linear-algebra

[wintry steppe]

it doesn't make sense to me

i just can't make sense of the questions

Yeah my teacher just said to ignore the eignevalues ddef, and just find where the matrix in not invertible

bruh

well your teacher's definition is wrong lol

so you should definitely ignore it

[teal grotto]

yea i thought they did what you said R2T2, but whats going on with the 1+lambda there lol

[wintry steppe]

imagine teaching a linear algebra class and messing up the definition of "eigenvalue"

[wintry steppe]

So then for part b does my work look good?

how did you get span (1, 1)?

That part I messed up on, it shouldn't be there

But besides for that is everything else good?

[teal grotto]

uh

[wintry steppe]

part a is correct

part b is not

this is fine but i don't know how you went from this to span (1, 1)

Than what would the span be?

Idk how i would find the span from this

wait

i might be high

give me a moment to make sure im not lying to you lol

okay, nevermind, the image will indeed be span (1, 1). for the image is the span of the columns of A, and it's clear that that's span (1, 1)

i got my rows and columns mixed up, sorry

you're correct, but a few words of explanation of how you went from the picture i posted, to the span being span (1, 1), would help

Okay, 👍 how would I explain it though?

"the image is the span of the columns of A"

short and sweet

Also would I need work to show how I got span of [1,1]?

[wintry steppe]

if you quote this then there's basically no work to be done

if you wanted to prove it directly

show that every Tx is a scalar multiple of (1, 1), and that every scalar multiple of (1, 1) is of the form Tx for some x

the amount of detail you should put into this really depends on what you've learned so far

ok thank you so much

no problem, and sorry about the confusion

i got my rows and columns mixed up earlier this week too

[patent wigeon]

Hi I am a beginner at vectors. I came across this problem

the answer says "|D| is the distance"

but distance from what?

it does not make sense to me.

[wintry steppe]

origin

i think

[patent wigeon]

I do not think that is true.

[wintry steppe]

me neither, i'm checking it now

[patent wigeon]

I mean a plane is (x-x0, y-y0,z-z0) dot (A,B,C)

so isn't d just -(Ax0 + By0 + Cz0)

what is "distance" the textbook is referring to?

[wintry steppe]

$|D| = |(A, B, C) \cdot (x_0, y_0, z_0)|$, where $(A, B, C)$ is normal to the plane, and $(x_0, y_0, z_0)$ is some point of the plane. if $|(A, B, C)| = 1$ (for simplicity), then $|D|$ equals the length of the vector projection of $(x_0, y_0, z_0)$ along the normal $(A, B, C)$. hence, $|D|$ is the distance in which we have to move the plane through the origin $${Ax + By + Cz = 0}$$ in the direction of the normal $(A, B, C)$ to reach the original plane. (the sign of $D$ is meaningful, but here it's asking for $|D|$ so we can ignore that.)

[stoic python]

R2T2 ✓

[wintry steppe]

it's been a really long time since i've thought about this kind of stuff

at least 3 years

[patent wigeon]

hmmmm, thank you very much

I will look over it

and I think you're right.

it makes sense! thanks

[wintry steppe]

👍

this argument also works in two dimensions, if you replace plane by line

easier to draw convincing pictures then

[spare crystal]

i think this is it..?

[stoic python]

Frank

[spare crystal]

oh, should have noted that omega(v', w' - w) = 0 by antisymmetry

is this tilde omega map basically the bilinear version of

[teal grotto]

they do be looking really similar

[wintry steppe]

let me think about it

@spare crystal you have the right idea, but your middle line is a little off

i think

it doesn't feel right

but yeah, writing v as v' + v - v' and w as w' + w - w' is good

the terms with either v - v' or w - w' (or both) in the arguments of \omega vanish, leaving omega(v', w')

i wouldn't be too surprised if this was a special case of that, in a way. since $\omega\colon V \times V \to \bR$ is alternating and bilinear, it factors through $\wedge^2V$ to a linear map $\wedge^2V\to\bR$

[stoic python]

R2T2 ✓

[wintry steppe]

maybe the kernel of this is related to the null space i defined earlier?

maybe not

[thorny hemlock]

how does one carry out a change of coordinates in the projective space?

specifically here

they move 1,1,1 to 1,0,0

not sure how

[half ice]

(X1, Y1, Z1) = (X, Y - X, Z - X)
I imagine. Then X = 1, Y = 1, Z = 1

Am I missing something because projective space? Haha

[thorny hemlock]

no

one sec

Is it just that we solve for all the m's

that would put them in terms of X' only?

or we choose coefficients for a convenient transformation?

[placid bone]

can you explain me this

[lavish jewel]

this fits better in #advanced-analysis

[thorny hemlock]

<@&286206848099549185> could someone help me out please?

[covert shadow]

what is the difference between a redundant system and a square system of linear equations?

[hollow finch]

someone else more qualified can correct me, but it looks like theyre doing a substitution x=Px'.
this is sometimes used to diagonalize things. like if A=PDP^-1 then the substitution x=Pu can turn x'=Ax into u'=Du once you multiply by P^-1.

cant say for sure though

[hollow finch]

what is a redundant system? like a system with linearly dependent equations?

[covert shadow]

i'm not sure

that's what i'd like to know

[lavish jewel]

according to google, it's a system in which at least one equation is "redundant", i.e. linearly dependent

idk what you call a "square system of linear equations" either

that's kinda weird terminology

[hollow finch]

redundant kind of makes sense i guess

square system of linear equations i assume means like Ax=b where A is square/nxn

[lavish jewel]

maybe, yeah. but idk if they also imply that the system is solvable with that name

[covert shadow]

yea

so if i have a mxn system, there could be a redundant equation in there

[hollow finch]

yeah

[covert shadow]

good

[hollow finch]

theyre not mutually exclusive though

[covert shadow]

do you know what are fill ins in an upper triangular matrix?

after QR factoring the coefficient matrix and applying linear transformations to them, i'm supposed to get fill ins

but i have no idea of what they are

and how to reduce the amount of them in order to save computation time

[lavish jewel]

sounds like it means it alters the structure

you start with an upper triang mat, but the decomposition can yield 2 dense matrices with all entries nonzero

[covert shadow]

i dont understand what you mean

you mean the Q and R matrices are dense?

[lavish jewel]

they could be, yeah

[covert shadow]

what does that mean? being dense...

[lavish jewel]

wait, lemme re read what structures QR uses

[covert shadow]

Q = orthogonal matrix (orthogonal to what?) R = upper triangular matrix

[rain skiff]

Orthogonal to nothing

[covert shadow]

[lavish jewel]

it has orthonormal columns

Q^T Q = Q Q^T = I

(and rows)

[rain skiff]

Orthonormal matrix is also a term used for it

[covert shadow]

oh

i se

ok

[lavish jewel]

which linear transformations do you mean here?

applying transformations to Q and R as in multiplying them by some other generic matrices?

because multiplying R by another generic matrix will in general not be upper triangular

[covert shadow]

sorry

not linear transformations

i meant the fill ins that appear in the upper triangular matrix that is a product from the QR factoring of the coefficient matrix

coefficient matrix is a sparse matrix in it's augmented form

i have no idea what those 'fill ins' are

[lavish jewel]

what do you mean by augmented form

[covert shadow]

i'm asked to symbolically determine the amount of fill ins in the upper triangular matrix.

[covert shadow]

Ax=b

augmented(A)=[A|b]

[lavish jewel]

i don't see how augmenting it would affect the sparsity, but ok

the matrices Q and R are in general not sparse

[covert shadow]

it doesnt

i'm saying the matrix A, which i want to QR factor, is sparse

[lavish jewel]

yep

in general, Q and R aren't sparse

that's what they mean by "fill in"

and the opposite of a sparse matrix is a dense matrix

a matrix with all/almost all entries nonzero

[covert shadow]

but how to symbolically determine the amount of 'fill ins' in the R matrix?

i still don't understand what is meant by 'fill in'

[lavish jewel]

it refers to nonzero entries in R that were originally zero in A

[covert shadow]

@lavish jewel do you know matlab scripting? i have a script here that would probably help me understand what i'm asking you if i were to understand what the script does

[lavish jewel]

i do know matlab, but i'm about to go to sleep

[covert shadow]

ok

[lavish jewel]

https://hal.inria.fr/hal-00932882/document

you can just read page 6 in that for a quick intro

but the idea is that A is sparse because it has a lot of zeros

[covert shadow]

THANK YOU

[covert shadow]

yep

[lavish jewel]

fill in refers to the factor matrices having nonzeros where A or A^TA has zeros

[covert shadow]

ohhh

[lavish jewel]

as to how exactly find the worst case fill in, i don't know the algorithms, but they should be a quick google search away for common decomps like QR, LU, and cholesky

[covert shadow]

it's okay

i get it now

thank you so much!

[covert shadow]

this is exactly the type of answer i was looking for

[covert shadow]

assuming this is the matrix R

would inputting the command 'sparse(R)' reduce computational times?

it's the upper triangular matrix after applying Tinney-2 ordination

[lavish jewel]

you'll have to read the flavors of sparse matrices in matlab

[willow ibex]

Is the MIT LA course by Gilbert strang recommended or is there something much better out there? I had a really bad lecturer this semester and honestly didn't understand anything.

[covert shadow]

wym flavors

[lavish jewel]

some of them do sparsity by indexing rows or columns with nonzero elements

neither of those help here

[covert shadow]

why?

[lavish jewel]

because all rows and columns have nonzero elements

[lavish jewel]

i like strang's content myself

[covert shadow]

but i would have to store a 1400x1400 matrix in memory

that's quite a lot

[lavish jewel]

1.) that's why you look for a matrix representation that is not ONLY based on either row or column indexing, and 2.) that's still pretty small

you'll have to check what exactly matlab's sparse function is doing

the more common ones in python only throw away rows or columns, as far as i recall

[covert shadow]

i know it uses the incidence matrix to index the non-zero elements

is that right? lol

[lavish jewel]

maybe, i don't remember lol

[covert shadow]

also

the size of the matrix is relative

for electrical engineering applications (my case) i need an application that can work in near real time with an updating coefficient matrix

so optimization is crucial

[lavish jewel]

ah, real time stuff

[covert shadow]

yep

[exotic wedge]

could someone explain why row operations don't change linear independence without using using the notion of rank?

[lavish jewel]

row operations are the same as adding a quantity to both sides of an equation

[thorny hemlock]

Thanks for replying but I figured it out thanks

[lavish jewel]

it's still the same relationship between both sides of the equation

[teal grotto]

suppose $A$ is an $m$ by $n$ matrix. Row operations on $A$ can be thought of as multiplication on the left by $m$ by $m$ invertible matrices, called elementary matrices. If $E$ is an $m$ by $m$ invertible matrix and $v_1,...,v_n$ are the column vectors of $A$, then $E(v_1),...,E(v_n)$ are the column vectors of $EA$. to keep things brief, if the first $r$ column vectors of $A$ are linearly independent, then $E(v_1),...,E(v_r)$ are also linearly independent (since injective linear transformations take linearly independent sets to linearly independent sets) thus preserving the linear independence of the column vectors of $EA$.

[stoic python]

coycoy

[stoic python]

Frank

Frank

[exotic wedge]

thanks

[livid notch]

how do i prove that these are not a system of generators of R^4?

[lavish jewel]

put them as columns of a matrix and do gauss jordan?

[inner tree]

reduce it to row echolon form

[viscid kernel]

Or rows

[inner tree]

if the rank <4 it doesn't span R4

[gray dust]

you already asked this. a solution was given in the book #help-9 message

[inner tree]

U can prove using any of this

😂

ez way is find det

if det is non zero it spans R4

[native rampart]

Det is usually bad

Because it's so annoying to compute the det

[inner tree]

depends on dimention

[native rampart]

Gauss Jordan is better

Always

[lavish jewel]

yeah, me no likey, but indeed, several ways to show it

[inner tree]

I'm just saying there are multiple ways

do what's convinient for u

and don't mind my spellings

you don't need to do gauss jordan, gauss elimination will do

[viscid kernel]

Not really, if you reduce a matrix to an upper triangular matrix, the determinant is just the multiplication of the elements of the main diagonal.

[inner tree]

u just need it in row echolon form

don't need to reduce it to reduced row echolon?

oh nvm

[inner tree]

in this case

det will reduce to a 3x3 sub matrix

[viscid kernel]

Also ( correct me if Im wrong ) if the determinant of a submatrix ( so a matrix in the original matrix ) is 0 then the det of the main matrix is 0.

Im not sure about this tho.

[inner tree]

I don't think so, but I'm unsure

[viscid kernel]

I dont know how to search this up in google lmao

[inner tree]

that's essentailly saying if a minor is zero, det is zero

which is not true

[viscid kernel]

Yup

[viscid kernel]

Oh

😦

[inner tree]

yeah

[lavish jewel]

yeah, that sounds like you could apply it recursively

so any matrix with a 0 would have det 0

[teal grotto]

its just not even true in trivial cases. two by two identity matrix has two minors that have determinant zero while the two by two identity has determinant one. similar situation with any n by n identity matrix

[solemn tusk]

wwhat does this mean

does it mean,

$\frac{\partial}{\partial x^{\mu}B_{\nu \rho}}$

?

[stoic python]

LolLol

[wintry steppe]

this notation is common in physics. i think it means $$\partial_\mu B_{\nu\rho} - \partial_{\nu\rho}B_\mu,$$ but i could be wrong

[solemn tusk]

what eta

do u mean rho

[wintry steppe]

typo

[stoic python]

TTerra

[solemn tusk]

ah alright thanks bro

[wintry steppe]

don't take my word for it

[solemn tusk]

wait

what would partial_{vp} me then

[wintry steppe]

you should ask in the physics server

[solemn tusk]

oh fair

[wintry steppe]

differentiate wrt nu and then rho

ive seen this notation used in general relativity

[solemn tusk]

Oh alright so its shorthand for $\partial_v\partial_p$

[stoic python]

LolLol

[solemn tusk]

im studying ssomething similar

im doing quantumgravity

im just not used to physics notations

[wintry steppe]

the physicists will have an answer for you lol

[solemn tusk]

cause im tryna use math for physics and it gives melike a panic attack

[wintry steppe]

i just gave a wild guess as to what it means

[solemn tusk]

nah im sure ur answer is right

its for string theory

specifically quantum grav so u r correct

:p

[wintry steppe]

check #old-network

physics server there

it's either that or some kind of sum of permutations of the indices on the brackets

[marble sierra]

Guys I'm new here and I have doubts in a specific question could someone help me?

[crude falcon]

just ask @marble sierra

[marble sierra]

I have this vector space

With the operations defined by

and need to determine the null vector of V

[dusky epoch]

is that a vector space at all though?

it doesn't even look closed under addition

(0, -1) + (0, -1) = (0, 1) as defined here, and (0, 1) is not in V.

[marble sierra]

The exercise stated that it is a vector space that's what confused me

[dusky epoch]

then the exercise lied to you and should be skipped and reported to your professor.

[marble sierra]

ok thank you very much i will talk to my teacher about

[hasty wasp]

hello

i am trying to find the limit of a sequence

or to prove that a sequence has a limit, rather

i don't really understand how to do it

[wintry steppe]

you're probably in the wrong channel then

[hasty wasp]

oh

what kind of math is it

(what channel should i go to?)

[wintry steppe]

more fitting channels would be #proofs-and-logic, #calculus, and #advanced-analysis

[hasty wasp]

oh thanks

[odd kite]

@solemn tusk $$\partial_{[\mu}B_{\nu \rho]} = \frac{1}{3!}\left(\partial_{\mu}B_{\nu \rho}+ \partial_{\rho}B_{ \mu \nu} + \partial_{\nu}B_{ \rho \mu} - \partial_{\mu}B_{\rho \nu } - \partial_{\rho}B_{\nu\mu } -\partial_{\nu}B_{\mu\rho} \right)$$

[stoic python]

Timon

[odd kite]

it's called the anti symmetric part of the tensor

https://en.wikipedia.org/wiki/Ricci_calculus#Antisymmetric_or_alternating_part_of_tensor

part of Ricci calculus notation

(in case it's not clear $\partial_{\mu}B_{\nu \rho} = \frac{\partial}{\partial x^{\mu}}B_{\nu \rho}$)

[stoic python]

Timon

[teal grotto]

is that it? what have you tried?

[wintry steppe]

ty for the correction, i was close enough lol

i figured it was something like a commutator of indices but that makes sense too

[umbral birch]

how would u solve this?

[chrome obsidian]

Nothing. I'm not confident I completely understand this.

[last iris]

that's not linear algebra, unless you don't know how to compute linear maps

[chrome obsidian]

I know that the answer is no, but I don't know why

I'm probably just being stupid, though

[nocturne jewel]

what are the requirements for a subset of a vector space to be a subspace?

(ie what does subspace test say)

[chrome obsidian]

It has to be a space

That's it, right?

[nocturne jewel]

...

[chrome obsidian]

That's what my text says

[nocturne jewel]

looking for a bit more, like what you do for subspace test

or naive test

[chrome obsidian]

Check that it's closed on addition and multiplication

[nocturne jewel]

and one more thing

[chrome obsidian]

I may be forgetting this

[nocturne jewel]

naive test:

1. ?
2. Closure under vector addition
3. Closure under scaling

what special vector has to be in the set for it to be a subspace?

[chrome obsidian]

Oh. Zero vector?

[nocturne jewel]

yes

[chrome obsidian]

That's what I thought

[nocturne jewel]

0 has to be in the subspace

is 0 in that set?

[chrome obsidian]

What's the set? P2?

[nocturne jewel]

the set you're determining if it's a subspace or not

[chrome obsidian]

Or am I checking if it fulfills the condition?

[nocturne jewel]

${a_0+a_1t+a_2t^2|a_0+2a_1+a_2=4}$

[stoic python]

Mosh

[chrome obsidian]

right

[nocturne jewel]

is 0 in that?

[chrome obsidian]

So you're asking me if 0 + 2(0) + 0 = 4?

[nocturne jewel]

yes

[chrome obsidian]

No

[nocturne jewel]

then 0 isnt in the set

so it cant be a subspace

[chrome obsidian]

So it's not a subspace

[nocturne jewel]

yes

[chrome obsidian]

It's that simple?

[nocturne jewel]

yep

[chrome obsidian]

Okay

I was seriously overthinking this

[nocturne jewel]

yeah

now if the 4 was a 0 you'd have to check the closure conditions

[chrome obsidian]

Right

[nocturne jewel]

but here it's simply 0 isnt in the set

[chrome obsidian]

Does the zero vector follow from being closed under multiplication?

[nocturne jewel]

?

we already know the existance of the 0 vector due to it being a subset of a vector space, so either 0 is included or not included in the subset

if it's not in the subset, we cant call that subset a subspace

[wintry steppe]

yes, if the set were closed under multiplication, it would contain 0. for if x is any element of the set, then 0x = 0 must be in the set

actually checking that the space contains 0 is always unnecessary

but it's quick enough and usually works that it's included in "subspace tests" anyways

[teal grotto]

how do you change you're nickname on the server

[wintry steppe]

have active role i think

[teal grotto]

how to obtain said role

[wintry steppe]

be active

[teal grotto]

who determines how active i am lol

[nocturne oracle]

me

[stoic python]

Tim O'Brien

[forest quiver]

So this is what I did

but I feel like this is too elementary

How would I solve this with a matrix?

I am trying pretty hard to stop using baby algebra

I tried doing it with matrix stuff and it was just complete nonsense

[weak needle]

Well, you can look at the determinant of {1,h,3,6} when it is zero, the matrix is not invertible

[forest quiver]

Sorry I don't know determinant yet

I just started today

I am slow learning alone :(

[weak needle]

your method works, so that’s all that matters

[teal grotto]

what is R/2 ?

[forest quiver]

yeah but I could have done this with the normal algebra I learned this year

I want to do new stuff

[forest quiver]

I am pretty sure it means "the set of real numbers minus 2"

Like all real numbers except 2

[torpid portal]

should be $h \in \bR \setminus {2}$

[stoic python]

111211211111221312211

[forest quiver]

Ok

My bad

[weak needle]

Well, for this problem you can’t get too highbrow, but another way of looking at it maybe is that when the two coolumns of the matrix are linearly dependant, you can’t get every value when you multiply by some column vector, the answer will just be a constant times one of the column vectors of the matrix

[forest quiver]

Columb vector is i hat and j hat

?

I don't know

[weak needle]

Column, like the column of a matrix

[forest quiver]

I don't have the best understanding of this

But I will learn it I guess

I know linear dependence is when two vectors can only make one line

[weak needle]

Yeah, i think it will make sense later

[forest quiver]

when multiplied by scalar

I am not picking up how columns work

I know rows are equations

[weak needle]

The column of a matrix, say {1,3} in our case, is a vector

[forest quiver]

Hmm

That is

Scary

So the columns are vectors

and the rows are line eqs?

[weak needle]

{h,6} is the other column vector

Even more fun: the rows can be a vector too

[forest quiver]

Wait but this is so complicated

I don't know if I can learn this on my own

[weak needle]

No don’t worry about this for now

[forest quiver]

Ok

Have you ever studied without a teacher and friends?

[weak needle]

When did you start linear alg?

[weak needle]

Yes

[forest quiver]

I started like 1 week ago watching videos on khan academy

But there wasn't enough practice

so I downloaded a textbook pdf

that was today

I like the idea of matrices though

[weak needle]

I see, then you don’t have to worry at all. This will all be there a little later. For now I guess you have to stick to the usual methods to do things though

[forest quiver]

OK got it

Something that excites me

I learned that you can put any data points in a matrix, like 3 variables describing one thing

and you can graph it to get a visual representation

that is super useful because a lot of times there are multiple variables that go into something, like damage, attack speed, and health in a video game

I hope I am not just saying nonsense; this is just waht I picked up today

[teal grotto]

yea linear algebra is really cool. im glad you're excited to learn about it. its one of my favorite subjects too

[weak needle]

Not at all, this sounds like exactly the sort of thing linear algebra is good for

[umbral birch]

this is a question from my linear algebra course

[wintry steppe]

mniip (owner) has a bot that tracks activity. it depends on the channel. for example, #chill messages count less towards the activity roles than questions channels

[nocturne oracle]

what about activity in #hentai

[weak needle]

Tracked by god

[forest quiver]

Really?

I am doing Boolean Algebra for my Extended Essay over the summer

And I had no idea that it had something to do with matrices

I am just not a good researcher

Im pretty much terrible at everything I do XD

[nocturne oracle]

:c dont say that. ur doing great 🥺

[forest quiver]

https://tenor.com/view/scooch-raccoon-chubby-chonk-absolute-unit-gif-14789220

Me every day

[teal grotto]

wow. that is one fat raccoon

[solemn tusk]

thanks!

[solemn tusk]

is there a geometric interpretationof this @odd kite

pls 😿

like er for any

like what does the antisymmetric part MEAN.

[onyx palm]

I am supposed to find a Diagonalmatrix D and a Transformation Matrix P for the regular Matrix A for which I know all the Eigenvectors and Values

I know that I can use a matrix made up of the Eigenvectors with A = PDP^-1

but now i need to find one that fullfills A = PDP^T

so i gotta find P so that P^T = P^-1

[dusky epoch]

define "regular matrix"?

[onyx palm]

inversible

[dusky epoch]

just invertible, so all eigenvalues are nonzero, and that's it?

[onyx palm]

yup

[dusky epoch]

hmm

i have a feeling that a decomposition into PDP^T may not always be possible, but don't quote me on this

[onyx palm]

ok so this is the thing https://cdn.discordapp.com/attachments/732592586803380314/856190750853562398/unknown.png

[dusky epoch]

ah, so A is symmetric!

[onyx palm]

mb

[dusky epoch]

then its eigenvectors for different eigenvalues will be orthogonal.

[onyx palm]

always? and for all eigenvectors?

[dusky epoch]

yes. it can be proven by simple algebraic manipulation.

for P, all you need to do is normalize your eigenvectors so they have length 1.

[onyx palm]

and then i will have a matrix where P^-1 = P^T ?

[dusky epoch]

yes

you will have an orthogonal matrix

[onyx palm]

hhhhm neat, I dont understand why but thats good enough

[dusky epoch]

if you are ok waiting around 10 minutes i can walk you through the proof of what i mentioned

[dusky epoch]

this

[onyx palm]

sure

another question. If I have a matrix with just orhogonal vectors in it I can do P^T = P^-1 always?

[marble lance]

You need orthonormal columns

So must be unit vectors too

[onyx palm]

ok so length 1

My eigenvalues are 6 and 2

my iegenvectors are

$\vec{x}= \left(\begin{array}{c} 1\ 1 \end{array}\right)$
$\vec{x}= \left(\begin{array}{c} 1\ -1 \end{array}\right)$

[stoic python]

Mergintaim

[dusky epoch]

these should both be divided by sqrt(2) to normalize them

[onyx palm]

yeah was just about to say that

does the order play an importance?

lets say my d is
6 0
0 2

then the eigenvector for 6 would be at first column as well?

[marble lance]

Yes

[onyx palm]

sweet

[dusky epoch]

the order must match between P and D

[slender patio]

Could someone explain Gauss reduction of quadratic forms to me ? In particular, I can't see the common point between the "classic version" (where you try to write q(x,y,z) as a sum of 3 or less squares) and the matrix version (where you basically do a gaussian elimination on both rows and columns on the matrix of the quadratic form)
It's bilinear algebra but I guess this is the right place anyway

[torpid wedge]

how do you integrate in regards to 2 variables?

[dire thunder]

#multivariable-calculus

[torpid wedge]

thanks

[forest quiver]

Yo guys what is the point of linear transformations?

It took me a couple of re-reads to understand

But I fail to see how it's useful

[lavish jewel]

wdym what's the point? you want like real world examples or something from the mathematical viewpoint?

[nocturne jewel]

rotation, shearing, reflection, cryptography

derivatives, integration

[onyx palm]

Most (real) Systems of any kind, if you are into engineering

[nocturne jewel]

Schrodinger eqn in its simplest form

[lavish jewel]

stuff like calculus operations and certain flavors of differential operators that show up differential eqs are linear

[native rampart]

If you want to do physics or engineering, you will have to understand linear transforms

[solemn tusk]

what is hodge star operator

[forest quiver]

Yes but how exactly?

Sorry to drag this on

[native rampart]

Ask the physics people

[forest quiver]

And linear transformations

The important part

is the movement on the vectors righht?

it's not what happens to the xy plane

does that make sense?

[native rampart]

xy plane is xy plane

[forest quiver]

I am watching a 3b1b video on it

and he keeps having the xy plane move with the vector

and it's lowkey confusing

[lavish jewel]

the movement itself is not important

only where the vectors start and where they end

[onyx palm]

@forest quiver lets start super easy at an average level. Lets say you have some kind of Blackbox system that has 2 inputs and 2 outputs. If its linear you just need two Measurements to be able to formulate an equation for the relationship between the inputs and ouputs. This is because if you have linear systems you can use superposition because in the definition of linear it says f(x_1+x_2) = f(x_1)+f(x_2) and f(c *x)=x * f(x).
And superposition is every engineers wet dream. With this you can make a super complex problem insanely easy because you can just calculate the system for a specific input, then calculate for the next input and then just add them together.

[nocturne oracle]

lol this isnt starting "super easy"

[onyx palm]

Ok lets say you have this magic box

and lets say you put some voltage on the input

you will get some kind of output

and you do this again, for another set of inputs

So far you only know two pairs of inputs and outputs

But because the magic box is a linear system, or a linear transformation if you want to call it in math terms you actually have enough information to calculate the output for every possible input combination

imagine, if you will the inputs and outputs here as vectors

so for the inputvector (6,4) you get the output (2,-1), and for the inputvector (2,3) you get the output (3,2)

because we know this system is linear and the vectors (6,4) and (2,3) are linearily independant we actually have a base for our 2D-input space here

this is pretty neat because now we can write an equation for the output vector in the form of:

$a \cdot ( \begin{array}{c}6 \4 \end{array} ) + b \cdot ( \begin{array}{c}2 \3 \end{array} ) = \vec{O}$

[stoic python]

Mergintaim

[forest quiver]

Before I get on my computer to get a better read at this, could you explain simply what exactly a linear system is?

[onyx palm]

A system that fullfiills these properties f(x_1+x_2) = f(x_1)+f(x_2) and f(c *x)=c * f(x)

[forest quiver]

And v_1 , v_2 are the base vectors of a vector ?

[onyx palm]

no v1 and v2 are just the inputs

[forest quiver]

Oh ok like a multi variable function

I see

[onyx palm]

6,4 and 2,3 are the base vectors of the linear transformation that is this system

the system is something "real" in the world. The linear transform is the mathematical model for it

[onyx palm]

yeah but for it to be linear it has to follow these rules. for example x^2 or e^x would not be a function that you would see in a linear system

[forest quiver]

Yeah and\
$\begin{bmatrix}
a\
b
\end{bmatrix}$\
Is where the vector ends up post-transformation right ?

[stoic python]

Tim O'Brien

[onyx palm]

you can however pick a small part of x^2 and say its approxximattly linear in this sector with the approximation function blabla

a and b are just the scaling factors of the base vectors

you know how base vectors work?

[forest quiver]

Yeah

[ a b ] is the new vector

Let me read all this on my computer give me 10 mins

[onyx palm]

sure, [a,b] would be the new vector if you split the vectors to get the standard base vectors yes

so instead of $\begin{bmatrix}
6
4
\end{bmatrix}$\

[stoic python]

Mergintaim

[onyx palm]

you would have
$\begin{bmatrix}
1
0
\end{bmatrix}$\

[stoic python]

Mergintaim

[forest quiver]

This all makes adequate sense to me

Except for this part

The writing the equation

[solemn tusk]

I think you mean c * f(x) 🍭

[nocturne jewel]

they wrote the O vector in terms of a different basis

that's all

since bases are not unique to the vector space

[hoary flicker]

can anyone explain this to me?

[nocturne jewel]

run through the axioms and find one that doesnt work

[hoary flicker]

for example it wouldn't follow closure under addition

right?

[nocturne jewel]

closure isnt a distinct axiom

[teal grotto]

o plus doesn’t have an identity

[wintry steppe]

addition isn't commutative

[wintry steppe]

[dreamy iron]

that was a lot of fun. you got any more?

[hoary flicker]

is that sarcasm :(

[hollow finch]

probably not but thats just my guess

[wintry steppe]

How would you go about proving that if W is orthocomplement of U then U is orthocomplement of W, where "orthocomplement" defined as:

orthocomplement $U^\perp := {\vec{v}\in\mathbb{R}^n\mid\vec{v}\cdot\vec{u}=0,,\text{ for all },,\vec{u}\in U}$ where $U$ is a subspace of $\mathbb{R}^n$

[stoic python]

Orangus

Commander Vimes

Commander Vimes

Commander Vimes

[wintry steppe]

yes, that's where I'm lost
edit: wait no, W = U^perp by definition
I misread your answer

[stoic python]

Commander Vimes

[wintry steppe]

?

[stoic python]

Orangus

Orangus

[dire thunder]

wait lol

[stoic python]

Commander Vimes

Suppose $u \in U$. We want to show that for all $w \in U^{\perp}$ we have $\langle v, w \rangle = 0$

Commander Vimes

Commander Vimes

Commander Vimes

[dire thunder]

for converse inclusion

@wintry steppe

[stoic python]

Commander Vimes

[wintry steppe]

I can't use that, as that's what I need to prove next

in my problem

from U^perp^perp = U

[dire thunder]

hmm

you may want to prove then uniqueness of orthogonal complement prolly

[teal grotto]

@wintry steppe Let $V$ be an inner product space with inner product $\langle\cdot,\cdot\rangle$ and $\dim V=n$. Recall that if $U\subseteq V$ is a subspace of $V$, then $$U^{\perp}={v\in V:\forall u\in U:\langle u,v\rangle}$$ and $$(U^{\perp})^{\perp}={v\in V:\forall w\in U^{\perp}: \langle v,w\rangle=0}$$
If $u\in U$, then we have $\langle u,v\rangle =0$ for all $v\in U^{\perp}$, which by definition, implies $u\in (U^{\perp})^{\perp}$.
For the converse inclusion, let $\mathbf{u}=(u_1,\dots,u_r)$ be an orthonormal basis of $U$ (exists because of Gram-Schmidt). Extend this to an orthonormal basis $\mathbf{u}'=(u_1,\dots,u_r,u_{r+1},\dots,u_n)$ of $V$. Then consider the map $f:V\to V$ given by $$f(v)=\sum_{i=1}^{r}\langle v,u_i\rangle u_i$$

Hints would be to show that $U=\textnormal{im}(f)$ and $U^{\perp}=\textnormal{ker}(f)$. Then replace $U$ with $U^{\perp}$ and repeat the same process to get that $\textnormal{ker}(f)=(U^{\perp})^{\perp}$ and $\textnormal{im}(f)=U^{\perp}$. You should be able to conclude, with rank nullity that $\dim U=\dim (U^{\perp})^{\perp}$ and it should complete the proof, since $U\subseteq (U^{\perp})^{\perp}$.

[stoic python]

coycoy

[wispy thicket]

A:R^3 -> R^3

I have gotten the orthonormal basis B, what does the last part mean?

[marble lance]

A represents a linear transformation that takes in component vectors w/r/t the standard basis for R^3 and gives out component vectors w/r/t the standard basis for R^3. I think [A]_B^B is the matrix that represents the same linear transformation but w/r/t the basis B you found.

So if f is the transformation,
$$[f(x)]_B = [A]_B^B [x]_B$$

[stoic python]

Lunasong the Supergay

[marble lance]

I might be talking shit, but idk what else it could be

[dire thunder]

yes

it looks like they want you find A wrt basis

[wispy thicket]

so I just apply the transformation to basis B?

[marble lance]

And then write the output in terms of B

[wispy thicket]

I see thanks, I've only seen the notation with different basis

[dire thunder]

j-th column of matrix is image of j-th basis vector

[delicate oxide]

Why is Option 2 correct, when it is clearly mentioned in the Question that Vector Space is non-zero.

[native rampart]

Do you understand what a nonzero vector space is?

[limber sierra]

let a = 1.

[dire thunder]

2. is true for any space tho

[limber sierra]

yes.

thats why 2 is correct.

[crude falcon]

some off topic but I passed Lineal Algebra and wanted to shout out to everyone helping in this channel, you helped me a lot !

[marble lance]

@delicate oxide nonzero vector space just means the vector space has nonzero vectors. It still has the zero vector. All vector spaces must have a zero vector.

[north anvil]

Is the following true?

If a matrix $A$ is invertible, then it has an inverse matrix, $A^{-1}$.

[stoic python]

Shen

[lavish jewel]

isn't it part of the definition?

a matrix A is invertible if there exists a matrix B such that AB = BA = I

and B is referred to as A's inverse

[north anvil]

I was just about sure this was the case, but I figured additional clarification would make me feel better - appreciate that

[nocturne jewel]

yes, existence of the inverse is the quintesential point of determining invertibility

[wintry steppe]

that is the definition

[lavish jewel]

that is the Tteppa

[marble lance]

The matrix has columns T(1,0,0) T(0,1,0) T(0,0,1)

Yes?

Are you trolling me?

Oh okay, dw

[marble lance]

This works for any linear transformation

If T: V -> W and V has a basis {b1, ..., bm} and W has a basis W = {w1, ..., wn} then your matrix has columns [T(b_i)]_W where [ ]_W means the coordinate vector in terms of W

So here you just have the canonical basis

So you just check what each of the basis vectors get mapped to

T(1,0,0) = (2,0,0)
T(0,1,0) = (0,1,0)
T(0,0,1) = (0,0,1)

[torn hornet]

in the standard basis you can also think of this as multiplying T with the identity, and the resulting vectors are the columns of T*1

[marble lance]

Then the matrix is
$$\begin{pmatrix} 2 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1\end{pmatrix}$$

[stoic python]

Lunasong the Supergay

[marble lance]

Yes

[marble lance]

To explain this, the i-th column of AB is A* i-th column of B. So let T be represented by matrix A. Then AI = A where I is the identity. So you can get the i-th column of AI (which is A) by multiplying A by the i-th column of I, which is the same as finding T(i-th basis vector)

[wispy thicket]

I managed to do a

but in here, do I also have to prove that T is an isomorphism, or just that the inner products are equivalent, or both?

[teal grotto]

both. but bijectivity and linearity are almost immediate, so there’s that

[rose umbra]

Be A nXn matrix , if A-3λ row are linear independant, what can you say about A?

is it Diagonalizable ? represent invertible matrix ? zero is not eigenvalue , or none of those?

[teal grotto]

what are you're thoughts on this ?

[rose umbra]

i think none

[teal grotto]

why do you say that ?

A - mu I is certainly invertible, where i just take mu = 3 lambda.

[rose umbra]

why?

[teal grotto]

its rows are linearly independent

[rose umbra]

oh wait it should be "represent invertible linear transformation "

[teal grotto]

i might be making myself look like quite an a**hole right now. you are right. it is none. sry about that

[weary spire]

if i want to show that two vectors are parallel then i have to show that they are collinear right? i.e one has to be a multiple of the other, but for which a is this the case here? if i let k = -2, then i have that the first vector is equal to (2, -2a, 4), but this contradicts the other side

[rose umbra]

nah its my fault i forgot important note..

[nocturne jewel]

cannot read your writing

[weary spire]

its k * (-1, a, -2) = (4a, 1, 4)

[nocturne jewel]

use cross product instead of scalar multiples

$a,b\in\mathbb{R}^3 \ a\parallel b \iff a\cross b =0$

[stoic python]

Mosh

[teal grotto]

i dont think such an a exists. you have solved for k, since you have to have -2k=4 which gives k = -2. then from the second row you need -2a = 1 so a = -1/2. but from the first row you have 2=4a so a = 1/2...

[weary spire]

yeah this is what i came across aswell

[weary spire]

sorry, we didn't cover cross products yet

[nocturne jewel]

ok then yeah, check if you get a solution to those 3 equations, if not then no a exists

[wispy thicket]

Is it possible to get a matrix representation for T here? I'm stumped on how to start with this one

[wintry steppe]

V is infinite dimensional (most likely) so no

also it's not even obvious what a basis of V should be

what are you stumped on? can you recall the relevant definitions (linear, surjective, injective, and eigenvalue)?

[wispy thicket]

I can recall the definitions, I'm just not sure how to solve this infinite dimensional

This is my first time encountering it

[wintry steppe]

what about V possibly being infinit dimensional is stopping you from checking those things?

[wispy thicket]

V would contain f(1) and I can't find a T(f)=f(1)

and i dont know how get eigenvalues if I can't make a determinant

[wintry steppe]

eigenvalues still make sense even in settings where determinants don't

they're just scalars c such that Tf = cf for a nonzero f in V

you could probably assume that this is true, and use it to find out what the eigenvalues are

[teal grotto]

wait why? a basis would just consist of functions f_i: N->R where f_i(k) = 1 if k=i and 0 if else

[wintry steppe]

i'm currently sleep deprived

need one like this but with bags under the eyes

[teal grotto]

lol

[wintry steppe]

wait but this isn't a basis

f(n) = 1 for all n can't be expressed as a finite linear combination, for example

[teal grotto]

it’s an infinite dimensional vector space tho

[wintry steppe]

ok

[teal grotto]

oh. it has to be a finite linear comb? even in an infinite dim setting?

[wintry steppe]

this is the distinction between a "hamel basis" and a "schauder basis" iirc

[teal grotto]

oh. i will go research those then

[nocturne jewel]

ham

[wintry steppe]

i realize i sort of threw words at you and didn't really address your question

if you only want to consider hamel bases then no, it's not a basis, but i guess that something similar to that would form a schauder basis

[wispy thicket]

it is not injective because of the upper limit converging?

[wintry steppe]

the moral of the story is that for infinite dimensional spaces you shouldn't work with bases

(at least if you're not in a banach/hilbert)

[teal grotto]

i was going to say… that isn’t entirely true. i could see some things getting messy with convergence of infinite series otherwise tho

[wintry steppe]

its not injective because it has a non trivial kernel

im not sure what you mean

im going to make the question a lot easier for you and tell you that V is just the space of convergent real valued sequences, and that the linear map T is just the left shift

[wispy thicket]

Yeah since they converge there's essentially f,g, f!=g such that T(f)=T(g)

[wintry steppe]

you can make it simpler: T is injective if and only if Tf = 0 implies f = 0. can you think of a non zero f which gets sent to 0?

[wispy thicket]

I'm not sure what a non zero f would even be, but I cannot think of one

[wintry steppe]

T maps a sequence (a, b, c, ...) to (b, c, ...)

[wispy thicket]

oh then (1,0,0,0,...) would be mapped to 0

[wintry steppe]

yes

that works

[wispy thicket]

so for surjective I need to prove that for every sequence (b,c,d,...) there exists (a,b,c,d,...)?

[teal grotto]

yea. so given a sequence a_n=a1,a2,…, in V, we can define a new sequence b_1=1 and then b_k=a_k-1 for k>=2. then where does T map b_n to

[warm kite]

Does anyone know how to find these eigenvalues? Or is more info needed

[teal grotto]

@warm kite try plugging in lambda = a

this should be a root of the characteristic polynomial of L - lambda I, so you can factor the cubic into a linear factor and a quadratic. the zeros of the quadratic along with 'a' will be the eigen values of L

[warm kite]

@teal grotto why can I do that?

[teal grotto]

do what

[warm kite]

Set lambda to a

[teal grotto]

the eigen values of L are the zeros of the polynomial you found

[warm kite]

Yeah I know

[teal grotto]

if you evaluate that polynomial at lambda = a, you will get zero

[warm kite]

Wait but why would I want that

[teal grotto]

because you can factor the polynomial to find the other two zeros, if they exist in the real numbers

[warm kite]

But when lambda is a I get that the polynomial becomes 0

How will I factor it

[wintry steppe]

[teal grotto]

um. this is called polynomial long division. if you have a polynomial p(x) and you know that p(a)=0 for some a, then there is a polynomial q(x) such that p(x)=(x-a)q(x)

[warm kite]

ahhhhh I see what you mean

So would I divide by a basically

[teal grotto]

basically, yes

[wintry steppe]

by x - a

[teal grotto]

@wintry steppe have you slept yet lol

[wintry steppe]

my disdain for do carmo's poor writing keeps me going

[teal grotto]

lmao, not to be creepy, but i saw that on mse. looks like a tough problem

[wintry steppe]

it's not tough it's just stupid

lol

do carmo constantly dodges small technical points like the one im confused on

[teal grotto]

oh. well, it looks tough to me because i just dont understand anything that was written on the page

but that seems really irritating

what is your math background if i may ask?

[wintry steppe]

it's a wonderful textbook to read if you want the idea of something, but if you want to get into the gritty technical details (i do) then it's not so great

just finished my third year of an undergraduate math degree

[teal grotto]

oh cool, i just finished my first year. you seem really knowledgeable tho

[warm kite]

Sorry to bother but how do I solve this?

[teal grotto]

the quadratic equation ?

[warm kite]

@teal grotto yeah, like how would I get values for lambda

[teal grotto]

well, use the quadratic formula of course

[warm kite]

Oh wait I see

[wintry steppe]

(lambda + a/2)^2

u can complete the square

factor it in your head

[teal grotto]

wait so what goes wrong, or are the key differences, if i use a schauder basis vs a hamel basis ?

other than not being able to write some vectors as a finite linear combination of basis vectors with the schauder basis, is there anything else that is different?

[wintry steppe]

hamel basis makes sense for an arbitrary vector space, schauder basis is for a normed vector space and allows for infinite "linear combinations" provided they converge

axiom of choice implies that every vector space has a hamel basis

[teal grotto]

right. with zorns lemma

[wintry steppe]

same thing xd

a concrete example of a hamel basis is {1, x, x^2, ...} for R[x]

as for schauder

[teal grotto]

right. so hamel basis can be infinite but each element can be written using only finitely many basis elements

[wintry steppe]

yup

[teal grotto]

hmm. schauder basis for space of real sequences would just be the the one i provided above then

those feel like good examples to remember the distinction by then. thnx

[wintry steppe]

for a concrete example of a schauder basis, consider \ell^2, the space of square summable sequences. then {e_1, e_2, ...} is a schauder basis that isn't a hamel basis

something like that

every hamel basis of a normed vector space is trivially a schauder basis, but as you've just seen the converse is false. that should make the distinctions a bit clearer

[teal grotto]

ig they give different conditions for linear independence and spanning as well

[wintry steppe]

right, you have to be careful for schauder

you ask that every element of the space be representable uniquely as a convergent "infinite linear combination" of the basis vectors

that unique part in italics is the linear independence analogue ig

[teal grotto]

ah ok. cool

[wintry steppe]

does anyone have any idea abt this

Hey guys I’m stuck on a hw problem

Pls ping

For this I have an answer ^ just not sure

[restive raft]

send it

[wintry steppe]

Matrix 2 x 2

Top row 1 0

And 0 -1 bottom

Our textbook has it different in different spots

Also can u verify this bc I wasn’t sure if it’s 2 x3 or 3 x 2

[restive raft]

2 rows, 3 columns. So a 2x3

[wintry steppe]

So the way i have it is right?

[restive raft]

yeh looks gud

[wintry steppe]

One more question

If u don’t mind

for this matrix I got when h is not equal to -4 it’s consistent

m conflicted on if this is right or if it’s all values will make it consistent

Does anyone happen to know? ^

nvm not necessary^

[violet river]

does anyone wanna help me with Exponential Functions, Features of Exponential Functions and Graphs, Solving with Exponentials, Analyzing Quadratic Functions homework?

I will pay

It's for my course online and im struggling

[vocal prairie]

You are requested not to offer money for getting your homework done. Read #rules .

Also, read #❓how-to-get-help to get help with these topics in an appropriate channel.

[delicate oxide]

I am trying to derive this formula, but unsuccessful yet:

I understand it is to do with "determinants".

The numerator is the product of adjacent sides of the parallelogram, and the denominator is the "determinant" of the vectors of the parallel lines.

[signal badge]

do you know the dot product?

[lavish jewel]

i think cross product is what they're looking for

[delicate oxide]

Anyways, got it. Thanks.

[drowsy flower]

stuck on this question....

I don't think what I wrote in the middle paragraph is wrong....

But I am not sure how to do the last one since I am not given any explicit mapping? Would the last one also not have enough information in this case?

[rose umbra]

remind me what is it ?

[lavish jewel]

norm of some kind

[dusky epoch]

poorly typeset $\nrm{v}$ probably

[stoic python]

Ann

[rose umbra]

ah probablly ty

[wintry steppe]

Can anyone help me out with this

[dusky epoch]

here's a hint: think about geometric series

this might lead you to a formula for (I-A)^-1

[wintry steppe]

Okay ty

[broken sun]

Hello. Any $2 \times 2$ unitary matrix with determinant $1$ except $\pm I$ can be described as a nontrivial $3 \times 3$ rotation matrix?

[stoic python]

MathPhysics

[wintry steppe]

@dusky epoch can guide me?

[native rampart]

Hint (1-x)(1+x)=1-x^2

[exotic wedge]

If the column operations on matrix A are identical to row operations on A^T, why don't column operations preserve the rank?

[dusky epoch]

they do preserve the rank. who's telling you they do not?

[dusky epoch]

buncho dragons gave a hint.

[wintry steppe]

Yeahh i got it thanks @dusky epoch and @native rampart

[exotic wedge]

oh, the answers online xd

[broken sun]

<@&286206848099549185> Hello. Any $2 \times 2$ unitary matrix with determinant $1$ except $\pm I$ can be described as a nontrivial $3 \times 3$ rotation matrix?

[stoic python]

MathPhysics

[rose umbra]

T(0) != 0 therefore T isn't linear transformation. (is this true also in case of complex numbers above R and C like in this example?)

[dire thunder]

yes, any linear map should map 0 to 0

[rose umbra]

ty

[wintry steppe]

Hi

How are you?

I'm studying linear algebra from zero

and right now I'm at the gaussian elimination part (and also Gauss-Jordan)

And I'd like to see a proof of that

Especifically, the case where you multiply a row by a number

I don't know why you can do that

And I'd like to know if there's a proof

[native rampart]

Why do you think you shouldn't do that

[wintry steppe]

No, it's not that

[native rampart]

You want to know why elementary operations work?

[wintry steppe]

Yeah

Especifically, the one where you multiply a row by a number

The book I'm reading says that it works because if we consider 3 elements $a,b,c$ from a field $K$, the idea is based on the fact that $a = b \Longleftrightarrow ac = bc$

[stoic python]

Obamid

[wintry steppe]

I don't understand how that proves that you can multiply a row by a number and then the 2 matrices are equivalent

[native rampart]

Let's start with something simple

You know how to solve linear equations?

[wintry steppe]

Yes

[native rampart]

Consider the following systems of linear equations:
2x+y+z=0
3x+2y+4z=0
5x+3y+z=0

and
2x+y+z=0
x+y+3z=0
5x+3y+z=0

These 2 should have the same solutions

Because you just subtracted equation 1 from equation 2 in 1 st system to get 2nd system

[stoic python]

Buncho Dragons

Buncho Dragons

[native rampart]

You say these 2 matrices are row equivalent

Because the corresponding linear systems can be converted to the other form by doing some perfectly legal operation(In our case system 1 -> system 2 Do eqn 2-> eqn 2- eqn 1
system 2-> system 1 Do eqn2 -> eqn2+ eqn1)

Elementary row operations are defined such that they convert a matrix into another row equivalent one

[wintry steppe]

And when you multiply a row by a number?

[native rampart]

Multiply by inverse to get back original system

That's assuming inverse exists

That's why multiplication by 0 is not elementary

[wintry steppe]

So I can understand that with lines, yeah?

[native rampart]

Well,with linear equations

I guess you can make some Intuition with lines

[wintry steppe]

In R², you would have that 2x + 4y = 5 is the same as 4x + 8y = 10, right?

[native rampart]

Yed

[wintry steppe]

So that's why in gaussian elimination you can multiply a row by any rational

Okay, thanks

:)

[native rampart]

You can multiply a row with anything that has an inverse

So, if you were talking rings in general,that means you can multiply only with units

[warm kite]

Can someone please explain reducible matrices to me

This makes absolutely no sense

[wintry steppe]

that's quite the definition

[spare crystal]

It means you can split the indices $1, \dots, m$ into two sets $S_1$ and $S_2$ such that any entry $a_{i, j}$ in the matrix where $i \in S_1$ and $j \in S_2$ is zero

[stoic python]

Frank

[spare crystal]

at least i think lol

[wintry steppe]

does this come with any examples?

[spare crystal]

heres an alternate definition

https://math.stackexchange.com/questions/1342022/definition-of-reducible-matrix-and-relation-with-not-strongly-connected-digraph/1342135

[wintry steppe]

to no one's surprise it's just a hard way of saying something simple

[spare crystal]

The permutation just rearranges the indices so S_1 = 1, ..., k and S_2 = k, ..., m

[spare crystal]

I'm trying to prove that if $T$ is a linear operator over a complex vector space and has an orthonormal eigenbasis, then $T$ is normal (one direction of the complex spectral theorem). This stack exchange post says that if $v_1, \dots, v_n$ is an orthonormal eigenbasis of $V$ with eigenvalues $\lambda_1, \dots, \lambda_n$, then $T^*v_i = \overline{\lambda_i}v_i$. Does anyone know how I can show this?

[stoic python]

Frank

[spare crystal]

https://math.stackexchange.com/questions/2307642/complex-spectral-theorem-proof

[split slate]

any hints please

[wintry steppe]

1. there's a typo in the displayed equation, the middle term should be |lambda_i|^2 v_i

2. show that |(T* - \bar{\lambda_i}I)v_i|^2 = 0

[spare crystal]

oh hmm lemme try that

[wintry steppe]

i think that'd work

[spare crystal]

i think i have to somehow use the fact the v_i's are orthogonal because otherwise it wouldn't be true but im not exactly sure how to incorporate that

[wintry steppe]

ah you're right probably. i thought about it a bit more and for my proof to work you need to assume T is normal (which is what you want to prove...)

you can probably show that the inner product of (T* - \bar{\lambda_i}I)v_i with each v_j is zero

[spare crystal]

ohhh wait I think I can show the inner product of T*v_i with each v_j is zero

0 = <lambda_j v_j, v_i> = <Tv_j, v_i> = <v_j, T*v_i>

Which means v_i is an eigenvector of T* which I think is most of the work

yeah i think that works

[wintry steppe]

the first equality isn't necessarily true

[spare crystal]

but v_1, ..., v_n is orthonormal

[stoic python]

TTerra

[wintry steppe]

if i = j the first equality won't necessarily hold

[spare crystal]

oh right, but if it holds for all j ≠i then that shows that T*v_i is orthogonal to all v_j where j ≠ i, which is all you need

[wintry steppe]

ah yeah

that's right

and works too

[spare crystal]

oh this is a nice way too

[wintry steppe]

to finish your argument off, $T^*v_i = cv_i$ for some scalar $c$. since $\langle v_i, v_i \rangle = 1$, $c = \langle T^*v_i,v_i\rangle=\bar{\lambda_i}$

[stoic python]

TTerra

[wintry steppe]

so orthonormality is, as you pointed out, very explicitly needed here

neat

[weak needle]

Wait, I know this is probably wrong but don’t we immediately get T to be normal just by saying that is is diagonal wrt to the basis {v_i}?

[wintry steppe]

you'd have to know that the adjoint of T has matrix equal to the conjugate transpose of that of T (wrt orthonormal basis) and the proof of that fact probably follows from exactly the same work done here

[weak needle]

Wait T* is just the conjugate transpose right?

So what work exactly do I have to do

[wintry steppe]

T* is the adjoint of T. with respect to an orthonormal basis, it's matrix is the conjugate transpose of that of T

if one knows this already then there's no work to be done, as you say

[spare crystal]

oh i was trying to show it without this, yes if you use that its immediate

[weak needle]

wrt any basis, not just orthonormal right?

[spare crystal]

no it has to be an orthonormal one

take the identity operator, and write it as a matrix w.r.t. something thats not orthonormal. you get something that's not symmetric

[weak needle]

Oh yeah, that’s true

[wintry steppe]

wait i don't get it. the matrix of the identity operator with respect to any basis is going to be the identity matrix

[spare crystal]

LOL oops yeah thats not a valid example

[wintry steppe]

[spare crystal]

I suppose you can take something like $$\begin{bmatrix}1 & 0 \ 0 & 2\end{bmatrix}$$ and write it with respect to some other basis

[stoic python]

Frank

[wintry steppe]

a random example i pulled out of my ass is T(x, y) = (x + y, y) and the basis (1, 0), (1, 1)

i just wrote down the first operator i could think of lol

[spare crystal]

ah thats easier. for some reason i was trying to come up with a normal operator

[wintry steppe]

there's a tiny bit of computation to be done so it's not immediately clear that this example works

or if it works i could have made a computation error

[spare crystal]

i think it works, I got the original operator has the identity matrix but the adjoint doesn't have the identity

[wintry steppe]

if the original operator had the identity matrix then it'd be the identity operator on C

[spare crystal]

that'd be correct

what about $$\begin{bmatrix}1 & 1 \ 0 & 1\end{bmatrix}$$ for $T$ and $$\begin{bmatrix}0 & -1 \ 1 & 2\end{bmatrix}$$ for $T^*$

[stoic python]

Frank

[wintry steppe]

that's what i got

[spare crystal]

ok whew i can still do math

[wintry steppe]

computations are good for you

[weak needle]

Anyway, thanks for the clarification, I figured I was going wrong somewhere

[teal grotto]

is it true that every invertible matrix with entries in {0,1} have a real eigenvalue?

was curious because I came across a problem which asked to show that an n by n invertible matrix with entries in {0,1} can have at least n ones and at most n(n-1)+1 ones.
the number n(n-1)+1 happens to be the dimension of the space of n by n matrices with v_0 as an eigen vector for some non-zero v_0 in R^n

[broken sun]

Hello. Any $2 \times 2$ unitary matrix with determinant $1$ except $\pm I$ can be described as a nontrivial $3 \times 3$ rotation matrix?

[stoic python]

MathPhysics

[broken sun]

<@&286206848099549185>

[torn hornet]

what did you try?

[broken sun]

I have no idea about it.

[torn hornet]

well lets start by explicitly writing out what elements of 2x2 unitary matrices with determinant 1 look like?

[broken sun]

Ok

You are are waiting for my answer?

[torn hornet]

yes

i am expecting you to explicitly telling me what these matrices look like

[broken sun]

[a b \ -b* a*] with |a|^2 +|b|^2=1

[torn hornet]

mhm

now what about 3d rotations

[broken sun]

about what axis?

[torn hornet]

any

(a better way to talk about this btw would be using quaternions, are you familiar with those?)

[broken sun]

Only definition

[torn hornet]

i see, so unit quaternions are precisely rotations in 3d

i.e things of the form a+bi+cj+dk with a^2+b^2+c^2+d^2 = 1

now heres the hint, can you think of a way to represent i,j,k as 2x2 complex matrices?

(its related to unitary matrices)

[broken sun]

[i 0 \ 0 -i], [0 1 \ -1 0], [0 i \ i 0]

[torn hornet]

mhm, now use this idea to write your SU(2) matrices (the 2x2 unitary matrices with det=1) as unit quaternions

[broken sun]

How?

[torn hornet]

Break the SU(2) matrices as linear combination of those matrices and the identity

[broken sun]

A linear combination of unit quaternions is a nontrivial 3 * 3 rotation matrix?

It must not be true; their ranks are different.

[torn hornet]

The unit quaternions are the rotations in R^3

the 3x3 matrices are just a way to write them

so if you show each element of SU(2) can also be written as a unit quaternion then you have shown its a rotation in R^3 and you are done

Anyhow notice

[broken sun]

Why are unit quaternions rotations in R^3?

[torn hornet]

oh let me find a good source on it

https://graphics.stanford.edu/courses/cs348a-17-winter/Papers/quaternion.pdf

try reading this @broken sun , rotations in 3d are almost always written as unit quaternions

[broken sun]

@torn hornet Thanks. So, to answer my original question, I have to first show that rotations are quaternions, which is somewhat complicated?

[torn hornet]

I think you could go without it, but those would probably be disgusting algebra bashing with 3x3 rotation matrices. This is probably better and would help intuition

[broken sun]

Ok. Thanks.

[torn hornet]

(btw if you want some good detailed expositions on this, John Stillwell's "Naive Lie Theory" chapter 1 is very good to explain this)

[silk ether]

Does swapping rows during elementary row change the sign of a matrix?

[limber sierra]

if you mean the sign of its determinant

yes

each "swap performed" multiplies the determinant by -1

you can verify this by computing the determinant of a permutation matrix P [it'll be -1 if its just a single swap] and recalling that det(AP) = det(A)det(P)

[rose umbra]

is there a quick way to check whether A and B are diagonalizable matrices ?

[lavish jewel]

you'll have to check the eigenvalues and eigenvectors

i can't think of a faster way than trying to do it

[rose umbra]

alright wanted to make sure I dont miss anything

[torn hornet]

they are not btw

well B isnt*

basically you know the eigenvalues are 1,1,2 right, you just need to confirm 1 has geometric multiplicity 2

[rose umbra]

yea

[twilit anvil]

i thought u can just check if they are symmetric