youre good to go

[round coral]

you gotta go fast here, you know!

[blissful pagoda]

but it changes my answer from {(1, 8, 4, -6),(0, 1, -4/3, 5/3)} to {(1, 0, 43/3, -58/3),(0, 1, -4/3, 5/3)}

this is because i performed one last row operation to make sure the leading 1 was the only nonzero entry in its column

[hollow finch]

there are infinitely many possible bases for any given vector space except the zero vector space

[blissful pagoda]

is here a way i can tell from just eyeballing it

*there

[hollow finch]

wdym

[round coral]

yeah, for that do it tell RRE form, then it is mostly obvious

[blissful pagoda]

mhm but i would like to reduce as much work as possible as you guys said so if i know when my rows and columns are linearly independent before actually reducing all the way to rref that would be good

determinant?

System of rows of square matrix are linearly dependent if and only if the determinant of the matrix is equals to zero.

but that only works for square matrix

[hollow finch]

doesnt work if you dont have a full set of vectors yeah

often you can eyeball it

but theres no strict formula for at what point it becomes obvious of course

[round coral]

it depends on how messy your matrix is to begin with

if it is too messy, and you are lazy, use symbolab

[hollow finch]

i prefer wolfram alpha tbh. i hate symbolab

not that it matters

[round coral]

yeah any site, I use both for different works

[gray dust]

there are infinitely many possible bases for any given vector space
{0}

[hollow finch]

😦

[blissful pagoda]

I feel like I must have missed something because the only way I can think of testing linear independence/dependence is to see if a row = 0 has a nontrivial solution lol

Other than of course having a row be a multiple of another row

right?

[hollow finch]

actually there is a relatively pointless benefit to getting rref though now that i think about it. for an mxn matrix A of rank r then if you put the rows of the rref matrix in an rxn matrix R and the columns the rref matrix told you to choose for your coulmn space in an mxr matrix C then A=CR. not sure why youd do that but there you go.

row reduction always shows linear dependence

row operations are just replacing rows with linear combinations of all the rows. so if you get rows of zero thats because you found the linear combination that cancels out one of the vectors.

and row operations also preserve column relationships, and its much easer to see in a row reduced matrix

[silk dragon]

https://i.imgur.com/UP0OHp4.png I understand how to get i but can someone explain ii in a little more depth for me?

[round coral]

you need to find the change of basis matrix

@silk dragon

[silk dragon]

so do i use what I got in part i? @round coral

im a little confused as to what exact operations i need to do

[round coral]

you have to do this , e_ [T]B . B [T]B . B [T]_e

where e_[T]_B simply means that you are sending the standard basis vectors to the chosen basis that you have B

[edgy orbit]

Why can you tell that one of the eigen values is 2 just from look at this

[shrewd mortar]

because the det is even

jk

but uhh usually think about how trace and det relate to eigenvalues

[edgy orbit]

trace is the sum of all eigenvalues and det is the product of all eigenvalues

but how does that help with this

[dreamy iron]

from Axler 3.B, page 67.....this is an iff statement....idk why he's only asking us to prove this in one direction.

[shrewd mortar]

[wintry steppe]

[shrewd mortar]

just do it yourself if you really want to

[silk dragon]

i like this one.

its how im feeling rn.

[edgy orbit]

oh i meant how do you get eigenvalue of 5 not 2 from that matrix

[dreamy iron]

okay....i mean fair enough. yeah, i'll do that.

am i missing something, is he asking that it be proved in one direction only? or is that language meant to say prove it in both directions?

(there's a parallel convo going on about trace of dets, i'll just prove it both ways. dont need this to take up air time.)

[wintry steppe]

you're thinking about this too hard lol

[dreamy iron]

I am a ninny.....comes with the territory.

[wintry steppe]

yess

im done with linear algebra

that is all

b-but linear algebra is easy, right? haha

the class i took was

but typing matrices onlinne is not

i will say that the second level of linear algebra looks daunting

[gray dust]

what's lvl 2

[wintry steppe]

applied is super easy but abstract idk yet

i think thats like proof-based stuff for my university

[tame mural]

@silk dragon This is probably out of bounds for your question, but

The answer I got was M = [[1, 0], [0, 0]]

[stoic python]

meow

[tame mural]

The non-trivial vectors that would satisfy the problem are...

[1, 0] and [0, 1]

M [1, 0] = [1, 0]

[1, 0] + [1, 0] = [0, 0]

So there are in fact non-trivial vectors where Mx=0, Mx=x, and Mx=-x

I don't believe there's an inner product space in this world

But it is a vector space

this is silly idea?

[native rampart]

You can always make a vector space into an inner product space

[tame mural]

But if you try to

maybe I'm mistaken

for this world

there are vectors which are orthogonal to themselves

hence it would have to be a degenerate form

[native rampart]

Orthogonal as in (v,v)=0?

[tame mural]

yeah

<V, V> = 0

u need example?

I can grab one I was thinking of earlier

[native rampart]

Go on

[tame mural]

A = [0, 1, 1], and <A, A> = 0

Again, for F_2^3

Oop

[native rampart]

Nvm

[tame mural]

So you agree?

Not every vector space should be an inner produce space

they are <strict> subsets

right?

[native rampart]

Inner products kind of exist,only for R or C

mb

[tame mural]

yes because in this world

we cannot have ordered field, I THINK

that was my reasoning so far

so yeah, for that guys question

I think it was possible

There is Mx = 0, Mx = -x, and Mx = x

For non-trivial choices of x

[tawdry yacht]

can someone help me with this?

[wind yacht]

#prealg-and-algebra

This is not the type of linear algebra this channel refers to

[wintry steppe]

schoolgebra

[wintry steppe]

technically linear algebra

[frosty vapor]

see pins

[coral ferry]

for 1) would the best way to prove be by showing that the null space is trivial?

[viscid kernel]

@coral ferry well Idk since the sum of the identity + A can make that new matrix invertible since if A itself is not invertibale, I guess.

[coral ferry]

sorry but i don't follow

@viscid kernel

[wintry steppe]

It's written here that ${(a,b,c) \in \mathbb{C}^3 : ... }$

[stoic python]

LINEAR_ALGEBRA_GUY

[viscid kernel]

So, if the nullspace is trivial ( isomorphic ) it means that the determinant of that ( nxn ) matrix is not zero, which means the dimension of the nullspace is zero, this means that A is invertibale. But if itself is invertable, isnt there a posibility that the new matrix ( sum of Identity and A ) itself is not invertibale ? Thats what Im wondering

[wintry steppe]

But I thought $\mathbb{C}^3$ is a 6-tuple, since we need 2 coordinates for each part?

[stoic python]

LINEAR_ALGEBRA_GUY

[wintry steppe]

At least that is what someone on here told me

That $C^3 = R^6$

[stoic python]

LINEAR_ALGEBRA_GUY

[wintry steppe]

In terms of tuples

[tame mural]

Or more like this? [[R, R], [R, R], [R, R]]

[wintry steppe]

oh ok

So for example, in (a,b,c) in C^3, we have that a = z + wi?

[tame mural]

You mean that a is a complex number? yeah

C^3 just means a list of 3 complex numbers...

[coral ferry]

@viscid kernel oh yea A might or might not be invertible

i don't know if A being invertible will help with this proof tho

[dusky epoch]

@wintry steppe in C^3 each coordinate is a complex number

who told you C^3 was equal to R^6? who told you these two were the same?

they are (informally) similar, and are in fact isomorphic as real vector spaces but they should not be blindly identified

[coral ferry]

is there a way I can simplify $ (I- A)(I + A)^{-1} (I - A)^{-1} (I + A)

did i do that right

oops i didn't get the bot right

$(I- A)(I + A)^{-1} (I - A)^{-1} (I + A)$

[stoic python]

snoopy

[coral ferry]

ok oof bot works

yea so is thre a way to simplify this?

where I is the identity matrix

A is askew symmetric

[dusky epoch]

these should all commute

[coral ferry]

if only i could switch the second and third terms

really?

why's that

[dusky epoch]

I+A and I-A commute

their product is I-A^2 either way

[coral ferry]

ah ok

is there a name for the property

so i can just search it up or smth

nevermind I bruteforced it

thank you for your help

[raw magnet]

what does this theorem says? I have read it but I don't understand it

Starts from the part : .... then assigning to each linear ....... until the end
Also what is that symbol that looks like L, L(V,W) is that kernel or something?

[dreamy iron]

L(V,W) is the set of all linear transformations from domain V to codomain W.

Even more than that, a proof will show that L(V,W) is a vector space in its own right, so it is sensible that it may be used as a domain of another linear transformation, which is your example is called M.

T lives in L(V,W) actually. Do you can sensibly say something like:

[stoic python]

ninnymonger is a physics main.
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[wintry steppe]

This theorem says that linear maps can be written as matrices and vice versa, that is $\mathcal{L}(V,W)$ is isomorphic to $M_{n \times m}(\mathbb{F})$.

[stoic python]

bacon prince

[wintry steppe]

isomorphic means there exists a bijection.

[raw magnet]

so it basically says that I can write a linear transformation as matrix or vice versa, am i right?

[dreamy iron]

If that’s true, then that seems like a powerful theorem.

[raw magnet]

yes i think it is very powerful though because we can actually write a linear transformation into matrix

so we can compute the linear transformation with matrix property..?

[wintry steppe]

well it is used without proving in high school and mid school, so yeah

[acoustic path]

kids learn about isomorphisms in mid school and high school?

[ashen sinew]

we do

not mid school though wf

[wintry steppe]

they use the matrix of a linear map to solve systems of linear equations in high school though

[acoustic path]

oh

[soft burrow]

yes, $\exp: (\bR,+) \to ((0,+\infty), \cdot)$ is an example of one

[stoic python]

derivada.schwarziana

[wintry steppe]

maybe you'll see examples of isomorphisms

but if you go up to the average middle schooler and ask them to give you an isomorphism

i doubt they could

unless they're one of the prodigies on this server

[edgy orbit]

Does an SVD only work when the matrix is positively definite

[ashen sinew]

we learn that in 12th grade

groups and rings

[edgy orbit]

what do you do if it is not positively definite

give up and cry?

[hollow finch]

SVD works on all matrices afaik

oh i think maybe i get your question

A^T A will always be positive semidefinite

[edgy orbit]

oh that makes sense

so you don't have to worry about A^T A not having a PD right

[hollow finch]

yeah symmetric matrices never have defective eigenvalues

and theyre all real

[edgy orbit]

what is the difference between semidefinite and definite

[hollow finch]

i believe definite means strictly nonnegative while semidefinite means including zero

but im checking that

[edgy orbit]

yeah

[hollow finch]

then yeah there you go

you can pretty much always believe in symmetric matrices

they never really let you down

[edgy orbit]

does that also guarantee orthogonal columns.

if you do A^T * A

[hollow finch]

indeed. the eigenvectors of symmetric matrices are always orthogonal

or at least you can get an orthogonal basis out of any eigenspace

[edgy orbit]

ok that makes sense. Thanks

[hollow finch]

np

[night citrus]

eq 1.3.18 says the matrix of the inverse transformation is the inverse of the transformation matrix, right?

[hollow finch]

The matrix of the inverse transformation is the inverse of the matrix of the transformation yeah

[night citrus]

alright, thanks

[magic plover]

Would anyone be able to look over a proof I did?

I already completed it, just need someone else to verify I didn't do something stupid

[round coral]

maybe, put it up here

[hollow finch]

yeah thats good

but it can be more concise using the sigma notation

$e^A=\sum_{n=0}^\infty \frac{A^n}{n!}=\sum_{n=0}^\infty \frac{(PDP^{-1})^n}{n!}=\sum_{n=0}^\infty \frac{PD^nP^{-1}}{n!}=P\left(\sum_{n=0}^\infty \frac{D^n}{n!}\right)P^{-1}=Pe^DP^{-1}$

[stoic python]

nix

[steady fiber]

smh even a middle schooler could give the trivial isomorphism

[wintry steppe]

[steady fiber]

one of "send everything to itself" or "send everything to the identity"

ez

I'm sure middle school me could give one of those

[hollow finch]

Every time someone tells you the entries of a matrix theyre basically doing an isomorphism

"The matrix is a,b,c,d"

[wintry steppe]

ew, choices of bases

🤢

do things coordinate free smh

[steady fiber]

smh that's too hard

[wintry steppe]

i signed up for a course with 1/2 slots

[tame mural]

in person?

[wintry steppe]

oh wait this is thge linalg channel

nvm

[steady fiber]

you need to know manifolds for coordinate free stuff

I don't even know what manifolds are smh

well I kinda do but not really

[hollow finch]

Pretty sure thats abstract algebra, but essentially a group is a set of elements with an associative law of composition, an identity element, and an inverse for every element.
Think of nxn invertible matrices under multiplication (the general linear group). Matrix multiplication is associative. Multiplying by the identity matrix doesnt change any other matrix. And any invertible matrix has an inverse such that when theyre multiplied together you get the identity.

[sharp shell]

I've to find the real parameter a such that their are infinity solutions and a a such that x + y + z is the smallest possible natural number

I've tried using the ranks of the coefficients matrix and row echelon form

but Idk what to actually do with it

[tame mural]

I think of groups as integers under addition but with optional commutativity

[hollow finch]

thats also a mighty fine group

i just prefer the general linear one because it isnt commutative. i worry someone will assume all groups are if theyre just given the integers first. thats just me though

[night citrus]

is there a difference between a R^2 vector and a R^n vector with only the first two entries actual values and all else 0?

[hollow finch]

That is a subspace of R^n which is "isomorphic" to R^2

That's the word to articulate the way in which those two vector spaces are related.

[night citrus]

makes sense. kinda like how coplanar lines can be extracted from 3D and mapped on a plane

[limber sierra]

yeah formally speaking theyre different things, but theres a function that "relates" them in the "obvious" way

and for all intents and purposes

you can interchange one and the other

and be fine

just dont try and add a vector from R^2 to a vector from R^17 or whatever

(but you can apply the function to make the first vector its "equivalent" from R^17 instead, and then add them as expected)

[hollow finch]

That function being both surjective and injective

[night citrus]

is the function you speak of a matrix?

[limber sierra]

it's linear, so you can write it as a matrix

[hollow finch]

You can represent it as a matrix yeah

[limber sierra]

it's just the map $\begin{bmatrix}a\b\end{bmatrix} \mapsto \begin{bmatrix}a\b\0\0\\vdots\0\end{bmatrix}$

[stoic python]

IsolSingPts(X)

[hollow finch]

Which is very easy to invert (just switch the direction of the arrow basically)

[limber sierra]

might be a good exercise to think how you'd write this as a matrix

[night citrus]

my original question came from an exercise asking for a transformation from R4 to R2 actually

came up with a rightside multiplication by a 2x4 matrix

[hollow finch]

I know a professor who calls this the fundamental theorem of linear algebra

Every n-dimensional vector space V is isomorphic to R^n

Yeah that's exactly right

[night citrus]

you both have been very helpful, thank you

[steady fiber]

there's a bunch of "fundamental theorem of linear algebra"

different people call it different things

not one of the more widely accepted fundamental theorem names

[hollow finch]

Of course. I personally find it the most important and fundamental but I wouldn't say it is absolutely.

[limber sierra]

if i had to choose a "most fundamental"

it'd be that the different definitions of basis coincide

[hollow finch]

Interesting

[wintry steppe]

ok george

.

[limber sierra]

🤔

[wintry steppe]

🤨

[limber sierra]

i think george's answer would be "the axiom of choice"

and he wouldnt be wrong

[spare crystal]

whats a good intuition for the transpose of a linear transformation in the context of dual spaces?

[slow scroll]

its just the regular ol' dual map

[spare crystal]

wdym by that

[native rampart]

If (Af)(x)=f(Tx),for all functionals f and all vectors x ,A is transpose of T

[slow scroll]

btw, by dual map i just mean if you have a linear map $T : V\to W$ you get a map $T^* : W^* \to V^*$ given by $\varphi \mapsto \varphi \circ T$

[stoic python]

kxrider

[slow scroll]

in the context of matrices from Rn to Rm, T* is the transpose Drake is talking about

[spare crystal]

hmm i guess the transpose just feels kinda arbitrary still

idk

[native rampart]

Shouldn't it be V* to W*

[spare crystal]

no T* goes from W* to V*

[native rampart]

mb

[spare crystal]

which i agree makes things weird

[terse halo]

Weirder than a midget giving a parrot a back rub?

[spare crystal]

yeah

i can visualize one, i can't visualize the other

[native rampart]

Consider g(x)=f(Tx) ,g is obtained by fixing the functional f and applying a transform on the vector. Now,Can g be obtained by fixing the vector and applying a transform on the functional? That kind of seems like a natural thing to ask

[blissful pagoda]

Very dumb question but |v| what does this mean?

||v|| means norm of vector v

[hollow finch]

magnitude of the vector v

[blissful pagoda]

Isn't that ||v||

[hollow finch]

both notations are used

[blissful pagoda]

Oh

[hollow finch]

i prefer |v|

[blissful pagoda]

Okay i was gettng confused lol

So the dot product of a vector and itself is its magnitude/length/norm^2?

[spare crystal]

yeah

[hollow finch]

the magnitude/length/norm is more generally defined as the square root of inner product of a vector with itself

the dot product isnt the only way to define distance 🙂

[spare crystal]

right so I get this but it feels rather arbitrary compared to how non-arbitrary of an operation the transpose does to a matrix

idk maybe i just need to let it sit and do some problems involving it

[blissful pagoda]

@hollow finch The dot product of u = {u_1, u_2, …, u_n} and v = {v_1, v_2, …, v_n} is the scalar quantity u * v = u_1*v_1 + u_2*v_2 + … + u_n*v_n

this is what i have

[hollow finch]

yeah thats one definition for an inner product

[dusky epoch]

may i suggest using something other than {} for containing coordinates in a vector

even () or [] would have been better

[blissful pagoda]

Thank you for catching that normally I write ()

because {} is sets lol

[wintry steppe]

well depends on what you mean by length

dot product is more general

[dusky epoch]

nah dot product refers to the "standard" product

inner product is what's more general

[wintry steppe]

Why is $\mathcal{P}(\mathbb{F}) = \mathbb{F}[x]$?

[stoic python]

LINEAR_ALGEBRA_GUY

[wintry steppe]

it was @empty copper I think?

[dusky epoch]

P(F) and F[x] are... different notations for the same thing? if i understand you correctly

does that answer your question

[wintry steppe]

Isn't that the powerset, though?

[dusky epoch]

not unless it's specifically been stated as such

[wintry steppe]

So they're just using mathcal P for fun???

[dusky epoch]

your book should have introduced both notations

[wintry steppe]

Even though that always represents the powerset?

[dusky epoch]

"always"

notations can mean different things in different contexts

[wintry steppe]

@dusky epoch For example, $\mathbb{F}_2[x]$ means polynomials with coefficients in F mod 2, right?

[stoic python]

LINEAR_ALGEBRA_GUY

[dusky epoch]

in F_2

but yes

[wintry steppe]

Yeah well look at this

[dusky epoch]

i mean like

is there anything else you wanna ask besides notational complaints

[dusky epoch]

ok sure

that's a notation

good enough for their purposes i suppose

[wintry steppe]

[dusky epoch]

you don't use powersets often in LA, and even then

there's a different notation for 'em which doesn't clash with this even apparently

if you want you can write sth like $\mathbf{F}[x]_{\leq m}$ for the same space! notation really doesn't have as much weight as you're putting into it!

[stoic python]

Ann

[dusky epoch]

what matters is that both you and your audience understand what stands for what

[wintry steppe]

How do I show that $\mathbb{F}[x]$ is infinite-dimensional?

[stoic python]

LINEAR_ALGEBRA_GUY

[dusky epoch]

show it has arbitrarily large LI subsets.

[wintry steppe]

LI?

Linearly independent?

[dusky epoch]

linearly independent

yes

show that for any n there exists a linearly independent set of n vectors in F[x]

[wintry steppe]

How do you do it without using linear independence?

[dusky epoch]

what do you mean

what's your defn of an infinite dimensional space?

[wintry steppe]

Using span?

[dusky epoch]

what's your defn of an infinite dimensional space?

no definition no proof

[dusky epoch]

and what's the defn of a finite dimensional vector space?

one that has a finite spanning set?

[wintry steppe]

A vector space is called finite-dimensional if some list of vectors in it spans the space.

[dusky epoch]

okay then you can show that no finite list of polynomials spans F[x]

[wintry steppe]

finite

[dusky epoch]

you might find it handy to make use of the notion of degree

[wintry steppe]

?

finite list

[wintry steppe]

What about degree?

[dusky epoch]

which F[x] comes with

[wintry steppe]

Polynomials $P(\mathbb{R})$ can be spanned by $\beta = {1,x,x^2,\dots}$

[dusky epoch]

this is an infinite list, and being spanned by an infinite list says nothing

also we're not considering the space of polynomials up to a certain degree, we are considering the space of ALL polynomials

[stoic python]

bacon prince

[wintry steppe]

Now I'm confused

[dusky epoch]

i was responding to bacon prince there, sorry

anyway

i gave you a hint there.

your roadmap for the proof is:

consider a finite list of polynomials in F[x]. show that it cannot span all of F[x].

to show a list does not span a space, one good strategy is to produce a vector in the space which is not in the span of the list.

[wintry steppe]

Yeah seems impossible since even something like f(x) = ax can reach every point in R^2

For different values of a

?

[dusky epoch]

??

that has nothing to do with this

the graphs of polynomial functions don't matter at all here

[wintry steppe]

Why not? Isn't the span what points we can reach?

[dusky epoch]

in F[x], a polynomial is a point

[wintry steppe]

What??? How?

[dusky epoch]

"point" and "vector" are more or less synonymous in linalg

and a vector is just something that lives in a vector space

and here your vector space is F[x]

its elements - its vectors - are polynomials themselves

[wintry steppe]

Okay... how would you represent them graphically?

but what is F[x]

The set of polynomials with coefficients in F

oh

?? What did you think it was

I had my doubts

[wintry steppe]

What did you think it was?

[dusky epoch]

Okay... how would you represent them graphically?
you would not

not all vector spaces have dimension 2 or 3

and not all vector spaces admit a clear graphical representation

[wintry steppe]

I did think it was polunomials over a field, but it was weird as the notation I use is $P(\mathbb{F})$

[dusky epoch]

especially not ones over fields like F_p

[stoic python]

bacon prince

[wintry steppe]

@dusky epoch F_p is easy to represent graphically as discrete points?

[dusky epoch]

well good luck trying to make a graphical representation for something like F_7^100 lmao

thatll be a lot of points

[wintry steppe]

It could be done by a computer..

don't

[dusky epoch]

L_A_G what are you looking for here exactly

[wintry steppe]

Can you explain this proof

[dusky epoch]

what proof

that F[x] is inf-dim?

[wintry steppe]

I don't understand it and I wanted a different one

[dusky epoch]

this is the proof i had in mind

which part do you not understand here?

[wintry steppe]

I understand up to the first Thus

[dusky epoch]

so you understand that everything in the span of your list has degree ≤ m?

[wintry steppe]

Yes

[dusky epoch]

bacon prince

this is wrong

not all polynomials are monomials

and also youre taking way too high a degree there

the former is more serious than the latter

anyway youre not helping

@wintry steppe what is the degree of z^(m+1)?

[wintry steppe]

m + 1

[dusky epoch]

so, do you understand why z^(m+1), a polynomial of degree greater than m, cannot be in the span of your list?

[wintry steppe]

No because as I said even something like ax can reach any point in the plane for different values of a, and single variable polynomials are only in F^2

[dusky epoch]

what the hell are you talking about

we don't give a shit about F^2

we're not working in F^2 we're working in F[x]

every point - every polynomial - in the span of your list has degree m or less

z^(m+1) has degree m+1

m+1 > m

therefore there's no way z^(m+1) can be in the span of your list

how is this not obvious?

[wintry steppe]

We don't give a shit about F^2? But do you disagree all polynomials of one variable are in F^2?

[dusky epoch]

yes i disagree, because a polynomial shouldnt be identified with its corresponding polynomial function

or its graph

[wintry steppe]

Don't understand this sentence. What are you talking about?

[dusky epoch]

alright let me start over

let's take our field to be the real numbers just for the purposes of this

and take the polynomial x^3 - 2x

[wintry steppe]

ok

[dusky epoch]

the polynomial x^3 - 2x can be considered as a function and graphed, giving this curve

this curve is a set of points in R^2

some of the points on this curve include (0,0), (1,-1) and (2,4)

agree?

[blissful pagoda]

may i ask something or would i be interrupting

[dusky epoch]

you would be interrupting

ask in a free questions channel

[wintry steppe]

I agree yes

[dusky epoch]

however

this does NOT make the polynomial x^3 - 2x itself a point in R^2.

[wintry steppe]

I never said that it did

[dusky epoch]

single variable polynomials are only in F^2

all polynomials of one variable are in F^2?

[wintry steppe]

Yes?

I meant that every point of any single variable polynomial is in F^2

[dusky epoch]

see this is something that has nothing to do with F[x] as a vector space

you're trying to think of polynomials in terms of their graphs

[wintry steppe]

No this is what I am thinking

Each element in F[x] has a graph

And if we pick like z^(m+1) from F[x]

Then every point in the graph of this polynomial can be reached by ax, for example, for different values of a

[dusky epoch]

Each element in F[x] has a graph

yeah, again

nothing to do w/ F[x]

[wintry steppe]

But the way to think of the span is the points that each element can reach when multiplied by a scalar.

[dusky epoch]

bad wording

you're working in F[x] not in F^2

your points aren't points in F^2

your points are polynomials

your points are elements of the vector space youre working with

for example, x^3 - 2x is in the span of {1, 5x, 3x^3 + 9}

[wintry steppe]

Why is it in the span of that?

[dusky epoch]

$x^3 - 2x = \frac{1}{3} \cdot (3x^3 + 9) + \paren{-\frac{2}{5}} \cdot 5x + -3 \cdot 1$

[stoic python]

Ann

[dusky epoch]

x^3 - 2x is expressible as a linear combination of 3x^3 + 9, 5x and 1

and in that image above i did just that

[wintry steppe]

oh

ok..

[dusky epoch]

if you want, i can explain how exactly the proof's argument works with this particular list

[wintry steppe]

Yes please

[dusky epoch]

the value of m as defined in the proof is 3 for the list {1, 5x, 3x^3+9}

thus the proof argues that {1, 5x, 3x^3+9} cannot span R[x] because x^4 is not in span{1, 5x, 3x^3+9}

[wintry steppe]

Alright thank you ma'am

[errant cedar]

Give an example such that L is onto and one to one and linear

when i say p(x) = x

i think it's satisfied

[native rampart]

How is that a map from P[F] to F^inf?

[errant cedar]

its all i know

about the question

[errant cedar]

???

[wintry steppe]

there should be a fairly natural choice of map to take

write out an element of each side

[errant cedar]

ho

how

[wintry steppe]

write what an element of P(F) looks like and ask yourself how you can turn that into something in the set on the right

if you think hard enough there's a very simple choice of map

[errant cedar]

PF looks like ax+b

[wintry steppe]

maybe you can do the simpler case of, say, nth degree polynomials and F^(n+1)

are you sure?

[native rampart]

What about x^2?

[errant cedar]

x^2 is not in P^2(f)

or do i know false

[wintry steppe]

and that's not what you're working with

you're working with P(F)

presumably the set of all polynomials

how can you even construct a function from P(F) if you don't know what it is?

that's why i said write out an element of it ;)

[errant cedar]

i wrote

i couldnt figure it out

why its not ax+b

or 3x+5

[wintry steppe]

can you show us your definition of P(F)

[errant cedar]

the set of polynomails in max 1 degree

[wintry steppe]

can you show us the definition of P(F) given by whatever you're studying from

[errant cedar]

the definition is far away

hard to find

[wintry steppe]

yeah you still haven't given a definition of P(F) so i don't know what to say

id assume it's "all polynomials with coefficients from the field F" but i want you to verify that yourself

(where "polynomials" refers to polynomials with all but finitely many coefficients zero)

maybe check the back of the book? is there a list of symbols/notation?

[round coral]

@errant cedar It would be better if you see the previous chapters where he tells about polynomials as vector spaces and the standard basis of them

In 3.34, he already expects you to know all that.

[wintry steppe]

what book?

[round coral]

it's axler

[wintry steppe]

ty

it has dimension 3 x 1

[round coral]

rank depends

on your matrix

[wintry steppe]

so a 3x1 matrix is a linear map from a 1 dimensional space to a 3 dimensional space. if you write out rank nullity, you'll see that
rank = 1 - nullity
in which case it's either 0 or 1

(one way, not the only way, to see it)

the more straightforward way would be to just notice that the image of a 1 dimensional space will be of dimension <= 1

both work

when presented with something in finite dimensions (or with matrices) i like to see what rank nullity tells me, it's very powerful

maybe a technique worth keeping in mind

[round coral]

do you really need rank nullity to see it, it is just 1 column of matrix and that is for sure linearly independent so rank =1. Or am I wrong?

[errant cedar]

yes But P(F) is in form ax+b yes

?

[wintry steppe]

nah you don't need rank nullity, i gave an elementary way to see it after using rank nullity. it could be zero (eg all entries zero)

[round coral]

yeah, I get it

[wintry steppe]

rank nullity is baby's first isomorphism theorem

sure, post it here

[errant cedar]

you are giving always

but say pls something

[errant cedar]

what are u thinknig about that

if u give an example

i can understand maybe

[round coral]

@errant cedar go to 2.12 definition

[errant cedar]

OOOhh

wtf did i think until now

[wintry steppe]

could you take a screenshot of the question? feels like this might be entirely trivial lol since the third and fourth lines seem to imply that the null space of B is just the image of A (and image = column space...)

[errant cedar]

absoulutly

[wintry steppe]

yeah i think just like, ||the null space of B is the column space of A,|| so you're done

i think you're mistaking the size of a vector and the dimension of a vector space

B is a 3 by 5 matrix so the elements in its image/column-space will be 3 x 1, but it's null space will consist of 5 x 1 (i.e. vectors with 5 entities)

and the dimension of that will mean something different than the size of the vectors in it

[errant cedar]

Let V be a finite-dimensional and W an arbitrary vector space over F.
and L ∈ L (V, W). Show that there is a subspace U of V such that U∩kern (L) = {0}
and image (L) = {Lu | u ∈ U}.

[errant cedar]

x^n+...+x^2+x+1

is it work

[magic light]

So I have V = R[x] the space of all polynoms with one variable and coefficients from R
U = {p(x) in V | p(0) = 0 }
now, it's clear to see that this means that this is all the cases where the last coefficient isn't 0

so anx^n + ..... + a1x + a0

but how do vectors from V or U look like? I'm a bit confused on that

are the vectors just the coefficients?

like [an, a(n-1)...., a1, a0] ?

similarly, how would a linear transformation look like from V->U?

[wintry steppe]

they are literally just the polynomials. "vector" in this case means "element of a vector space"

[west spade]

The vectors are just what you say they are - polynomials. They're only sets of coordinates with respect to some basis... though a good exercise is to try and figure out exactly which basis you're subconsciously referring to when you reference those coordinates

[tame mural]

all polynomials can be uniquely defined by their coefficients

[half storm]

"It's clear to see that this means..."
I think you mean where the last term is zero.

[magic light]

really?

[tame mural]

Wait, is that not true?

I'd like to know @_@

or did I forget to mention

variables and coefficients

[magic light]

I mean it doesn't seem true but it'll help me solve my problem

lol

[magic light]

yeah,

So basically, I'm tasked with proving that U and V are isomorphic

from my understanding, if I can find a T s.t it's bijective, that proves they're isomorphic

is using the identity function, except in the last term, meaningful here?

so T(f) = { f(an), f(a(n-1))... ,f(a1), 0 }

so for T(3x^2 + 2x^1 + 3) = 3x^2 + 2x + 0
this would work since p(0) = 0 as needed for U

the issue is that it's not bijective since 3x^2 + 2x^1 + 3 = 3x^2 + 2x^1 + 4, as an example
it would work for anything in the last term since I'm mapping it to 0

am I making sense here at all? >_>

Going to assume I was coherent;
so I need to find a way to make the last term meaningful, I tried to multiply everything by the last term but:
T(3x^2 + 2x + 3) = 3(3x^2 + 2x ) = 9x^2 + 6x
T(9x^2 + 6x + 1) = 9x^2 + 6x ... still not bijective

[half storm]

p much if you find a linear transformation that is a bijeciton then by defintion it's isomorphic.

Go ahead keep typing

I'm just reading

[magic light]

I have another idea

[half storm]

Yea you're making sense

[magic light]

which is to map everything to the the derivative of n+1 divided by n+1

[half storm]

I mean one easy way of showing this is just showing that the two spaces are of equal finite dimension

But you're probably going to find that they're not.

Which is why you're having trouble of finding such an isomoprhism; it doesn't exist.

Do you see why?

[magic light]

the assignment tells us that there is one

so maybe I didn't write something correct?

https://i.imgur.com/2wsFLkP.png
https://i.imgur.com/6Obvn6U.png

"prove U and V are an isomorphism"

[half storm]

I'm struglging to see there is. Beacuse if I'm understanding what you're underlying sets are correctly then we have that $R[x] = { p(x) | p(x) = a_1 x + a_0 , a_1, a_0 \in R }$

[stoic python]

TheDon

[half storm]

like if that's the case that subspace has dimension two because a basis for that space is $\beta = {x, 1 }$ and for the other space all you need is $\alpha = {x }$

[stoic python]

TheDon

[half storm]

If that's those are the sets that you're talking about, then there can't be an isomorphsim between them.

But maybe I'm misunderstanding the sets your talking about.

[magic light]

R[x] is all polynoms with one variable

there can be x^2

[half storm]

All polynomials of one variable

That's a huge set

It's basis is infinite dimensional

There's no way there can be an isomorphism.

[magic light]

it's basically all polynoms like a0 + a_1x^1 + a_2x^2 ... + a_nx^n

[half storm]

Yea, there's no way.

[magic light]

or maybe without a0 i'm not sure

[half storm]

Now if you want to show that there is a homomorphism between the two of them, then that's a possibility and you've already given it.

That wouldn't make any meaningful difference.

[magic light]

it says clearly isomorphism

[half storm]

The situation is still the same.

[magic light]

OK, so these are my instructions:
"There is a very natural transformation:
https://i.imgur.com/PpNL2Rn.png

[half storm]

There's a theorem that states that two vector spaces of finite dimension are isomorphic if and only if they are of equal dimension.

Oooooooh

nvm i completely misunderstood the sets.

This might be doable.

[magic light]

"...But this transformation isn't injective"

[half storm]

Yea it 100% is.

[magic light]

they instruct us to "fix" it

I have an idea

[half storm]

I know you can because the bases are both of countable infinite dimension.

So you could say that by a theorem; finding an explicit isomorphism is more difficult.

[magic light]

basically...
T(3x^2 + 2x + 2 ) = (3x^3)'/3 + (2x^2)'/2 + (2x)' = 3x^2 + 2x + 2

actually that's not good

since p(0) != 0

Instructions say: "Try 'fixing' the transformation so that it'll 'remember' the free variable, and won't 'delete' it"

[magic light]

on this

[half storm]

hmmmm

[magic light]

yeah, hard.

[half storm]

I see

Well that helps a decent amount.

I think.

[magic light]

you see my previous issue?

[half storm]

Yes.

[magic light]

^

I tried to find a way to make a0 significant or something unique

[half storm]

Try this $T(a_nx^n +a_nx^{n-1} + \dots + a_0) = \frac{a_n}{a_0}x^n + \frac{a_{n-1}}{a_{0}} x_{n-1} + \dots + \frac{a_1}{a_0}x$ if $a_0 \neq 0$ and just what they gave you earlier if $a_0 = 0$

[stoic python]

TheDon

[half storm]

That might work but I literally just came up with that on the top of my head

[magic light]

I tried that

[half storm]

Dang

[magic light]

so what if a0 = 0? then it is undefined

[half storm]

No

you just redefine it

you define it for polynomials where the last term is nonzero and then just make it the regualr map that they gave you for when it is 0

[magic light]

OK, I thought of that a little but

I got stuck trying to prove that this is indeed unique

[half storm]

That what's unique?

[magic light]

Well, basically that T(f) = T(g) => g=f

because otherwise it's not a bijection

[half storm]

Even if you proved this, it wouldn't show it's a bijection because you have to show that it's surjective too.

[magic light]

yeah, but it's a start

Is Ker T = {0} sufficient for bijection?

[half storm]

YES

That's exactly what I was gonna say

lol

Provided that T the map is a linear transformation

[magic light]

Just beforehand

technically

do I need to split T into a0 = 0 and a0 != 0?

[half storm]

Yes

you define it for when a_0 is zero and whne a_0 is nonzero.

[magic light]

Like T = { (something) a_0 = 0, (something else) otherwise

[half storm]

You need to make sure this transformation is linear first because even going about trying to show that the Ker = {0} because that'll be a valid proof method if the function is lienar

[magic light]

yeah, I'll prove it's a transformation

[half storm]

yea remember I said that you define it to be the transformation where you divide the last coefficient if a_0 !=0 and just the way that they said was a "natural" linear transformation for the case when a_0 = 0.

[acoustic path]

merry christmas my linear algebra people

[magic light]

OK, so hold up

[half storm]

M christmas 😄

[magic light]

I need to prove KerT = {0}
so that means that x = 0?

[half storm]

I'm probably gonna go out of my way to check that's a linear map because I have a feeling off the top of my head that it might not be.

yes

[magic light]

I feel like that has to be wrong... hmm let me think

wait

[half storm]

No that's right.

[magic light]

p(x)=0 is by definition

of being in U

no no we're making a mistake

[half storm]

no p(0) = 0 is the conditon

not p(x) = 0

[magic light]

we need to prove T(f) = 0 iff f(x) = 0?

[half storm]

Yes

This definitely holds.

So injectivity taken care of.

[west spade]

for which transformation?

[magic light]

@west spade it's so hard to explain simply lol

[half storm]

#linear-algebra message

[magic light]

OK hold up

1. assume f(x) = 0
   the T(0) = 0
   that's one direction

2. assume T(f) = 0

[half storm]

Yes

[magic light]

so in other words assume (anx^n + ... a1x1) / a0 = 0

[half storm]

For the case in which a_0 !=0 yes

[magic light]

a0 is always non-zero so we can get rid of it

[half storm]

It's always nonzero.

yes

by assumption

[west spade]

that's not linear, consider x+1 and x+2 as your polynomials

[magic light]

so anx^n + ... ax = 0.. isn't that just our function?

[half storm]

It's no the original funciton

it's missing the a_0 term on the end

[west spade]

and it's certainly not bijective

[half storm]

That's what I was worried about

The linearity is what concerned me

[west spade]

an indirect way to show it's not linear is that only one polynomial maps to 0, but the map isn't injective

if it were linear that couldn't be possible

[magic light]

Yeah I got a contradiction

[magic light]

why is this impossible?

[west spade]

if a linear map has two inputs that output some vector, you can do some arithmetic to transform that into two inputs that output 0

so the set of vectors that map to 0 will tell you all you need to know about injectivity

[magic light]

mm yeah

[west spade]

(this is all for finite dimensions)

[wintry steppe]

@winged jay

[winged jay]

.

[magic light]

well, I'm trying to think of what linear transformation can help keep the constant

that doesn't work sadly @half storm

[half storm]

Maybe multiplication would be the next thing I would think of off the top of my head

[west spade]

what exactly are you trying to find here?

[half storm]

I thought it might not

which is why I said go back and check for linearity

[magic light]

this is the nutshell

[west spade]

a linear map from the space of polynomials degree n to itself?

[magic light]

not to itself

V-> U

where U is all p(x) in V s.t p(0) = 0

in other words the constant is zero

[half storm]

The set of all polynomials into the set of polynomials such that p(0) = 0; basically the set of polynomials with no trailing term.

[west spade]

what about it? that image is cut off

[magic light]

there's text explaining what V is thats all

[west spade]

and you're trying to show...

[magic light]

I'm trying to show U and V are isomorphic

[west spade]

where U is

[magic light]

the images here don't load for you?

U = { p(x) in V : p(0) = 0 }

V = R[x]

[west spade]

got it, my b

[magic light]

basically, I need to find a transformation

that's bijective between them

there is a transformation where you just keep everything except the trailing member, set to 0

[west spade]

got it

[magic light]

but then it's not bijective because, for example T(3x + 3) = T(3x + 2) = 3x

[west spade]

luckily in this case you're handed two bases to biject

forget about the polynomial structure, just strip the problem down

[half storm]

You have two countable bases so by a certain theorem you know that their isomorphic. But given that the bases are countable there's probably a way to find a bijection between the bases themselves that can lead you to the right answer.

so basically what @west spade said.

[magic light]

I don't mind doing something different, it's just that I'm a little inexperienced here; I have instructions in the question

[magic light]

^

I'm a little confused though

T(f) = { f'(an + 1) / n+1, f'(an-1 + 1)/n, .... f'(a0 + 1) }
maybe I'm writing it wrong but basically

OK, I don't understand something

3x^2 + 5x + 3 is not in U, right?

oh I get it

[half storm]

Right

[magic light]

T(3x^2 + 3x + 5) = 3x^3 + 3x^2 + 5x < why wouldn't that work?

[half storm]

It could

[magic light]

basically up all powers by 1

[half storm]

I'm going thru my head and I think it might work

[magic light]

T(5) = 5x
T(5x) = 5x^2

[half storm]

Wrong section

Complex variables; the channel below this one.

[rustic panther]

Sorry

[half storm]

👍

I'm worried about linearity still though

[magic light]

Is integral not a linear map?

I mean it's not exactly integral

but it's the same concept

[rustic panther]

On the tablet

[half storm]

But it's not integration so it may not work.

[magic light]

Prove T(f) = 0 <=> f(x) =0, f(x) = anx^n + ... a1x + a0

side 1: assume f(x) = 0, T(0) = 0 by def

side 2: assume T(f) = 0, then anx^n+1 + ... a0x = 0

if x=0 => then clearly f(0) = 0
if x != 0 we can divide by x once =>
ax^n ... + a1x + a0 = 0 => f(x) = 0

[half storm]

So if you take two polynomials one of degree m and another of degree n, apply this map:
$p(x) = a_nx^n + a_{n-1}x^{n-1} + \dots a_1x + a_0$ and $q(x) = b_mx^n + b_{m-1}x^{n-1} + \dots + b_1x + a_0$
What happens?

[magic light]

@half storm we just need to prove aT(f) = T(a * f) and that T(f + g) = T(f) + T(g)

I think that works

[half storm]

What you typed seems to check out for me that if it's a linear map then it's definitely injective

[magic light]

the only thing I need help with is how to explicitly state that function?

[half storm]

You should show that it works first. But you've already stated what the functoin is

[stoic python]

TheDon

[magic light]

I don't think I wrote ite correctly mathematically

[half storm]

For any $p(x)$ $\implies T(p(x)) = xp(x)$