linear-algebra

[quartz compass]

I guess I'd try to make a plan of attack by looking at the first row

seems to have terms of (a+b+c)^3 in them

going to need some kind of polynomial identity in mind

the truth is, someone started with some formula, and then they just messed around with it until they got this meaningless thing

and now it's your problem

I think if you work one piece at a time, make the a+b+c part and get that to work, it would eventually yield

[wintry steppe]

I have no idea man, I tried tons of shit, and adding row2 and row3 to row1

and it gets real real weird

[vast torrent]

Maybe use the formula already known for vandermond for the lhs

?

Pi(xj-xi)

[wintry steppe]

🤔

[sinful heron]

what is this property for cross product

[quartz compass]

@sinful heron antisymmetry

[sinful heron]

the vectors are oppositely symmetrical?

[quartz compass]

antisymmetry just means reversing the order makes it negative

[sinful heron]

oh

[vast torrent]

I'd've said anticommutativity

[quartz compass]

"oppositely symmetrical" doesn't mean anything to me

that works too

[wintry steppe]

so the LHS =
(d-c)(d-b)(d-a)(c-b)(c-a)(b-a) ?

[vast torrent]

Zoran maybe it would be easier to replace the lhs with that formula

Yes

Not sure if that helps

[wintry steppe]

🤔

let me take a crack at it with that try

I think I'll just go to bed, take it on fresh next morning

[sinful heron]

gn man

in this question, what are the vectors n1 and n2?

[vast torrent]

normal vector

[sinful heron]

to each respective plane?

[vast torrent]

ya

[sinful heron]

i see thx

if i take the cross product of these two vectors i should get the vector of the line that's parallel to the intersection line for the two planes

[quartz compass]

yep

[uneven bloom]

@wintry steppe The RHS is homogeneous of degree 6 (like the LHS) , so you just need to conclude that it is 0 if any two of a,b,c,d are equal.

[north sierra]

https://cdn.discordapp.com/attachments/442470054978650122/646194557117005825/unknown.png

what do the sets actually mean?
so there's a vector such that a + b + c = 2 i dont really understand what that means
i know how to answer the question but like i dont get the notation

[dusky epoch]

{object : condition} is the set of all objects for which the condition is true

i.e. for example

$\begin{bmatrix} 4 \ 7 \ -9 \end{bmatrix} \in W_7$, because $4+7+(-9) = 2$

[stoic python]

Ann:

[dusky epoch]

while $\begin{bmatrix} 1 \ -4 \ 2 \end{bmatrix} \notin W_7$, because $1+(-4)+2 \neq 2$

[stoic python]

Ann:

[dusky epoch]

(W_7 here means the set given in exercise 7)

[north sierra]

okay makes sense, thank you!

[marble bone]

Hey, if we are given a matrix M = diag(d1, d2, d3, ..., dn), does the "diag" notation imply that M is a diagonal matrix or does this notation just tell us about the diagonal entries?

context:
"A,D exist within the set Mnxn(R) and D is a diagonal matrix. Let D = diag(d1, d2, ..., dn) for some d1, d2, ..., dn that exist within the set R \ {0}, show that D is invertible and that D^-1 = diag(d1^-1, d2^-1, ..., dn^-1)"

is the D^-1 implied to be a diagonal matrix? It would seem so according to my intuition. But I'm not entirely convinced.

[cold ravine]

Anybody willing to help me with an explanation? I think I'm good on what the actual answer is but putting into math terms is a little harder for me.

Tha basic gist I can muster is that you end up with the same set of vectors back because when you take the projection of one vector onto the other you'll get zero.

So when you subtract your projection from the second vector you just end up with that vector as your result.

Does that make sense?

[carmine terrace]

https://i.imgur.com/pl2pjZq.png

anyone know how to find d?

I'm having problems finding it

[cold ravine]

As a side note to my original request, this is what I have written down.

[twilit terrace]

just want to clarify some things about notation

in the above picture, when it say T(v1), do they mean the 1st column of T (that's it) or that T acting on basis vector v1 (i.e T v1 = T(v1)) gives me the 1st column of T

[half ice]

T(v1) is the transformation acting on v1

Which is a basis vector here @twilit terrace

[twilit terrace]

i tried doing it that with some numbers tbh(should i post it here?) but the results don't match with T acting on v1

hmmm

[half ice]

Doing what with some numbers? This is a definition

Or do you mean that theorem below?

[twilit terrace]

the top part

[half ice]

Hey, nice excel work

[twilit terrace]

^^

[half ice]

Yes, they aren't the same.

Note that v1 in the top is not v1 in the bottom

[twilit terrace]

yup

i was using 2 different bases

i know that the Ts would be different in terms of physical meaning(same element, different bases), but was trying to show that in the 2nd case, T(v1) = 1st column, which i couldn't get it

[half ice]

T(v1) = (2,1,3)

In the second case. If your math is right.

I feel you might be trying to suggest that both cases should be the same. Why?

[twilit terrace]

erm...not sure if i can word it well,
I mean, given T(v1) = 1st column of T,
so, i just try to prove for myself?

and stumbled here

i was almost convincing myself that T(vi) is just a way of calling the ith column of T rather than T acting on any vector, but now, i am confused again

[wintry steppe]

if v1, v2, v3 are the standard basis, then T(vi) = the ith column of the matrix representing T

this is a bit more fundamental than you would think

[coral sage]

Hey can anyone explain how principal component analysis works to a 2nd semester linear algebra student

I want to understand word2vec better

[sinful heron]

I'm confused as how to go about this question please help

[minor galleon]

@sinful heron get the vector line equation

u have P=r + kv where v is the direction vector

work out requirement for k such that distance is 3

then add n subtract

[sinful heron]

okay ill try the vector line equation but I also don't quite understand what it means by distance 3

does it have to do with the length of the vectors

[vast torrent]

@wintry steppe did you figure out the vandermonde thing

[wintry steppe]

yeah

Well, a mate of mine did

[vast torrent]

Can i hear how to do it

[wintry steppe]

It's a lot easier if you go at it from the right side to the left side

By manipulation you get a triangular matrix with all the factors that the Vandermond matrix has

[vast torrent]

Ah

[wintry steppe]

And the manipulation is: all rows - first row; factorize what you can (on the first step you're able to factorize (d-a)(d-b)(d-c))

And then expand along the first column

At least that's how I found it easier to work with

[vast torrent]

Ah so you did use the formula (c-b)...(b-a)

[wintry steppe]

Ye ye

We know the LHS = (d-a)...

So let's see if it's the same for the right side

Is basically what my mate told me when he figured it out

Still an awful fucking problem

😠

[carmine terrace]

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

does any1 know how they get 4(sqrt26)/13?

I've been trying to figure out how to find the shortest distance for a long time and this is the part I think I'm getting stuck on

[clever cedar]

No sorry

They just used length formula

[carmine terrace]

thanks That was my first thought but I used the wrong vector somehow...

[carmine terrace]

Repost as I posted it wrong

https://i.imgur.com/QTW0kLc.png

I'm trying to find shortest distance and point Q, IDK what I'm doing wrong

[opal plaza]

kinda struggling with where to start with this question could use a push in the right direction ❤️

[quartz compass]

well first off, what are the dimensions of the matrix for T?

can only be two possibilities, 3x4 or 4x3 right, which is it? @opal plaza

[opal plaza]

umm it would have to be 4x3

so I'd have to figure out what T is?

1 1 0
0 0 0
2 0 1
0 1 1

Would it be the first two colums of T, since one row is all zeros?

[clever cedar]

@carmine terrace still need help?

[carmine terrace]

@clever cedar I was able to figure it out. Thanks for offering the help

[clever cedar]

nice, np

that question was on my exma last friday

exam*

so make sure u know it :p

[carmine terrace]

I did it so many times I have it engrained in my memory.

[wild pagoda]

Can someone explain the wording of question 8. I'm confused about the definition of e|x=7(f)

[slow scroll]

@wild pagoda it evaluates a function at x=7

[wild pagoda]

don't all functions equal f(7) when evaluated at x=7

[slow scroll]

consider f(x)=2x a function from R to R

[wild pagoda]

ok

does that not satisfy it?

[slow scroll]

f(7)= 14 which is not 7.so not all functions equal 7 when evaluated at 7

[wild pagoda]

ooo

I misunderstoof

so it's the set of f such that f(7) = 7

[slow scroll]

No, e is the linear functional that evaluates any function at x=7

So like e(x^2) = 49 for example

[wild pagoda]

O

ok

I get it

[coral sage]

hey, would someone mind reading my proof to make sure it's sound

i'm too tired to read it myself and have it make any sense

here it be, out of Axler's linear algebra done right

this is literally the longest and hardest proof ive ever written so im like nearly certain it's bunk

[dire delta]

i think ive got this one down intuitively but im just checking

I want to make a convolutional neural network that scans over the elements of a matrix where the matrix is of form

\begin{bmatrix}
a_1_1 & a_2_1 & a_3_1\
a_1_2 & a_2_2 & a_3_2\
a_1_3 & a_2_3 & a_3_3
\end{bmatrix}

[stoic python]

The Red Team Is Online:
Compile Error! Click the reaction for details. (You may edit your message)

[dire delta]

where the rows and columns can be arranged arbitrarily

what is the complexity of doing the 2x2 convolution over every arrangement of row and column for an nxn square matrix?

Without any optimizations my naive answer is (n!)^2 which is... completly unfeasible but since I'm after the 2x2 convolutions specifically I'm trying to figure out how much complexity can be removed through symetries

its proably something basic I've forgotten / not realised I can apply here though thank you in advance for any help

[wintry steppe]

In triangle ABC let P and Q be points such that B->P = 1/3 B->C and A->Q = 1/3 A->C.
S is the intersection of lines AP and BQ.
Find A->S as a linear combination of A->B and B->C

Graphically it seems to look like
A->S = 2/3 A->B + 2/9 B->C

But I can't figure out how to get there 'normally'

[frosty kraken]

If S is a finite subset of Rn
and Span S=V, then S is always basis for V

what would be the answer to that t/f question?

[empty copper]

Well... I'd assume yes, obviously, unless I'm missing something

[dusky epoch]

you are

[empty copper]

ohh basis

Yeah some elements of S might be linearly dependent on each other

Which isn't allowed to happen in a basis

gotcha

[dusky epoch]

ew wording

[empty copper]

engineer wording

[frosty kraken]

ty thats what I was thinking but wasn't sure

[wintry steppe]

hi, im trying to find if a system of linear equations has an infinite number of solutions, the equations evaluate the same when i sub one value of a into them but differently when i sub the other value of a into them, also i only came up with one value of b

have i done it right?

sending pic

i misread the question sorry pff

the question asked for which pairs of a,b does it have infinite solutions..both pairs won't work

[sour spear]

Can you send a picture of the equations

@wintry steppe

[wintry steppe]

x-ay = 1
ax-4y = b

is that what you mean?

[sour spear]

So you need to find the combinations of a and b resulting in infinite solutions, yes?

[wintry steppe]

yeah

i found out it's a=2, b=2 but i only got one value for b anyway, but for a i got 2 or -2

idk if i should have 2 values for b

[sour spear]

I think you would have 2 b values.

Rearrange the first equation in terms of x and plug that into the second equation.

If a is 2 or -2, you'll still have the y terms cancelling, yielding a = b

[wintry steppe]

so the pairs of solutions are a=2, b=2 and a=-2, b=-2 ?

[sour spear]

Yes

[wintry steppe]

im struggling to do that rearrangement, this is what i got so far

x = 1-ay (first eq. rearranged)

2(1-ay) - 4y = b

2-2ay-4y = b

[sour spear]

In the second line, keep it as a(1 - ay) before plugging 2 in for a.

Oh wait

1 + ay = x

In your first step

[cobalt tartan]

If we have $[L]_B = P^{-1}[L]P$, where $P^{-1}$, $[L]$ and $P$ are matrices, can we do
$P^{-1}[L]P - \lambda I = P^{-1}[L]P - \lambda P^{-1}P = P^{-1}([L]P - \lambda P) = P^{-1}([L] - \lambda I)P$?

[stoic python]

Liria ^(;,;)^:

[cobalt tartan]

Or do matrices not work like that

[thorn egret]

im trying to relearn everything in linear algebra so i can actually understand anything else my professor defines

can someone help me with this definition

read this in english

[quartz compass]

what it's basically saying is they have some operation that takes two things from V and maps them to 1 thing from V

and specifically that map is (x,y) |-> x+y

that make sense or still not completely clear yet @thorn egret

[thorn egret]

i mean i get all of it from the beginning

im not completely new to this

but im trying to figure out if there's a clear formal meaning to all of this, like even the ","

[sonic osprey]

There's no formal meaning to the comma no

The rest of it, yes there are formal meanings to it

[thorn egret]

i know that | means "such that"

[sonic osprey]

The : means that + is a function

Then it describes the domain and codomain of this function respectively

[thorn egret]

yeah i get that

the reason i want to relearn all of this is because

when i get into the real rigorous stuff

everything is defined as an ordered n-tuple

a "vector spaces over a space F" is defined as an ordered 3-tuple V = (V, +, *)

such that : (image above)

also im aware its not called a 3-tuple but my english is really bad

[sonic osprey]

3-tuple is what you would call it

Yeah depending on what linear algebra you're exactly trying to learn

You might not need all the formalism

[thorn egret]

"The relation ρ on a set A is every subset of cartesian products
A^n = A x A x ... x A"

i have no idea what this means

[sonic osprey]

I mean yeah

If you're actually trying to learn this stuff, don't really look for linear algebra stuff

Look for some intro to proofs stuff

[thorn egret]

basically all of my courses are divided into theory and practice, i get half the points from a practical exam and the other half from an oral theory exam on the whiteboard in front of my professor

[wintry steppe]

@sour spear thanks

[thorn egret]

i don't know if i have to learn all of this or not

[vast torrent]

linear algebra is sometimes the course students are introduced to basic proof concepts

but seeing as the vector spaces you first learn about are cartesian products of fields numbers, you should know what the cartesian product of sets means

and a relation is a subset of the cartesian product

that's a definition, not a theorem

[thorn egret]

thanks

[sonic osprey]

But yeah, looking for linear algebra resources on this isn't really what you want

Pick up an intro to proofs book

Like How to Prove It by Velleman

for information on this type of stuff

[thorn egret]

you're right

i just spent so much time trying to figure out the difference between a ring and a field

and i realized we will never mention rings again in linear algebra

lol

[sonic osprey]

Yeah that's what I mean

A lot of this formalism isn't actually required or useful

[quartz compass]

probably the main takeaway for rings in linear algebra to remember is you can add and multiply matrices, but sometimes you can multiply 2 nonzero matrices together to get the 0 matrix, and so you don't always have inverses to matrices meaning it's not a field

[thorn egret]

🤔

[vast torrent]

a matrix can be singular and not nilpotent

@quartz compass

[quartz compass]

so what

[vast torrent]

Merosity

probably the main takeaway for rings in linear algebra to remember is you can add and multiply matrices, but sometimes you can multiply 2 nonzero matrices together to get the 0 matrix, and so you don't always have inverses to matrices meaning it's not a field

Nov 20, 2019, 9:53 PM

just not sure waht you meant by that

[quartz compass]

just saying one example to help him see a difference between a ring and a field cause he doesn't seem to really remember

[vast torrent]

oh okay

[quartz compass]

if he's gonna forget it anyways, might as well say that one

[thorn egret]

last one, how do i read this and what does it mean

[vast torrent]

cant you have fields of finite characteristic? or am I thinking of rings

[quartz compass]

in a matrix? yeah

[thorn egret]

i mean specifically R^4[X]

i know what to do with it when i see it, i just dont know formally what it really is

[vast torrent]

maybe it means degree 4 or less polynomials in the variable X with real coefficients?

I'm guessing

actually no

I think what it means

is the set of 4-tuples, each of which is a polynomial in the variable X with real coefficients

so an example of an element of R4[X], if my guess is correct

would be (1+x^2, x^5, 1-3x+5x^11, 0)

[thorn egret]

the first thing it asks is to prove that V is a subspace of R^4[X]

so R^4[X] is a vector space, right

[vast torrent]

yes, an infinite dimensional one

[thorn egret]

wym infinite dimensional

[vast torrent]

know what a basis is

[thorn egret]

yeah

[vast torrent]

any basis of it has infinitely many things in it

[thorn egret]

oh ye

the 2nd thing it asks is to find dim(V)

we learn how to do all of these things in practice but i dont really pay attention in class when we explain theory

[vast torrent]

if it's the first class of linear algebra they will just want you to prove its infinite. or maybe whether or not it's countable or uncountable

if you haven't seen "countable" or "uncountably infinite" before then they just want you to show its not finite dimensional

but I'm not 100% sure that's what the notation means so Id prefer it if someone could confirm my guess

[sonic osprey]

notation of whta

[vast torrent]

$\mathbb R^4[X]$

[stoic python]

gfauxpas:

[vast torrent]

I guessed it's 4-tuples of polynomials with real coefficients

[sonic osprey]

Oh this

This usually means polynomials up to and including degree four

At least in a linear algebra context

[vast torrent]

oh, for that I've seen $P_4(\mathbb R)$

[stoic python]

gfauxpas:

[vast torrent]

But if R[X] means polynomials with real coefficients than R4[X] should mean what I said

oh well

[quartz compass]

yeah R^4[X] makes me uneasy like they're talking about polynomials of the form a+bx+cx^2+dx^3, I prefer the other notation

[vast torrent]

That's what galois says it means in lin alg

[quartz compass]

no he said something different

I put a degree 3 polynomial

cause there are 4 real numbers

[vast torrent]

Oh i see

Well the first term is like x⁰

[quartz compass]

lol

[vast torrent]

No joke

[quartz compass]

I know

I don't know what you think you're explaining to me, I was just expressing my discontent with that notation

I don't need a response really

[thorn egret]

i just always thought of R^4[X] as all vectors (u, v, w, t)

[quartz compass]

wew

that's pretty much what we agreed it isn't @thorn egret

[thorn egret]

lol

[quartz compass]

if it's a degree 4 polynomial it would be like vectors of the form (a,b,c,d,e)

[thorn egret]

we literally don't even mention polynomials in this part of the course

[quartz compass]

ok I'm out of this channel for the next hour good luck lol

[vast torrent]

S[X] means polynomials of one variable

[quartz compass]

sort out what the notation should be from your book or teacher

[vast torrent]

That's what [X] in set notation means

You're right that there's an obvious isomorphism between polynomials with real coefficients and tuples of reals

[silver ore]

can I ask a very very quick question

What does W^V even mean

[thorn egret]

oh here we go

lol

[vast torrent]

Functions with domain V and codomain W

[thorn egret]

oh

[silver ore]

is that all

[thorn egret]

really

[vast torrent]

Yes

[silver ore]

I feel stupid for worrying

[vast torrent]

L(V,W) means those functions for which + and . Are linear

Note that the order is counterintuitive

Domain V, codomain W

Is W^V

[silver ore]

yeah I knew

just the W^V part

[vast torrent]

K

[silver ore]

they had the explanation of the L(V,W) thing above

but not the W^V thing

its 10:30pm now

gotta sleep

[vast torrent]

Bye

[silver ore]

Bye

[half sentinel]

Could I ask a question?

I did part a but part b is kind of tripping me up

I first thought about doing a geometric series, but doing that proof of allowing that is hard

[vast torrent]

No you don't want to do that

multiply both sides by id-T should be easier

@half sentinel

Show that id = ... (id - T)

[half sentinel]

really?>

that's it????

[vast torrent]

Well to me that would be easier

To show

[half sentinel]

makes sense

then how would you do part c

would it be multiplying (s-t) to both sides of the idv?

[vast torrent]

Well if you derived what operator L satisfies L(S-T) = id

Then L = (S-T)‐¹

It's not always easier, but it's equivalent

[half sentinel]

oh

that was it

man

[vast torrent]

That's what inverse means

I'm sure there's more than one way to do it

Terminating geometric series for inverses of nilpotent operators you can prove by induction if you want

Where 1/(1-T) is (id-T)‐¹

So that would be another way

Good practice if you're bad at induction 👍

[half sentinel]

i;ll try doing it that way again

thanks for the help!

[vast torrent]

Np

[clever cedar]

when it says T is induced by the matrix A

does it mean that the linear transformation for T occurs because of A?

[vast torrent]

Actually smores you can multiply both sides by id-T and still use induction

@half sentinel

It means T maps x to Ax

And S maps x to Bx

[clever cedar]

right on, i see

tyvm

[vast torrent]

Np

[clever cedar]

what does it mean for $T: {R}^{n}\rightarrow{R}^{n}$ to be linear

[stoic python]

mike0x81:

[clever cedar]

i understand that T is a linear transformation

but what does it mean for it to simply just be 'linear'

[sonic osprey]

the same thing

[clever cedar]

oh thank you

is there a difference between $\mapsto$ and $\rightarrow$

[stoic python]

mike0x81:

[clever cedar]

in the context of linear transformations?

[vast torrent]

the regular arrow means domain:codomain

the arrow with a flat tail means element of domain: element in codomain mapped to

$f: \mathbb R \to \mathbb R \times \mathbb R, x \mapsto (2x, x+1)$

[stoic python]

gfauxpas:

[vast torrent]

and linear can be different than linear transformation

consider the indefinite integral

it's linear, but it's not a transformation

[clever cedar]

idk what indefinite integral is

soz

[vast torrent]

didnt take calculus?

lemme think of another example

[clever cedar]

currently in calc 1

[vast torrent]

well you know what the definition of a function is, right

[clever cedar]

yh

[vast torrent]

if something isn't a function, like for example it's not defined for all value sin the domain

it can still be linear, in that when it's defined, it's linear for those elements for which it's defined

or, if something isn't a function because it's many valued

it sends one thing to two things

it can still be linear, but it's not a transformation

do you want an example

[clever cedar]

hm okay gonna take me a sec to swallow that

appreciate the detail

[vast torrent]

let's say there's the price function, that tells you how much something sells for online

then it might be linear, because price(2 pizzas) = 2 price(pizza)

but the price function isn't really a function, because not everything has a price

[clever cedar]

right

sorry how does this relate back to transformations

ive lost track

[vast torrent]

a transformation is a functio

n

something can be linear and not be a function, so then it's not a linear transformation

[clever cedar]

oh i see

[vast torrent]

you'll see an example of something that's linear and not a function because it's many valued at the end of calc 1

called the indefinite integral

it sends functions to a family of functions

[clever cedar]

interesting,

i hope i get to learn it

appreciate ur help

[vast torrent]

there's always khan academy

np

[paper egret]

Just to clarify, if you're trying to do a change of bases from some arbitrary base B to the standard basis, all you gotta do is take your original vector from B and multiply it by that basis to get the coordinate in standard basis

i hope that makes sense lol

[paper egret]

also, can anyone here clarify what it means for a transofmration to be a basis B
$[T]_B$

[stoic python]

MemesPlease:

[north sierra]

hi

is that good first steps?

[half ice]

@paper egret
I usually see [T]B refer to the matrix constructed by the basis vectors after they've been transformed

[paper egret]

hmmm ic ic...

i actually have an explicit definition, but for some reason can't make some sense out of it

$[T]B = P{B \to E} [T]E P{E \to B}$

[stoic python]

MemesPlease:

[paper egret]

oh wait the arrows go reverse direction <-

[half sentinel]

@opaque sun , we have (A − λI)x = 0

so we want to find x, in the case of a 2x2 matrix, x = (x, y)^T

we can then find the values accordingly

you can do that as well

right?

x = (x,y)^T is just the same thing but now like a column like that picture

[sour spear]

Solve for x in terms of y and z from that equation

[half sentinel]

right

[sour spear]

And then make 2 eigenvectors from the coefficients of y and z

I.e y * (-1/2 1 0)

Imagine that's vertical. I don't know the code for the bot

If you think of the answer like x = -0.5y -z, y = y, and z = z

Then the eigenvectors are the coefficients of y and z

[paper egret]

free variable

[sour spear]

^

Solving a matrix/system of equations gives you x from the first equation, y from the second, and z from the third

But the y and z equations are all 0s

So y and z can be anything

Those 2 in the picture are right (it's your equation times 2 to get rid of the 1/2)

You want to split up the coefficients into y * (matrix of y coefficients) + z * (matrix of z coefficients)

When I get back to my computer I could type it out and it would probably make a bit more sense

(-1/2)
( 1 ) * y
( 0 )

(-1)
(0) * z
(1)

If I were on my computer it'd be prettier. I'll be back on it in 10 minutes or so.

@opaque sun

x = -1/2y - z

Which you got from the (1 1/2 1) matrix with the bottom 2 rows as all 0s

and the coefficient matrices would be your eigenvectors

[half sentinel]

Hey guys, could you help me out with this problem

[sonic osprey]

what are you stuck on

[half sentinel]

so I cannot really see the proof

especially the hint

[sonic osprey]

Have you tried doing the proof

[half sentinel]

Well yes

[sonic osprey]

Okay, and what did you do

[half sentinel]

so i did n1 of (x,y) and then to (y,z)

and added the sums

[sonic osprey]

what's n1

what's z

[half sentinel]

of which i got an inequality that the addition of (x,y) and (y,z) to be greater than sum of (x,z)

n1 is norm 1

and z is some other vector

and I feel like that's not what it's asking for

[sonic osprey]

Sure that's the triangle inequality

[half sentinel]

wait

[sonic osprey]

But that's the triangle inequality for metrics

[half sentinel]

right

[sonic osprey]

It says in the parentheses

What they want

[half sentinel]

sorry if it's a dumb question but would the first step be taking the norm of itself then?

[clever cedar]

for a surjection does the Transformation function have to be of the same R^n as the vector X it transforms onto

[sonic osprey]

I mean

Read what it asks you to show

Write it out

[clever cedar]

im not sure who ur talking to 😦

[sonic osprey]

not you

[clever cedar]

😿

[sonic osprey]

the answer to your question is no

[clever cedar]

oh cool ty

[half sentinel]

hi @sonic osprey okay, so I eventually got that (x,x) for || x || 1 becomes 0, and that's why it does not satisfy the parallelogram law, correct?

i meant | | x | | 1

[clever cedar]

is surjective and injective mutually exclusive?

[terse mirage]

no?

because Bijective functions exist

[clever cedar]

idk what bijective is

just learning about surjective and injective now

didnt mean to offend :p

[terse mirage]

sorry i didnt realize

[clever cedar]

no problem

[terse mirage]

bijections are defined to be maps which are both surjective and injective

[slow scroll]

bijective <=> invertible <=> columns of matrix form basis for target space <=> image of transformation is the target space

[terse mirage]

^

also trivial kernal

[clever cedar]

oo wow ty

[slow scroll]

ah yes, trivial kernel

[clever cedar]

the reason i ask is this

[slow scroll]

and nonzero determinant

[clever cedar]

it says if rank is n then its injective

but rank is m then surjective

so thats why

i asked

[slow scroll]

well if rank(T) = m then that means the column space / Im(T) has dimension m. Well, the smallest subspace of R^m which has dimension m is R^m itself. Therefore T is surjective Does that make sense?

[clever cedar]

Yeah I believe so, seems straight forward

[slow scroll]

In other words, since Im(T) = R^m, every element in the codomain gets mapped to, i.e. Tx = b has a solution for all b in R^m

[clever cedar]

ahhh okay

thats more clear

ty

[lapis current]

Can someone help me solve this

<@&286206848099549185>

[uneven bloom]

Wrong channel. This is not college math.

[lapis current]

Sorry

[uneven bloom]

Go to #prealg-and-algebra

[clever cedar]

what is an x-shear

[feral mountain]

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

[clever cedar]

appreciate it

[brittle orchid]

Hi, I was wondering which of the following ebooks is a recommended resource to understand fields and vector spaces?

• Linear Algebra Done Wrong
• Elementary Linear Algebra
• Linear Algebra Third Editiion
• Linear Algebra with Applications
• Abstract Algebra
  I'm looking for a text which is a little noob-friendly 😄

[feral mountain]

• Linear Algebra Done Wrong

oh it is a thing

[brittle orchid]

yup, starting with that, in fact 😛

[wintry steppe]

hi, im trying to solve using the Gauss method, im not sure where i've gone wrong here

for each row operation, can we only do it so that it makes a 0 or a 1?

[brittle orchid]

what have you done in the second operation btw

[wintry steppe]

added rows 2 and 1

[brittle orchid]

my bad I didn't realise you wrote the operation above the matrix

wait

if you added rows 2 and 1 how did you get 0

for the first element of the second row

[wintry steppe]

ahh

i was too focused on keeping it 0, i forgot to add to it

but then the operation is wrong

[brittle orchid]

tbh my rre/gaussian elimination isn't particularly good as I'm pretty new to it myself xD

I was just having a read of this idk it might help you

http://www.math.jhu.edu/~bernstein/math201/RREF.pdf

[wintry steppe]

ok thanks

[half ice]

@brittle orchid
Linear algebra done wrong is a great book, but may be pretty technical if you've never done anything technical. If you're really new, take an easy run through an applied linear algebra book. 3b1b can be a great complement to this

Afterwards, come back and get the details right

[minor galleon]

Ok does anyone have any idea how to start finding curves that are invariant under a 2d matrix transformation?

[timber minnow]

sounds like an eigenvector no?

with eigenvalue 1?

[half ice]

Curves? 2d transformation? What space are you working with?

[minor galleon]

It’s just a student at a class I’m a supervisor over has posed a question I said I would get back to him

Basically Cartesian transformation matrix a 2x2 one

[timber minnow]

wat?

[minor galleon]

How would I go about invariant curves in a way that doesn’t really involve eigenvectors stuff cuz they haven’t learnt it yet

We have invariant lines but what about invariant curves

[timber minnow]

you need to be more explicit...

[half ice]

"curves" is an odd term here. A space has vectors in it. If you do mean vectors, davidL just hit the nail on the head

These are the vectors where T(v) = v

[timber minnow]

if you want to "be more general" (?) you can say eigenfunction, with eigenvalue 1

[minor galleon]

No I literally mean say y=x^2 is an invariant curve of the matrix transformation (-1, 0),(0,1)

We ofc have infinitely many solutions but I was simply wondering whether there are any similar things for other 2x2 matrix transformations

[half ice]

I think you may mean the vectors (x, x²) for some x?

This transformation is R² → R², correct?

[timber minnow]

I think he wants like L2 or smth

[minor galleon]

Yeah

[half ice]

Because there are spaces of quadratics, but you'd need a 3×3 matrix

[timber minnow]

I don't think he means like polynomial basis

I think he means from function space to function space

where he applies some operator

and the function remains unchanged

but like, these are just statements (?), I don't understand what his question is

[half ice]

But then that's infinite dimensional. Even worse, you'd need to be explicit about a basis

[timber minnow]

"can I explain this without eigen-shits" not well

it seems like you don't really know what you're looking for (?)

I would not try to explain this to other people, especially at a very reductive level

I probably don't have enough background to cover this well either. I haven't taken functional yet

@half ice he doesn't mean vector space

but yes, I agree

he needs to specify many things before he can get any kind of meaningful response

@minor galleon care to clarify?

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

this is what you're looking for

[half ice]

I do think you mean R² → R², since you have a 2D matrix. all eigenvectors form a subspace, which in this case implies one of the following:

• They don't exist
• They are a line through the origin
• They are the entire plane (think of "do nothing")

[timber minnow]

I think they are just inconsistent enough that we can both be right XD

[half ice]

Nothing wrong with more info

[timber minnow]

I'm talking about the eigenfunctions of some linear operator (i.e. matrix) on some function space

but they specified 2D cartesian matrix, so ???

that implies what you said

[wintry steppe]

i have 3 vectors that create a subspace of R4, how do i calculate its dimension?

[gray dust]

cram the vectors into a matrix, row reduce, count the pivots

[wintry steppe]

the matrix is irregular

[dusky epoch]

what do you mean by irregular

[wintry steppe]

i mean i can row reduce but at least one of the vectors is linearly dependent

[gray dust]

you mean the set of vectors is LD

[wintry steppe]

uh

yes

[gray dust]

so the subspace dim < 3

[wintry steppe]

ok so the last row is all 0

thats it?

[gray dust]

cram the vectors into a matrix, row reduce, count the pivots

[north sierra]

the dimension is just the number of vectors in the basis right

[wintry steppe]

may i ask why its not 3 but < 3?

[north sierra]

cause its dependent

[wintry steppe]

o

[north sierra]

so itll ❤

[gray dust]

dim of subpsace=minimum number of lin indep vectors from that subspace needed to form a basis for that subspace

[north sierra]

< 3

i see

[wintry steppe]

its not possible for 3 vectors in R4 to create a 3d subspace is it?

[gray dust]

why do you say that

[wintry steppe]

ok maybe this is specific to my example but when you row reduce the last row will always be 0 in a 4x3 matrix no?

[gray dust]

think outside row reducing

the dim of the subspace is 3 if the set of those vectors is LI. consider the span of a single vector, then an LI set of 2 vecs, then an LI set of 3 vecs

[wintry steppe]

im having a hard time imaginging all this stuff

[gray dust]

3b1b's LA series has nice visuals

[wintry steppe]

right

[brittle orchid]

@half ice Thank you

Btw, could someone explain what's going on with the dot operator thing here

[wintry steppe]

@gray dust but even so, theres no way 3 vectors with each 4 elements can be linearly independent, no?

[gray dust]

As an easy example, pick any 3 of the R^4 standard basis vectors

[vast torrent]

@brittle orchid this is called a Cayley Table and it's a generalization of multiplication tables from elementary school . Which entry is bothering you

[brittle orchid]

the alpha and alpha+1

[vast torrent]

Have you seen finite groups before

[brittle orchid]

I think not

[vast torrent]

So in this context

This is like a set of 4 "numbers"

That you can multiply and add to get other "numbers"

[brittle orchid]

yea

but what does the dot operator at the top left even do

[vast torrent]

So alpha and alpha +1 are two of those elements

This table is tellung you precisely what dot does

[brittle orchid]

what's the relationship between each row, column and entry

[vast torrent]

(a+1).(a)=1

It's telling you how dot is defined

Because there are only 4 elements

You only need 16 boxes to tell you what dot does

[brittle orchid]

Right

So it's not a standard operator so to speak?

The table describes the dot instead of the dot describing the table?

[vast torrent]

Unless you have another description of dot from somewhere else

[brittle orchid]

I thought the dot acted as a modulo operator

for some reason 😕

[vast torrent]

Well maybe it does for F4

Do they give you a table for +?

[brittle orchid]

Btw what's the difference between F subscript n and F superscript

Sorry for spam but I'm just showing you what I've come across so far

[vast torrent]

It's fins

Fine

[brittle orchid]

I mean the way it is described it acs as the modulo operator doesn't it?

[vast torrent]

Sure

[brittle orchid]

and if that's the case for integers n >= 2, then why isn't that the case for F4?

[vast torrent]

So basically what you're discovering is

Fp is isomorphic to Z/pZ

That symbol means Integers modulo p

P is prime

[brittle orchid]

isomorphic? xD

Please don't take anything for granted, assume I'm an absolute idiot 😄

[vast torrent]

"Basically the same"

Iso=same, morphic=shape

[brittle orchid]

ah ok

[vast torrent]

Two fields are isomorphic if they're "basically the same"

[brittle orchid]

ah ok

So um, I was given a homework sheet with 4 vectors in F4, and I had to decide whether or not they're linearly independent when the field = rational.
I did that by reducing to RRE form and showing that all lambda values = 0

But I kinda don't get where any of this is going

I mean I get how to reduce to RRE but I have no idea why I'm doing so, nor what it really means for something to be linear independent, linear dependent or even linear

and I've especially got no clue of how to determine whether or not something is linearly dependent when F = F5

Since I'm conceptually really shaky

[vast torrent]

lambda values? is that the same as eigenvalues?

[wintry steppe]

the way i understood it is that if 2 vectors are linearly dependent then one is a scalar multiple of another

[brittle orchid]

um sorry

the coefficients?

[vast torrent]

wait so the vector space is F_4 and the scalar field is Q?

[brittle orchid]

I meant those lamba values

sorry about that

[wintry steppe]

those are scalars

[vast torrent]

there are two equivalent ways to think about linear independence

at least 2

one is that the only solution to

$\lambda_1 v_1 + \cdots \lambda_m v_m = 0$

[stoic python]

gfauxpas:

[vast torrent]

is if all the scalars are 0

the other way to think about linear independence is

[brittle orchid]

yup, I proved that all the scalars are zero

[vast torrent]

a set {v1,v2,...,vm} is l.i. if and only if no vector is a linear combination of the other vectors

so for example, this set is not linearly indepdent. give me a sec while I write it

[brittle orchid]

Hmm

[vast torrent]

$\left{ { \begin{bmatrix} 1 & 2 & 0 \end{bmatrix}, \begin{bmatrix} -1 & -1 & 1 \end{bmatrix}, \begin{bmatrix} -1 & 0 & -2 \end{bmatrix} } \right}$

pretend they're columns

[brittle orchid]

yea

[vast torrent]

if you notice, the first one minus two of the second one is the third one

[brittle orchid]

yup

[stoic python]

gfauxpas:

[vast torrent]

so the set isn't linearly independent

if I call those vectors x1, x2, x3

then the only scalars c1 x1 + c2x2 + c3x3 = 0 are c1=c2=c3=0

the two things I said are equivalent

[brittle orchid]

I don't understand what this has to do with the LI: if you notice, the first one minus two of the second one is the third one

[wintry steppe]

should be +2r_2 btw

and then v3a3 is 2 not -2

[vast torrent]

oh thanks

yeah I meant that

[brittle orchid]

really

[wintry steppe]

what hes trying to say is that the 3rd vector is a disguised 1st vector (?)

[vast torrent]

yeah the entry of v3 is 2, not minus 2

my bad

typo

third entry

the third vector can be written in terms of the first two vectors

[brittle orchid]

yea

[wintry steppe]

so its a linear combination of the other 2

[brittle orchid]

oh that's what it means to be a linear combination of something else?

[wintry steppe]

the linear combination is the set of all vectors you can reach by scaling a set of vectors i think

[vast torrent]

yes, you can write it in terms of the other ones with scalar coefficients, uniquely

[wintry steppe]

aka the span

[brittle orchid]

ngl I feel so dumb right now, like when you're introduced to the atomic model for the first time and you're studying this stuff but have no idea why or what it actually means

[vast torrent]

the span is the set of ALL linear combinations

[brittle orchid]

ie you have no tangible concept of it

[wintry steppe]

oh oops

[vast torrent]

linear combination means a (finite) sum with scalar coefficients

[brittle orchid]

So, what exactly does $F_5$ mean and what does $F^5$ mean

[stoic python]

WhiteBerry:

[half ice]

I see above they're giving you Cayley tables to describe fields. I understand why you're confused, these normally are explained in a seperate course. To note, the things in a Cayley table are not vectors

[brittle orchid]

I feel like the subscript and superscript are used interchangably (they're obviously not) but I can't find anything explicit in the notes for all of the actual notation and terminology

[vast torrent]

yeah im kind of surprised theyre showing it to you in linear algebra

[half ice]

F5 is the field of 5 elements. That is, this is 5 "things" on which you can define addition, subtraction, multiplication, and division

[brittle orchid]

is that sub or super

[vast torrent]

$F^5$ often means $F \times F \times F \times F \times F$

[stoic python]

gfauxpas:

[half ice]

This structure happens to be unique. There is exactly one field of 5 elements

[brittle orchid]

So when they say: Consider the following vectors in $F^4$

[vast torrent]

up to isomorphism, I briefly explained to him before Kaynex

[stoic python]

WhiteBerry:

[vast torrent]

$F_4, F^4$, or ${F_4}^4$?

[stoic python]

gfauxpas:

[brittle orchid]

The way I wrote it xD

$F^4$

[stoic python]

WhiteBerry:

[vast torrent]

do they say what F is

[brittle orchid]

May I just show you the question?

[vast torrent]

sure

[half ice]

"When F = Q"
"When F = F5"

Replace F with those, when proceeding

F⁴ means the vectors from the field F that contain 4 entries

[vast torrent]

in other words, FxFxFxF

[half ice]

So part A is asking if the vectors are LI over Q⁴

[brittle orchid]

Oh I see

[half ice]

This is pretty technical for a lin alg course, congrats if you're keeping up

[brittle orchid]

I only started attending lectures last week smh

😂 😂

Literally haven't handed in a single piece of homework

So now I'm grinding to catch up lol

Anyway so in the case F = F5

what does that even mean like if you're to try and understand the meaning of it

[half ice]

For a very simple explanation, F5 is the integers mod 5

[brittle orchid]

Sorry for dumb questions and I sincerely greatly appreciate your patience ^

[half ice]

So for example 3 + 4 = 2

[brittle orchid]

yea

[half ice]

Or 2×3 = 1

[brittle orchid]

yup, makes sense

and how do you check for LI over a field Fn

[vast torrent]

if they're tuples

you make a matrix

[brittle orchid]

yea

[vast torrent]

and then use that linear independence of the vectors holds iff linear indepdence of the reduced row echelon form holds

iff means if and only if

[brittle orchid]

What does the matrix look like and why? 😅

[vast torrent]

matrix looks like the column vectors glued together side by side

[brittle orchid]

I mean for F5 what would the matrix look like

[vast torrent]

how many vectors are you given and what's F

[brittle orchid]

4 vectors

[vast torrent]

5 rational numbers in each one?

[brittle orchid]

in F^4

[half ice]

Same thing, the only thing that will change is how you reduce

[vast torrent]

as to why it works

think about the sort of things you do when you row reduce

if, say, the vectors were related by

v1 = 2v2 - v3

and you replace v3 with 2v3

then v1 = 2v2 -1/2(2v3)

[brittle orchid]

Right

[vast torrent]

doesnt change the fact that they're not independent

certainly switching the order of the variables doesnt matter either

[brittle orchid]

I didn't mean why does RRE work

I meant why are we trying to solve this system of eqns

[vast torrent]

because you want to know if the vectors are linearly independent, so you're trying to turn the matrix into something where the answer is obvious

[brittle orchid]

Ahh

[vast torrent]

if you get $\begin{bmatrix} 3 & 1 & 0 \ 1 & \frac{1}{3} & 0 \end{bmatrix}$

[stoic python]

gfauxpas:

[vast torrent]

well then you can stop

because obviously the second vector is 1/3 the first plus the last

[brittle orchid]

Yeah

so that would prove that theyre independent

[vast torrent]

so you dont even need to finish row reducing if you see that one is the linear combination of the others if all you're asked to do is show independence or dependence

[brittle orchid]

I see

[vast torrent]

but if you completely row reduce it should be even more obvious whether theyre independent

[brittle orchid]

yeah

so when the scalars are = 0 the vectors are linearly independent?

else they can be expressed as a linear combination of..?

[vast torrent]

if the only solution to c1v1 + ... + cnvn = 0

is when c1=c2=...=cn = 0

[brittle orchid]

yeah

[vast torrent]

then and only then v1, v2, ..., vn are linearly independent

[brittle orchid]

Thanks a lot btw

I really appreciate it

[vast torrent]

it's worth thinking about why rather than just me telling you

np

about why those two ways to describe lin.ind. are equivalent

[brittle orchid]

I think I should read a bit of a book for a change

[vast torrent]

khan academy is good but he deals pretty much only with R^n rather than things like Q^n

but R^n and Q^n behave similarly

[brittle orchid]

Ah okay

brb lecture ;D

[wide warren]

anyone good in derivative, graphs and things like that in an early uni level ? 🙂

[quasi vale]

[wide warren]

So, I got f(x)= -2x^4+4x^3+10

With the question, Find the derivates Largest point

(English is not my first language as i am studying and trying to translate from swedish)

I used graphs, i derivated it and i got the largest point to be a point in (1,4)

BUT I am not sure how to get to this point and prove that the point is (1,4) on paper

without a graph

[quasi vale]

w/o a graph, you'd just differenciate the function, set the derivative = 0 if you're looking for maximum/minimum points

[wide warren]

W/o a graph its -8x^3 + 12x^2 = 0

Isnt it ?

because -8x^3+12x^2 is the second derivate

first derivate*

[quasi vale]

Yeah, then solve for x.

[wide warren]

when i did it i got (0, 1.5)

And it doesnt add up to the graph

[quasi vale]

How'd you do that, can I see the working?

[wide warren]

using mathway

for help atm

[quasi vale]

sec

[wide warren]

But more or less I had -2x^4 + 4x^3+10, I derivated this into -8x^3+12x^2

I put the -8x^3+12x^2 into the mathway and put it into graph mode, this gave me 5 points
X,Y
-1,20

0.0

1/2, 2

1,4

2,-16

[quasi vale]

Oh.

[wide warren]

the biggest point was 1,4

[quasi vale]

The question is to fine the derivatives' highest point, I get it.

[wide warren]

yes

[quasi vale]

That means to find the maximum/minimum points of the derivative function.

So what you have to do is take the derivative again, set it to 0, then solve for x.

[wide warren]

yeah, sorry my english isnt to good when it comes to translating school stuff

[quasi vale]

All good.

Take the derivative of -8x^3+12x^2.

[wide warren]

So i get f''(x)=-24x^2+12x

[quasi vale]

Set it to 0, then solve for x.

No, should be -24x^2+24x

[wide warren]

ah yeah

wait

nvm

so the largest value is 0, 0.5

[quasi vale]

How?

When you solve for x using -24x^2+24x, you get x=1.

[wide warren]

0,1**

[vapid coyote]

why is this in linear algebra?

[wide warren]

Because i dont know the english equalivement

for what its called in swedish

[quasi vale]

#calculus come hre

[vapid coyote]

probably calculus

linear algebra is about vectorspaces and linear functions

[wintry steppe]

so i have a transformation from R2 to R3
when it asks me to show the map as x |--> Ax, then it wants me to find the matrix i need to multiply the initial matrix with?

[brittle orchid]

Why are these two equations referred to as separate things?

Aren't both equations saying the exac same thing?

[wintry steppe]

the x are unknown scalars while a1, ... an are in the field K

at least thats how i would read it

[vast torrent]

First one is with constant coefficients, second one with variable coefficients

[wintry steppe]

when it asks me to show the map as x |--> Ax, then it wants me to find the matrix i need to multiply the initial matrix with?

am i just being a dumbass about this

[ionic steppe]

Could someone explain how they went from 3+6i+4i+8i^2 to that next step?

[wintry steppe]

i^2 = -1

therefore $ 3 + 8i^2 = -5 $

[stoic python]

yami:

[ionic steppe]

Thank you 🙂

[vast torrent]

@ @wintry steppe context?

[wintry steppe]

linear transformation

https://i.imgur.com/1P0IqJw.png

asking me whether F can be shown as x |--> Ax and if yes, to determine A

[vast torrent]

Oh, yeah, find the matrix M such that F(x)=Mx

Or A whatever

Or A whatever

Linear functions can be written in terms of matrices so if F is linear then it can be written as a 3x2 matrix

Linear functions can be written in terms of matrices so if F is linear then it can be written as a 3x2 matrix

[wintry steppe]

f is linear

[vast torrent]

Then A exists

[wintry steppe]

the kind of transformation where they dont get multiplied with each other is generally linear no?

[vapid coyote]

what do you mean by that?

[wintry steppe]

the kind of transformation that F is

only using scalars and addition/subtraction

[vast torrent]

If each entry is a linear combination in the variables

[vapid coyote]

that one is trivially linear

it doesn't make a difference if you write that as matrix or directly

[vast torrent]

F(x,y) =(sinx,siny) is not linear

[vapid coyote]

??

[vast torrent]

F(x,y) =(sinx,siny) is not linear

F(x,y)=(xy,y²) also not

[vapid coyote]

why is the text red

[wintry steppe]

yeah tahts what i meant @vast torrent

[vast torrent]

So the key is that to create a matrix for a linear transformation you only need to know how it acts on a basis

Do you see why?

[wintry steppe]

no

[vapid coyote]

It's not necessary to write linear functions in matrix form

[vast torrent]

That's the assignment mophra

[vapid coyote]

just for the standard basis

or for a different basis?

[vast torrent]

Standard

But im explaining a concept

If a vector v in terms of a basis e1,e2,
..,em is

v=c1e1+...+cmen

Then T(v)= c1T(e1)+...+cmT(em)

See?

[wintry steppe]

yeah thats one of the requirements for the transformation to be linear

[gray dust]

this may look better bashed out in latex, fauxpas

[wintry steppe]

or at least looks like it

[vast torrent]

But do you see how T(v) is wholly determined by how it acts on e1,e2,..,em

[wintry steppe]

yeah

[vast torrent]

Im on mobile, it's annoying to use latex

[vapid coyote]

it really is

[vast torrent]

Anyway, can you write c1Te1+...+cmTem in terms of a matrix multiplied on a vector [c1,...,cm]?

[gray dust]

$v=c_1e_1+...+c_me_m\\$if transformation $T$ is linear, $T(v)=c_1T(e_1)+...+c_mT(e_m)$

[stoic python]

RokettoJanpu:

$v=c_1e_1+...+c_me_m\\\\$if transformation $T$ is linear, $T(v)=c_1T(e_1)+...+c_mT(e_m)$

[vast torrent]

That's the matrix you want

$T(v)=c_1T(e_1)+...+c_mT(e_m)=M[c_1 $ \cdots & c_m]^\top$

[gray dust]

oh my, i'm sorry you had to do that on mobile

[stoic python]

gfauxpas:
Compile Error! Click the reaction for details. (You may edit your message)

[wintry steppe]

oh transposed?

[vast torrent]

I copy and pasted most of it

That's just to make it fit

M times a column vector

[wintry steppe]

hmm idk the matrix looks like an arbitrary m x n in my head

[vast torrent]

Someone on pc write that as a column vector to not be confusing

[wintry steppe]

np i get it

[vast torrent]

Tia

[gray dust]

$T(v)=c_1T(e_1)+...+c_mT(e_m)=M\begin{pmatrix}c_1\\vdots\c_m\end{pmatrix}$

[stoic python]

RokettoJanpu:

$T(v)=c_1T(e_1)+...+c_mT(e_m)=M\begin{pmatrix}c_1\\\vdots\\c_m\end{pmatrix}$

[vast torrent]

Think of the row.column or column.row interpretation of mtx multiplication

And think of each T(ei) as columns

Bbl

[wintry steppe]

Think of the row.column or column.row interpretation of mtx multiplication
not sure what you mean by this, are you talking about how the rows and columns interact with each other when multiplied?

[gray dust]

The goal is to write the linear transform as just left-multiplication of the input by some matrix

[wintry steppe]

yeah so like some M x (x1, x2)

[gray dust]

Aye

[wintry steppe]

and in order to find M i have to find out how T(x) behaves on a basis

[half ice]

Seems like you get the strategy. What's the question?

[wintry steppe]

how to apply it

lmao

[vast torrent]

https://www.khanacademy.org/math/linear-algebra/matrix-transformations/linear-transformations/v/linear-transformations-as-matrix-vector-products?playlist=linear-algebra

[vast torrent]

have 18 minutes?

[wintry steppe]

sure

[half ice]

Okay, what's F(1,0)?

[vast torrent]

watch the video I linked to, he explains it

[wintry steppe]

-1

[vast torrent]

needs to be a 3-tuple

[half ice]

Remember, F takes in vectors from R² and returns vectors from R³

[wintry steppe]

so -1,0,0 then?

seems wrong

oh lmao

$ \begin{bmatrix}
0 & -1 \
1 & 3 \
3 & -5
\end{bmatrix}
\begin{bmatrix}
x_1 \
x_2
\end{bmatrix} $

[stoic python]

yami:

[wintry steppe]

this is actually pretty cool

[wintry steppe]

so i understand how to compute the kernel but im not sure where it goes wrong

take the matrix above, put it into RREF and then x1 = x2 = 0?

[vast torrent]

It's possible for the kernel to be {0}

In fact, there's a very important theorem

Know what an injective function is? Also called a one-to-one function

[wintry steppe]

yeah

[vast torrent]

If T is a linear function