linear-algebra

[novel needle]

yeah i understand basically in laymen terms, k will increase and eventually i and j will = k

and thoses terms become 0

[native rampart]

Yea,but there will be exactly one term which will be nonzero

[novel needle]

so the only term is eijkeijk where ijk are not the same

yeah i fully understand now, thanks :))

[primal fable]

nvm
my dumbass forgot trace you add, not multiply

[native rampart]

1. No

[wintry steppe]

So a Vector Space is an algebraic structure of n-dimensional vectors over a Finite Field F with a set of operations, satisfying certain properties?

[native rampart]

You are assuming
1)the dimension is finite
2) the field is finite

[wintry steppe]

Yes

And what about no. 2?

[native rampart]

Need not be finite

R^n is also a vector space

[wintry steppe]

So no. 1 should be true?

[native rampart]

(Z/2Z) is a vector space over (Z/2Z)

[nocturne jewel]

any R-vector space with dim(V)=n will be isomorphic to R^n

[frank hedge]

stuck iwth part (a)

do we have to get parametric equations somehow ?

[quartz compass]

nope not necessary

first, what is the equation of the xy plane?

[wintry steppe]

👍

[fiery pike]

Determine the image of the point P(1, 3, 2) under the orthogonal reflection with respect to the plane
with equation x+ y+ 2z = 3 via the matrix formalism and afterwards check your result geometrically.

[wintry steppe]

Is this proof okay?

[coral temple]

Hello, I'm trying to prove that if $u, v$ are vectors in $\mathbb{R}^n$, so that $\left<u,v\right>>0$, then there is a symmetric positive definite matrix $A$ so that $Au=v$. Any hints on how to do this?

[stoic python]

Porphyrion

Porphyrion

[zinc timber]

@coral temple think about a LT s.t. Au=v similar to projection

[coral temple]

Okay, you can do a composition of a dilation and an orthogonal projection, so that Au=v. But the resulting matrix is only positive semi-definite. Is there a way to make it positive definite?

[zinc timber]

yea actually

as a hint say you have a diagonal [1 1 0 0 0]

you can add ϵ [ 0 0 1 1 1] to make it pd

starting with $A=\frac{vv^T}{v^Tu}$

[exotic vector]

$A = span{(1,1,1)}$ what is $A^{\circ}$ (the orthogonal of $A$).

[stoic python]

Salah

[zinc timber]

note that eigen values are ||v||²/⟨u,v⟩ and 0 with multiplicity n-1

so it's diagonalizable

[exotic vector]

sorry guys if i disturb

[zinc timber]

now pick orthonormal basis and add the P\eps P^-1 thing and that might work

@exotic vector there's an easier way and there's a hard one

[mystic dagger]

a canonical basis is an orthonormal basis for any given dot product?

[zinc timber]

I prefer the easy one

[zinc timber]

by canonical basis, I only know e_i's

[exotic vector]

yep show me

[zinc timber]

but sure you can always find one given any inner product

[zinc timber]

like since you are in lR³, A\perp has dim 2

so you can pick any 2 LI vector which is orthogonal to (1,1,1)

and then conclude A\perp is the span of those 2

@exotic vector

[exotic vector]

LI?

[zinc timber]

Linear Independent

[mystic dagger]

for this inner product
and the basis is the canonical basis from vector space of the the 2x2 matrices

[exotic vector]

ok but the issue we just cover what the perpondicular is and we don't know yet its dim

[mystic dagger]

i mean, no matter how you define the inner product, it's always orthonormal if it's the canonical basis of that vector space, right?

[exotic vector]

I think we gotta do it the hard way

[zinc timber]

[zinc timber]

i don't think that's what canonical means

or it could be

idk

count me out

it depends on what you want 'canonical' to mean

[exotic vector]

the canonical basis of $\bR^2$ is ${(1,0),(0,1)}$

[stoic python]

Salah

[exotic vector]

you can say it's the obvious basis

[zinc timber]

@coral temple does that work?

[coral temple]

I didnt fully understand the construction

[mystic dagger]

${ \begin{pmatrix}
1 & 0 \
0 & 0 \
\end{pmatrix}, \begin{pmatrix}
0 & 1 \
0 & 0 \
\end{pmatrix}, \begin{pmatrix}
0 & 0 \
1 & 0 \
\end{pmatrix}, \begin{pmatrix}
0 & 0 \
0 & 1 \
\end{pmatrix} }$

[stoic python]

leonardomoura

[mystic dagger]

this is the canonical basis for the vector space of the 2x2 matrices

idk the proper definition, i just see as the simplest basis

for the usual inner product, it's orthonormal

but i wanna make sure it's orthonormal for any given inner product

[zinc timber]

we start with $A=\frac{vv^T}{v^Tu}$

[coral temple]

Ok

[zinc timber]

since Au=v

but it's not +ve definite

since it's eigen values are ||v||²/⟨u,v⟩ and 0 (n-1 multiplicity)

is this clear?

[coral temple]

Yes

Wait no

OK yes

[zinc timber]

ok so can you see why A is diagonalizable

[coral temple]

Yeah

[zinc timber]

then let's say A= P Λ P' be the evd

then Λ= diag[1, 0,0,..0]

[coral temple]

Right

Wait, not 1

[zinc timber]

yeah not 1

[coral temple]

Something

[zinc timber]

mb

replace it with 1

anywau

take E= diag[0, 1,1..1]

then B=P ΛP' + PEP' then we get B= A+PEP'

it's+ve def as all eigen values are 1

ok let's add 2 and not 1

does it work?

or anything

[coral temple]

Yeah

Wait why is it symmetric?

[zinc timber]

P is orthonormal

[coral temple]

Oh ok

[zinc timber]

since A is symmetric

fine now?

[coral temple]

Right, really cool!

Thanks

[zinc timber]

I would have written a matlab code to verify it but feeling lazy

[exotic vector]

@zinc timber dude do you have any resource to learn duality and bilinear and quadratic forms?

[zinc timber]

ok there's a problem with my sol, it not longer has Bu=v @coral temple

because it doesn't guarantee u to be a eigen vector

[coral temple]

Right right

[zinc timber]

for that I think we can choose P to be an orthonormal basis with first entry u

gram schmidt

[coral temple]

But P has to have eigenvector columns no?

And u isn't one

[zinc timber]

yes we are not taking the same P anymore

just making sure that our modification does not mess up Au

for that we are choosing an orthonormal basis starting with u

[zinc timber]

textbooks are good enough, like LADR kr LADW

[coral temple]

Wait so how is the matrix defined this time?

[zinc timber]

it's not explicit this time ig

[exotic vector]

send me the amazon link please

[coral temple]

Well how is it defined anyway? Explicit or not, I don't fully understand

[zinc timber]

just google

[spare widget]

Linear algebra done right/wrong

[zinc timber]

no like say b=[u, e1 e2 .. en-1] s.t. it's a basis

[exotic vector]

ah yeah i have it thank you

[zinc timber]

then we use gram schmidt/ QR

then choose the Q to be our P in the previous calc

i.e. u and some basis of u\perp

[coral temple]

What? What's QR?

[zinc timber]

QR decomposition

it's basically Gram schmidt but matrices

[coral temple]

Hmmm

[zinc timber]

it's much more convoluted this time

there should be a simple one

A + (I-uu'/u'u)

how about this one?

[coral temple]

It's still symmetric so that's nice

[zinc timber]

ye just removed the whole basis part

[coral temple]

It should work I think

[zinc timber]

feels like so

[coral temple]

Yeah it works, because the extra factor is pd except for span u, and there the matrix A is pd

[zinc timber]

yes

[mystic dagger]

it's asking for the eigenvectors of that matrix, i don't know how to solve it by hand without taking hours on this question, any ideas?

[lavish jewel]

it's not asking for the eigenvectors

it's asking you which of the given vectors are eigenvectors

the latter is a lot easier to check: just plug them into the definition

sadly, if there is no evident pattern, it does mean you need to do 5 matrix-vector multiplications

[tough oxide]

If two subspace are complementary they are necessarily a direct sum ?

[gray dust]

this is in fact the definition of complement

[tough oxide]

Alright

And how do you find the complement

[gray dust]

theres no such thing as THE complement

a complement can be found by basis extension

[mystic dagger]

but i would need their eigenvalues right? so anyway i would have to find the characteristic polynomial?

[lavish jewel]

you don't

you only need to test the definition

it doesn't matter what the eigenvalue is

[mystic dagger]

but the definition of eigenvectors doesn't depend on eigenvalues?

[lavish jewel]

it doesn't depend on it

Ax = lambda x

it doesn't matter what lambda is

[mystic dagger]

oh

[lavish jewel]

if x is only scaled by A, then it is an eigenvector

[mystic dagger]

A is the matrix i have

[mystic dagger]

i see

thanks

[wintry steppe]

oh look, it's "death"

@meager harness what are you stuck on bro?

[meager harness]

idk how to do the whole thing

😂

for a) i was thinking of just doing a matrix like with f(a1) down

[wintry steppe]

you can also use the help channels if ya want

[meager harness]

something like that

[blissful vault]

Looks good to me. You encoded T with respect to your two chosen basis lists.

[meager harness]

do u know how to do part b

[mossy coral]

do you know how to find the image?

[meager harness]

its just the span of the column space

right

[mossy coral]

yep

[meager harness]

how do i get the column space tho

[mossy coral]

set your matrix to 0(so add a 0 column), row reduce. The rows that has a pivot are the column space

[meager harness]

but like the matrix has up to degree n so i feel like row reducing it

is gonna be nasty

[mossy coral]

sorry I am new to linear alg as well so im not quite sure. But i feel like there is an abstract way to do this

like you can reason it without actually row reduce

[meager harness]

lol its alright

thank you tho

[grim leaf]

"if we have three vectors in R3, and we made a matrix where each row was one vector, if that matrix can be row-reduced to an identity matrix, then the three vectors form a basis for R3"

is that right?

[stoic python]

totally not anamono

[blissful vault]

My initial thought is to show that M(T) (your encoding) is invertible, since that would show that M(T) is surjective. M(T) is invertible follows if its determinant is non-zero.

[meager harness]

wouldnt the determinant be 0 if any a1 = ...an

[native rampart]

Just show this is invertible

[halcyon spindle]

Hmm, shouldn’t the column of the matrix be those vectors? If it is then your statement is right.

[native rampart]

https://en.m.wikipedia.org/wiki/Vandermonde_matrix#:~:text=In linear algebra%2C a Vandermonde,Macon and Spitzbart (1958).

This whole exercise is basically proving the vandermonde matrix

smh

[native rampart]

nvm it's Lagrange interpolation

You can always find a polynomial f such that f(x_i)=a_i for i in {0,1,2...n-1}

a_i should all be distinct for this to work

[grim leaf]

this solution I found online seems to use the vectors as the rows

here is thm 12

[native rampart]

Yea that's equivalent to saying "You can get (1,0,0) ,(0,1,0),(0,0,1)" from your row vectors with just linear combinations

Since every elementary operation is just some linear combination of row vectors

[grim leaf]

and those span R3 and are linearly independent

therefore it makes basis of R3?

[native rampart]

Yea

[grim leaf]

dope

[lime zinc]

any hint ?

[wintry steppe]

Suppose A_n is a sequence that converges to A. You have to prove that A_n^T A_n converges to A^TA

[rotund verge]

is it ok to ask

linear programming here

[lavish jewel]

here, multivar calc and computational math should all be ok

[rotund verge]

okie

Aling Amy is a reseller of black t-shirts, and is lucky to have found a supplier who
offers her wholesale quantity discounts. Everytime Aling Amy purchases 1-100 shirts
in a transaction, she pays Php 20 per shirt. If she purchases 101-200 shirts in a
transaction, she pays Php 18 per shirt. If she purchases 201 shirts or more, she pays
Php 17 per shirt. Aling Amy can only purchase a maximum of 300 sh

does this problem violate certainty assumption of linear programming?

[dusky epoch]

what is the certainty assumption, in your own words?

[rotund verge]

all the coeffcients of the decision variables, constrabts and the rhs are constants and known?

and that it will not chanhe

[dusky epoch]

okay great

and are there any coefficients here that have some uncertainty or error bars attacked to them?

[rotund verge]

idk yet how to constract the variables tho

[mellow island]

hello can i seek help for my basic geometry activity here?

[subtle walrus]

probably try #geometry-and-trigonometry or a help channel

[wintry steppe]

Pls check anyone

[icy totem]

hellp

[dusky epoch]

there are ten problems here. which one(s) do you need help with?

@icy totem

[icy totem]

all of them , I'll compare own solutions

[dusky epoch]

what do you mean?

you already have written up solutions to these problems and want to know whether or not they are correct?

and to do that you're trying to make someone else do the same problems for you?

[icy totem]

I just learned about the subject. I'm solving questions, but I want to be sure if I'm solving correctly. Unfortunately I don't have the solutions or answers.

[icy totem]

6,7,8,9,10 questions priority

[tranquil steeple]

Ask your teacher.

Since you have solved them, hand them in and get the grade you deserve.

[icy totem]

I couldn't solve them

[spare widget]

try checking the definition for symmetric matrix for 6

for 7 try using (A+B)^2 = A+B and the fact that A^2 = A, B^2 = B (expand (A+B)^2, careful with the non-commutativity of matrix products)

for 8) a)try proving that I = (I-A)^{-1}(I-A) = (I+A+A^2)(I-A) and I = (I-A)(I+A+A^2)

8b) is clear

and you can try to guess what the form of 8c) is

9. compute the determinant of the matrix, a matrix is regular when the determinant is non-zero

10. use A * A^{-1} = I to figure out the inverse, it is regular when det !=0, try to figure out what the determinant is of a diagonal matrix

[stark quiver]

how to construct linear map T given subspace S, such that kernel of T is equal to S?

[subtle walrus]

what have you tried?

[stark quiver]

creating a matrix that will map basis of S to zero

[subtle walrus]

ok and whats the issue there?

that idea should work

[spare widget]

If A is a representation of T and B is a basis for S, then AB = 0.

What I would have done is solve B^Tv1 = 0 looking for any v1!=0 that solves the potentially underdetermined system. Then add v1 as a last row to B^T forming B1^T. Then solve B1^Tv2 = 0 requiring v2!=0, etc. Once the null space of Bk^T becomes {0} you are done.

This should produce vectors that are orthogonal to the column vectors of B and that are orthogonal with each other.

[neon lake]

The exercise asks to discuss the system depending on the value of lambda, but I can't seem to get it right doing echelon form. Also, the solutions are done employing the determinant method instead of the one I mentioned

[lavish jewel]

i'm under the impression you could reach the result either way

[neon lake]

The critical values according to solutions are -1 and -3/4

When I reduce the original matrix/system I get this, from which I don't know how to extract these critical values

[lavish jewel]

it's asking for which lambda the system has some number of solutions, right?

[neon lake]

yep

This reduced form is correct since WolframAlpha got the same one

[lavish jewel]

what is it asking you for, exactly? which lambda make the system have unique sol?

[neon lake]

By discuss, it means that depending on what value lambda takes, then if the system has one unique solution, either infinite solutions (with 1 or 2 degrees of freedom) or it is impossible.

[lavish jewel]

all right

[hexed urchin]

in this problem I set up the general polynomial a + bx +cx^2 +dx^3 and put it into the matrix and substituted 1+c,2+c or 3+c for x depending on where it was in the matrix. I then simplified it expected to get the matrix [p(1), p(2), p(3)] + [c,c,c] where I could then move on to scalar multiplication to see if it is a linear transformation. Did I do something wrong?

my figure did not simplify to anything workable when I did that

[mossy coral]

Is there a calculator where I can plug in a transformation and a input and output basis, and it will give me a matrix?

[gentle pumice]

do your lecturer not give examples of how to do these kind of stuff?

you can find alot of examples from youtube.

[grim leaf]

just want to check my understanding

[stoic python]

totally not anamono

[slow scroll]

This could be kind of confusing.

Lets say your basis $B$ is $v_1, v_2$. Then in general,
$$[\alpha]_B = [v_1 ; v_2]^{-1} \alpha$$
where $[v_1 ; v_2]$ is the matrix whose columns are the vectors in your basis. Here we have,
$$\begin{bmatrix} \cos \theta & \sin \theta \ -\sin \theta & \cos \theta \end{bmatrix} =\begin{bmatrix} \cos \theta & -\sin \theta \ \sin \theta & \cos \theta \end{bmatrix}^{-1} = [v_1 ; v_2]^{-1}$$

[stoic python]

kxrider

[grim leaf]

i see

i think

ah yeah ok

so what is the significance of multiplying the inverse of the basis and the vector?

[knotty saffron]

i dont understand why simply putting pi/2 into the function proves linear independency.

[soft burrow]

there's a theorem that says if the Wronskian of two functions is nonzero on an open set, then the functions are linearly independent on that open set

you can take them to be nonzero except at isolated points, which is what happens in this case

the Wronskian vanishes exactly at points k\pi for k in Z, but these are isolated points so sin(x) and sin(2x) are linearly independent as differentiable functions on R

[soft burrow]

so this doesn't really suffice, you have to check the Wronskian at each point in the domain

[lavish jewel]

if the columns were lin dep, the determinant would be 0 for all values of x

choosing a few values of x cleverly lets you check that this is in general not true

[slow scroll]

the idea is that v1 and v2 are written in terms of some ambient basis. v1 in its own basis is (1,0) and v2 in its own basis is (0,1). Therefore the change of basis matrix to go from B to the ambient (e.g. standard) basis is
[v1 v2] since v1 v2 = v1 and v1 v2 = v2. Thus, to go the other direction, we take the inverse of [v1 v2]

[grim leaf]

i see, thank you

[slow scroll]

npnp

[grim leaf]

is there a formal name for the expression $[\alpha]_B$?

[stoic python]

totally not anamono

[grim leaf]

this is all i have in my notes

[slow scroll]

"alpha in the basis of B"

[grim leaf]

gotcha

[lavish jewel]

coordinate vector in a basis, too

pretty much as it says there

[grim leaf]

kk ty guys

[knotty saffron]

thanks I understand now 🙂 Is the picture I uploaded wrong ? shouldnt it be W(sinx,sin2x) = the rest?

[soft burrow]

you're right

[knotty saffron]

I feel like I want to evaluate at pi/2 and 0 - would that be correct?

[dusky wadi]

could you say $\det M = \lVert\bigwedge\limits_{v \in\mcl{C}(M)} v\rVert$ where $\mcl{C}(M)$ is the column space of M

[stoic python]

all functions are alison

[soft burrow]

evaluating at two points isn't enough - you want to check that sin(x) and sin(2x) are linearly independent on their domain (R)

so you have to check that W(sin(x), sin(2x)) is nonzero for all x in R, except possibly at isolated points

x=k*pi, for k in Z are isolated points

[grim leaf]

in solving part b, where did the b3 = 2b1 go?

also i dont understand this step-by-step progression

actually i think i get the last three statements

just dont know what happened to b_3=2b_1

[mossy coral]

Late response, but I know how to do it. Just want a fast way to double check my work

[gentle pumice]

ahhh

it's okay

typically you can calculator at then end of what you want to find it.

[native rampart]

You just eliminate b3 and replace with b1

[grim leaf]

then where does 2b1 go?

[native rampart]

All vectors will be of the form
x(1,0,2,0)+y(0,1,0,0)+z(0,0,0,1)

If you substitute and compare ,you get x=b1,y=b2 and z=b4

So it's b1p1+b2p2+b4p3

[grim leaf]

i see

[wintry steppe]

What does the C stand for here?

[pseudo cairn]

Can someone help me with this question ? I will send my working now

Frankn is this channel taken ? I’m not sure how the subject specific channels work

[wintry steppe]

Does c have some predefined value which I'm not aware of? Because as I currently understand the problem, it has not been assigned any value.

[wintry steppe]

occupied/unoccupied does not apply to subject channels.

issa freeforall

[pseudo cairn]

Got it, are we allowed to post linear alg questions on the help chats too ?

[wintry steppe]

Yes

[pseudo cairn]

Thanks

I just needed to know whether I am doing the right thing here. I think I’m doing what my lecturer suggested but I’m finding it counterintuitive

Please ping me if u can help

[zinc timber]

not linear algebra tho

[zinc timber]

anchor points

nah I'm not sure anymore, ignore

[wintry steppe]

Ah

It's Ac = f

SO they are just stand-in variables

[zinc timber]

yes

coeff of cubic spline, but idk which

[weak root]

hey guys can yall help me out????

A sprinkler system is used to water two areas. If the total water flow is 920 L/h, and the flow
through one sprinkler is 85 % as much as the other, what is the flow of each?

[lavish jewel]

try a help channel or #prealg-and-algebra

[wintry steppe]

Is any Vector space subspace of itself?

[teal grotto]

yes, similar to how any set is a subset of itself

[wintry steppe]

Thank you 👍

[kindred hinge]

Is this the channel to make r the subject

[lavish jewel]

hmm?

[spare widget]

that would probably be #computing-software

if you mean R as in the R project for statistical computing

[normal void]

This is not adequate proof, right?

[stoic python]

FantaSkink

FantaSkink

[normal void]

This is the original function that we have to show is linear

[spare widget]

why not

You should do alpha * B + beta * C for linearity though

Theh take out alpha out front, and beta out front

Remember the definition of linearity

L(x+y) = L(x) + L(y), L(alpha x) = alpha L(x)

Or more concisely L(alpha x + beta y) = alpha L(x) + beta L(y)

In fact if you vectorize w_{ij} and O_{ij} you can write the kernel application as p = W o

o = vec(O), p =vec(P)

And W a suitable matrix with coefficients made up of the w

You can always flatten a 2d arrays into a 1d one using [i][j] -> [i *m + j] (here I assume i in [0,m-1], j in [0,n-1])

It's clear that matrix-vector multiplication is linear

[dusky epoch]

\times, not X

[chrome radish]

how can I do this?

[spare widget]

find x such that A x = (0 0 6)^T

[chrome radish]

you can't do RREF?

[spare widget]

idk what rrefis

[chrome radish]

row echelon form

[spare widget]

You can

[chrome radish]

to show that it only has one solution?

[spare widget]

you're supposed to solve the augmented system not show that it has only one solution

Find x such that Ax = b

[chrome radish]

yeah, but if you reduced it to row echelon form and showed that the last column consists of only zeros. Then <0, 0, 6> are in R(T), but if the last column has a 1 at the end then it's a contradiction, because 0 isn't equal to 1, right?

@spare widget I solved the following systems of equations: $\begin{cases} 4x+y-2z-3w=0 \ 2x+y+z-4w=0 \ 6-9z+9w=6 \end{cases}$

[stoic python]

John doe

[chrome radish]

and got x = w, y = w and z = w

[spare widget]

it's supposed to read 6x -9z +9w = 6

Not 6 -9z +9w = 6

[chrome radish]

oh yeah

meant 6x

[frigid kettle]

Would someone mind suggesting a beginner level Linear Algebra course I felt hard to solve problems on Computer Vision and Image Processing course of mathematical and physical underpinnings..

[lavish tusk]

The LCM of two numbers is 15 times of HCF. The sum of HCF and LCM is 480. If both number are smaller than LCM. Find both the numbers.

can anyone help me with this problem

[dusky epoch]

phi isn't even a linear map

[wintry steppe]

?

[lavish jewel]

the +2 and +1 terms make the transformation affine

[grave kettle]

how is the current definition of linearity motivated

[spare widget]

Study some simple linear spaces, e.g. line passing through the origin, plane passing through the origin

Notably if you take two vectors from those, then linear combinations of those stay within the space

[grave kettle]

rephrased my q

[spare widget]

By generalizing what I just mentioned

You take the same property and try to see how it works with other spaces than R^n

Then you come to the conclusion that finite dimensional real vector spaces are isomorphic to R^n

and note that the theorems that you have can thus be generalized

It's a standard approach to take properties of something known and look for conditions such that some of those properties hold elsewhere

It took some time to get from Descartes to modern linear algebra, if you'reinterested in how the concept evolved I would suggest reading history of mathenatics

[spare widget]

Maybe this will help you: https://youtu.be/kYB8IZa5AuE

[grave kettle]

i watched that like 3 times

even the last episode, mentioning the question of why lin transforms are linear

i remember you sending that

[grave kettle]

i don’t think he answered that

[lavish jewel]

if you think about it geometrically in R^n

[spare widget]

He shows geometrically what linear transformations do

[lavish jewel]

an expression of the form a v + (1-a) u, with a real between 0 and 1, defines a line segment between v and u

[spare widget]

L(ax + by) = aL(x) + bL(y) is just the algebraic counterpart.

[lavish jewel]

then we call a transformation "linear" if it satisfies that T( a v + (1-a) u) = a T(v) + (1-a) T(u), which is now a line segment between T(v) and T(u)

this generalizes to more abstract vectors though, which is why the axiomatization is so useful

[spare widget]

Already for R^4 it's probably harder to imagine geometrically

However the algebraic version remains the same

You basically take a concept that you have some intuition about and know what it means for some spaces and try to generalize it to other spaces which you may even be unable to imagine geometrically

This is a standard approach and doesn't apply only to linearity

Taking a concept and checking the conditions necessary so that it applies to other settings

[fringe fjord]

Doesn't this define the class of transformations usually called "affine" rather than the "linear" ones?

[lavish jewel]

i jumped a few steps and took artistic liberties here and there, recalling that the person asking the question has already asked it before and was dissatisfied will all the explanations we provided before

i thought i had been careful to only use consequences of homogeneity and additivity though, do tell me what i missed :x (admittedly i make mistakes and miss stuff often)

[fringe fjord]

Your identity is true for all linear transformations, but not everything that satisfies it is a linear transformation -- consider e.g. translations.

[lavish jewel]

ah, certainly, yes

[fringe fjord]

In fact I'm not sure there is any compelling reason why the word "linear" has ended up attached to the property we know and love under that name. It just happened, and the word is now everywhere and far too late to second-guess. So students just have to deal with the fact that f(x)=5x+7 is a "linear function" in high school, but stops being "linear" in higher math ...

[lavish jewel]

yeah, chromium might be too hung up on that. last time we had used "nice" or "well behaved" and they didn't like it though

[spare widget]

inb4 homogeneous coordinates translations 😂

on a serious note, in plenty of engineering papers people use linear and affine interchangeably and it's clear from the context what they mean

off the top of my head they call the affine elements in FEM linear

[grim leaf]

does anyone have some good resource that just really dumbs down the algebra of linear transformations? i watched 3b1b's video but it's more conceptual than algebraic

[spare widget]

Just solve some matrix systems I guess

[grim leaf]

i'm struggling with things like proving something is a linear transformation, finding conditions that would make some vector be in a null space/range/rank

i guess but i'm just trying to understand the connection between the algebra and the concept of linear transformations

[spare widget]

For those you need to know the definitions and then verify those

Being in the null space means that the map sends it to 0

For example imagine a projection onto a line through the origin in 2D

Then infinitely many points get projected to (0,0)

All vectorscorresponding to those are in the null-space

[grim leaf]

ic

[spare widget]

Now consider the range it is made up of all vectors that you can get as output

So that would be the whole line

in the above example

[grim leaf]

ic

[spare widget]

finally your column rank is the number of linearly independent columns and the row rank is thenumber of linearly independent rows

Imagine a rectangular matrix

In 3D say you have a plane

And vectors v1, v2, v3 are in that plane

[grim leaf]

ye ik what column and row rank are, i don't know what "rank" alone means

as in "What is the rank of T?"

[spare widget]

The column and row rank are equal

[grim leaf]

oh true

[spare widget]

So the rank is the number of linearly independent rows/columnsof your matrix

Geometrically the meaning is clear

Say you are in 3d

And your matrix maps vectors to a line

scratch that

[grim leaf]

ah ok

[spare widget]

Assume your matrix maps 2D vectors to a plane in 3D

[grim leaf]

e.g. R2 -> R3?

[spare widget]

$\begin{bmatrix} \vec{v}_1 & \vec{v}_2 \end{bmatrix} \begin{bmatrix} x\ y \end{bmatrix} = \vec{v}(x,y)$

[stoic python]

criver

[spare widget]

Where v1 and v2 are linearly independent vectors - theyform a basisfor the plane

It's clear then that the matrix has rank 2

You may not be as lucky, andsomeone may give you a frame - more than 2 vectors, but at least two are linearly independent

Then the rank is still 2, but you have something like this

$\begin{bmatrix} \vec{v}_1 & \vec{v}_2 & \ldots & \vec{v}_n \end{bmatrix} \begin{bmatrix} x_1\ x_2 \ \ldots \ x_n \end{bmatrix} = \vec{v}(x_1, x_2, \ldots,x_n)$

[stoic python]

criver

[spare widget]

The rank being 2 means that still results in a plane

Because n-2 of the vectors are linearly dependent, i.e. you have infinitely many ways to write the same point

If the rank was 1 then it would have been a map to a line through the origin

If the rank is m then it is a map to an m-dimensional hyperplane

[grim leaf]

ohhh i see

[spare widget]

Whether youspan that hyperplane with m vectors or with more results inyou havingabasis or a frame/overcompletebasis

Thenice thing about abasis isthat a point can be represented inaunique way

[peak plinth]

For my row reduced thingy I got this, and now I have to find the basis for the eigenvector

But I’m so confused

[spare widget]

This is not thecase for frames, but sometimes thisis an advantage: e.g. noisy data

[grim leaf]

i see

thank u

[grim leaf]

what's the significance of defining i =/= q and i=q as two separate things?

[slow scroll]

The idea is that you want to enumerate mn distinct linear transformations and show that they are linearly independent. This is basically like listing out the matrices $\begin{bmatrix} 1 & 0 \ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 1 \ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \ 1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \ 0 & 1 \end{bmatrix}$

[stoic python]

kxrider

[grim leaf]

ah i see

[solid leaf]

Has anyone here worked through “Linear Algebra Done Right” by Sheldon Axler? If so, was there any place to find solutions for the many times he asks you to prove something? I’m not sure how to start some of these or if I’m doing them correctly.

[slow scroll]

you can always just ask here. I think quite a few people here are probably familiar with axler

[solid leaf]

Oh ok great

In that case

I’ve been stuck on this for quite some time

I understand the actual theorem now i think

But I don’t understand how to do the proofs

How to prove that F^S is a vector space over F

[slow scroll]

you have to check each axiom for a vector space. For example, you would have to check that F^S has an additive identity

[solid leaf]

Oh so just the 10 axioms

[slow scroll]

yeah. A few of them should be pretty clear like commutativity for example f + g = g + f

[solid leaf]

Like closure under addition and scalar multiplication?

When I first did this earlier this year those were the big two ^

[slow scroll]

sure, we need our operations to be well-defined. However this is pretty much claimed in 1.23. f + g is defined to be the function (f+g)(x) = f(x) + g(x), so we are pretty much automatically closed under addition for example

[solid leaf]

Ok yeah

I figured I was just making sure I’m thinking of the correct axioms

One thing I’m confused about In this case

I understand that like fundamentally in math to prove something is incorrect you only need one example

In this case, how would I actually prove for all examples

Idk how to word this to explain what I mean

[slow scroll]

this is not exactly true. It depends on the statement. the negation of the statement "for all x, P(x)" is indeed "there exists x such that not P(x)," but not all statements are of the form "for all x, P(x)"

like if I asked you to prove "there exists a real number x such that x^2 = -1" is false, you would have to show that for all real numbers x, x^2 is not -1

[solid leaf]

Could I not square root both sides and show that x=i and then say that there are no real numbers x that are i?

[slow scroll]

My point is that any proof of this is going to require some kind of manipulation to an arbitrary real number x. For a proof, it would obviously not suffice to say "Counterexample: 2^2 = 4 != -1. qed."

so proving that a statement is false is not always just "providing a counterexample"

[solid leaf]

Oh yeah that I understand I think

So my question is like

How do I do this proof

Without using a counter example

[slow scroll]

like the example i gave?

[solid leaf]

Like for what you said would this be an acceptable statement proving your statement

[slow scroll]

showing that x=i is not showing that x is not real unless you've already proven that i is not real.

[solid leaf]

Ok sorry I’m not trying to sound like

Idk what the word is

But

Isn’t that the definition of i?

Like it’s not a real number

By definition

[slow scroll]

eh, i is just a solution to the equation x^2 + 1 = 0.

[solid leaf]

Ok I’m not gonna argue I assume you’d know better lol

But like let’s say I want to prove that

Uhh

T(x,y+x) is a linear transformation

I can just take a vector and prove it using two variables and the four axioms that define a linear transformation

[solid leaf]

How can I do that with this

[slow scroll]

do you mean T(x,y) = (x,y+x)?

[solid leaf]

Yeah sorry

Forgot the notation lmao

I can just take a vector (x1,y1) and prove that

What would I use instead of like

x1 or y1 to prove my thing

Just an arbitrary f(x)?

[slow scroll]

Alright, so we know linear means T(v + w) = Tv + Tw and T(cv) = cTv for v,w in V and c in F
vectors in F2 look like ordered pairs (x,y). So you can write v = (x1, y1), w = (x2, y2).
Then v + w = (x1 + x2, y1 + y2) and cv = (cx1, cy1) by the definition from above, and you can check that T(v+w) = Tv + Tw and T(cv) = cTv hold

[slow scroll]

also about this, you can formally "adjoin" i to the real numbers, so it seems like it should not be real, but in technically speaking, there could be some funky way to combine real numbers together to get back i. For real numbers though, you can show that x^2 >= 0, and so i can't be a real number since i^2 < 0.

[solid leaf]

Ohh

Interesting thanks

[solid leaf]

And this is verifying the

First thing I sent

That F^S is a vector space over F?

[slow scroll]

no, this is verifying that T is a linear transformation

[solid leaf]

Ok yeah I was gonna say that didn’t make sense for the first thing

I understand all of that

Everything we’ve done so far

But I still don’t understand like

[solid leaf]

To verify this

I need to verify that F^S is a vector space

First off

Can I ignore the

Over F part

Or does that actually require more verification

Than just proving if it is a vector space

[slow scroll]

the "over F" part just means that your scalars come from the field F. When we define c(x1, y1) = (cx1, cy1), we want to make sure that cx1, and cy1 actually make sense. That's the only thing about the "over F" part

[solid leaf]

OH

ok thanks

That cleared a bit up

So what I’m actually confused about now is what I can use to verify it. Obviously I need to use vectors but can I just use variables or do I actually need to use functions in the vectors? And if I had to use functions how would that work?

If you need more clarification I can try to send an example

[slow scroll]

im not sure what you mean by "variables" here

do you mean like (x1, x2, ..., xn) vs a function f : S --> F?

[solid leaf]

I think so yeah

Like F^S is a space of functions right

[slow scroll]

yes. and so is F^{1,..., n} = F^n = {(x1, ...., xn) : xi is in F for all 1 <= i <= n}

[solid leaf]

Wait

What

So like

R^2

Is a space of functions?

[slow scroll]

yes in a sense. We normally just write (x1, x2) for an element of R2, but this the same as the function f : {1,2} --> F defined by f(1) = x1 and f(2) = x2. This is how we extend the cartesian product of sets to products of infinitely many sets

so for example showing that F^S is a vector space for any set S shows that R2 is a vector space for S = {1,2} and F = R

[solid leaf]

Ok this is gonna take me a minute to wrap my head around

Quick question about what you said

Are the 1 and 2 important

Could I say like

[slow scroll]

honestly not sure if ive answered this question though

[grim leaf]

https://linearalgebras.com/ you may find luck here

didnt mean to interrupt just thought id drop that

[solid leaf]

Lol I have that already thanks though

[grim leaf]

ah kk

[solid leaf]

It only has solutions to the exercises

So @slow scroll could I say f:{3,2} —> F defined by f(3)=x1 and f(2)=x2 is the same as (x1,x2)

Like did the 1 matter at all or was it just an arbitrary value

I know this isn’t normal this is just to help me understand

[slow scroll]

If you are consistent with this convention, then sure. yea, Rn can be identified with functions f : S --> F where S is any n-element set. But typically we like {1,2, ..., n} because we can say f(1) is the first coordinate, f(2) is the second coordinate, and so on.

[solid leaf]

Yeah ok

But technically the 1 and 2 don’t actually matter, they could be any number as long as the functions are defined correctly?

It’s just easier to go in order

[slow scroll]

yes

[solid leaf]

Ok

Thanks for helping me understand the

Well

All of that

Lmao

[solid leaf]

So

For this specifically

I just prove it using vectors

Say like

(x1,y1,…)

But what dimension vectors would I use

Cause if it was say F^4 ik it’d be 4 dimensional

But idk how to do it with the S

Let me resend the problem real quick

[slow scroll]

you can't order the coordinates the same way you do with F^n. If f : S --> F, you just know that f(x) is the xth coordinate, and that we want to add coordinate-wise: (f + g)(x) = f(x) + g(x)

like for F^n, (x1, y1) + (x2, y2) := (x1 + x2, y1 + y2) is the same as saying (f + g)(1) := f(1) + g(1) and (f + g)(2) := f(2) + g(2)

[solid leaf]

Where like f(1) = x1

And so on?

[left grail]

heyy whats up linear algebra peeps
im in cryptology right now and i learned about the hill cipher, and one aspect of it is calculating the inverse of a matrix in mod 26 with something called reduced row echelon form

now heres the thing
i have never taken linear algebra in my life before and did a crash course on matrices before learning the hill cipher and i feel so lost

my main problem is just knowing the steps to calculating the inverse in mod 26 and know how to proceed from each step

[slow scroll]

yes

[solid leaf]

So basically you can always define a vector as a function? This is still a weird concept to me

[nocturne jewel]

vectors are just stuff in the vector space that arent scalars

functions, coordinates, sequences, matrices, etc are all vectors if you look at their spaces

[solid leaf]

Lmao maybe I need to better understand what a vector is

[slow scroll]

elements of F^S are functions, and F^n is canonically identified with F^{1,2,...,n}

[nocturne jewel]

F^n is FxFxF...xF

[slow scroll]

a vector is just "an element of a vector space" though, and we don't always think of vectors as functions

[nocturne jewel]

ie the n-tuples with entries from F

[slow scroll]

later on, you'll see that every vector space is isomorphic to F^S for some set S, so it does come full circle lmao

[solid leaf]

Yeah so like if n is a set that’s what’s confusing me

[nocturne jewel]

It's just notation

like how $\int f(x)\dd{x}$ is just notation for all functions that differentiate to $f$

[stoic python]

Mosh

[nocturne jewel]

$\mathbb{K}^S:={f|f:S\to\mathbb{K}}$

[stoic python]

Mosh

[solid leaf]

Oh so it’s the notation for all functions that map S onto F

Ok

[nocturne jewel]

yes

It's just notation

[solid leaf]

Just wondering is this topic

Like the idea of F^S usually in an introductory lin alg course

If you know

[nocturne jewel]

like how $\mathbb{K}^n:={(k_1,...,k_n)|k_i\in\mathbb{K}}$ is also notation

[stoic python]

Mosh

[solid leaf]

Oh ok

[nocturne jewel]

F^S and abstract vector spaces is usually 2nd half of a LinAl course

[solid leaf]

Ohh ok

[nocturne jewel]

1st half is in concrete Euclidean space

[solid leaf]

I’m currently in an intro Lin alg course but it’s year long so

[nocturne jewel]

so yeah, 2nd term you should see this

[slow scroll]

When S is an infinite set, the vector space that example 1.24 defines is infinite dimensional. Intro lin alg courses usually don't leave finite dimensions

[solid leaf]

I decided to self study Acker’s book

Axler*

[solid leaf]

Ok yeah cause I was gonna say I’ve never seen this stuff before

Most of the stuff we’ve learned so far has used a lot of determinants and matrices but I liked his idea of not using determinants

But it’s definitely harder

[meager harness]

idk how to do part c and d

[meager harness]

@atomic oasis

[atomic oasis]

lmfao

is AM2 = AM1 guys

Oh this is algebra sorry

[grim leaf]

how did they get the A11 -> A11-4A21 stuff? i assume it has something to do with the matrix B, but i can't draw the connection

[nocturne jewel]

Compute T[A_11]

then write that matrix in terms of the basis vectors

[grim leaf]

ohhhh

why is $b_1\alpha_1 + b_2\alpha_2 + b_3\alpha_3 \ne 0$?

[stoic python]

totally not anamono

[blazing moss]

The alphas are a basis. If one of the b’s is not zero, and the three vectors are linearly independent, the resultant cannot be 0

[grim leaf]

ohhh righttt

[wintry steppe]

Hello, for $x \in \mathbb{R^{n}}$ can I always write $x=\sum_{i=1}^{n} <v_i, x> v_i $ ?

[stoic python]

kaz

[nocturne jewel]

if {v1,...,vn} forms an orthonormal basis of R^n

[wintry steppe]

yeah

regardless of scalar product?

[nocturne jewel]

Well whether the basis is orthonormal depends on which IP you chose

[wintry steppe]

oh alright thanks

[nocturne jewel]

orthogonality is determined by the IP

[wintry steppe]

yeah yeah

[thick rain]

can someone please give me the Texit commands for linear algebra please

[thick rain]

Thank you

[amber osprey]

Ax,Ay = choice(CR[0,:], CR[1,:])

how can I do something like this I want to randomly choose points but if I do

    Ax.append(choice(CR[0,:],))
    Ay.append(choice(CR[1,:]))

I call choice twice which will not give me corresponding x and y

[undone hedge]

hello

[wintry steppe]

I still don't understand how the set {(1 0),(0 1), (2 3)} is linearly dependent. Like I understand the notion when exactly the given set of vectors is linearly dependent or independent, but consider it as a linear combination:
(a b) = a(1 0) + b(0 1) + 2(2 3)

(a b) will only become zero vector when a and b become zero, but the equation will then become:
(0 0) = 2(2 3)

Which is not correct

And I know that (2 3) can be represented as a linear combination of vectors (1 0) and (0 1)

But I just can't get my head around it

[lavish jewel]

i think you misunderstand the definition of linear (in)dependence

you are rather looking for the existence of not all zero a,b,c, so that (0 0) = a(1 0) + b(0 1) + c(2 3)

by subtracting c(2 3) and dividing by -c both sides, and after some substitutions, this is the same as asking whether there exist not all zero numbers m and n so that (2 3) = m(1 0) + n(0 1)

which is what you said is possible

so this means the set of vectors is lin dep

this here: (a b) = a(1 0) + b(0 1) + 2(2 3)

has nothing to do with linear independece of the set of vectors

[wintry steppe]

Thank you 👍

[radiant viper]

idk where to post this

but

[halcyon spindle]

#prealg-and-algebra

[wintry steppe]

Hello guys

I have a question about matrices

What am I doing wrong here?

[spare widget]

How did you get 1 in a22

3 * 2 - 8 = -2

Similarly for the rest

[ember pagoda]

Guys, in my clg, we just started linear algebra, so any resources for that?

[halcyon spindle]

https://textbooks.math.gatech.edu/ila/.

Interactive linear algebra.

[ember pagoda]

tnx

[wintry steppe]

So according to my understanding now:
To prove any set of vectors to be linearly dependent, we can do so:

1. If the given scalars a, b, c are not all zero
2. We can represent a vector from that set of vectors, by the linear combination of the other vectors in that same set, where m and n are not all 0.

[lavish jewel]

for 3 vectors, sure. it generalizes to however many you want (sort of)

[wintry steppe]

Thank you once again

[lavish jewel]

the usual definition you find in books is that the vectors ${v_1, v_2, \dots, v_n}$, are linearly dependent if there exist constants $c_i$ not all zero such that $\boldsymbol{0} = \sum_{i=1}^n c_i v_i$, where $\boldsymbol{0}$ is a 0 vector

[stoic python]

Edd

[wintry steppe]

Yes

Ah so there is also a non trivial solution of the system such that we get the zero vector, showing linear dependence of the given set of vectors

[lavish jewel]

precisely

this is equivalent to saying there is a nontrivial solution to $0 = Vc$

[stoic python]

Edd

[lavish jewel]

where V is the matrix whose columns are the v_i vectors, c is a vector of the c_i

so if a nontrivial c exists, the matrix V has a nontrivial kernel

[wintry steppe]

Oooh

[lavish jewel]

idk if you've seen rank-nullity yet

[wintry steppe]

Well I haven't, but I've heard about trivial/non trivial kernels

[lavish jewel]

ok. well, you can ignore it for now, it'll pop up eventually

[wintry steppe]

👍

[grave kettle]

https://math.stackexchange.com/questions/4409972/motivation-for-definition-of-linearity#comment9225292_4410075

can anyone help answer this

[blazing moss]

Everything is locally linear. In multivariable calculus you use a matrix called the jacobian, which tracks the partial derivatives of of a function and allows you to change your system of coordinates, as this set of derivatives described everywhere how the function is changing

[grave kettle]

have you seen what i’ve been exposed to

[blazing moss]

In polar coordinates in single variable calculus, rdtheta being a viable differential is due to the same reason

[lavish jewel]

chromium, no one can help you

[grave kettle]

why not

[lavish jewel]

because we already tried

there is nothing special about the name. people noticed these properties appeared fairly often in different fields, and so they sought to come up with a description that covered all of them

and that's all

[native rampart]

Didn't linear algebra come into existence because physicists needed a convenient way to model physical phenomena

[grave kettle]

i’d like to know the specifics

[lavish jewel]

that's one thing, but one of the main motivations behind it was war and the assignment of resources

[grave kettle]

like a story

[lavish jewel]

dude, go read and look it up yourself then

we already gave you the history a while back

did you just forget it or ignore it?

stop wasting people's time here

[grave kettle]

here?

[native rampart]

Send a link

[grave kettle]

neither

[lavish jewel]

scroll back to the beginning of the month

when you asked the same thing and several people gave you examples, definitions, and historical context

[native rampart]

In this channel?

[lavish jewel]

yes

chromium has been asking the same thing over and over and rejecting everyone's answers

all month

[blazing moss]

Better that than shudders… studying more linear algebra

[grave kettle]

ok so

maybe just drop this for now?

[native rampart]

Funny I will try once

[grave kettle]

and see how linear transformations apply

then can see its use

[native rampart]

You know what a line is?

[grave kettle]

yea

[native rampart]

You know what a linear equation is?

[grave kettle]

ax + by = c

[native rampart]

Not that alone

[grave kettle]

i’m not sure what answer you expect

[native rampart]

ax+by+cz...=d

So you know what elementary row operations are?

[lavish jewel]

https://en.wikipedia.org/wiki/Linearity https://en.wikipedia.org/wiki/Linear_system https://en.wikipedia.org/wiki/Linear_programming
https://en.wikipedia.org/wiki/Linear_algebra
https://mathworld.wolfram.com/LinearAlgebra.html
you can read these and the references in each

they'll point you to books, papers, etc

[grave kettle]

i haven’t touched matrices

[native rampart]

Ok,so you can't expect an answer

[lavish jewel]

they want a motivation behind the name

which doesn't really exist

as for the field itself, name aside, it's just something that happened to pop up time and time again

if you want more details, read the links i sent and continue the search yourself

linalg includes problems in assignment of resources, solutions to differential equations in physics, and can be abstracted to be applied more generally as well

some people just saw many problems and realized they were all the same at a more abstract level

[native rampart]

Linear algebra originated as an abstraction of way of solving linear systems I think

[spare widget]

touch matrices then, once you solve enough problems you may understand what this is about

[native rampart]

You can't talk about foundations of linear algebra without linear equations and matrix algebra

[grave kettle]

don’t matrices come from transformations

[lavish jewel]

you would know this if you would read your linalg book

[native rampart]

No,matrix algebra existed pre Linear algebra

Matrices originated as a convenient way of bookkeeping variables in linear equations

[limber sierra]

perhaps itd help if, instead of 'linear', you imagined it said 'particularly-easy-to-work-with' or similar

[lavish jewel]

we already explained it like this at the beginning of the month, nami

[grave kettle]

and i’d like to know how

[lavish jewel]

it's a lost cause

[spare widget]

solve some problems, and you'll understand it

take something from analytic geometry, e.g. intersection of planes in 3D

[limber sierra]

most applications of LA are reframing questions about continuous functions as about linear ones

[spare widget]

solve several problems and it will become clearer

no wonder nobody's explanation made sense to you since you haven't touched matrices

[limber sierra]

or actions of generating sets as bases but thats beyond you

[blazing moss]

Find a parabola that goes through the point (2,5) and has its vertex at (-1, -4).

Not as easy as a linear equation?

Boom, there you have it.

[grave kettle]

can i have one

just as an example

[spare widget]

find the intersection of planes in 3D

[limber sierra]

open your textbook

[spare widget]

$\vec{n}_1 \cdot (\vec{p}-\vec{o}_1) = 0 \ \vec{n}_2 \cdot (\vec{p}-\vec{o}_2)= 0 \ \vec{n}_3 \cdot (\vec{p}-\vec{o}_3) = 0$

[stoic python]

criver

[spare widget]

where n1, n2, n3 are the normals of the 3 planes, o1, o2, o3 are points respectively on each of the planes

studying the solutions of the above naturally leads to matrices and determinants

you get multiple cases depending on the rank - e.g. all 3 planes are parallel and there is no intersection or they coincide, 2 planes intersect in a line, or 3 planes intersect in a point

https://conf.math.illinois.edu/~lfolwa2/PlaneIntersections.pdf

[limber sierra]

heres another example

open a derivative practice pdf

you will apply linearity of the derivative many times

you might know it as the two rules, 'sum rule' and 'constant multiple rule'

collectively these rules are linearity

try doing derivatives without them

it might take a while and a lot of limits.

[grave kettle]

i see many problems showing some transformation is linear, then tell us to use linearity properties

but i can’t find ones that emphasise the easy-to-work-with-ness of linear transformations

[spare widget]

consider A *x = b, there are plenty of methods for finding a solution (even if the system is incompatible we can find the least squares "solution")

[limber sierra]

i mean... thats the point

once you have linearity properties you can use them

theyre quite handy

[spare widget]

now consider a1(x) = b1, ..., am(x) = bm where a1, ..., am are arbitrary nonlinear functions

this is a pain to solve

[limber sierra]

i dont know what youre after

"use linearity properties" IS emphasizing the easiness to work with them

thats literally the entire point

[spare widget]

probably motivation to study linear algebra 😛

[limber sierra]

do you just think we shouldnt bother having a name for this topic?

[grave kettle]

so it turns out these properties are ultra useful

[limber sierra]

i mean, if they care this much about motivation, it seems weird to me that starts with a text that introduces abstrac tlinearity before it introduces matrices or vectors

read an engineering LA text instead if you want motivation

¯_(ツ)_/¯

[spare widget]

^

you should start with something simple

[limber sierra]

or alternatively read a group theory text and realize how much nonlinear stuff sucks

[spare widget]

I recommend analytical geometry

[limber sierra]

and then get to the chapter thats like

"oh yeah actions of generators are like vector spaces"

and realize it makes every problem 10000x easier

[grave kettle]

generators?

[limber sierra]

you can ignore that quip, was semi-sarcastic

im referring to the more "mathematical" (specifically an algebraist's) motivation for linear algebra

[spare widget]

I can't recommend you an analytical geometry book unfortunately since the one I used is from the 60s

[grave kettle]

hm so

[limber sierra]

namely that it's often best to study structures using their generating sets, and for most generating sets we just look at them as a vector space

[grave kettle]

new question

why is matrix multiplication the way it is

[limber sierra]

the generating set analogue in linear algebra itself is a basis

[lavish jewel]

it is defined the way it is because it represents linear combinations in a succinct way, and makes function composition simple

[spare widget]

read the first chapter of hoffman and kunze

they explain it in there

should be in the first few pages

[blazing moss]

Read your entire linear algebra book

[limber sierra]

there are many many ways to frame an answer to this yeah

lmao

the naive first year's approach is that it just so happens to coincide with function composition

which is the "why" but not really the "how"

[languid sphinx]

Do you want to generalise it or just perform it

[blazing moss]

And Google your question

[limber sierra]

the "how" comes from, i suppose, looking at how matrices act on just column vectors

and thinking "how do i generalize this"

[languid sphinx]

I think function comp is a lot more general though?

[limber sierra]

and then noticing that it coincides with function composition as desired

[limber sierra]

linear function composition

[spare widget]

probably too advanced for him

[limber sierra]

where the dimensions work

[languid sphinx]

oh

That's so constrained

[spare widget]

he should just read the first pages of a practice oriented linear algebra book

[lavish jewel]

wdym that's so constrained

[limber sierra]

do you have a less "constrained" explanation?

[lavish jewel]

that basically means that whenever the multiplication is defined, it is a composition

so, always

[blazing moss]

How convenient!

[lavish jewel]

this whole thing is a failure to google and to read a book, let's not clog up the channel

it's like the 3rd or 4th time this month

[languid sphinx]

I mean it's very constrained on the composition

I think of functions as mapping from arbitrary places

so iterating different functions map to/from even more arbitrary places

functions can be anything you want them to be

[lavish jewel]

hmm?

and what do you call "iterating different functions"

[spare widget]

well yes matrix multiplication is composition of linear stuff

[grave kettle]

so matrices (can) come independently of vectors

wikipedia said multiplication originated from composition of linear maps

[grave kettle]

how do i get the composition if i don’t get linear mapping at all lol

[spare widget]

by reading the first few pages of Hoffman and Kunze

[grave kettle]

that’s fields

[spare widget]

read 1.2 and 1.3

1.3 is literally page 6

[languid sphinx]

composition?

[spare widget]

if you have questions once you read it and try to solve the problems, then that would be much more fruitful. Conversely if you're unwilling to read a few pages then I don't think people here can help you

[lavish jewel]

precisely

and at least in the finite dimensional case, you can pick a basis and define a matrix for your transformation

so the two are equivalent

saying it is "limited" is weird because the limiting component there is not in the matrices

[grave kettle]

when do we multiply matrices outside the context of linear transforms

[native rampart]

Pointwise multiplication is a well defined operation

[spare widget]

whether you interpret those as linear transforms or not is up to you

a matrix is just a representation, so what specific meaning you attach to it is application dependent

[native rampart]

https://en.m.wikipedia.org/wiki/Hadamard_product_(matrices)

[spare widget]

I don't think they are asking for more products

To not be as vague I can give you some examples: You can write image blurring as W * v where the matrix W contains some discretisation of a blurring kernel, and v is an image in vector form. Of course this is a linear transform, but thinking in terms of rotations and scaling is not very useful here.

On the other hand in computer graphics you may want to scale and rotate some object in which case you will have R * S * p, where p is a point of the object, S is the scaling matrix, and R is the rotation matrix

For discretisations of PDEs you may have something like:

Cu + (I-C)W^m u = f, where W is some discretisation of the Laplacian

[flint pivot]

does discretisation mean some kind of approximation/loss of precision?

[spare widget]

while all of those constitute linear maps, it's not very useful to think of those in the usual geometric setting of a linear transformation

[flint pivot]

I should say imply not mean

[spare widget]

in the above case it means going from the continuous to the discrete setting

e.g.

$\partial_x u(x,y) \approx \frac{u(x+h,y) - u(x,y)}{h} + O(h)$

[stoic python]

criver

[spare widget]

it's how you get linear equations from linear PDEs

[flint pivot]

I'm not too familiar with PDEs but is that a numerical matrix method?

[spare widget]

the standard ones are the finite element method, the finite volume method, and the finite difference method

they produce a system of linear euqations out of linear PDEs

e.g. the finite element method projects the continuous problem on a finite dimensional subspace

[flint pivot]

does it involve matrix operations or is that useful after it's in linear equation form?

[spare widget]

you get linear systems once you discretise

then you apply numerical solvers to solve those linear systems

[flint pivot]

got it

[spare widget]

it's a similar story with variational methods

and you also have a ton of linear algebra in machine learning

[flint pivot]

ah yes that I know

[small cosmos]

Can someone please help me with number 5

[halcyon spindle]

what are you thinking of to approach it?

[small cosmos]

I don't even know how or where to start

[thick rain]

You will get help faster in one of the helper channels, so maybe try that.

[spare widget]

start with the conditions for something to be a subspace

(check that it's closed under scalar multiplication, vector addition, and that 0 is in it)

this type of question is really meant to have you practice verifying those conditions

[summer topaz]

Could anyone please help me with this problem ? My teacher said it is a vector space but didn't explain why and im very confused, maybe i just didn't know how to check the axioms...?

[blazing moss]

Check inverses, Identity, closure, etc

[grave kettle]

or just notice that it’s a plane lol

planes are subspaces

[summer topaz]

Thank you

[wintry steppe]

Hello. Could I please get some help with part c please

[spare widget]

compute the adjugate of X

and then multiply it with X

I suppose [A] should actually read [X] where [X] = det(X)

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

note that the above is true for general matrix X in R^nxn

then adj(X) X = X adj(X) = det(X)I

[wintry steppe]

Thank you very much

[wintry steppe]

Is it that 'I' is the identity matrix?

[spare widget]

yes

[wintry steppe]

Okay, thank you so much

[spare widget]

in your exercise I think they want you to explicitly compute the adjugate

and then perform the matrix multiplication

[wintry steppe]

Oh okay

I believe so too

[spare widget]

I think you can prove it in the general case too though by using the laplace expansion

[wintry steppe]

Okay, well, we haven't learnt that as yet haha

[spare widget]

because when you dot column i of the adjugate with row i of the matrix, that's really the laplace expansion along row i

so you get the determinant

for row i and column j (i!=j), you can again from a determinant through the laplace expansion that has linearly dependent elements -> 0

so the general proof isn't even that hard

[wintry steppe]

Ok, thank you very much. Sorry for just responding

[cold sky]

Sorry, what does this mean?

[nocturne jewel]

Ortho complement of U

[cold sky]

Thank you

[nocturne jewel]

Most likely

[cold sky]

What exactly is the orthogonal complement of a Matrix?

Or a subspace rather?

[nocturne jewel]

If $U$ is a subspace of an inner product space $V$, then $U^\perp:={v\in V|\langle u,v\rangle=0\forall u\in U}$

[stoic python]

Mosh

[nocturne jewel]

For example, if you have a plane that passes through the origin in R^3, the orthogonal complement will be the span of the normal vector

@cold sky

$U={[x,y,z]^T\in\mathbb{R}^3|ax+by+cz=0}\implies U^\perp=\operatorname{span}{[a,b,c]^T}$

[stoic python]

Mosh

[limber umbra]

how do you use LU decomposition to find the eigenvectors and eigenvalues for this matrix?

[cold sky]

Thank you @nocturne jewel

[zinc timber]

why LU?

[limber umbra]

The actual question asks to use matrix decomposition to find the eigenvectors and values. One of the people I asked used LU.

[zinc timber]

[teal grotto]

LDU maybe? diagonals should be eigen values iirc

[spiral osprey]

So for math induction, if I am given a question asking to find the formula for "a^1,a^2,a^3...."
Is there a way for me to solve this without just knowing that the formula is a(a^n -1)/(a-1) ?
Or is this just a formula that is needed to be memorized?

[teal grotto]

do you mean a^0 + a^1 + a^2 + a^3 + ... + a^n?

[spiral osprey]

Im given a problem like this

so I am trying to find a way I can find the formula without google

but it feels like I just need to have the formula memorized for something like this

[teal grotto]

look at (1 + a^1 + ... + a^n) - a(1 + a ^1 + ... + a^n)

[native geode]

Find det(M-xI) = 0 Where M is the matrix and I is the identity matrix. x is the eigenvalue. Now do null(M-xI) and those give you the eigen vectors for each eigenvalue

[wintry steppe]

Could someone please help me with this question?

I have worked it 2 ways but I am not sure if my working will worth 5 marks

I tried combining it now

[teal grotto]

det(A) could be either -1, 0, or 1. the second way is not right. just because AB = 0 doesnt mean A or B = 0.
there is really no way to know precisely which one it is, since the zero matrix, the identity matrix, and the identity matrix times -1 all satisfy A^3 = A

[wintry steppe]

ohhh

Okay, thank you!

[wintry steppe]

when i read a textbook section, i feel like i understand everything said and examples, but when i get to the homework problems for the first time i can only solve a few by myself without looking at the solution. what can i do to fix this and learn linear algebra from a textbook most efficiently?

the beginning concepts were easier and i could do most if not all of the hw problems by myself but when i got to subspace/basis/span etc i keep on confusing concepts

the solutions are easy to understand as well when i read them and seem obvious, i just have a hard time coming up w one myself when looking at a problem and im thinking it might be because the concepts arent concrete in my mind or something else

[wintry glen]

Getting to more abstract concepts for the first time is definitely difficult. Since you say coming up with a solution yourself is the difficult part, I'd recommend reading the examples with more scrutiny. Focus on the thought process, the why behind each step. Why did they do this step? Oh, it's to verify this subspace axiom. Why did they reduced this matrix to RREF? Oh, it's to identify the pivot columns so that the dimension and a basis for the subspace can be found. When you do this enough, you have a good idea of the procedure to follow. Then, when you solve problems, you should hopefully recall the procedure. Once you do that, then how to carry out the procedure, like to show that the sum of the vectors in the subspace also belongs in the subspace or to reduce the matrix to RREF to identify the basis vectors of the column space, should come more naturally to you.

[wintry steppe]

that seems like really good advice, that made me realize i dont really focus on the conceptual why/thought process behind each step in the examples as much, ill definitely keep that in mind, thank you!!

[spiral osprey]

Yo for determinants, I recall there was some method that when solving if you multiple a row or column by a number that you also have to multiple the entire determinant by that number at the end.
what I dont recall is when this was used.
is this for any column multiplication? Only with Triangular matrix thereom for solving determinants? Or is it used in cofactor exspansion

[teal grotto]

this is just a property of the determinant as a multilinear function

[spiral osprey]

property for which rule tho or is it in general for column swaps

[teal grotto]

for any column (or row) multiplication.

[spiral osprey]

sorry jargon making my brain fry exam thursday done way too much studying thank you ❤️

[icy totem]

could you look at three questions?

[native rampart]

What have you tried

[wintry steppe]

What's the difference between the notations $F_n[\mathbb{R}]$ and $F[\mathbb{R}]$ for polynomials?