linear-algebra

[little crater]

ie inner product

[quartz compass]

yeah, I guess I skipped a step there by using the transpose (AB)^T=B^T A^T

[little crater]

yeah and then that theorme

[quartz compass]

yeah so focus on what U^T U is

[little crater]

okay

U^TU is identity

because they are orthogonal

or no?

[quartz compass]

yeah

[little crater]

okay

[quartz compass]

an normal

[little crater]

oh

[quartz compass]

the diagonal is their magnitudes squared

[little crater]

the normal part matters at all for this though?

[quartz compass]

err normalized*

meaning unit length

[little crater]

well orthonormal matrix

[quartz compass]

right, well I'm saying U^T U is the identity not just because they're orthogonal but because they're also normalized

[little crater]

I am a bit confused are orthogonal matrices always square

[quartz compass]

the identity matrix has 1s down the diagonal

[little crater]

because this says mxn

[quartz compass]

the vectors are orthogonal which makes zero

yeah it will give you an nxn identity matrix

[little crater]

Like by definition this is what my book says.

[quartz compass]

so here, what you should realize is if the matrices A and B contain column vectors, then A^T B is the matrix of all their dot products

so if you look at every entry on the diagonal, you are dotting a column vector with itself to get 1

and everything off the diagonal are 0 because they're orthogonal

[little crater]

this is right? I am just picture cases of orthogonal matrices that are not square

[quartz compass]

the fact that it's not square doesn't matter here

you could fill it out with more vectors to make it a square matrix if you wanted

[little crater]

well sure but I just wanted to know if my definition of "orthogonal matrix" is wrong.

[quartz compass]

that definition is right

[little crater]

so this isn't an orthogonal matrix?

[quartz compass]

no but it has orthonormal columns

[little crater]

oh okay

[quartz compass]

err I guess there is a problem sorry

[edgy shale]

hi, so I learned a few days ago that $\mathbb{R}^n$ is the set of functions such that $f : [1, ..., n] \to \mathbb{R}$. So if we have $\mathbb{R}^{[0, 1]}$ it is the set of functions over domain [0, 1] that maps to $\mathbb{R}$. Let say we have $\mathbb{R}^2$ or even $\mathbb{R}^3$, what exactly does the the elements in the space really mean? Like for $\mathbb{R}^2$, what is the meaning of the vector $[0, 2]$ or $[1, 4]$ and the case of $\mathbb{R}^3$ what is the meaning of the vector $[1, 4, 2]$?

[quartz compass]

no there's not, I was thinking of something else

[stoic python]

ModernNoob

[little crater]

okay I think I got it, thanks!

[quartz compass]

cool

[edgy shale]

from the $\mathbb{R}^{[0, 1]}$ example, i thought perhaps the elements in $\mathbb{R}^2$ such as $[0, 3]$ represents the domain of the set of functions that map to $\mathbb{R}$, but i feel like this is incorrect because i couldn't relate this when dealing with $\mathbb{R}^3$

[stoic python]

ModernNoob

[hexed urchin]

I never figured out this question if anyone could take a look ;_;

[copper pelican]

It’s not really a basis (since in R2 you can only have two linearly independent vectors at most), but you are right in saying that the range is R2

So using the standard basis should be fine

[hexed urchin]

thanks! : )

[little crater]

@quartz compass sorry to ping but what is so important about an orthogonal matrix?

I mean I know what it means to be orthogonal but idk the specific chartersitics of an orthgonal matrix other than U^TU = I

feel like this should be easy

but I must be missing something

[copper pelican]

That identity should suffice to prove orthogonality of UV

Apply it to the matrix UV and think about how the transpose operation works with products

And then use orthogonality of U and V

[little crater]

okay I think I got it

[sudden cedar]

https://stats.stackexchange.com/questions/63143/question-about-a-normal-equation-proof machine learning perspective right here (on why orthogonality is important for the normal eq i.e. a determinate method to optimize cost where cost is a polynomial regression function) p.s. basically, at least, it's really important for an extremely common ML algorithm

[knotty tiger]

would anyone mind helping me out with #help-14 🥲

[little crater]

wait I think the U^TU = I will only work if U is orthonormal

[copper pelican]

Hm wiki says an orthogonal matrix is the same as an orthonormal one (which is also what I learned)

[little crater]

oh....

See I knew I wasn't sure on "orthogonal matrix"

[copper pelican]

I dont think matrices with just orthogonal columns are studied as much

[little crater]

okay

[copper pelican]

yeah an orthogonal matrix should have orthonormal columns

[little crater]

Ik it seems dumb but what makes it okay to say (UV)^-1(UV)=I

by this I mean what makes it okay to even assume the inverse exist?

[copper pelican]

Oh it is a fact that the product of two invertible matrices has an inverse, but in a sense you don’t really need that here

[little crater]

Here is what I did and I assume this is what they would want

[little crater]

True true

[copper pelican]

Hm that seems fine enough

[little crater]

I wrote on the side about

U^TU=I, U^T=U^-1....

since they were given to be orthongal matrices

but I should add the comment about the product of invertible matrices

[copper pelican]

Imo it’s slightly cleaner to do it from bottom to top, since all we need is to check whether (UV)^T (UV) is the identity

But it should work with your comments too

[barren crystal]

something wrong with EdX or am i missing something?

the answer to first problem is (-3, 6) but like how lmaoo

[green trench]

uhhhh

what

like [0,-12] + [0,-12] + [0,-12]???

how

[barren crystal]

some chrome extension was messing up the rendering, realized it later on lol

[green trench]

lol

[dusky epoch]

something's wrong

[glacial mantle]

where should i start if I want to learn linear algebra

and what is the fastest and most efficient way of learning it

[copper pelican]

there's a list of recommendations in the pinned messages in #book-recommendations

[languid otter]

im having trouble with algebra can somone help

[wintry steppe]

if you ask a question, someone could help, but if you don't, no one can

[dim epoch]

orthogonal transformations preserve lengths and angles so |x| = |Ox| for an orthogonal matrix O, same with scalar products, <x,y> = <Ox,Oy>, that's maybe another interesting thing when talking about characteristics of orthogonal matrices

makes them important for a lot of stuff in numerics, another nice thing is the way they act on unit spheres

[wintry steppe]

brandon: see "real spectral theorem"

[winter pond]

Does anyone know how could this be done?

[wet stratus]

can you find a nonzero vector v such that T(v)=0 ?

[winter pond]

No

The thing is I missed my class and I can’t catch up with knowing what invective and surjective is 😭😭😭

Only cuz my lectures aren’t recorded

[gray dust]

the definitions can be googled

so can theorems relating them to linear maps

[dim epoch]

and then you can think about what that implies for linear maps

could be a nice exercise after missing class to get into it

[winter pond]

I only got the part where injective is 1 to 1 and surjective is one to many

[dim epoch]

so how can you refute injectivity?

[winter pond]

If the nullity of the transformation isn’t 0?

[clever totem]

we have been given that
$\mathbb{R}[x]/\mathbb{R}x$ gives the complex field
we now have to show that
$\mathbb{C}[x]/\mathbb{C}x$ is not a field
any ideas?

[stoic python]

~Martin

[dim epoch]

think about why that is the case

and try to refute injectivity with that reasoning

[dim epoch]

why this refutes injectivity i mean by that

[winter pond]

I’m still not getting it

[dim epoch]

have you looked up the definition of injectivity

[winter pond]

All I’m getting is “To test injectivity, one simply needs to see if the dimension of the kernel is 0. If it is nonzero, then the zero vector and at least one nonzero vector have outputs equal 0w implying that the linear transformation is not injective.”

[dim epoch]

i think you might have better luck with that in #groups-rings-fields

[dim epoch]

so do you understand what's written there

this only makes sense if you know the definition of injectivity

[wet stratus]

can you find two elements $a,b\in\bC[x]/(x^2+1)$ such that $ab=0$? then this can't be a field.

[stoic python]

Denascite

[clever totem]

thats what im looking for as well as examples where a*b=1 does not hold but i cant seem to find any

oh wait

i think i got it

thx for the tip

[hard drum]

Since x^2 + 1 splits over C, we can also use CRT to show this ring is isomorphic to C x C, which I find perhaps more insightful

[wet stratus]

yes but tbf, to show that CxC is not a field you also look for zero divisors

[hard drum]

Yes, exactly

Or use the fact x^2 + 1 certainly isn't a prime element of C[x], many ways to do it

[quartz compass]

a polynomial of degree d in a field can have at most d roots, but z^2+1 has at least 4

[winter pond]

The transformation is injective if only kernel is 0?

[dim epoch]

yes but why is that the case

why is a transformation whos kernel isn't only 0 not injective, that's what i tried to get out of you

[little crater]

Yeah I am seeing that in the book. Today (yesterday) was rushed and my prof skipped around sections and we skipped stuff like orthogonal projections. Trying to piece together the bits of information he selected from the course because he felt they were important before the course ends

which made it hard to figure out how to get an orthogonal vector given my 2 basis vectors

which I needed for this diagonalization exercise

[dim epoch]

why would one skip orthogonal projections

[little crater]

I guess because we have 2 days left in his class

[dim epoch]

jesus

[little crater]

so it is at the point where he has cherry picked topics

[dim epoch]

orthogonal projections are neat

[snow shore]

I have a problem where I'm given that T^10 = T^8 and that T is a linear transform from the unitary space V to V.
And I have to show that T is self adjoint.

Now I already figured out it's eigenvalues are 0,1 and - 1.

And if I look at the solution, out of that they immediately concluded it was a normal operator/matrix.
I get that if the eigenvalues of a normal matrix are real then it is hermitian.

But why does T need to be normal for it to have only eigenvalues 0 1 and - 1?
I would think that having real eigenvalues is not equivalent to being self afjoint?

I guess im trying to ask: if T is an endomorphism of a unitary space, then how does it having as eigenvalues 0 1 and - 1 imply it is self afjoint?

[fringe fjord]

$T=\begin{bmatrix}0&1\0&0\end{bmatrix}$ satisfies $T^{10}=T^8$ but is not self-adjoint. You need more assumptions.

[stoic python]

Troposphere

[snow shore]

Thats what I thought as well but it was a question on an old exam

It was stated as:

Oh damn

Nvm

In the question it was already stated that T is normal

I just didn't notice and thought T was a random endomorphism 😩

[raven badger]

Is this correct? Where that rotating thingy is the tensor operation

And the v's and w's are basis vectors

[fringe fjord]

Yeah.

[raven badger]

#swag

[raven badger]

I can write a tensor product as a nXm matrix or as a n*mX1 vector.

Is this somehow analogous to superposition?

[winter pond]

Can someone explain me this?

[raven badger]

I don't suppose, that you know whether or not p(x) is a linear function?

[winter pond]

It is.

It is indeed a linear function

[uneven raptor]

Is it not a polynomial of degree at most 2?

[hard drum]

I assume P2 are the polynomials of degree <= 2 and hence not necessarily linear

Lol

[copper pelican]

Yeah it looks like the polynomials of degree 2

[winter pond]

Oh my bad

[copper pelican]

Also note most of the answers have x^2

[winter pond]

Wait I think I get an idea so will it be A

If not I got the wrong idea

[copper pelican]

I don’t think it’s A

[hard drum]

Which polynomials p have p(-x) = -p(x)

[raven badger]

Have you tried writing P2 as a vector itself?

[hard drum]

P2 is a vector space not a vector

[copper pelican]

One way to do it is to find the matrix representation of T and find the null space of that matrix

[grave garden]

Hifi guys

[uneven raptor]

Check if the polynomials in each option is even in the kernel

[grave garden]

So M is a diagonal matrix ?

[winter pond]

Wait I don’t even know the dimension

[hard drum]

No, M could be any matrix

[winter pond]

I’m so bad at this 😭😭

[hard drum]

And I wouldn't worry about writing T as a matrix honestly. Just think about which polynomials have p(-x) = -p(x) and then you'll realise what the kernel is, from which it's easy to find a basis

[copper pelican]

Yeah it looks like it’d be easier to just do what potato says

[winter pond]

Okay lemme try that

[uneven raptor]

M could be upper or lower triangular

[raven badger]

t

my internet just went

swag

[grave garden]

By its element in the diagonal coincide with its eigenvalues I take it as diagonal?

[grave garden]

Ohhh

[hard drum]

I meant more that the definition extends to all possible M in Mn(C)

[grave garden]

D1 is just [a] right ?

[uneven raptor]

Yea

[grave garden]

Been trying to study linear algebra for awhile just so I can solve this problem

Today I try to see how far I can go with my current knowledge hahaha

So guys, what with the same multiplicities do ?

[slender yarrow]

you got it ?

[slender yarrow]

☠️

[stoic python]

JohnL

[past barn]

I'm assuming this was meant to be done through guesswork, but I wanted to try this approach for fun, but I am not getting the right values. Is it supposed work?

[slender yarrow]

there's some factoring you can do in the middle two equations

that could probably help

[past barn]

Yeah I just put it into a calculator but the values I got didn't match up with the answer sheet

[slender yarrow]

wdym ?

[past barn]

I used wolfram to solve for the simultaneous equations but I got complex values

And the answer sheet had a simple matrix, I think 0, 1 / -1 0 if I remember correctly

[slender yarrow]

the factoring I suggest could help you solve the system by hand though

if you factor the two middle equations, you get something like $$\begin{cases} a_{12}(a_{11}+a_{22}) = 0 \ a_{21}(a_{11}+a_{22}) = 0 \end{cases}$$

[stoic python]

aPlatypus

[slender yarrow]

you can split cases, looking at whether a_12 or (a_11 + a_22) is 0 (for the first eq)

@past barn

but if a_12 or a_21 is 0, you'd get that the square of a number is -1

so necessarily a_11+a_22=0

[past barn]

ah yeah that makes a lot of sense

[slender yarrow]

you could probably continue from there

if you want all possible matrices of course

[past barn]

yeah i think i can manage, thank you for the help

[distant schooner]

if a vector v exists in the basis, it must be an odd polynomial. for polynomials of degree 2 or less, that leaves only {x}, so the answer should be D

also i just realized that this is from many hours ago my bad lol

[slender yarrow]

yeah and you didn't reply to the poster of the question :/

oh well

[copper pelican]

Yeah my point was that this implies P2 is the space of all polynomials with degree at most 2

But in hindsight this is unclear from my statement

[sage gale]

Is there another way to come to this conclusion besides finding this counterexample?

[wintry steppe]

you could find another counterexample

[vague crane]

^

[sage gale]

lol

[vague crane]

in general the way you show something false is to either prove its negation or find a counterexample

[wintry steppe]

these are the same thing

[vague crane]

not always

[wintry steppe]

by finding a counterexample you are proving the negation of a "for all" statement

[vague crane]

ig finding a counterexample is a subset of proving the negation

[wintry steppe]

yep, that's what i wrote

[vague crane]

yeah i was agreeing

tbh idk why i typed that much instead of just saying like

[wintry steppe]

i was agreeing with your agreement

[vague crane]

ohh

very cool

thank you for agreeing with my agreement

[pallid rampart]

Hmm

[sage gale]

cheeky interaction

[little crater]

okay Im confused. How does the the residual vector fit into this

Like how do I compute the residual vector?

[sudden cedar]

extremely simple. y(obs) - y(predicted).

in this case, X*B -> y(predicted).

[little crater]

okay

wait

so I have this worked problem I did

You are saying i need to plug in my found B back into x

[sudden cedar]

yes

[little crater]

and then how does that fit into my actual model?

see I have this

[sudden cedar]

you can view it as an additional column in a matrix with the coefficient 1

[little crater]

but idk how I plug the residual back into that equation

[sudden cedar]

slap a 1 at the end of your B vector, and a new column to X

[little crater]

why did the work example not do this?

[sudden cedar]

are you trying to just find the error term, or do you want to add it to your column to make a specific solution?

[little crater]

well I just don't really know if I even have the right equation

i didn't use the residual

[sudden cedar]

it fits right there with bo and b1x

it's another polynomial term, e

y = bo + b1x + e describes any one point

[little crater]

okay

[sudden cedar]

the bo + b1x is model described, and the e is what it couldn't fit

[little crater]

Oh

so if there is an e?

we can do that

I guess I am just confused when my equation I am modeling needs this e term as a correction factor or whatever

[sudden cedar]

if the e is there it stands for the error. Some texts usually just show it so you know it's there, but you don't try to model it.

no, it's not correction, you don't put it there - it's calculated

not a lot of people add an error term to their model. you can.

[little crater]

well I am missing with the desmos and plotted my computed model and it seemed off

and idk if it has to do with the residuals

[sudden cedar]

that looks good to me. e1 is your error.

i.e. a residual

[little crater]

okay but I guess I mean what did desmos graph in green and why is it off of what i graphed

i just don't know if I made a mistake or desmos is doing something different

[sudden cedar]

it looks like you graphed two things?

[little crater]

the orange curve is mine

the green curve is what desmos computed

[sudden cedar]

well the orange one is fucked

x1 x1 x1

[sudden cedar]

your x's are unlabeled

[little crater]

I am not asking desmos for a regression I just put it back into the model

ie

[sudden cedar]

I don't use desmos, so I can't say for sure what's going on there.

[little crater]

I just plugged my b terms in

do you know a site I can test these regressions on

like that will compute them with a specific model and poitns?

points*

[sudden cedar]

I dunno, but desmos can definitely do that

it's probably worth figuring out what's up

[little crater]

well that is what it was doing but idk if I screwed up

[sudden cedar]

I'd google "what does ~ mean desmos"

in stats it symbolizes "distributed as"

[little crater]

that is how you setup the regression

https://help.desmos.com/hc/en-us/articles/360042428612-Nonlinear-Regressions

This is what I used

thats tragic because the other one I computed seemed to give the same result as desmos

😦

[sudden cedar]

I would trust desmos' computation as long as you understand the syntax

the y ~ syntax is correct for your purposes

I'd scrap the orange. Might be doing something weird, I dunno

[little crater]

Well it is just practice but the thing is inputtting a 8x3 matrix is annoying and I don't know how to easily do it with wolfram

wolfram

like creating variables

[sudden cedar]

I think desmos is probably the right tool for you, unless you want to get into matlab / R

there's a bit of overhead there that might be more frustrating

[little crater]

oh I mean on the computation end

I have been plugging theses matrices into a matrix multiplication calculator and then replunging them in.

is there not a way to define variables in the wolfram thing and then do the computation from it

[sudden cedar]

I dunno, but I would guess so. I'd also have thought desmos could do that.

I don't want to waste your time, but if you've got it to spare and you're in STEM stuff it'll save you time in the long run to just learn R or python as soon as possible.

[little crater]

well I know some python and I think I will be learning R for this EDA course I plan on taking in the fall

what libraries do people use in python?

just math plot lib or something?

or I guess that is a big vaue

vague*

[sudden cedar]

for what you want, you'd write maybe five lines of code in R and a few more in python.

ah, starting out, you can get by really readily with just base R or base python + scipy

scipy.stats is an important library that basically contains a lot of basic functionality

makes python more like R. maybe also just start out with pandas.dataframe, also makes python more like r

if you're going to be taking a course in EDA later, r might be a better bet

it blasts python hard when it comes to ease of graphing

[little crater]

just have to pass my math stats course starting next week . Seems like it will be interesting but never did stats behind the basic highschool/ap stats stuff (plug into calculator and no clue what is really going on)

[sudden cedar]

r is also imo a little more beginner friendly. You won't wedge up against any object-oriented-tact-required moments

so you can just focus on whatever math you're supposed to be doing. I love it, I spent a lot of the last two years doing it professionally.

I'm a huge fan of r + ggplot for starting off with any dataset

good luck 2 u mann

[little crater]

thanks

[old minnow]

R is really nice because of the objects it provides too, its easy to manipulate matrices, vectors, and dataframes

[sudden cedar]

yeah. And honestly Rstudio is an amazing IDE, you don't have to configure emacs or pycharm or w/e for it.

[little crater]

i think that is what my syllabus for that EDA course says we will use.

at least this is what I am interrupting from this

okay it does say it

shame my linear algebra class is over after today. Kinda rushed and somehow are doing this linear models stuff even though we skipped projections

so it has been very "algorithmic"

[old minnow]

you might find PCA fun (principal component analysis)

[sudden cedar]

he said he skipped projectionns

[little crater]

well the class did but I tried looking into it as much as I can within 1 day

[sudden cedar]

I'm not a mathematician but I tend to think of it as basically a PCA thing, which you can just read about and grock conceptually. And then have R package do it for you!

pca is really cool though

[old minnow]

Darn, PCA is one of the best ways u can integrate LA and stats

[sudden cedar]

yeyp

[old minnow]

its possible to self study projections though

[sudden cedar]

I went to a seminar once on it, being in bioinformatics

[little crater]

well how much knowledge of projections do I need to know

[sudden cedar]

it's cool. Very useful for weird dimensional spaces like expression along whatever metabolic pathway

not a lot at all, for pca

pretty much just the basics

I'm not sure what more advanced uses would be. I really think it's a domain-specific thing that you could pick up pretty easily, too. Conceptually projections are very intuitive, any physics you've had of decomposing things into x/y components will get you mostly there.

[old minnow]

if you ever need to efficientlt represent something from higher dimensions but with a lower dimensional vector, it would come in handy

[little crater]

yeah I understood what it means but I will say some of the computation of just using y hat = (y dot u1)/(u1 dot u1) u1

and I get what it is saying somewhat but still need to look into it more

I know there is some relationship between the dot product and trig stuff

and I get the concept of it is like a "shadow" casting down

[sudden cedar]

yeah you get it. For the details, https://www.stat.cmu.edu/~cshalizi/uADA/12/lectures/ch18.pdf

[little crater]

and using orthogonal projections allows you to get that y-y hat with the shorted distance

which has to do with minimizing .

[sudden cedar]

key thing is just it finds vectors* to represent dimensions in R with which the dot products of the sample vectors of information are maximized

[little crater]

that seems logical

well makes sense

[sudden cedar]

yeah. like if you could just say x axis is now x = sqrt( y ) and you're looking at hyperbolic data

it's cool for compression

in the best possible case, you get such a neat fit you can just leave a little note describing the axis and delete all the x data. So basically, a ton of data becomes x = sqrt(y)

and you get rid of every single x value in your data.

other cool stuff like the edge of a square, you can just say well principal component is dotproduct(x,y).

note though - don't think these simple cases are ever actually used, it's normally high-dimensional and noisy data.

[little crater]

Seems interesting

[sudden cedar]

yeyp sorry to ramble. it's neat.

[normal loom]

true or false question

not sure how to approach

[little crater]

Well can you transform A into the identity

And B into the identity matrix of size nxn.

And elementary row operations are reversible

[normal loom]

ohh oko thanks

so A can always be transformed into identity matrix?

if rank is n, that is

[little crater]

Well given it is full rank

[normal loom]

ic

[blissful vault]

Yes. Since A is full rank and square, it is invertible. All invertible matrices can be reduced to the identity by elementary row operations.

[wintry steppe]

Can I ask why row operations are a thing? How do they not change a determinant?

to answer the latter question, they will change the determinant, depending on the operation you do. elementary row operations come from left multiplying by elementary matrices, which have determinants c, -1, and 1, corresponding to the cases of multiplying a row by c, interchanging two rows, and adding a row to another, respectively

det(EA) = det(E)det(A) dictates how the determinant then changes

Ohh! I thought it didn’t change them! Like putting it into rref didn’t change rot for some reason

[keen sierra]

if that were true, then every invertible matrix would have determinant 1

since they can all be row reduced to the identity

[bright vapor]

Hi, do you know upper bounds on |AB^{-1}|_{op} for A, B symetric positive definite matrices ? (operator norm)

I know there exists upper bounds on |A^{1/2}B^{-1/2}|_{op}...

[grave garden]

Guys, how do I prove that f is not surjective ?

Since it is endomorphisme I try to prove that it is not injective

$f^2=-id\newline (f^{-1} \circ f)\circ f = f^{-1}(-id)\newline f(x)=f^{-1}(-x)$

[stoic python]

Potato

[grave garden]

I play with it for awhile and get sth like this which doesn't look quite useful ?

[dusky epoch]

did you crop something out here?

[grave garden]

It is in Frenglish XD

[dusky epoch]

[grave garden]

Some are French term

[dusky epoch]

franglais at its finest

okay so to phrase this in less language-mixed terms

Let f ≠ 0 be an endomorphism of R^3 such that f^3 + f = 0. Show that f is not surjective.

[chilly helm]

hey is this where i go for geomertry?

[dusky epoch]

no. #geometry-and-trigonometry

[chilly helm]

ok thx

[dusky epoch]

@grave garden for this i would think about eigenvalues

[grave garden]

I see

[dusky epoch]

∅ ≠ {eigenvalues of f} ⊆ {roots of X^3 + X}

[grave garden]

This is some review exercise from past lesson before eigenvalues stuff, so I thought eigenvalue is not need here

[clever totem]

sry to interrupt
$V=\mathbb{C}^n,f\in End(V)$
let $\mathbb{C}_f^n$ be a $\mathbb{C}[x]$-Modul
such that
$p.v=p(f)(v)$

i lack a bit of intuition how this works. for example what happens when i multiply 2 elements of this modul?

[stoic python]

~Martin

[dusky epoch]

this is a module not an algebra

so you can't multiply together two vectors

[clever totem]

how do i then show that another module is a submodule if i cant multiply?

[dusky epoch]

you can multiply elements of your module by polynomials

[clever totem]

why though?
the elements of my module are polynomials after all

[dusky epoch]

no, the elements of your module are vectors in C^n

[clever totem]

but it is a C[x] module

[dusky epoch]

sure is, but its elements are not polynomials

its scalars are polynomials

[clever totem]

ohhhh

thx

do you know hoe to understand this structure p.v=p(f)(v)?

[dusky epoch]

f is an operator on V. do you understand what the notation p(f) refers to?

[clever totem]

i think p is a polynomial and i input f

for example p=x+1 then p(f)=f+1

[dusky epoch]

p(f) = f+e

where e is the identity operator

[clever totem]

is identity op the neutral operator or the null operator? we havent used this term since i study in german

ah ok the neutral i think

[subtle walrus]

whats the german term?

[dusky epoch]

einheits...something?

[clever totem]

neutrales element

[subtle walrus]

neutrales element, yes

[dusky epoch]

i know identity matrix is Einheitsmatrix

or something

[clever totem]

yep

[subtle walrus]

ye thats correct

[dusky epoch]

so Einheitsoperator? idk

ich spreche kein Deutsch

[subtle walrus]

in general groups its called neutrales element

[dusky epoch]

what do you call the 1 in a ring?

[subtle walrus]

eins

[clever totem]

neutrales element as well

[subtle walrus]

Einheit is unit

[clever totem]

or eins

[dusky epoch]

i mean the operator e which satisfies ev = v for all v in V just to be clear

that's called the identity operator

[clever totem]

the thing where im stuck is what should i do if i want to show that U is a sub module
generally i want to show that c.u is in U

[dusky epoch]

anyway, p.v = p(f)(v) should be taken at face value really

the module as a whole is encoding the action of your f on the space

yes

so in this case it's enough to show that U is a subspace of V as a C-vectorspace and also that U is f-invariant

from this it'll follow that U is closed under module multiplication by any p

[clever totem]

we have actually given that U is a sub vectorspace and that f(U) lies in U

[dusky epoch]

so you have proved those two already?

[clever totem]

i didnt have to they were given haha

[dusky epoch]

oh

well then

if you know U is a C-subspace of V and that for all u in U, f(u) in U

then it should not be hard to show that for all u in U, any C-linear combination of f^k(u) where k in N is also in U

[grave garden]

Guys, if a linear map is neither injective or surjective, is there sth we can say related to its diagonalization ?

[grave garden]

So the second question here is asking to prove f is not diagonalizable

From one we know f is not injective and surjective

[quartz shale]

I'm having problems proving that an affine application of a vector subspace is an affine space. The question itself goes as such:
Let E be a vector space, not necessarly euclidean. Let φ : E → E be a linear application and F a vector subspace of E.
Let a ∈ E and Ta : E → E a translation associated with the vector a.
Consider the affine application λ: E → E defined by λ = Ta ○ φ.
Prove that λ(F) is a affine space.

My first idea was using the definition of a canonical affine space brought by the sum of a vector and a subset of a vector space E, which is just what λ does. Does anyone have a better idea on how I should prove this?

(If this is better suited to a help channel, please do tell me and I'll move over there)

[ionic gate]

how do you read this? is it "matrix a augmented b" or "augmented matrix of a and b?"

[dusky epoch]

matrix A augmented by B?

or just write it and point at it tbh lmao

[ionic gate]

lmao okay ty

[unreal thistle]

why is the answer B? arent only options A and D consistent?

[edgy shale]

Answer A spans $R^2$ because the third component is always 0 which basically means it spans a plane

[stoic python]

ModernNoob

[edgy shale]

Answer D is incorrect because the third vector is linearly dependent on the first vector. Vector 3 = -2 (Vector 1)

I guess another way u can think about it is u write the basis in a matrix and find the determinant. You will see that the determinant will be 0 for A and D which means it’s degenerate and will not span R^3. (I’m pretty sure lmk if there’s something wrong)

[copper pelican]

Diagonalizability isn’t really related to injectivity/surjectivity

For example, the zero matrix is neither injective nor surjective, but it is diagonal

The identity is bijective and also diagonal

The criterion for bijectivity is whether or not 0 is an eigenvalue of the matrix

[grave garden]

In my case what should I do ?

I dunno that matrix associates with it

[copper pelican]

On in this case f needs to be diagonal already

Think about what happens when you cube a diagonal matrix

[grave garden]

U get back a diagonal right

[copper pelican]

Yeah

To be precise, what happens to each element

[grave garden]

They are cube too

[copper pelican]

Yeah

Now add f to the cube; what do you get

For each entry

[grave garden]

x³+x

[copper pelican]

Yep

Now each of those must be equal to 0

What does that force x to be

[grave garden]

0 and ±i

We can't take i right since it is R³ ?

[copper pelican]

Yep

So what does that tell you about f

[grave garden]

Hmmmm

[grave garden]

We have a zero matrix associate with it ?

[copper pelican]

Yeah

Except remember it’s necessarily surjective

(From part a)

So can f be diagonal?

[grave garden]

It need surjective ?

You said it isn't right

[copper pelican]

Well when we assumed that f is diagonal, we find that it must be the zero matrix

But remember that f is necessarily surjective

Is the zero matrix surjective?

[grave garden]

I dunno ?

I can't see it 🤔

[copper pelican]

Well what does it mean to be surjective

[grave garden]

For all elements in its range we can find an element in the domain associates with it

[copper pelican]

Yeah and what’s the range of the zero matrix

[grave garden]

Hold on, what do you mean by that ?

What's the mapping

[copper pelican]

Well it corresponds to the map that sends everything to 0

[wintry steppe]

you should absolutely make sure you understand the correspondence between matrices and linear operators if you don't right now

[grave garden]

Are we talking about f(0) ?

[wintry steppe]

f(0) is always zero when f is a linear map, so it's not clear what you mean

[grave garden]

Like this

I dunno what is the range/domain of it since i dunno what mapping we use for it

[wintry steppe]

if A is an m by n matrix with entries in a field F, then it corresponds to the map F^n -> F^m given by matrix multiplication, x -> Ax

[grave garden]

Ohhh

Now i see it

XD

[wintry steppe]

(and conversely, any linear map F^n -> F^m is given by such a matrix)

(choosing bases of more general vector spaces identifies your spaces with F^n and F^m and lets you do the same thing)

[grave garden]

So with zero matrix its range is 0

🤔

[wintry steppe]

that is right

[grave garden]

And it is surjective ?

[wintry steppe]

what do you think?

what does surjectivity mean?

[grave garden]

I think yes

[grave garden]

I think this one

Since all elements in the domain is mapping to 0, which is the only element in the range

It is surjective

[wintry steppe]

this is not what surjectivity means. this is a trivially true statement

the range is by definition the set of things you get by mapping elements in the domain

you would like to replace "range" by "codomain" in this post

then you will have the correct definition of surjectivity

in this case, you're considering the zero matrix as a map F^n -> F^n (using the correspondence i wrote earlier), so the codomain is F^n

now that we've fixed your definition, is this surjective?

[grave garden]

Still think it is surjective

[wintry steppe]

why?

it's not surjective, but if you think it is, some misconceptions need to be cleared up

[grave garden]

Its codomain only has one element right ?

[wintry steppe]

you are confusing "range" with "codomain"

[grave garden]

Ohhh

Yeahhh it is not

[wintry steppe]

the codomain is F^n, the range is {0}

[grave garden]

Got it now

[wintry steppe]

[grave garden]

Back to the problem where is this come from ?

The condition that it is surjective

[copper pelican]

Oh does part a say it’s surjective or not surjective

[wintry steppe]

n'est pas surjectif

[copper pelican]

Ahh so my bad

[grave garden]

Not surjective

[copper pelican]

I think it’s simplest to go with the original question statement then

[copper pelican]

Where it says f must be non-null

[grave garden]

This is my proof for that

[wintry steppe]

en français

[grave garden]

Frenglish XD

[copper pelican]

(I’m assuming not nul means not null right)

[grave garden]

f ≠0 I think

[copper pelican]

Yeah

And we’ve found f must be zero if it’s diagonal

[sudden cedar]

pas toi aussi

[grave garden]

Hold on I have probability quiz 💀

Be back guys

[copper pelican]

Good luck

[sudden cedar]

luck

[wintry steppe]

nitpick: why does such a real eigenvalue/eigenvector exist? once you've got this, this is a good proof

(one always does exist, i just think you should mention why)

added "real" so no one says "go to the complex numbers"

[velvet moss]

is it true that all transformations which take a vector in one dimension of size n and bring it to another dimension also of size n

are surjective?

i want to say its not true

[wintry steppe]

no, see the above conversation. the zero map R^n -> R^n provides a counterexample

[velvet moss]

how come I see many books

proving surjectivity by providing such an argument

[wintry steppe]

what you wrote
"transformations which take a vector in one dimension of size n and bring it to another dimension also of size n"
doesn't say anything about every vector being reachable by the transformation (which is what surjectivity means)

i don't understand how what you wrote constitutes an argument for anything, it basically just says "linear transformation from R^n to R^n"

[velvet moss]

I understand that surjectivity means that an element in the codomain can be assigned to some value in the domain

its just that sometimes I see arguments ( though I may be seeing them wrong ) somehow proving the surjectivity of a transformation

simply by implying that the the dimension of the image is equal to the dimension of the domain of T

I dont know, im probably just misunderstanding their proofs

or I have some deeper lack of knowledge on how they are proving it 🤷‍♀️

[wintry steppe]

surjectivity is equivalent to the dimension of the image being equal to the dimension of the codomain of T, but in the special case you wrote that the domain and the codomain of T have the same dimension, what you wrote is good

the reason this works is the following: every n-dimensional subspace of R^n is equal to R^n

if you have a linear map T: R^m -> R^n (i am no longer restricting to the case n = m) and you can show that image(T) has dimension n, then you have image(T) = R^n, which is precisely what it means for T to be surjective. this is the argument you are probably seeing

[velvet moss]

oh I think I understand now

thanks so much, that was annoying me for the longest time

[wintry steppe]

[sudden cedar]

I'm trying to understand the implications of being able to write any solution to a system of equations as a linear combination of a vector in the kernel and the particular solution

is this equivalent to saying that any solution can be composed like a transformation to be centered at 0 across all dimensions* in the codomain + an additional transformation?

[grave garden]

Since the mapping is R³ then its characteristics polynomial must be degree of 3 so at least one real root must exist

What do you think ?

[wintry steppe]

very good

[wintry steppe]

So I have a question about the wronskian. I know it’s used to determine linear independence of functions?

Are wronskian ever nonsquare matrices? Like a 3x2 matrix?
Could someone shed some light on this for me please 🙏

[grave garden]

Generally, the expression in 4 always hold true

But how do I prove it in this case ?

[wintry steppe]

no, it isn't always true. if it were true in general, there'd be nothing to prove for your problem

by rank-nullity, dim(ker + im) = dim(ker) + dim(im) - dim(ker \cap im) = dim V - dim(ker cap im), so to show there's a direct sum you only need to show that ker cap im is {0}

ker(f) cap im(f) = ker(f restricted to im(f)), so it's the same thing as saying that f restricted to its image becomes injective (think of e.g. projections, a nice example for which this is true)

maybe these remarks will help you get started

[grave garden]

Ohh nooo

[grave garden]

So dimV = dim(ker) + dim(im) is always true right ?

[wintry steppe]

that is rank-nullity

yes

[grave garden]

I thought that implies V = ker (+) im

[wintry steppe]

no

not at all

V is the direct sum of subspaces W and W' if W + W' = V and W cap W' = {0}

the latter condition ensuring every element of V decomposes uniquely as a sum of elements from W and W'

[grave garden]

I see

You saved me right there

Who know how many crimes I'm going to commit if not realizing that

[wintry steppe]

the police will hunt you down

[grave garden]

If f restricted to its image then it is bijective right ?

Ohh wait

I feel like it is surjective ?

[wintry steppe]

after restricting the codomain as well, sure

[grave garden]

You say it it injective after that right ?

Why is that

[wintry steppe]

...

[grave garden]

For some reason I can see it is surjective

[wintry steppe]

this is the first fact everyone learns about linear maps

injectivity is equivalent to trivial kernel

[grave garden]

Yes yes I know that one

[wintry steppe]

so if the kernel of f restricted to its image is zero, f restricted to its image is injective

[grave garden]

Ohhhhhh

[wintry steppe]

f restricted to its image is a map im(f) -> V. it is injective. once you restrict the codomain, you get a map im(f) -> im(f). this is injective, and a linear operator, so it's also bijective (or more easily, restricting the codomain of a map to its image gives you something surjective trivially)

[grave garden]

I was talking about the general mapping where if their codomain restricted to just their image then the function would become surjective

Btw is it true tho ?

[wintry steppe]

i wanted to make sure you were aware of this

there is a difference between
f restricted to its image
and
f restricted to its image, with the codomain restricted to the image as well

the first would be injective in the case we are talking about, the second would be bijective because it is injective (first case) and surjective (we have restricted the codomain)

[grave garden]

Hmmm i thought f restricted to its image mean E->Im(f)

I get the definition wrong

[wintry steppe]

i should have been more specific, by "restricted to its image" i meant "restricting the domain to its image"

that is what i have meant all along

[grave garden]

I have another question

What if the image of f is not in the domain at all ?

[wintry steppe]

then you're not looking at a linear operator at all

[grave garden]

🤔

You mean I miss that f is endomorphisme right ?

[wintry steppe]

yes

[grave garden]

I see

So it works for this case

[grave garden]

What is tr(f) ?

[wintry steppe]

trace

[grave garden]

So trace of linear map is a thing

[wintry steppe]

of an endomorphism, yes

[grave garden]

Ohh so it exists in an endomorphisme only right ?

Honestly I never heard of a trace of a linear map

So I'm being skeptical about it a bit

[wintry steppe]

the trace of an endomorphism V -> V of a finite-dimensional vector space is defined by picking a basis of V and taking the trace (i.e. sum of diagonals) of the matrix representation. there may also be a more high-brow definition related to the fact that the trace is the sum of the eigenvalues, but i don't know off the top of my head

matrix representations to two bases are related by conjugation by the change of basis matrix, and trace is invariant under conjugation, so the result is independent of the choice of basis (i.e. you get a well-defined trace)

[leaden nexus]

hello if i have a sesquilinear form over a linear space V, does the scalar multiplication still work normally as in: (k*a,0) = k(a,0) resp. (0, a*k) = (0,a)*k?

[wintry steppe]

if your form is sesquilinear then it must conjugate the scalar when pulled out of one of the components

which one that is is up to your conventions

although what you wrote is technically still true, since putting 0 into one of the components will make both sides vanish

[leaden nexus]

Ok true, but if i the only information i have that its over a linear space V can i assume that real numbers is whats its mapping to?

so the conjugation would still be 1 right?

[wintry steppe]

if V is a real vector space, sure

[leaden nexus]

Ah ok

[wintry steppe]

not all vector spaces are over R. C^n's dot product (z, w) -> \sum z_i \bar{w_i} is sesquilinear

you must conjugate the (complex) scalars when you pull them out of the second component

[leaden nexus]

Ok i see

so proving that (a,0) = (0,a) = 0 can be done by
(a, a+(-a)) = (a, a) + (a,-a) = (a, a) - (a,a) =
(a+(-a), a) = (a, a) + (-a, a) = - (a, a) + (a,a) =
= 0 only if the vector space is real

[wintry steppe]

where did you use the fact that the vector space is real (it doesn't have to be) in this argument?

the complex conjugate of -1 is -1

[leaden nexus]

the (-1) would be a scalar right?

wait

lol ok so this works for complex and real

[wintry steppe]

yep

[leaden nexus]

okayy, thank you

[floral monolith]

。

[grave garden]

So TTerra, i was able to prove that (f², f) spam Im(f)

( is it correct tho ? )

But cannot prove that it is linear independent

Do you have sth to feed me ?

Gonna send it once more

Now I'm at the last question

[grave garden]

So in my case tr(f)=0?

We dun take the complex root right

[velvet moss]

Im having trouble with proving this one

[versed holly]

Which topics do I have to master before I start learning Linear Algebra?

[velvet moss]

I also want to clarification on what the author meant by "The matrix of T relative to the pair B, B'"
do they mean a matrix which takes input B as coordinates and applies to T then converts to B'

or the opposite

its pretty ambiguous imo

[wintry steppe]

this is correct. you can see it by writing f in the basis (f(x), f(f(x))); the diagonal entries will be zero, so the trace will be zero.

i have not yet looked at the rest of your questions

[grave garden]

I'm pretty sure you can start it right away

[wintry steppe]

if you are just learning about matrices and systems of equations, nothing except for elementary algebra.

if you are learning about vector spaces and linear maps in general, then you should be familiar with the basics of proof writing

i say should for you can pick it up as you go, if you are strong enough

[velvet moss]

i would argue you could learn about linear maps before proof writing

just learning their computations

[wintry steppe]

you could, but probably not very deeply

linear algebra is rather light on theoretical details, which is good, but you can only get so far into the subject without knowing how to read or write a proof

[grave garden]

You can learn everything in an engineer way ( no rigorous proof, no asking why, just number and how to use it )

[wintry steppe]

the depressing way

[velvet moss]

.-.

[wintry steppe]

also, i specified vector spaces and linear maps in general

in general

[versed holly]

Thank y'all

[wintry steppe]

the columns of the matrix will be the coordinate vectors (relative to B') of the images of the elements of B

[velvet moss]

oh ok ty

[wintry steppe]

#linear-algebra message
see my post here (for the definition for linear operators, at least. you can figure out how to generalize)

[velvet moss]

i guess the proof is pretty straightforward from there

[versed holly]

So regarding proof writing, how can I make sure that I could write a proof coherently?
For sure, if I'm not gonna write proofs in this topic I'll need to do so in other topics regardless of the fact that proof writing is genuinely important tho

[wintry steppe]

write proofs and have people who know how to write proofs criticize them

[versed holly]

Fair enough

[wintry steppe]

simple things, like basic induction, contrapositive, contradiction, and all that. at least be familiar with these

[versed holly]

Oh I have to check these for sure, I've just graduated from high school and the curriculum didn't involve anything more advanced than basic set theory which I learned back in the 9th grade. Either way, thank you very much!

[wintry steppe]

some people shun "proof books" that teach skills such as these, but they are really good for making sure you know them

i read "how to prove it" by velleman after graduating highschool before university and really benefitted from it

at the very least peek through something like that and make sure you're familiar with most of the content

[versed holly]

Oh interesting, I shall look for it up. I had printed a book regarding proofs as well called "proofs" by jay Cummings before my finals, I shall give it a try

[dusky epoch]

who are those people

[wintry steppe]

daminark

people who think you should suffer in your first proof based math class

etc.

Can dot product of two non-unit vectors, not normalised and the angle between them is acute, be greater than 1?

[velvet moss]

yea

[wintry steppe]

(2, 0) dot (2, 0) = 4 is greater than 1, and both of these fit your crtieria

the angle is 0, surely that counts as acute?

Yes

[velvet moss]

[dusky epoch]

(2,0) and (2,0.1)

if you don't want a zero angle

[wintry steppe]

Lol

[remote valley]

How would I go about correcting a vector of acceleration values from a tilted accelerometer using the vector of a gyroscope which gives the offset from uh "ground"?

[grave garden]

@wintry steppe Hellooooo

Sorry for pinging

So hmmm I think for quite awhile and arrive at this

Assuming there exists two element $x_1, x_2$ such that $f(x_1)=f(x_2)=a$ and $x_1 \neq x_2$. Then there could exist two elements $y_1, y_2$ such that $f(y_1)=x_1$ and $f(y_2)=x_2$. So the mapping f restricted to its image is not guarantee to always be an injective

[stoic python]

Potato

[grave garden]

What do you think ?

[wintry steppe]

sure, i guess

you're just restated the definition

(for linear maps, it's really better to think about injectivity in terms of elements going to zero. you will save a lot of writing)

[clever totem]

i already asked this yesterday and i think i semi understood it, but i need help with writing this down:

[wintry steppe]

this is some very strange formatting

[dim epoch]

did you figure this out btw?

oh lol

didn't see you asked about it again

[wintry steppe]

why does it remain to show that U is f-invariant? give the details

you're avoiding writing down a precise argument which makes me think you don't actually have one

[clever totem]

well you are partially right

[wintry steppe]

so you already know U is closed under addition, since it's a subspace of V

you now need to show that it's closed under the action of C[x] on V

fuck i keep hitting enter too early

[dim epoch]

i actually have the same problem so i'm just gonna be lurking here

[wintry steppe]

so for p(x) in C[x] and v in U, you need to argue that p(x) . v = p(f)(v) is in U. why is this?

[grave garden]

So hmmm now what should I do for that then ?

[wintry steppe]

yes, it's because U is f-invariant. but in detail please

[clever totem]

p.v=p(f)(v)
f(v) is in U
i have to show that p(f)(v) is in U
i would say that p(f) is a linear combination of f to some degrees
since f(u) is in U for every u
f^n(u) should also be in U

and a linear combination of element in U should also be in U

[wintry steppe]

sounds about right

[clever totem]

great ^^

[dim epoch]

[wintry steppe]

i would just write out "p(f) is a linear combination of f to some degrees" in symbols, though

then go "each summand (scalar)*f^n(v) is in U because v is in the f-invariant U, so the whole thing is"

this completes the proof

[wintry steppe]

stop worrying about interpreting the definitions of things and just try to prove what you want for your original question (the f^3 + f = 0 one)

i gave you a few ways to do so in that long post involving rank-nullity and stuff

i'm going to sleep, so i'm not going to respond if you ping me again

[dim epoch]

good night

[wintry steppe]

it is 9 am for me

[dim epoch]

[wintry steppe]

quite far from night time

[dim epoch]

that's an interesting sleep schedule

[wintry steppe]

i need to get on a plane very early in the morning tomorrow

might as well ruin it now and not feel like crap during the ride

[dim epoch]

fair enough

[past barn]

How would I approach this?

Forgot to provide context, here it is:

[snow flint]

Suppose $dim V \geq 2$. Show that the set of normal operators on $V$ is not a subspace of $\mathcal{L}(V)$

[stoic python]

الله أكبر

[snow flint]

CAn anyone give a hint🥺

ik that i should show its not closed under addition but idk how

[little crater]

what does L(V) mean (asking because I have never seen that before)

[copper hinge]

The set of all linear maps from V to V.

[little crater]

I never even seen "normal operators" either

I must be missing out

@snow flint what book are you using in your course?

[past barn]

Yep this helped i think I was mostly just struggling a bit to understand all the definitions but now I got it, thank you

[zinc timber]

is that supposed to be ij?

[past barn]

Yeah it should be "Let I_jj..."

[snow flint]

linear algebra done right

normal operator are operators that commute with their adjoint

[velvet moss]

what’s an intuitive reason for why trace(AB) = trace(BA)

[zinc copper]

classmate asked this question: what's the point of looking at congruent matrices when these arent symmetric?

is there any use to it?

[zinc copper]

there's a much more general fact about the characteristic polynomial of BA in terms of AB

i wouldnt call it intuitive though (at least i dont find it particularly intuitive)

i mean there's a more or less natural proof method to showing it but i dont find the result particularly intuitive

[golden mulch]

that fact is not at all intuitive

depends on your level of intuition of course

but when i first checked this fact my intuition drew me sideways

[wintry steppe]

this depends on your level of intuition on the trace of a linear transformation

[golden mulch]

that's what im sayin xD

[wintry steppe]

why is the trace well defined for a linear transformation?

thats a good question to start with

[golden mulch]

yeah other than that i found a paper which illustrates that fact

[wintry steppe]

like somehow the sum of the diagonal entries of a matrix is independent to choice of basis

how come the diagonal sum has this property, rather than say the off diagonal?

or summing the entries of the top and bottom rows

or the outer square entries of a matrix

none of those sums are stable under change of basis

[golden mulch]

no elements of diag(AB) are the same for (BA) given that A!=B

so yeah

unstable sums

all around

also given that B!=A^-1 i guess

but yeah very unintuitive

[wintry steppe]

the intuition is that the trace is some kind of generalized dimension counter for a linear transformation

[golden mulch]

as dimension counter for a linear transformation is very intuitive xD

[wintry steppe]

the same way that the dot product calculates some kind of generalized perpindiculatiry

yeah for a projection map U the trace of U is the dimension of the image

and for non projections there is some additional factor, the way for non unit vectors there is a factor in the dot product not corresponding to the angle

and finally why the dimension is related to the diagonal entries has to do with the dual space, since these components are unchanged by taking the transpose and the dual space has the same dimension as the origional v space

[golden mulch]

tr(AB)=∑i(AB)ii=∑i∑jAijBji=

=∑j∑iBjiAij=∑j(BA)jj=tr(BA)

here is what i found

[wintry steppe]

lmao very simple algebraic proof extremely subtle geometric intuition

linear algebra for ya

[golden mulch]

i hate linear algebra

so yeah @velvet moss

a couple of approaches to your problem above

my question is that is there ANY way to find the permutation matrix in PA=LU decomposition without going through GEPP

my guess is no but it would save a shid ton of time if there is

[clever totem]

Let $R=K^{n\times n}$ be the Ring of $n\times n$ Matrices over a field $K\newline$
interpret $R$ as a (Left)module over itself$\newline$

[stoic python]

~Martin

[clever totem]

I am looking for a decomposition of this module into a direct sum of indecomposable sub modules

I found this in my script:
R is a euklidian ring and M a free R-module
then there are indecomposable elements $f_1,\dots ,f_s \in R$ and $t,n_1,\dots ,n_s \geq 0$
such that
$M \cong R^t \oplus R/(f_1^{n_1})\oplus\dots\oplus R/(f_s^{n_s})$

[stoic python]

~Martin

[velvet moss]

lol sorry guys, I didn’t realize the “intuition” for that fact about the trace would be so esoteric

[quartz compass]

a surface level intuition is that the trace is multiplying the left side of the matrix with its own right side, so AB and BA both look the same when "wrapped around" making tr(AB)=tr(BA)

[idle atlas]

Why do we have that matrices over a field are nilpotent if and only if its characteristic polynomial is x^n?

[dusky epoch]

which direction are you interested in?

charpoly x^n => nilpotent, or nilpotent => charpoly x^n?

if both, which one would you like explained first?

@idle atlas

[idle atlas]

Both, but I think charpoly x^n => nilpotent is easier

@dusky epoch thank you

[dusky epoch]

charpoly x^n => nilpotent is obvious by cayley hamilton

for the other direction, note that a nilpotent matrix cannot have eigenvalues other than 0: the presence of such an eigenvalue would mean that its eigenvector would never be annihilated by any power of your matrix

[idle atlas]

Yes

then?

[dusky epoch]

so all eigenvalues of our matrix are 0

[idle atlas]

If the field is C I get why polychar = x^n but what if it is not the case, for instance for R?

[dusky epoch]

you can pass to the complexification of your vector space i think

[idle atlas]

The what?

[dusky epoch]

nevermind

well ok like. the complexification of a real vector space V is a complex vector space in which the vectors are formal sums of the form v + iw where v, w in V

with multiplication by complex numbers defined in the obvious way

but. hm.

i actually can't think of an argument off the top of my head why nilpotent implies charpoly x^n in a non alg closed field

[idle atlas]

Ok np, thank you for your help anyway! @dusky epoch

[wintry steppe]

you would use generalized eigenvectors i believe

wait hmmm

passing to the algebraic closure might work

[dusky epoch]

that's what i tried to do but then faltered @wintry steppe

[wintry steppe]

if you're nilpotent in an algebraically closed field, then the entire space is the generalized eigenspace corresponding to 0, so jordan form gives you the charpoly you want

(this reasoning might be circular since existence of jordan form uses some facts about nilpotents, but let's ignore that)

so does passing to the algebraic closure preserve what we want here? hmmmm

how do the characteristic polynomials of the operator and of the operator extended to the algebraic closure relate?

i am in a super long airport line so i have some time to think on this

i want to say they are the same. i think that is true

the determinant det(T - tI) should not be changed so long as t lies in the base field, other than now giving a splitting polynomial once we pass to the algebraic closure

so i think this argument works

if i have made any blunders then let's blame it on my lack of sleep

maybe you can use some arguments about generalized eigenspaces to just cut out the whole "passing to the algebraic closure" bs

[wintry steppe]

i have overcomplicated it. if T is nilpotent, its minimal polynomial divides x^k for some k, so is of the form x^l for some l. but the minimal and characteristic polynomials have the same irreducible factors (i think. same roots at the very least), so the characteristic polynomial is x^dim V

@dusky epoch @idle atlas

this is the shortest and simplest i can come up with, that doesn't use anything about algebraic closure

the interrogation by airport security has got my mathematical juices flowing

[idle atlas]

@wintry steppe The answer I got in #groups-rings-fields is the following:
Nilpotent implies that the matrix satisfies x^m = 0 for some m, so its minimal polynomial divides this so must itself be of the form x^k.
All factors of the characteristic polynomial appear in the minimal polynomial by working in the splitting field of the characteristic polynomial and using https://math.stackexchange.com/a/101284
Then use the fact that x is the only irreducible divisor of the characteristic polynomial, because it is the only irreducible divisor of the minimal polynomial

Approximately

[dusky epoch]

how do we know minpoly and charpoly share same irreducible factors over a non closed field

@wintry steppe

[wintry steppe]

great question. i don't know off the top of my head

[dusky epoch]

[wintry steppe]

i have always taken this as fact

my la class assumed most fields were closed anyways

[dusky epoch]

[wintry steppe]

i think that it remains true anyways

[dusky epoch]

R is totally closed bro i promise

[wintry steppe]

assumed most fields were closed (for jcf and min poly purposes)

R is closed though

(in its topology)

[wintry steppe]

yeah, that's pretty much my argument. didn't realize you'd posted in #groups-rings-fields, so sorry for the ping

some googling convinces me it works, ann

[dusky epoch]

[severe magnet]

For finite and infinite dimensions, are the eigenvectors of a self adjoint operator the same as for it‘s operator exponential?

[zinc copper]

@dusky epoch there’d a few nice proofs here: https://math.stackexchange.com/questions/825848/showing-that-minimal-polynomial-has-the-same-irreducible-factors-as-characterist

[zinc copper]

is this true? Im trying to take an alternate route to an exercise problem and it would really work out well for me if this was true. Let $A$ be an $\mathbb{C}^{n\times n}$ matrix and $\lambda$ be the minimal eigenvalue of $A^TA$. Then $\sqrt{\lambda}\leq \operatorname{det}(A)^{\frac{1}{n}}$

[stoic python]

𝓛ittle ℕarwhal ✓

[zinc copper]

god im a fool this is trivial

nvm me

[still socket]

The matrix system feels like a video game

[golden mulch]

why do we need to find the interpolating polynomial if the function is well defined

the answer is x^4, no?

i mean it's literally through this function how the knots and values are defined

usually we find the polynomial for ill-defined functions, distinct (x_i, y_i) pairs for i = {0, 1, 2, ... n}

[dusky epoch]

no

a polynomial interpolation through four nodes will give us a polynomial whose degree is at most 3.

[golden mulch]

but why do we even need that

[dusky epoch]

also, imo there is value in seeing what a polynomial interpolation procedure gives you when the points to be interpolated come from a known function.

[golden mulch]

if the function f already describes the relation

[dusky epoch]

that way, you know whether your interpolation scheme works as intended.

[golden mulch]

how would you know that

if you're getting a smaller degree polynomial

[dusky epoch]

know what

[golden mulch]

know that it works as intended

how does a given well defined function contribute to taht

[dusky epoch]

ok let's try going back to the basics

an interpolation method is an algorithm that takes in a bunch of points and outputs a function that passes through those points

how it accomplishes this is entirely beside the point

[golden mulch]

yessir

[dusky epoch]

...don't call me sir.

but ok.

presumably, in whichever use case the interpolation method is intended for, its input points won't come from a formula-defined function.

[golden mulch]

it's a phrase i know you're non binary and i respect that

[dusky epoch]

yeah and i don't like being told "yessir" because it always sounds like people call me "sir" and i don't like that

but anyway

[golden mulch]

ok sure, got it

[dusky epoch]

in whichever use case the interpolation method is intended for, its input points won't come from a formula-defined function.
but we can TEST it on those that do.
and for a certain class of "nice" functions, we should have that taking a function, sampling it and then interpolating the samples should give us back what we started with

in the case of polynomial interpolation on n nodes, the "nice functions" are polynomials of degree n-1 or lower.

[golden mulch]

sorry what do you mean by "sampling"

im from Poland and we might use a different terminology

[dusky epoch]

sampling as in taking some points from the function's graph

as was done in your problem

[golden mulch]

the f_i's or y_i's per se

okay

[dusky epoch]

the x_i and y_i.

[golden mulch]

reading and trying to compile now

[golden mulch]

so you're saying that the way a given function contributes to our ability to check the result, is the single fact that deg(p(x)) <= deg(f(x))

[dusky epoch]

hnbnffng.d/

[golden mulch]

sorry?

[dusky epoch]

i don't know how to answer your question.

[golden mulch]

ok np

[fierce relic]

If you identify a triangle as a 6-tuple in R^6, the dimension of the subspace of only congruent triangles is 3

I don't understand why

Is it because we need three pieces of information to identify a unique triangle?

and so the remaining 6-3 are being used to differentiate them

but then by what characteristics are the remaining pieces of information differentiating the congruent triangles by

[vague crane]

.>

did you really just

[fierce relic]

?

[vague crane]

boi i saw that

don't even pretend

[fierce relic]

srry ;-;

I'm desperate

i've been trying to figure this out for the past 6 hours

[vague crane]

#❓how-to-get-help

[fierce relic]

oops

well could you help me?

[vague crane]

i have no idea what a triangle in r^6 even looks like sry

[fierce relic]

no its just

each point is 2 numbers

identifying it

and then you just stick them together

[fringe fjord]

I think he means a triangle in the plane, with the coordinates of all the vertices jammed into a length-6 vector.

[fierce relic]

yes this ^^

[fringe fjord]

But a class of congruent triangles won't make a subspace of that.

[fierce relic]

ic

But then that makes it even more confusing