linear-algebra

[thorny hemlock]

T: V1 x ... Vm -> V1 + ... + Vm

[wintry steppe]

wait

[thorny hemlock]

like that

[wintry steppe]

okay first off

in the original problem, V_1 + ... + V_m doesn't even make sense, since the V_i's aren't necessarily subspaces of some larger space

i initially misread, forget about the previous message

so neither of 3.77 or 3.78 apply here

one way you could do the problem is by proving the contrapositive statement

another way to do it is to ||show that each V_i is isomorphic to a certain subspace of the product||

a third way to do it is to just furnish a basis of each V_i

pick your favorite

[sick trail]

is it true that |Av|/|v| >= minimum eigenvalue of A

if so how do you prove it?

[thorny hemlock]

why doesnt V1 + ... Vm make sense? owo

[wintry steppe]

because adding vector spaces only makes sense when all of the summand spaces are subspaces of some larger space

[thorny hemlock]

V are vector spaces over F

[wintry steppe]

so? you can't make sense of adding two elements of completely different vector spaces

try adding an element of R^2 and a polynomial for me

[thorny hemlock]

ok

i see

[wintry steppe]

so all the summands have to be subspaces of some bigger one

so that the addition of their elements makes sense

[thorny hemlock]

i understand

so you recommend tryna prove the contrapositive ?

[wintry steppe]

i think the easiest way to do it is the spoilered one

there's an "obvious" injective linear map from each V_j into V_1 x ... V_m

[thorny hemlock]

where do i go from there tho

[wintry steppe]

and since V_1 x ... x V_m is finite-dimensional, all of its subspaces are

[thorny hemlock]

ok

[wintry steppe]

well maybe the contrapositive way is the easiest since it avoids any funny linear map stuff

personal taste

[thorny hemlock]

T: Vj -> V1 x ... x Vm such that T(0,0,...,vj,...,0,0) vj is in Vj ?

is that the obvious linear map

[wintry steppe]

you might wanna adjust that a little

but i think you have the right idea

[thorny hemlock]

hmm

how does that show that Vj is a subspace of V1 x .. x Vm ?

i mustve forgot

[wintry steppe]

it won't be a subspace of the product but it will be isomorphic to one

and if a vector space is isomorphic to a finite dimensional one...

[thorny hemlock]

doesnt the linear map have to be surjective also to be an isomorphism ?

[wintry steppe]

good point

yes, it does, but note that i said "isomorphic to a subspace of V_1 x ... x V_m"

and not all of the product

since your linear map T(v_j) = (0, ..., v_j, ..., 0) (which im going to assume you meant to write) is injective, you can define an inverse linear map from the image of T back to V_j, proving you have an isomorphism of V_j with a subspace of the big product

[thorny hemlock]

oooh

thanks

[wintry steppe]

For each $1 \leq j \leq n$, define a map $\iota_j \colon V_j \to \prod_{i=1}^n V_i$ by $$ \iota_j(v) = (0, \dots, v, \dots, 0), $$ where $v$ is in position $j$. The map $\iota_j$ is an injective linear map, so it follows that $V_j$ is isomorphic to $$\iota_j(V_j) = {0} \times \cdots \times V_j \times \cdots \times {0}.$$ Since $\prod_{i=1}^n V_i$ is finite-dimensional, $\iota_j(V_j)$ is finite-dimensional. Since $\iota_j(V_j)$ is isomorphic to $V_j$, it follows that $V_j$ is finite-dimensional.

[stoic python]

TTerra

[wintry steppe]

now do all the other proof methods i suggested

[thorny hemlock]

lol

im not good enough :o

[wintry steppe]

suuure

i think that proving the contrapositive statement is very easy

if some V_j is infinite-dimensional then the product must also be

[thorny hemlock]

proof?

[wintry steppe]

u do it

[thorny hemlock]

lol

ok

[wintry steppe]

just write down what "infinite-dimensional" means

[thorny hemlock]

yeah

Vj has no spanning list of vectors

soooo product wont

thats it maybe?

[wintry steppe]

i think you dropped an adjective or two

[thorny hemlock]

basis?

[wintry steppe]

finite

unless by list you meant a finite list

[thorny hemlock]

yeee

[wintry steppe]

that's pretty much the idea

quite a bit easier than fussing around with linear maps in my opinion

[thorny hemlock]

the linear maps thing could be useful for other questions

[wintry steppe]

but i think it's good to see multiple ways of doing things

ya

also if you ever start messing around with like

universal property garbage

good to know i guess

[thorny hemlock]

whats that

[wintry steppe]

i think axler is kind enough to not throw that at you, though

a certain property that a space has that characterizes it

like

"if space X has the universal property, then any space Y with that property is (maybe uniquely) isomorphic to X"

you really don't have to worry about it right now

[thorny hemlock]

ooh interesting

[wintry steppe]

by uniquely isomorphic i mean, there's a unique isomorphism between them

which means they basically are the same thing

i swear i'm not an UGCT

[thorny hemlock]

is that a university :0

[wintry steppe]

lol

nah, undergraduate category theorist

it's a meme

[thorny hemlock]

oh lol

[thorny hemlock]

[soft burrow]

Nope. Consider $A=\begin{pmatrix} 1&1 \ 0&1 \end{pmatrix}$ and $v=(1,-1)$. The only eigenvalue of $A$ is $1$; it's a Jordan block. Then, if $\norm{\cdot}$ is the euclidean norm:
$$
\dfrac{\norm{Av}}{\norm{v}} = \dfrac{1}{\sqrt 2} < 1 = \lambda_{\mathsf{min}}
$$

[stoic python]

derivada.schwarziana

[soft burrow]

bounds like these are often in terms of the spectral radius, i.e. the largest eigenvalue in absolute value, since the operator norm induced by the euclidean norm has some relation to the spectral radius.

[sick trail]

hm ok

any idea where this inequality is coming from then?

you can take this lambda_min statement as given

[soft burrow]

no idea. maybe post it in #advanced-analysis with some background info?

[sick trail]

lol the notation isn't important i think

but i see

[wintry steppe]

what is beta

[soft burrow]

probably not, it's more in case the inequality follows some other way

cause I'm absolutely sure that's not true (in general) for any linear operator

[sick trail]

beta is smoothness of matrix, or upper bound on eigenvalue of hessian

the 1-alpha*beta statement is clear to me

perhaps it's because A is symmetric here?

so maybe the statement is true for all symmetric matrices or something 😂

[blissful pagoda]

if you were a linear algebra professor and this question had to be out of 10, how would you distribute the marks for each part of the question (closed by addition, closed by scalar multiplication, and standard matrix)

my linear algebra prof made it 2, 1, 7 for some reason is it just me or is that a bit unreasonable

[soft burrow]

the first two are sort of obvious and the latter is the important skill you should attain from a linalg course

I'd have done 2-2-6 just b/c I like even numbers lmao

[wintry steppe]

2,1,7 seems fine to me, maybe a bit skewed but checking linearity is the most straightforward thing in the world so it makes sense

imagine not doing 2.5/0.5/7

2.75/0.25/7

[blissful pagoda]

Damn

Okay then

[north sierra]

True/False

When u and v are nonzero vectors, Span{u, v} contains the line through u and the origin.

stuck on this.

[blissful pagoda]

@north sierra it's true

By the definition of span

At least I think

[north sierra]

can you say why?

like, is it cause the vectors u and v could be multiples of each other?

[blissful pagoda]

By the definition of span as in like "if every vector in a vector space can be written as a linear combination of vectors in a set S, then S is a spanning set of the vector space". So by the definition that would mean that you could form a line along u which will also pass the origin because you can have 0, 0 (it's a linear combination)

same method of solution just worded differently if that helps

[north sierra]

i still haven't learnt vector spaces so i don't really get your explanation.

but is the explanation in the picture because v =0?

[nocturne jewel]

span{u,v} = c_1u + c_2v. If c_2 = 0, then you get the line spanned by u. when c_1 = c_2 = 0, you get the origin

[north sierra]

but it says u and v are non zero vectors

@nocturne jewel

[nocturne jewel]

yeah

[north sierra]

but c_2 can still be 0 tho?

[nocturne jewel]

yes

[north sierra]

i see

ok

[nocturne jewel]

c_2 is the scalar

[acoustic path]

The scalar can be zero too

[north sierra]

trueee

okay

i get it now

[nocturne jewel]

{0} is always a subset of span

[north sierra]

thank you all

[nocturne jewel]

the span is just the set of the linear combinations of the vectors

[blissful pagoda]

Oh crap my bad

Imo vector space should be taught first but that's fair

[north sierra]

nah man you're good!!

[thorny hemlock]

how did they get that value of z ?

[spice storm]

They forgot to say Let Z = this ...

[thorny hemlock]

wdym

[spice storm]

In proofs usually when you want to introduce a variable you usually say "Let Z be this"

Z can be anything

[thorny hemlock]

is does say z is an element of F

[spice storm]

Z in this case is the the fraction

[thorny hemlock]

how did they get that fraction tho

wheres the motivation ?

[spice storm]

You assume a polynomial

[thorny hemlock]

degree 1 polynomial?

<@&286206848099549185>

[quartz compass]

does that argument really work, I'm thinking over the finite field $\mathbb{F}_q$ we have the zero function polynomial $f(X)=X^q-X$

[stoic python]

Merosity

[thorny hemlock]

idek

i dont understand this proof :/

[wintry steppe]

maybe F is supposed to be the reals or complex numbers

[thorny hemlock]

it is ...

[wintry steppe]

anyways the basic idea is that if a complex-coefficient polynomial has a nonzero term, then you can find an element where it's nonzero

that z happens to be one such element

then it's just inequality pushing

[round coral]

also an important thing , even if you take the finite field, F_p and your polynomial is z^p - z , I don't see any problem, why the argument doesn't work

[wintry steppe]

(what's the absolute value or modulus of a coefficient if it's from a finite field, anyways?)

[thorny hemlock]

hm i dont get it still

[wintry steppe]

what step

[thorny hemlock]

the first step

defining z

[wintry steppe]

are you asking why axler chose that z, or why he's choosing a z in the first place?

[thorny hemlock]

wheres all the other z's

[wintry steppe]

🤨

so the latter

[thorny hemlock]

how does he know that z equals all those coefficients

[round coral]

he chose a z like that , so that he can use triangle inequality

manipulation

[thorny hemlock]

ok

so how is it used in the inequality?

yes

ok

so using triangle inequality

i would have expected

|a0 + a1z +a2z^2 + .... + am-1z^m-1| <= |a0| + |a1z^1| + |a2z^2| + ... + |am-1z^m-1|

[round coral]

could you please explain to me, a coefficient from the finite field F_p, will be from the equivalence classes 0, 1,2 ...., p-1, so the absolute values, will also be 0,1,2,...., p-1 , and if we take that polynomial over F_p. f(z) = z^p - z , [-1] = [p-1] , so aren't the absolute values of z^p and z , 1 and p-1 . Or am I wrong somewhere?

[thorny hemlock]

z^m-1 >= z^j

[spice storm]

@thorny hemlock he could have chosen X, y or Z or even S. He just chose Z

[thorny hemlock]

no :/

something with the definition of z probably

[stoic python]

slimvesus

[thorny hemlock]

yes

yes

OHHH

ok

ok

ok

i see

so

he defined z only so z^j <= z^m-1 will be true?

ok

what happens in the next step?

sub in $z|a_m|$ ?

[stoic python]

Yes

[thorny hemlock]

ye

yeah, idk why i struggled on that

ok

tysm!

[wintry steppe]

Yes is learning

[thorny hemlock]

yep

[wintry steppe]

don't sully that

[thorny hemlock]

sorry

[wintry steppe]

let me share this

http://www.math.toronto.edu/payman/mat247/main.html

many problem sets and whatnot here

[thorny hemlock]

[wintry steppe]

i might have posted it before

but it might be good to look at

[thorny hemlock]

nice

[wintry steppe]

it seems that some things go wrong if you try to define a notion of absolute value on a finite field, https://math.stackexchange.com/questions/2410561/how-to-show-the-only-absolute-value-on-a-finite-field-is-the-trivial-one

from wikipedia:

The trivial value is the only possible absolute value on a finite field because any non-zero element can be raised to some power to yield 1.

[thorny hemlock]

im gonna try other books after i finish axler

[wintry steppe]

trace and determinant on page 295

[thorny hemlock]

are determinants important? :o

[wintry steppe]

ya

axler avoids them

which is one reason the book is so divisive

(probably the reason)

[thorny hemlock]

i really like axler

[wintry steppe]

i think :sully-gun: would be a good emote

[thorny hemlock]

:sully-gun:

it doesnt work :o

[wintry steppe]

well it's not an emote

[thorny hemlock]

lmao

[latent ledge]

i don't have much experience using linear algebra, @wintry steppe would you mind tell me why

[quartz compass]

not using determinants at all is a mistake, but you don't really have determinants to use in functional analysis, I think that's really the angle he takes, but idk I never read LADR

[tame mural]

the determinant captures the notion of the change in volume

it's a super nice intuition to have

it's also related to eigenvalues

So if you think about eigenvalues, you mind as well also talk about the determinant

[wintry steppe]

the determinant tells you exactly when a linear operator on a finite dimensional space is invertible

which is rather important

[wintry steppe]

axler's an operator theorist iirc so that's not at all surprising

the determinant also gives you the characteristic polynomial so you can find eigenvalues

i think axler defines the characteristic polynomial that way lol

[quartz compass]

that's not always possible

$\dv{x}e^{kx} = k e^{kx}$

[stoic python]

Merosity

[quartz compass]

I'm talking about eigenvalues right now but there's no corresponding determinant

[wintry steppe]

kek

[modern glade]

what do you guys think about this book? https://www.amazon.in/Linear-Geometric-Algebra-Alan-MacDonald/dp/1453854932

[wintry steppe]

linear transformations on page 113

[modern glade]

covers a bit of clifford algebra

[wintry steppe]

pdf for those who wish to read more than the amazon preview

[modern glade]

i have so many pdfs

[wintry steppe]

🏴‍☠️

[modern glade]

which edition is this?

[wintry steppe]

there's no word on the edition in the pdf and the first few pages match the one on amazon

so i'm going to assume first (and only)

[round coral]

weird, he discusses inner product spaces before linear maps

[wintry steppe]

i saw the title for chapter 9 and i thought it'd be some like

actual representation theory

[native rampart]

Well, atleast It's not quadratic forms before vector spaces

[wintry steppe]

well i can't comment on the geometric algebra stuff since i don't know that

the standard linear algebra stuff seems to be there, although in a bit of a weird order

[quartz compass]

what the heck is this

[wintry steppe]

read the caption

[quartz compass]

if you told me this was from IUT i'd believe you

[native rampart]

Linear Transformations are not part of Linear Algebra? This guy has some issues

[wintry steppe]

where is that written

[wintry steppe]

🤨

i think it's done that way to introduce linear transformations with the point of view of geometric algebra

[modern glade]

the book is more geared towards geometric algebra

there isn't many resources for GA out there yet

[dire thunder]

hi @native rampart are you drake

[native rampart]

No,I am moonbears

[dire thunder]

bad

[smoky crystal]

i think thats fine, its oriented around vector spaces

then it talks about linear transforamtion

but tbh it really shouldnt be called linear algebra , it should be something like linear spaces or some shit

[native rampart]

Where are affine spaces,if you are going for ga?

[round coral]

there are also very limited proof based questions, mostly computational ones, that you can use method and solve. Isn't proof based techniques and approach needed in GA? I am quite ignorant on it , so don't know

[wintry steppe]

[hollow finch]

Is it true that all for all 2x2 nilpotent matrices N, col(N)=null(N)?

[native rampart]

Yes,You can show that explicitly

[hollow finch]

Of course. Just confirming I hadn't broke the law or forgotten anything lol

[wintry steppe]

it's okay to break the law, just don't get caught

[native rampart]

Well,No one understands the law

[hollow finch]

when you get too hand-wavey with your linear algebra proofs so ||the guards burst into the classroom to arrest you||

[dire thunder]

KRGHM

[opaque plover]

Hello, may somebody help me please. I need to show the equivalence of the following. I am using the determinant of the matrix and the subsets of Q, R and C to proof this. For II to I it's easy, but if I try the other way round from a complex matrix to Q I am stuck. Because if I have an invertible complex matrix in C, the determinant is obviously not (0,0). If I go now to R the determinant of the matrix can be 0, if the complex determinant is just imaginary. (I am not sure if this is valid!) So basically I can't find a valid proof that a complex determinant in C, which is not (0,0) is also not 0 in R.

[quartz compass]

you're doing what?

[opaque plover]

Trying to show the equivalence of i and ii, by going from i to ii

[pallid swallow]

What are you doing?

I think you are leaving out some context??

[spare tapir]

Can anyone link me to any seminar for matrix equations

I need for literature

i can't think of good introduction

[opaque plover]

I wrote down everything I thought of, I think I am just doing it completely wrong

[quartz compass]

GL_m(C) is just matrices with nonzero determinant, obviously there are ones with complex number determinant that aren't in GL_m(Q)

so if you're trying to show their equivalence, you're mistaken

[opaque plover]

yeah I just noticed what I am doing wrong, I need to show that an invertible matrix of Mat_m(Q) it is an element of GL_m(C) and GL_m(Q), i have to prove this equivalence

[rose umbra]

what does it means about the matrix

[pallid swallow]

symmetric? hey wait no, that isn't a matrix, that's a set

[fickle citrus]

The elements are symmetric

[marble lance]

It's the set of symmetric 3x3 matrices

with real entries

[rose umbra]

how do you read it?

that its symetric

[fickle citrus]

The A=A^T part?

[marble lance]

That's the definition of symmetric

If you don't know it, a symmetric matrix is one which is equal to its transpose

[rose umbra]

oh make sence

thanks

[celest remnant]

I apologize for a somewhat silly question, but i think i should ask it anyway. What's an intuition behind computing eigenvectors and eigenvalues of a matrix. I would glad to know about that from various perspectives(linear algebra, computer graphics, and statistics)

[pallid swallow]

It's probably not a silly question...

[celest remnant]

I thought it is))

[pallid swallow]

Graphically, 3b1b has a good video

https://www.youtube.com/watch?v=PFDu9oVAE-g

[celest remnant]

Thanks, i definitely will.

[pallid swallow]

In markov chain stuff, the steady state is the eigenvector with eigenvalue 1

Any eigenvector with eigenvalue 0 is in the nullspace

If a matrix is diagonalisable (similar to a diagonal matrix) then there's a nice formula for calculating powers

[celest remnant]

And what about svd of a matrix?

why to compute them there?

[pallid swallow]

Also, the matrix product $\prod_\lambda(A-\lambda I)=0$ where $\lambda$ ranges over all eigenvalues...

[stoic python]

Element118

[pallid swallow]

For singular value decomposition, it involves computing eigenvalues of $AA^T$.

[stoic python]

Element118

[fickle citrus]

There's one perspective of SVD that, if you arrange the eigenvectors by the absolute values of the eigenvalues, you get an approximation of the original matrix

That's useful in an engineering sense - it's the probably the 'simplest' form of Matrix compression if you will

And if you use the full SVD, you recover the original matrix

[celest remnant]

Yeah, i've heard about matrix approximation by means of computing eigenvectors and eigenvalues.

That was the only meaning of computing eigenvectors and eigenvalues until that moment. I just needed an elaboration in order to grasp it fully.

[fickle citrus]

Well honestly Eigenvec/vals are quite deep, it depends on how deep you want to go

My perspective of Eigenvalue/vect is that they are 'natural' parts of a matrix

[celest remnant]

So, you mean that they are some "building blocks" of a matrix that constitute it?

[fickle citrus]

In the sense that if you want to pick coordinates systems over some space, you'd probably use eigenvectors

As for building blocks of matrices, that comes from the rank of the matrix and how many eigenvectors you need I suppose

[celest remnant]

And what about computer graphics? Do eigenvectors/values have application in this field? For example, to represent some transformations or something.

[fickle citrus]

This says yes to me
https://en.wikipedia.org/wiki/File:Eigenvectors.gif

[celest remnant]

oh

[fickle citrus]

Although computer graphics is quite a big area by itself

Other than transformations/scaling

You need geometry

By geometry I mean like, angles and stuff

For that you can't just use linear spaces, but you need affine spaces

Anyway it's a little vague what you're really looking for, but if it's a big summary I think 3B1B is probably a good idea.

The thing about LinAlg is that it has been generalised way beyond matrices of numbers and vectors which are tuples-of-real-numbers

[celest remnant]

Ok, i'll do that right now. Thank you very much

[rose umbra]

If i have to find the basis B of 5 vectors u1 u2 u3 u4 u5 so that:

what does it means?

[rose umbra]

i mean I know to find the basis , but what this ^^ adds ?

[subtle walrus]

you have to find a basis only using those elements

[rose umbra]

so why does they give me the fifth vector?

[subtle walrus]

what

if you have 5 vectors, you may not include the 5th one in your basis

you can only use u_1, ..., u_4

[rose umbra]

@subtle walrus so I need to find the basis of u1...u4?

[subtle walrus]

that makes no sense

a set of vectors does not have a basis

[rose umbra]

sry i meant the span of their subspace

[subtle walrus]

maybe, i would have to see the full question

[rose umbra]

that not in english so one sec

[subtle walrus]

you have to find the basis of some (sub)space

but the basis may only include vectors u_1, ..., u_4

[rose umbra]

oh.. so what the way for the solution?

[subtle walrus]

check if u_1 is in your subspace, if yes you may add it to the basis

then check if u_2 is in your subspace, if yes and if it is linearly independent from the vectors you have already added, you may add it to the basis

and so on

[rose umbra]

all the vectors are in the sub-space already

i mean the question says that there is a subspace which made of span of (u1,u2,u3,u4,u5)

[subtle walrus]

then you just need a linearly independent subset of {u_1, u_2, ..., u_4}

so basically find a basis of that subspace that does not include u_5 as a basis vector

[rose umbra]

but there is only one basis

[subtle walrus]

then find that

[stone pebble]

any body here?

[round coral]

just ask the question

[rose umbra]

What the differences between eliminating vectors by row or columns?

[round coral]

you mean row reduction and column reduction?

[rose umbra]

yes

[round coral]

they are fundamentally different

ERO preserves the row space, ECO preserves the column space

second, Left multiplication by an elementary matrix represents ERO, while right multiplication by an elementary matrix represents ECO

[rose umbra]

wait i think it wasnt my question

for instance if i got 3 vectors , i can use gauss elimination by taking each vector and put it in a row, or in columns

like this

[round coral]

do you even know what gaussian elimination is?

you want to create a matrix, that is not what gauss elimination is for

and you have 2 vectors not 3

[rose umbra]

yes that just was another example, im just confused about the differences

for example when finding basis for the span of the vectors, sometimes I see the solve with the left matrix and sometimes the right

[round coral]

you have two vectors v = (1,2,3) and w= (4,5,6) , so the span of the two of them will be span (v, w)= { av +bw : a,b belonging to F the field}

so you can write it as span (v,w) = a(1,2,3) + b(4,5,6) or as a(1,2,3)^T + b(4,5,6)^T

is your question answered now?

@rose umbra

[rose umbra]

no , im not sure how to explain it

one sec i will give u an example of what i saw

[round coral]

yes, so what's the question

you have 5 vectors , what to do with them?

[rose umbra]

just a moment i writing in in Paint

those are 2 ways I see the vectors are put in matrix :

i forgot the =0

i know that when I do it on the left matrix I get either infinite or one trivial solution

and this is how I found the base of the span of the vectors

the sub-space of the span*

but i dont understand when to use the right one

[round coral]

Ok, I see, just write them as the left matrix( that is written by you on the left) , and after Gaussian elimination , row operations when you get to RE or RRE form, the zero rows will lin dependent, the rows with the echelons will be linear independent, and that will form the basis of this list of vectors,

you know how to do GE anyway, no?

[rose umbra]

yes i do

[round coral]

you can write them as the right matrix too, but then you will need to use elementary column operations, the linearly independent columns will then make the basis of your list of vectors

[rose umbra]

what is RE RRE?

[round coral]

RE means row echelon form , RRE means reduced row echelon form

[rose umbra]

and when do I use the right ?

[round coral]

you can use anyone you want to find the basis of the list of vectors you are given, if you use the left matrix , you have to use elementary row operations and if you use the right matrix, you will need to use elementary column operations

[rose umbra]

So when does it matter?

[round coral]

it depends on what you like. I am more comfortable with doing ERO rather than ECO, so I will write as the left matrix

[rose umbra]

i understand

thanks a lot

[winged axle]

If I multiply a certain row of my matrix by a paramater "a", I need to write the condition a not equal to 0, right?

[native rampart]

Yes,If you want an elementary row operation,it should be nonzero

[winged axle]

matrix of a system of linear eqautions*

[round coral]

yes if a is 0, then what is the purpose of the row operation

[winged axle]

Yeah I was sure about this, the problem is in the answer of this exercise where it lists the solution of the system for every value of the parameter, a=/=0 is not listed

[round coral]

can't understand what you are saying, could you put up the image?

[winged axle]

So I tried to reduce the matrix

and to do so

I multiplied a row by a

and I wrote the condtion a=/=0

[round coral]

I see, so upon solving, you got no solution right for a!=0 ?

[winged axle]

the case a!=0 is not listed in the answer of the exercise. I just have a!=-2 and a!=1

[round coral]

i think that can be verified only while doing row operation by own

when you will be doing ERO, you will encounter such a case for sure, if this answer is true

[modern glade]

How does the arrow head works on vectors? Does it change based on the quadrant? is that arrow always relative to the x-axis or y-axis?

[round coral]

you mean the arrows we use above writing vectors in physics?

[modern glade]

no, i mean geometrically

[native rampart]

There are non geometric vectors

Take polynomials for example

The arrow head is completely arbitary if you mean the physics thing

[modern glade]

i mean this

[nocturne jewel]

Geometric Interpretation is literally just an arrow. The tail is at the start point and where you end is where the head points

[modern glade]

don't these arrow tell you the direction?

[nocturne jewel]

The vector "equivalent" of say a point (1,1) just has the tip on (1,1) and the tail on (0,0)

yes

whichever way the arrow points is the direction the vector faces

[modern glade]

right

i guess i'm referrinng it to programtically

as it doesn't know where to point

i have to explicitly define a radian to point the correct way

[round coral]

you can easily find the direction of euclidean vectors

[modern glade]

yep

Math.acos(Vector.Dot(a,b)/(a.Length*b.Length))

Is it possible to find the angle of a single vector though?

[native rampart]

Is this java?

[nocturne jewel]

rectangular components?

[modern glade]

nah it's JS

building a simulationn

lol

[nocturne jewel]

Oh that's what you meant by the arrow head being different

[modern glade]

yeahhhh

[nocturne jewel]

that's just computer being stoopid

[modern glade]

that's what i thought too

so i guess somehow i need to define angles of this

i was thinking to take the angle of the vector relative to the y axis

[nocturne jewel]

If you know the co-ordinates of where the head lands it should be easy

[modern glade]

of course I do

[nocturne jewel]

Ok so then you can just use sohcahtoa to find the angle

[modern glade]

between x,y value?

[nocturne jewel]

If the vector is [a,b], then the angle it makes with the horizontal is given by tan(theta) = a/b

[modern glade]

and that's the measure of the angle it makes with the horizontal line right?

[round coral]

you can see it like this, if your vector is $v= a\hat{x} + b\hat{y} \Rightarrow \tan \theta = \frac{b}{a}$

[stoic python]

Kanishk

[round coral]

if it is a vector in 3D, you find the unit vector , direction cosines

[modern glade]

@nocturne jewel thnx

[rose umbra]

im trying to prove that something is wrong but i get its true.. can anyone tell me where i wrong?

red is what i try to proof, green is given

A and B are matrix ofc

[soft burrow]

I don't think it holds in general (consider A=B=I), maybe you're being asked to find a example of A and B such that AB=I but A^{-1}B^{-1} is not I.

(and I'm not even sure that's possible if A and B are invertible?)

[rose umbra]

im being asked if this ^ is wrong or true

given A and B and invertible

yea but it didnt work in my proof

[verbal chasm]

but it is true when AB = I

idk why you wrote the thing in green

[rose umbra]

but why doesnt my proof works?

yea they are

[verbal chasm]

how do you know

[rose umbra]

its given

[verbal chasm]

ok then the statement is true

why do you think it is false

[rose umbra]

I was teached the true statement is (AB) = B^-1A^-1

and not the opposite

yes

[verbal chasm]

the opposite is false in general but when AB = I, it does happen to be true

[rose umbra]

@wintry steppe i dont think that true

[verbal chasm]

it’s likehow (a + b)^2 = a^2 + b^2 is false in general, but true for a = b = 0

[rose umbra]

im being asked in general

[verbal chasm]

no you’re being asked in the case where AB = I

[rose umbra]

this one

[verbal chasm]

you are being asked to see if it is true in the special case where AB = I. in general, it is false

[rose umbra]

the general sentence is anyway when AB = I

miss click

[verbal chasm]

no the general case is when A and B are just any unrelated matrices (of dimension nxn)

can you just post the question?

[rose umbra]

the question is not in english

let me translate one sec

"If A and B are inversed matrixs so AB is inversed and :

[verbal chasm]

AB is invertible? or AB = I? these are different things

[rose umbra]

its the same aint it?

[verbal chasm]

no, X is invertible and X = I are different

[rose umbra]

i mean if A and B are invesed so AB is inversed

[verbal chasm]

i don’t think that AB = I is actually given it just seems to be a misunderstanding

and the language barrier doesn’t help

[rose umbra]

oh wait a second

[verbal chasm]

saying a matrix is invertible and saying a matrix is equal to the identity are different statements

[rose umbra]

they said A and B are invertible matrix

i assumed by mistake that AB is I

[verbal chasm]

yes

exactly

[rose umbra]

thanks brothers u helped me 😄

[thorny hemlock]

hm i dont think i get the part about T being unique

[verbal chasm]

which bit in particular? you understand what it means for the map to be unique?

[thorny hemlock]

Only one map that can do what T does ?

is that it?

[verbal chasm]

kind of

it means that any other map that has the same properties as required by the theorem is equal to T (functions are equal iff they agree for all inputs)

[thorny hemlock]

i see

How did they get to that conclusion tho?

[verbal chasm]

it supposes that there is some map with the required properties, and applies some theorems to show that it must be in that specific form

basically, all maps satisfying the properties are the same

[thorny hemlock]

which theorems?

it seems like just defining it on the basis seems to make it unique

[verbal chasm]

well it uses the homogeneity and addivity of T

let me think if there is a better way to explain it

[north sierra]

so do the columns of the matrix span R^3?

[wintry steppe]

another way to see it is to prove that if $T,T' \colon V \to W$ are linear maps with $Tv_j = T'v_j$ for each $j$, then $T = T'$

[stoic python]

TTerra

[wintry steppe]

the columns of that matrix are elements of R^4, how could they span R^3?

[thorny hemlock]

i see

[north sierra]

oh

I just thought since it has 3 pivot rows, it should span R^3 but not R^4?

[nocturne jewel]

Does it span R^4 if the 4th entry of the vector in the span will always be 0?

[north sierra]

what

4th entry of which vector?

[north sierra]

is this true ? @wintry steppe

[thorny hemlock]

[wintry steppe]

did you read what i said

[north sierra]

ya

[wintry steppe]

the elements of the matrix you posted have columns with four entries. the elements of R^3 have three entries.

so how can the columns of your matrix possibly span (let alone be in) R^3

[north sierra]

goood point

[thorny hemlock]

could someone help me prove this

[wintry steppe]

just write out both sides

in terms of the entries of A and C

[thorny hemlock]

ok

[soft burrow]

that's almost the definition of the matrix product, just in terms of dot product

[nocturne jewel]

it's almost like the title is "Matrix product"

[quartz compass]

not really a fan of this notation

it's like einstein summation notation without the indices, but the context seems like it's just introduced the concept so it seems weird to not outright put the summation

[pallid swallow]

with a dot instead of an index

[modern glade]

can cross* product work on 2d vector?

[native rampart]

You kinda can(with generalised cross products) , but axb will be 0 for all a,b

[round coral]

By definition the cross product of 2 vectors yields another vector which is in direction perpendicular to the vectors participating in the cross product.
But if you take a cross product of vectors in 2D, they give a scalar.

so, it is not like you can't do it, but it is useless and uninsightful, it is not defined for 2D as well

so it can be called an analogue of cross product which outputs a scalar

like for 4D, we have an analogue of cross product which gives a second rank antisymmetric tensor

[spice birch]

I am trying to use a 3x3 affine transformation matrix to define a 2D camera (similar to what OpenGL does for its camera, but in 2D space). I am currently trying to figure out how to implement something similar to a zoom mechanic by zooming into or away from a specified point. Where the point is relative to the camera's perspective (affine transformation matrix).
My idea was to take the original matrix, A, and translate A to the point p, scale it by the zoom factor z and translate it back by -1/z to get A'.

[native rampart]

You should use 2x2 matrix,in that case

[spice birch]

I need 3x3 for translation

[native rampart]

Ok

[spice birch]

The 3x3 matrix is used to represent translation, rotation and scale. Being an affine transformation, the last row is always 0, 0, 1. This technique is used to manipulate the draw transform for HTML canvas elements (https://www.w3schools.com/tags/canvas_transform.asp), which I am trying to use.

Albeit, I have a separate library to manipulate the matrix before I set the canvas' transform to perform extra operations of my choosing.

[spice birch]

So, the current attempt is:
$$
A' = \begin{bmatrix}1 & 0 & -p_x/z\0 & 1 & -p_y/z\0 & 0 & 1\end{bmatrix} \begin{bmatrix}z & 0 & 0\0 & z & 0\0 & 0 & 1\end{bmatrix} \begin{bmatrix}1 & 0 & p_x\0 & 1 & p_y\0 & 0 & 1\end{bmatrix} A
$$

[stoic python]

UnknownError

[quartz compass]

how's it working out for you?

[spice birch]

It looks wildly off. It certainly zooms, but the translation is all over the place.

[native rampart]

Translation is not a linear operator

[quartz compass]

it is in homogeneous coordinates like they're doing

[spice birch]

Well, after looking at the results, the translation appears to move to an offset that is not intended, then when zooming out the translation moved to x' = x / z^2

[quartz compass]

not sure exactly what is doing what here, I guess maybe just try flipping stuff, like translate by -p_x and -p_y first

or multiply by z rather than divide, without seeing specifically, it's hard to say

try to constrain your translation/scaling to something tiny to see if you can figure out what it's off by, that might make it less wild

for instance, small scale and large translation seems about ok, but small translation and large scale seems to break, just looking for things like this

[spice birch]

I am recording the results to a gif so that you can see what I am working with

lol, probably should not have recorded the gif in 4K

[quartz compass]

it's sort of hard to tell what's happening

I said all that stuff so you could figure it out yourself, not so I could try to figure out your problem

good luck, you're on your own lol

[spice birch]

I didn't mean to...

I'm sorry man

[quartz compass]

I'm not offended or going anywhere lol

[spice birch]

I guess I got caught up on "without seeing specifically, it's hard to say"

So, getting back to the math, when z approaches 1 from the math I shown above, A' should be nearly identical to A right?

[quartz compass]

yeah probably so

[spice birch]

When z = 1 that should be:
$$
A' = \begin{bmatrix}1 & 0 & -p_x \ 0 & 1 & -p_y \ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix}1 & 0 & 0\0 & 1 & 0\0 & 0 & 1\end{bmatrix} \begin{bmatrix}1 & 0 & p_x\0 & 1 & p_y\0 & 0 & 1\end{bmatrix} A
$$

[stoic python]

UnknownError

[quartz compass]

so when z is larger than 1, are you expecting things to look larger?

[spice birch]

The issue right now is when I have z near to 1, A' != A

And my intention is that it should be identical

[quartz compass]

well it should be unless you're computing the matrix product incorrectly

how's it look like to you if you don't put the 1/z scaling in the last matrix and just do plain translation?

[spice birch]

After looking over things, I found a typo and was not performing the operation correctly.

Instead of performing -p, I was performing 1/p.

[spice birch]

So after fixing the typo and running the zoom operation without scaling the last matrix's translation component, I get interesting results. Hard to really describe what the results really due from looking at the behavior. The scale does not center on anything in particular while zooming. When attempting to zoom in anything on the +x, +y quadrant, I get a drifting zoom attempting to center on something in the -x -y quadrant.

[quartz compass]

maybe you've flipped your translation vectors around backwards

usually I think of subtracting first to center, then adding back in the end

[spice birch]

Alright, this gives better results when A starts as the identity matrix. As soon as I translate or scale A, the center of the zoom will drift to an unintended position (not p)

Ooooohhh. So, thep was left in the wrong reference frame. I needed to get a corrected p such that it is in A's frame of reference. To do that I multiplied the vector p by the affine inverse of A. The operation now looks great!

[quasi frigate]

Guys is it true for determinants?

Because when you add two matrices, you add each element a_ij

it is just not intuitive for me that there are different rules for adding two matrices or determinants

This would be true for the matrices, right?

[acoustic path]

ye

[nova marsh]

Both are true.

[round coral]

for understanding why that works for determinants, you will have to study the theory behind them

[quasi frigate]

Thanks, maybe I will dive into it later

[round coral]

nonetheless, if you can understand, you can see this https://math.stackexchange.com/questions/2219163/sum-of-two-determinants

[nova marsh]

It is different because matrices correspond to linear applications while determinant correspond to a n-linear application (but I don't know if you've seen enough theory to underrstand the difference)

[round coral]

simple I would say that is true, because determinant is multilinear

[acoustic path]

hey why is this a direct sum if all the coordinates except in the jth slot are 0?

[quasi frigate]

Thank you all, It's seems interesting and I will read more about it

[acoustic path]

oh wait

is it cuz U1 has a nonzero coordinate in u1, U2 has a nonzero coordinate in u2, etc?

and so that comes out to be u1+u2...+un

right?

[winged axle]

Is it normal calulating the inverse matrix of a 4x4 matrix is such a pain? Im using the cofactor method

[round coral]

you can also use gaussian elimination

[winged axle]

so solving AA^-1=I ? Where I is the identity matrix?

[quasi frigate]

Exactly

[winged axle]

ok thanks i'll try

[quasi frigate]

If you could give me the matrix A, I could calculate this as well, and then we could compare results

[round coral]

You can always check these matrix inverses after solving if it is correct from symbolab also

[hoary osprey]

yes

[modern glade]

Find the coordinates of the vector (4,2,3) with respect to the basis {(2,0,0), (0,1,0), (0,1,1)} of R3

is the answer for this (6,0,0)+(0,2,0)+(0,3,3)?

[thorny hemlock]

could someone help me out on this?

[hoary osprey]

no

thats just the sum of 3 random vectors

not what the q is asking

[modern glade]

is it (6,2,3) then?

[thorny hemlock]

try and equate (4,2,3) as a linear combination of the basis

[modern glade]

4e1+2e2+3e3

that's the linear combination I get

suppose e1,e2,e3 is {(2,0,0), (0,1,0), (0,1,1)} respectively

[quasi frigate]

Hello, I have a question concerning matrices.
Is it true that the biggest element of the matrix of Cofactors of a matrix A, is also the minor of the matrix A?

[thorny hemlock]

2e1 + -1e2 + 3e3 = (4,2,3)

[modern glade]

where did you get -1e2 from?

[thorny hemlock]

just a scalar

-1 + 3 = 2

[modern glade]

is that artbitrary?

[thorny hemlock]

no?

2(2,0,0) + -1(0,1,0) + 3(0,1,1) = (4,2,3)

[modern glade]

oh, i get it!

-1(0,1,0) are you distributing here?

[thorny hemlock]

yep

-1(0), -1(1) , -1(0)

[modern glade]

(0,-1,0)

[thorny hemlock]

ye

[native rampart]

For if direction:
Let v be a vector in null T_1, then
S((T_2)v)=0,but this implies T_2(v)=0 since S is invertible,implying null T_1 is a subset of null T_2. Now,write T_2=(S^-1)T_1 and repeat the argument to complete the proof

[modern glade]

should that be -1 then? (4,-1,3)

[thorny hemlock]

For if direction:
Let v be a vector in null T_1, then
$S((T_2)v)=0$,but this implies T_2(v)=0 since S is invertible,implying null T_1 is a subset of null T_2. Now,write T_2=(S^-1)T_1 and repeat the argument to complete the proof

[modern glade]

@native rampart everytime you mention null i feel like you're referring to me

[stoic python]

Yes
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[quasi frigate]

Sorry, I need to refresh the question.
The rank is the biggest minor of the matrix A. So is the rank also the biggest element of the Cofactors Matrix (since it is built from minors of the A matrix)?

[thorny hemlock]

thanks

[native rampart]

For only if direction:
Notice that range T_1 and range T_2 have the same dimension,implying there is an isomorphism between range T_1 and range T_2,That isomorphism is your required S

[thorny hemlock]

ok

[thorny hemlock]

i dont get this

we are under the assumption that S is invertable and it brings T2 to T1

[native rampart]

Consider a basis {$e_1,e_2,...e_n,e_{n+1},...e_k$} such that {$e_1,e_2...e_n$} is the basis of null($T_1$)
Note {$T_1e_{n+1},T_1e_{n+2}...T_1e_{r}$} and {$T_2e_{n+1},T_2e_{n+2}...T_2e_{r}$} are linearly independent sets

[stoic python]

MoonBears-D-

[native rampart]

Now consider the map from range($T_1$) to range($T_2$):
$S(T_1(e_i))=T_2(e_i)$

[thorny hemlock]

er

isnt it easier

just to

S is injective

So nullS = 0

implies that T1 = T2

so nullT1 = nullT2

[native rampart]

That is for if direction

[thorny hemlock]

youre showing the forward direction?

ok

For the backward direction, is mine fine?

[native rampart]

I am showing invertible map S exists when null(T_1)=null(T_2)

[thorny hemlock]

ok

[stoic python]

MoonBears-D-

[thorny hemlock]

S is injective and surjective

[native rampart]

Yes

now note T_2=ST_1

[thorny hemlock]

ok but

we dunno if T1 and T2 are surjective

S is surjective for range of T1 and T2

but maybe not surjective in W ?

@native rampart

[native rampart]

You know S_(T_1)=(T_2) for elements {e_{n+1},e_{n+2}...e_r}

Note for other basis vectors(i.e,anything in {e_1,e_2...e_n})
\S_(T_1)e_i=0 and T_2(e_i)=0

[thorny hemlock]

i dont get how you know that T1 and T2 span W

[native rampart]

I don't think you need that

[thorny hemlock]

S cant be surjective otherwise?

S needs to span W to be surjective tho right? :o

oh wait

-_-

since S is operator, being injective implies its surjective right

[native rampart]

Yes

[thorny hemlock]

ah

i forgot that existed

i get it

[thorny hemlock]

why not consider this in the first place?

[native rampart]

Well, You need to construct a map explicitly

[thorny hemlock]

ok

for the opposite direction, S is injective so T1v = 0 when T2v = 0

[native rampart]

Yes

[thorny hemlock]

nice

thanks

[native rampart]

S is an element of L(W,W),so injective implies bijective

If S is a member of L(W,V),That need not be true

[acoustic path]

Suppose U1 and U2 are subspaces of V. Prove that the intersection
U1 intersect U2 is a subspace of V

umm they both share the 0 vector? which is a subspace of V?

is that correct

[soft burrow]

that's a good first step

[acoustic path]

should i prove after that 0 is a subspace?

of V

[soft burrow]

whether ${0}$ is a subspace of V or not is irrelevant really, what you should prove is that given $a,b\in U_1\cap U_2$, then 1) $a+b\in U_1\cap U_2$ and 2) for any $\lambda \in \mathbb F$ you have $\lambda a\in U_1\cap U_2$ (where $\mathbb F=\bR$ or whatever field you're working with)

[stoic python]

derivada.schwarziana

[soft burrow]

you should've seen a result like this already?

[acoustic path]

yeah

mhm okay i got it

thank you

[acoustic zodiac]

how can i find a complementary subspace without matrices?

[tame mural]

What information do you have?

[acoustic zodiac]

i know the original subspace

but that's pretty much it

[thorny hemlock]

hint?

am i just taking definition?

ST is injective so invertable and by definition TS=I also exists ? thats all?

operators?

It does for operators tho right

invertable, inverse exists

not an operator tho

oh

is ST also considered an operator?

cuz we can think of ST as a single transformation ?

ok i think im correct then

[steady fiber]

an operator is not some well defined term in math, it's just another way of referring to functions, often used when the map (another not well defined word) is from a domain that's not just some field

you can often mentally interchange operator/function/map

[wintry steppe]

just wait until you hear about normal operators!

[thorny hemlock]

since ST is invertable

from the second part

TS = I also exists?

yeah

ST = I

o

but

ok

[acoustic zodiac]

unrelated but i haven't been using the L(V,W) notation, what does it represent?

[thorny hemlock]

yeah im wrong

[acoustic zodiac]

ah gotcha

[thorny hemlock]

any hint?

[acoustic zodiac]

we've been using Hom() notation, pretty much the same thing right?

[thorny hemlock]

since ST is invertable, S and T are also invertable

hm

yep

got it

[thorny hemlock]

Not too sure how to prove the only if direction

the direction that we assume ST=TS for every S and prove T is a scalar multiple ?

the backward

ive done the forward

oh my bad

well T has to have the domain and codomain as S

thats enough? @ornate valve

[thorny hemlock]

ST S: V -> V

T must map something into V

and TS shows that T is V -> V

v1 = av2 for some a in F v1 v2 in V

@ornate valve

[grim delta]

how do i use the tex bot? ill try to explain

[wintry steppe]

put single dollar signs around math stuff and double dollar signs around math stuff you want displayed

[grim delta]

well yeah i know latex but

does it just auto do it for you

[wintry steppe]

try it

[grim delta]

this is a test (suppose $p$ is a prime and that $a$ and $b$ are elements in a field of characteristic $p$). Then $(a+b)^p=a^p+b^p$.

[stoic python]

skrub tub

[wintry steppe]

ya

[grim delta]

hehe okay

[native rampart]

That's true in a field of char p

(Assuming p is a prime integer)

[pallid rampart]

People who use p in number theory to mean anything other than a prime shouldn't be kept alive

[native rampart]

Applicable to ring theory too

[grim delta]

Proof.

Lets say that the field over which $V$ is a vector space is $F$ (elements of which are of course scalars). Since $V$ is finite dimensional, say with dimension $n < \infty$, we let $\mathcal{V}={v_1,v_2,...,v_n}$ be a basis for $V$.

\noindent ``$\implies$": Suppose $T$ is a scalar multiple of the identity, that is, $\exists c\in F$ such that, $\forall v\in V$, $T(v)=cv$. Now let $S\in \mathcal{L}(V)$ be arbitrary. Then, $\forall v\in V$, $(ST)(v)=S(T(v))=S(cv)=cS(v)=T(S(v))=(TS)(v)$. Hence, $ST=TS$.

\noindent ``$\impliedby$": Suppose $ST=TS$ for every $S\in\mathcal{L}(V)$. For each $i$, since $T(v_i)\in V$ and $\mathcal{V}$ is a basis for $V$, we let $T(v_i)=c_{1i}v_1+...+c_{ni}v_n$ for scalars $c_{ji}\in F$. Now define, for each $i$, the linear transformation $S_i\in\mathcal{L}(V)$ by $S_i(v_i)=v_{i+1}$ (the $i+1$ is taken modulo $n$ here, meaning we take $v_{n+1}=v_1$) and for $j\neq i$, $S_i(v_j)=0$ (recall that any linear transformation can be uniquely defined in this way: by its evaluation on each basis vector). For each $i$, then, we have the following string of equalities: $c_{ii}v_{i+1}=0+...+0+c_{ii}S_i(v_i)+0+...+0=S_i(c_{1i}v_1+...+c_{ii}v_i+...+c_{ni}v_n)=S_i(T(v_i))=(S_iT)(v_i)=(TS_i)(v_i)=T(S_i(v_i))=T(v_{i+1})=c_{1(i+1)}v_1+...+c_{(i+1)(i+1)}v_{i+1}+...+c_{n(i+1)}v_n$. By linear independence, this implies that $c_{ii}=c_{(i+1)(i+1)}$ and that $c_{j(i+1)}=0$ for $j\neq i+1$. Since this holds for each $i$, we let $c=c_{11}=c_{22}=...=c_{nn}$ and note that $c_{ji}=0$ whenever $j\neq i$. Thus, for each $i$, $T(v_i)=c_{1i}v_1+...+c_{ni}v_n=0+...+0+c_{ii}v_i+0+...+0=cv_i$. Finally, for each $v\in V$, since $\mathcal{V}$ is a basis for $V$, there exist (unique) scalars $d_1,...,d_n\in F$ such that $v=d_1v_1+...+d_nv_n$ and thus $T(v)=T(d_1v_1+...+d_nv_n)=d_1T(v_1)+...+d_nT(v_n)=d_1cv_1+...+d_ncv_n=c(d_1v_1+...+d_nv_n)=cv$. This shows that $T$ is the scalar $c$ times the identity and so we're done.

q.e.d.

sorry that took so long

[stoic python]

skrub tub

[winged axle]

Hi, isnt L(i+j) just L(i,j) ? Because they are linear independent

[half ice]

What's L, i, and j?

[winged axle]

yea sorry

L is the span

i is (1,0,0) and j is (0,1,0)

[gray dust]

L{i+j} is a line containing i+j. L{i,j} is the xy plane

[winged axle]

Oh right

[limber sierra]

for example

(1, 0, 0) will be in the span of {i, j} but not the span of {i+j}

[winged axle]

Thanks,

yes a span of {i+j} is L(1,1,0) for example

if i understood right

[limber sierra]

yes

i+j is (1, 1, 0)

[acoustic path]

can someone explain how he jusst factored out r?

how are (rf)(-x)=r(f(-x))

the same thing

nvm

[unkempt crater]

Does anyone know why we use choosing when doing the binomial expansion?

[acoustic path]

choosing?

[thorny hemlock]

pascals triangle

[acoustic path]

Oh

How did u understand that

[winged axle]

I have a subspace generated by (u1,u2,u3) and I want to find a basis, so I put my vectors in columns in a matrix. If I find rank=2, can I conclude that my subspace dimension is 2 and a basis is (u1,u2)? (Or alternatively (u2,u3),(u1,u3),(u2,u1),(u3,u1),(u3,u2) so basically 2 vectors in any order)

[marble lance]

But what if u2 is 2u1?

[winged axle]

Wouldnt reducing the matrix and finding rank=2 mean since 3-2=1 the problem is only one of the vectors? But yes I understand that I cant know which vector causes the set to be linearly dependent

So should I throw away one of the 3 vectors ad check again for linear dependence? Is this the only way? Cant I avoid this check just by considering the rank of the initial matrix?

[marble lance]

If you have 3 specific vectors, then unless the one is a multiple of one of the others, then you can throw any away

It's only if two of them are multiples of each other that you have to throw one of those out, but that should be very easy to spot

[winged axle]

yeah that makes a lot of sense

so if one is the linear combination of the other 2 it doesnt matter which one i throw away. But if one is multiple of another one I need to pay attention to throw away one of those two

[acoustic path]

can someone pls explain why axler wrote z^(m+1)

how does he conclude that no list spans P(F)

[hoary osprey]

its just an example

of an element not in the span of the supposedly spanning set

[acoustic path]

ohhh

did he wanna show that the list of polynomials with degree most m isnt convergent? cuz there can always be a higher degree?

[hoary osprey]

convergent?

[acoustic path]

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

so if the vector space isnt spanned by any list of vectors its inf dimensional?

[hoary osprey]

yea

[acoustic path]

ahah

thx again

wait tho if we have a set of polynomials of degree at most 3, is the vector space with polynomials higher than 3 considered inf dimensional

when compared to the set of at most degree 3 polynomials

[thorny hemlock]

Yeh

[hoary osprey]

well its not really a vector space to begin with

[acoustic path]

oh

true

[severe magnet]

u can say a vector space is infinite dimensional if there is an infinite number of linearly independent vectors in ur vector space

[stoic python]

Nabil

[acoustic path]

thank you 🙏

[tame mural]

One idea is that you can always go either way with derivatives and integration of a polynomial

so any polynomial can be expressed as a derivative, so it is surjective but not bijective

but any finite dimensional vector space mapped to itself which is surjective is also bijective

[acoustic path]

meow

[acoustic path]

meow meow

(that means ty in cat language)

[tame mural]

Does anyone have a proof which is even nicer, and has fewer requirements for knowledge?

[acoustic path]

i think nabils proof is good

i think they were talking about their own idea lol

not axlers

[tame mural]

How do you know a combination of lower degree polynomials can't equal a higher degree one?

do u use the fundamental theorem of algebra?

[acoustic path]

so a degree 4 polynomial can be considered inf dimensional?

cuz that sounds weird

oh sorry

the dimension of the vector space

[tame mural]

the set of polynomials are infinite-dimensional

[acoustic path]

the one with at most degree 3 polynomials

oh yea cause the +1

[stoic python]

slimvesus

[round coral]

Simple as told above, if a vector space can be spanned by a finite list of vectors , it is finite dimensional. an infinite dimensional v. space , it can't be spanned by a finite list of vectors

[tame mural]

I see

i see, i see

thx

[acoustic path]

is that what axler tried proving, using polynomials?

i think i get it now

[tame mural]

yeah, it's just that his proof was a little succinct

[acoustic path]

like fr get it

yea lol

We claim that P is infinite dimensional.

Suppose to the contrary that P is given by the span of k polynomials in P, p1,…,pk. Let m denote the maximum of the degrees of these k polynomials. Then x^(m+1) is a vector in P but it cannot be written as a linear combination of p1,…,pk because taking linear combinations of polynomials of degree at most m cannot give polynomials of degree higher than m. Hence, xm+1 is not in the span of {p1,…,pk}, which is a contradiction.

i think i understand this proof more

its better worded than axlers

[acoustic zodiac]

How would you prove that yhe orthogonal of the orthogonal of F is F?

[round coral]

What is F??

[wintry steppe]

🤨

[acoustic zodiac]

F as in a vector subspace sorry

[limber sierra]

youre trying to prove two sets are equal here, essentially

F itself and the orthogonal complement of the orthogonal complement of F

the typical approach to prove sets equal is to show they're subsets of each other

so you need to:

• show any vector in F is also in the orthogonal complement of the orthogonal complement of F

• show any vector in the orthogonal complement of the orthogonal complement of F is also in F

for the second part, it may be helpful to use the fact that a vector space is the direct sum of a subspace and its orthogonal complement (assuming you've already proven that)

which means that we can write any vector in the set as u + v, for u in F, v in the orthcomplement of F

[severe magnet]

if the question still stands: Assume u have a family of polynomials $(x^n){n \in I}$ and want to check whether $$\sum{n \in I} \lambda_n x^n = 0 \Rightarrow \forall n \in I: \lambda_n = 0.$$ Then u can recognize that $$\sum_{n \in I} \lambda_n x^n = \lambda_0 + x \cdot (\sum_{n \in I \setminus \lbrace 0 \rbrace} \lambda_n x^{n-1}) = 0$$ which has to go for all $x \in \Bbb R$. Setting $x=0$ implies $\lambda_0 = 0$ so following by induction u can conclude that all coefficients have to be $0$.

[limber sierra]

(note that this requires F comes from a "nice enough" vector space; this statement is not true in the general case!)

(it is true if, say, F is a subspace of a real or complex space though)

[stoic python]

Nabil

[acoustic zodiac]

@limber sierra Thanks!

[acoustic path]

Happy new years

[thorny hemlock]

happy new years

[gray dust]

hny

[rocky wolf]

Why use lambda?? Use a better greek letter

lol

[limber sierra]

lambda for scalars is quite common

[acoustic path]

isnt it cuz of eigenvalues

[wintry steppe]

eigenvalues...

[fallen karma]

appreciate this lil channel

[hollow finch]

if we have PA=AP then what can be said about P?

or what conditions on A are necessary to know anything substantial

bc yea if A=0 then P can be anything

[limber sierra]

in general theres nothing particularly nice you can say immediately

[modern glade]

it's not 'lil'

it's quite large

[limber sierra]

the result that comes to mind is that matrices commute iff they have a common basis of generalized eigenvectors

but... thats not really practically useful for much

a necessary condition for matrices commuting is being simultaneously triangulizable

(so for some matrix S, SAS^-1 and SPS^-1 are both upper triangular)

this isnt a sufficient condition however; it becomes sufficient if you "loosen" restrictions slightly

requiring the square of the commutator [P, A] to be 0

rather than just requiring [P, A] to be 0

also if A, P commute and we know both A and P are both diagonizable we know they can be simultaneously diagonalized

(so for some S, SAS^-1 and SPS^-1 are both diagonal)

but if A or P isnt diagonalizable then obviously this falls apart

im not really sure of any other notable facts

[hollow finch]

i see

[limber sierra]

(the "commuting matrices are simultaneously triangularizable" result is actually a special case of Lie's theorem on representations of solvable lie algebras over algebraically closed fields of char 0)

(but thats beyond the scope of #linear-algebra )

now if you know some "niceness" results about P and A

for example, that theyre matrices from C^(n times n)

(or similarly R^(n times n))

we know that this means we can write A as some polynomial in P

so A = p(P) for some p(x) in C[x]

(and vice versa)

actually wait i may be misremembering that slightly

let me think

huh yeah i just came up with a counterexample lmao

okay i must be forgetting some condition

ignore that last point then

my bad

oh

this requires the minimal and char poly of A being the same

iirc

or maybe it required that of P?

one or the other lmao

in any case, still not a very practically useful result AFAIK

maybe its used as a lemma somewhere though

[hollow finch]

i was trying to prove/derive that for a real $3\times 3$ matrix $Q=PDP^{-1}$ to be skew symmetric

where $D=\lambda\begin{bmatrix}0&1&0\-1&0&0\0&0&0\end{bmatrix}$ (which is also skew symmetric)

then $P$ must have orthogonal columns. implying that if $\vec{w}=\vec{u}+i\vec{v}$ is the complex eigenvector associated with one of the complex eigenvalues of $Q$ and $\vec{n}$ is a nonzero vector in the null space of $Q$ then $\vec{u}$,$\vec{v}$, and $\vec{n}$ are orthogonal.

Using $Q^T=-Q$ you can get to $P^TPD=DP^TP$. By brute force calculation it appears that $P^TP$ must be diagonal (with $(P^TP){11}=(P^TP){22}$), which does indeed appear to imply the column of $P$ must be orthogonal by extension proving the orthogonality of the vectors mentioned about since $P=\begin{bmatrix}\vec{u}&\vec{v}&\vec{n}\end{bmatrix}$ (or at least some scalar multiples of them).

[stoic python]

nix

[hollow finch]

so i guess im wondering if there was any way i could have known that P^TP had to be diagonal (assuming i didnt just mess up)

shoot sorry for the wall of text 😳

[jovial furnace]

Hello, happy new year. Might be a simple question, but: When finding the determinant for a 3x3 matrix that is multiplied by a fraction (in this case 1/5). Why do you take the third power of the five?

[trail dirge]

Taking 5 common from each row?

$|kA_{n \cross n} | =k^n|A|$

[stoic python]

usernamephobic

[trail dirge]

Also, I have a question, how many hadamard matrix of order 4 are there?