linear-algebra

[dusky epoch]

mmh

okay if you give me some time to get home i can write up a thing

[surreal loom]

the exercise starts in 30 minutes so I guess I'll get the solution by then 😃 but thank you very much anyway! awsm that you would do that for me

[pastel aspen]

@surreal loom well i know im late but every permutation in Sn can be written as the composition of a finite number of transposition if that helps

[solemn flax]

What is Cramer’s rule?

[dusky epoch]

it is an explicit formula for finding the sol of a consistent system

for a system Ax = b, with A an n by n matrix with nonzero determinant, the solution's coordinates x_i (i = 1, 2, ..., n) are given by the following formula

[plain fjord]

related to cofactor matrix stuff

[dusky epoch]

$x_i = \frac{\det(B_i)}{\det(A)}$, where $B_i$ is the matrix formed by replacing the $i$'th column of $A$ with $b$.

[stoic python]

Ann:

[dusky epoch]

@solemn flax

[plain fjord]

Pretty hard to understand without an example though

[solemn flax]

Okay

Thanks bread boy

[wintry steppe]

bread girl

[half swift]

smh

[ashen warren]

Hey does anyone here have an example on how to express a matrix multiplication over a set of basis?

[broken girder]

if A^2 is hermitian, then A is hermitian. true or false?

[half ice]

By contrapositive, if A is not hermitian, then A² isn't either. True or false?

[broken girder]

uhhhh

[half ice]

Odds are no, I'm trying to think of a counter example lel

But maybe there's a reason I'm not seeing

[broken girder]

we can always just take the 2x2 case and do it component-wise

and use properties of conjugates of complex numbers

[half ice]

Since (AB)' = B'A', (A²)' = (A')²

Assume:
(A²)' = A²
(A')² = A²

Since a square root of a matrix isn't unique, A is not hermitian

[broken girder]

oof

[half ice]

It's hard to use that to conjure a counter example, lel. Taking square roots is hard

[broken girder]

yeah...

ok so since im allowed to use Maple

it turns out that Maple has a way of finding square root of matrices

so yea ur right, the claim is false

@half ice

[broken girder]

uhh more linear algebra pasta

this question i feel is trying to get out of us that the restriction on A for it to be diagonalizable is that a = -2 and b= 0 (making A hermitian)

in other words, that if (a,b) =/= (-2,0), then A is not hermitian, and therefore that it is not diagonalizable.. but is this correct reasoning?

[timber solar]

Hellooo

alright so I have a question! I have 3 Coordinates in R2 that I need to transform into 3 other given Coordinates in the same Plane via affine transformation. How do I start with this one

[slow scroll]

I don't really know how to tell if two matrices are similar. Could someone guide me through the logic behind figuring it out?

[dusky epoch]

well you could try to use the definition of similarity

which given the matrices are 2×2 isn't that bad

or you could try to compute their characteristic polynomials. if they aren't the same then the matices are not similar

[slow scroll]

I don't know what characteristic polynomials are. The best justification i've been able to come up with so far is that they are both rank 2 🤷

[dusky epoch]

have you not worked with eigen-things before

eigenvalues, eigenvectors etc

[slow scroll]

nope

[dusky epoch]

hh ok

then all you have is the definition i guess

[slow scroll]

yep

[dusky epoch]

letting $A = \begin{bmatrix} 1 & 3 \ 2 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 2 \ 4 & 2 \end{bmatrix}$

[stoic python]

Ann:

[dusky epoch]

you want to see if there exists an invertible matrix $S$ such that $SAS^{-1} = B$, or equivalently $SA = BS$

[stoic python]

Ann:

[dusky epoch]

so i guess you can let $S = \begin{bmatrix} a & b \ c & d \end{bmatrix}$

[stoic python]

Ann:

[dusky epoch]

and write out the equation SA = BS

as a system of four equations in four variables

and see if it has a solution

[slow scroll]

oh alright that makes sense.

[dusky epoch]

actually wait.

have you encountered trace before

[slow scroll]

yes

[dusky epoch]

and so you know that if two matrices are similar then their traces are the same

right?

[slow scroll]

i was also asked to prove that on the problem set. Wasn't too sure what to do lol

[dusky epoch]

ah well that isn't too hard to prove

if you can prove what i think is called the cyclic property of the trace

namely, that tr(XY) = tr(YX) for any two square matrices X and Y

[slow scroll]

yea it gave that as a hint

[dusky epoch]

because then you can make the following very simple argument

$\tr(SAS^{-1}) = \tr(S(AS^{-1})) = \tr((AS^{-1})S) = \tr(AS^{-1}S) = \tr(AI) = \tr(A)$

[stoic python]

Ann:

[slow scroll]

omg of course. very nice.

[dusky epoch]

so with that in mind

if two matrices have different traces then they are definitely not similar.

and your two matrices have traces 3 and 2 respectively

[slow scroll]

ahh that makes sense

thanks for explainin ann c:

[wintry steppe]

I've got the following question:
For n quadratic matrices $K_i, \ K(x)=\sum x_i K_i$ we consider the cone $K={x\in\mathbb{R}^n: K(x)\succeq 0}$. Is it possible to find $x \in K$ such that K(x) is a rotation matrix, if we add $K$ to be pointed and $int K\neq \emptyset$?

[stoic python]

mathDE:

[lyric sphinx]

what does K(x) \geq 0 mean in that context?

that K(x) is definite positive?

[wintry steppe]

yes, $\succeq$ is supposed to mean positive semi definite

[stoic python]

mathDE:

[lyric sphinx]

the matrixes are all nxn?

[broken hawk]

uh, bit confused about the notation here. sum over x_i K_i is a row vector, so it can't be a rotation matrix

oh I see

x_i aren't vectors but the components of x

no assumption on the positive definiteness of the K_i?

[wintry steppe]

The Matrix size isnt fix since most cones have several representations with different dimensions

[broken hawk]

Also, how do you define rotation matrix?
identity, or a rotation of a 2D subspace and identity elsewhere?

[lyric sphinx]

tAA = I with det A = 1

[broken hawk]

that's an orthogonal matrix

[wintry steppe]

For example the lorentz cone in 3 variables has representations with 2x2and 3x3 mateices.

[broken hawk]

to me rotation matix is more specific

[lyric sphinx]

tAA = I is an orthogonal matrix

[broken hawk]

right, special orthogonal

[wintry steppe]

Orthogonal and keeping Scalar product

[lyric sphinx]

rotation matrix is the subgroup of these matrix with determinant 1

[broken hawk]

a special orthogonal matrix is the product of rotation matrices, but this is why I asked

cause not everyone uses this terminology the same way

[wintry steppe]

Orthogonal and det 1 is fine es well

So do you have any ideas? Since int K isnt empty and K is a cone some continuity argument can be used to show that there are x in K such that det(K(x))=1. But to show that one of these is orthogonal seems more comlicated.

[lyric sphinx]

im trying to see if $x \mapsto K(x) K(x)^T$ defined on the cone passes through identity

[stoic python]

oumid:

[lyric sphinx]

or is there is a connected path to I

from say 0

[wintry steppe]

Thats a idea, although im not sure how this could be seen easily.

Good*

[broken hawk]

I guess understanding the manifold structure of SO(n) would be useful here eh

[quiet crane]

hi guys this isnt a question about an assignment, but could anyone please send me an example of what linear algebra looks like?

[half ice]

Linear algebra is the study of vector spaces, and linear transformations between them

Basically, take vectors, take functions that preserve the structure of those vectors, what can you say about them?

@quiet crane

[quiet crane]

i may need help with this subject in the future, i just want to get a head start

[half ice]

Yeah of course, feel free to ask

[summer dagger]

For the definition of subspace when they say closed under vector addition can someone please explain closed ?

[slow scroll]

that means any vector in the space added to any other vector in the space is a vector in the space.

[summer dagger]

does this include negative scalars?

[slow scroll]

like scalar multiplication? Yea, vector spaces also must be closed under scalar multiplication

[summer dagger]

ty makes sense just need it reworded

[slow scroll]

np c:

[summer dagger]

@slow scroll with the save spent too much time on this one question instantly got it once you said that

[slow scroll]

glad i could help!

[covert dragon]

What’s the dif between this and normal algebra

[half ice]

Linear algebra is the study of vector spaces, and linear transformations between them

You will probably fairly soon learn what vectors are, lin alg is a lot of that.

[swift plinth]

Hey guys I have got A (1/4/-1) B (-2/1/-1) C (-3/-1/0) those dots are on a circle. I wanna find the center coordinates of the circle . anyone got an idea how?

[covert dragon]

Well I never did this

But the point should have the same distance between all points

Just that sry

Lol

[swift plinth]

or maybe I am wrong. The task says those coordinates I just gave you are positions of a planet who is orbiting a star. and I need to find the coordinates of the star

[covert dragon]

Well you could take the 2 furthest away

Connect them

Wait nvm

lol

[swift plinth]

well I drew it via geogebra

that doesnt work

[covert dragon]

They’d need to be away from the center point

Equally

[swift plinth]

exactly

thats not the case

[covert dragon]

Wait can you send how it looks I’m geigebra

I’ve never done this but I wanna have fun

[covert dragon]

Wtf lol

Maybe angles

?

Na

[swift plinth]

I have found some formulars for R^2

but I am not sure if they work for R^3

[little cairn]

Can someone help me with this seemingly simple problem?

[covert dragon]

Simple

):

[swift plinth]

I wish I had this problem

[covert dragon]

What’s a matrix

[little cairn]

I solved the one above and it seems very simple, but like.. the opposite

[covert dragon]

Is there a formula for circles in graphs

[little cairn]

What does that have to do with linear algebra?

[covert dragon]

No idea

[little cairn]

Did this 3 second google search help your problem @covert dragon

[covert dragon]

Yes I checked later

[glacial gulch]

@little cairn that column A gonna be a vector, yes?

[little cairn]

I assume by Col (A) it meant Columnspace of A, so it's going to be a set of vectors. at least I assume

considering the previous problem

[glacial gulch]

t in K?

[dusky epoch]

that's an R, i think

strange how it's not $\bbR$ or even $\mathbf R$

[stoic python]

Ann:

[glacial gulch]

@little cairn https://www.khanacademy.org/math/linear-algebra/vectors-and-spaces/null-column-space/v/null-space-and-column-space-basis will this help?

[little cairn]

Thanks @glacial gulch I’ll check it out

[glacial gulch]

@little cairn http://www.math.drexel.edu/~jwd25/LA_FALL_06/lectures/lecture5C.html Example 5

[stoic python]

wizz:

[glacial gulch]

should be your A unless I accidentally crossed my eyes while writing them down xD

[winter reef]

how do I check whether a point is in H where H is affine combinations of some vectors?

[glacial gulch]

@winter reef https://math.stackexchange.com/questions/985645/express-a-point-as-an-affine-combination-of-another-two-points3d-collinearity will this be of assistance?

[winter reef]

ty will check it out

[little cairn]

Thanks @glacial gulch that was a lot easier than I expected

I even left a proof at the end

[glacial gulch]

np, sry it took forever to get down to googling :D

[little cairn]

Because I felt like it was too easy

No problem lol

[steep marten]

Hi guys, so I have to find the orthogonal projection of (-1,1,2,2)^T on the subspace spanned by:
v1 = (1,1,1,1)^T
v2 = (0,0,2,2)^T
Which would be easy if these two were orthogonal themselves, but they're not. Is there a smart way to do it without v1/v2 being orthogonal?

[lethal raptor]

@steep marten Just use gram schmidt, works with any set

[steep marten]

So the subspace is preserved through GS and then it becomes ez

[royal fiber]

can anyone simplify these into real numbers?

i got eigenvalues: 4i, -4i

eigenvectorss: <1/2-i/2,1> and its conjugate

[dusky epoch]

you mean eigenvalues 4i and -4i

[royal fiber]

woops yeah

im having trouble simplifying c1_e^4i*(1/2-i/2)

[dusky epoch]

you're gonna get cos(4t) and sin(4t) eventually, i think

[royal fiber]

i got to c1/2*(cos4t-icos4t+isin4t+sin4t)

[lean aspen]

pls don't give answers

[royal fiber]

idk how to go from there

[dusky epoch]

ok wait

let's double those eigenvectors

<1-i, 2> and <1+i, 2>

[royal fiber]

mhm

[dusky epoch]

so one of your complex sols is <1-i, 2> e^4it

take the real and imaginary part of that

[royal fiber]

shouldnt all the values end up real

[dusky epoch]

the imaginary part of a complex number is a real number

$\Re(zw) = \Re(z)\Re(w) - \Im(z)\Im(w) \ \Im(zw) = \Re(z)\Im(w) + \Im(z)\Re(w)$

[stoic python]

Ann:

[royal fiber]

ah ok i did some reading

thanks for the help

my class hasn't actually gone over this yet

[swift plinth]

okay guys I have a ship starting point (50,12) moving in the direction (5,12). It moves with 26 km/h. I need to find the coordinates of the ship after one and two hours.

<@&286206848099549185>

[swift plinth]

okay problem solved: I needed the unit vector

[near edge]

I'm not understanding this. Is this a linear transformation? If T(xi) = axi + b. For two inputs we have T(x1+x2) = a(x1+x2) + b = ax1 + b + ax2 =/= ax1 + b + ax2 + b = T(x1) + T(x2). Maybe i'm seeing it wrong and what they call a linear transformation here is not exactly what I think it is.

[empty copper]

t!wiki affine transformation

[glass sluice]

📖 | ** https://en.wikipedia.org/wiki/Affine_transformation **

[near edge]

ye... but it says linear transformation :/ , not affine

[charred stirrup]

can anyone explain what a "span" is exactly? I have having trouble wrapping together the concept with the idea of linear independence and basis, and dimension

[brittle juniper]

the span of a family of vectors is the smallest (for the inclusion relation) vector space that contains that family of vectors

For example, if $v_1, v_2, v_3$ are vectors of $E$, it means span$(v_1,v_2,v_3)\subseteq E$

[stoic python]

Tuong:

[broken hawk]

https://www.youtube.com/watch?v=fNk_zzaMoSs&list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab watch this playlist, first two videos are all you need right now but it’s great

[brittle juniper]

you can prove the span of a family of vectors is the set of linear combinations of said vectors

[broken hawk]

so just watch it all

[brittle juniper]

Oh yeah those vids were nice

[broken hawk]

the first video is just a bit of setup, the second is where he explains span, basis and linear combinations

he puts it in a non-formal setting, so if you’re doing very formal linalg, treat this as examples

this is imo mandatory watching for anyone taking a first linalg course

and some of the videos are good to rewatch after you’ve worked with the material a bit imo, there are some things I found nice on a first watch but only really grokked after I had a much deeper understanding

like the video on duality

also honestly, span is weird to explain because once you understand it, the word is so obviously descriptive

[charred stirrup]

@brittle juniper @broken hawk thanks a bunch, coming to an end of my lin alg course and they dropping all this stuff on us

[broken hawk]

wait what

[brittle juniper]

must be tough

[broken hawk]

what did you do all year

even if the whole course is just matrix stuff, linear dependence is important there too

[charred stirrup]

@broken hawk we did matrixes, and learned about hyperplanes and a bunch of intersection stuff

point normal

[broken hawk]

ah so it was more vector geometry?

[charred stirrup]

yeah

[broken hawk]

rather than actual linalg

[winter herald]

^ thats a nice icon, roxas

[winter reef]

Any good source to learn affine stuff?

[chilly dragon]

i'm confused about something

prove that $ cos \theta = \frac{x^T y}{(| x | _2 | y | _2)} $

[stoic python]

Wed.:

[chilly dragon]

using the law of cosines

$ || y - x ||_2 ^2 = || x ||_2 ^2 + || y ||_2 ^2 - 2 || x ||_2 || y ||_2 \cos{\theta} $

[stoic python]

Wed.:

[chilly dragon]

now the problem is that

$ || y - x ||_2 ^2 = x^t x + y^t y - 2 x^t y $

[stoic python]

Wed.:

[chilly dragon]

i don't get how $ 2 ||x||_2 ||y||_2 = 2 x^t y $

[stoic python]

Wed.:

[chilly dragon]

because $ ||x||_2 = \sqrt{x^t x} $

[stoic python]

Wed.:

[broken hawk]

use $|$ instead of $||$

[stoic python]

Sascha Baer:

[broken hawk]

\|

[chilly dragon]

too much lines to replace ..

[broken hawk]

nah dw about that, just for teh future

looks better and plays better with discord

[vague musk]

To prove a subset is a subspace, do you first have to prove that the subset is a vector space?

[wintry steppe]

For vectors yes oof I'm wrong

[dusky epoch]

@vague musk no, you have to prove that your subset satisfies the definition of a subspace

[vague musk]

@dusky epoch yep you're right, thanks! turns out I didn't know what closed addition was and thought you had to do both for some strange reason

in the criterion*

[stoic python]

woopy:

[half ice]

@patent glacier
That's correct

[stoic python]

woopy:

[half ice]

Looks right

That's a dumb way for them to notate that but whatevs

[stoic python]

woopy:

[dusky epoch]

(0,0)

but eys

yes*

[empty copper]

Sure are~
Keep it up!

[ancient river]

english isn't my first language, is linear algebra the right channel for matrices?

[winter reef]

yes

[ancient river]

so uh, I'm studying economics and I think I understand how matrices interact with one another but I struggle a lot with how to lay them out from the text I have. (not sure if I should go in questions instead?)

I have a big pdf that says how to use them yet only half a slide adresses "production ratios" and everything I have to do start with that :/

[slender yarrow]

just wondering, are you french?

[ancient river]

yes 🥖

[slender yarrow]

ptdr c'est pas mal ça 🥖

[dusky epoch]

👀🥖

[sage mauve]

你好

[slender yarrow]

Γνοτι σεαυτον

[dull kettle]

Hello guys, is statement II true or false?

[lyric sphinx]

It's true you can show it by induction

[dull kettle]

can you elaborate please?

[lyric sphinx]

Do you know how the Vi's are defined?

[dull kettle]

every distinct pairs are orthogonal?

[lyric sphinx]

I mean the formula

[dull kettle]

oh gram-schmidt process? yes

[dusky epoch]

without getting too far into the details of the gram-schmidt process, you have that $v_j = u_j - \sum_{i=1}^{j-1} \alpha_i v_i$ for some constants $\alpha_i$ whose exact values are irrelevant

[stoic python]

Ann:

[dull kettle]

hmm

how does that make span ( u1, ... , uk) = span (v1, ... , vk)

[dusky epoch]

well, every v_j can be eventually expressed as a linear combination of the u_i for 1 ≤ i ≤ j

thus v_j ∈ span {u_1, ..., u_k} for all 1 ≤ j ≤ k, and as such span {v_1, ..., v_k} ⊆ span {u_1, ..., u_k}

does that make sense to you?

[dull kettle]

yes but how do you see that every v_j can be expressed as a linear combination of u_i for 1 ≤ i ≤ j

[dusky epoch]

this can be shown by strong induction

for v_1 it's trivial since v_1 = u_1 by def

and for the inductive step it's likewise obvious from that formula i wrote out

[dull kettle]

oh i see, you can use the same formula to prove that span (u1, ... , uk) is a subset or equal to span (v1, .. , vk) right? that would show that they are equal

[dusky epoch]

yes

[dull kettle]

thanks @dusky epoch !

[twin loom]

When applying gram-schmit to polynomials, is it not a problem that this inner product definition allows for <u|v> = 0, when u and v are not zero?

Or any inner product definition really

[quaint heart]

Yeah, when thinking about the inner product you should think of the dot product as an example

[twin loom]

What do you mean?

I know the dot product is an example of an inner product

[quaint heart]

The dot product is an inner product on R^n

[twin loom]

but thats only 0 if one of the vectors is 0

[quaint heart]

No

[twin loom]

OH

ok yah im dumb

its <u|u> = 0 => u = 0

[quaint heart]

Yeah

You can take sqrt(<u, u>) as a norm

So it has that property

[twin loom]

But what about in the problem I have?

I ended up with a division by 0

And I dont know what that means

[quaint heart]

The grant schmidt process works for a collection of linearly independent vectors

So none of your vectors should be the 0 vector

[twin loom]

The vectors I started with are all linearly independent tho

[quaint heart]

So check your work

[twin loom]

fuck i didnt square the last integral

god Ive been checking this problem over for the past half an hour

thanks

[dusky epoch]

this inner product definition allows for <u|v> = 0, when u and v are not zero?

that's called orthogonality

😛

[heady cypress]

Hey, I'm having a bit of trouble finding in my textbook how to find the number of inversions for a set of integers, it just shows sets and then it says "this is the number of inversions" but it isn't exactly clear how it gets that number

nm found the answer

or rather, found out how to do it

[hallow leaf]

Hello I am needing help on creating this problem into a matrix operations. NH3 + H2SO4 → (NH4)2SO4 I am not sure how to solve this using algebra.

[wicked trellis]

Can anyone help me with these two

[quaint heart]

For the first one do you know the rank nullity theorem?

[wicked trellis]

I don’t, I could look

[quaint heart]

Ie rank + nullity = n

In this case n is 5

For the second one just take 2 linearly independent vector in R^3

And use those to make a matrix

@wicked trellis

[wicked trellis]

So two vectors need to be linearly dependant?

[quaint heart]

Yep

[wicked trellis]

Thanks

[quaint heart]

Dim(column space) is rank btw

[limber wind]

I have a math assignment to solve some linear equation systems using "row operations", does that mean I'm supposed to use Gauss-Jordan elimination? Terms are translated from Norwegian and may be different from English terms

[steep marten]

Yes it does, bring it to row reduced echelon form if possible then you will have a solution

[limber wind]

That's the diagonal of 1's with 0s in the other columns?

[steep marten]

Aye

[broken hawk]

not necessarily on the diagonal

row echelon form is this: https://en.wikipedia.org/wiki/Row_echelon_form?wprov=sfla1

reduced is the second example given

[limber wind]

Thanks @broken hawk

Does it matter whether I follow the algorithm of reducing rows below before swapping etc?

I swapped because I would get a 0, but now I have an entire row of 0's

Tried the other way and got a full row of 0's anyway

[random trellis]

(for a matrix in rational canonical form) why is the largest invariant factor the same as the minimal polynomial?

[rare barn]

not really sure how to do this, the options are 0, -1, and 1

[slow scroll]

$A(A( v)) = A(\lambda v) = \lambda^2 v = 0$

[stoic python]

kxrider:

[rare barn]

got it, one more : if lambda is the eigenvalue of A, then is lambda also the eigenvalue of A^T?

<@&286206848099549185>

[dusky epoch]

you mean an eigenvalue in both cases

[rare barn]

yes, i figured it out just now too

it makes sense

[dusky epoch]

det(A - λI) = det(A^T - λI)

[rare barn]

yes

@dusky epoch this is another question (its true/false)... not really sure how to check this: Every square n×n matrix has n distinct eigenvalues.

[quaint heart]

Can you think of an example where this isn't true?

[dusky epoch]

this depends on what field the matrix is over

[rare barn]

i dont see where it wouldnt be true

what do you mean what field

[dusky epoch]

for your class, are your matrices always with real entries?

[rare barn]

yes

[dusky epoch]

do you allow the eigenvalues of a matrix to be complex nonetheless?

actually, you don't even need that.

a matrix may have all its eigenvalues be real yet fail to have n distinct eigenvalues

[quaint heart]

There's a very trivial example

That may have been a pun

[rare barn]

how though

[quaint heart]

What are the eigenvalues of the 0 matrix?

[dusky epoch]

alternatively, what are the eigenvalues of the identity matrix?

[quaint heart]

I guess that the identity matrix is a trivial example as well lol

[rare barn]

its just 0

and 1

[wintry steppe]

can anyone recommend a good book on linear algebra? something that's thorough and explained for brainlets like me

😄

[broken hawk]

are you looking for applied linalg (matrix algebra and vector geometry and the likes) or theoretical linalg (vector spaces and all the theorems)

[wintry steppe]

theoretical

although my curriculum is a bit mixed, but mostly theoretical

[broken hawk]

personally, I like Friedberg/Insel/Spence Linear Algebra

it’s very rigorous and imo takes the right approach of avoiding matrices where possible

but it may be too dry :P

slash not suitable for someone calling themselves a brainlet

[wintry steppe]

i dont mind it being dry tbh

[broken hawk]

(note, I’ve not read through all of it. I’ve read the beginning stuff and liked it when I was starting out)

I hear Linear Algebra Done Right and […] Done Wrong are both good

but I’ve not read them at all

[slow scroll]

ive been using "Linear Algebra Done Wrong." I think its pretty good

[broken hawk]

basically, look a bit through those books and pick whichever you like, I guess

hi potato

I see you 👀

[wintry steppe]

I will do that, thank you very much

😄

[broken hawk]

also, obviously, watch 3b1b’s series on linalg

gives all the right intuitions for the basic concepts

[wintry steppe]

i already did that lmao

[broken hawk]

do it again

:P

[wintry steppe]

lmfao

[broken hawk]

honestly, rewatching them when I knew more linalg actually taught me some things I didn’t really get the first time round

[wintry steppe]

yeah the visualisation component is definitely a plus when you are learning this stuff

[broken hawk]

yea, though you have to take care not to overgeneralize too much, cause not all spaces are ℝ² or ℝ³ :P

but then again there’s not that many “important” vector spaces

like, there’s $\mathbb{K}^n$ for some field $\mathbb{K}$ (in particular also finite ones), there’s $\ell^2$ and $L^p$, and maybe like the space of polynomials

[stoic python]

Sascha Baer:

[random trellis]

(for a matrix in rational canonical form) why is the largest invariant factor the same as the minimal polynomial?

[charred stirrup]

how legitimate is this strategy for finding the RREF of a matrix? I just need any sort of method or steps to follow that will eventually lead me to the correct answer

https://imgur.com/gallery/LzQ8R

[half ice]

-Pick the leftmost un-rref'd column
-Pick where the 1 should be
-Whatever's there, divide it out. This means dividing the entire row. There will be a 1 there now.
-Add/subtract the other rows until that column is zero.
-Repeat.

[charred stirrup]

oh shit

[half ice]

There's no "strategy", it's an algorithm. Computers do it

[charred stirrup]

you're hype as fuck

[half ice]

It always goes the same way

[charred stirrup]

THAT'S A NASTY TRICK

YOO

[half ice]

Lel, does that make sense?

[charred stirrup]

yeah you put it really well

i have a question though

[half ice]

Well, I guess there can be some strategy, but it's usually easy enough to just "follow the recipe". If there's a faster way, it's not much faster

[charred stirrup]

it's confusing but- it's redundant to say "interchange row 1 with (1/x) (row 1)" instead of "multiply row 1 by a scalar (1/x)" right?

they mean the same thing?

[half ice]

Yes, you are allowed to multiply a row by any scalar

So multiply in 1/x

I don't know what your teacher may want lel. I would accept "multiply row X by X"

[charred stirrup]

thank you

i like both your methods

you get the 1's taken care of first but they get the 0's taken care of first

[half ice]

I don't think it really matters, just find a way to do it that works for you

Perhaps by handling the 0s first you can avoid fractions? Idunno

[half ice]

I have no idea what you want to do with that statement

[slow scroll]

f o h is a function. You don't "solve for x"
also, #prealg-and-algebra

[wintry steppe]

ill delete my post im sorry kamen rider

[sand oak]

can anyone help me out

[slow scroll]

for the second one, the span of any set of vectors would be a vector space

for the first one, the $f(2) = 0$ just wants you to see that when you take the sum of two vectors, call them $f_1 (x)$ and $f_2 (x)$ and let that be equal to a new vector, $f_3 (x)$, then $f_3 (2) = f_1 (2) + f_2 (2) = 0 + 0 = 0$. Therefore the space described in problem 1 is closed under vector addition.

[stoic python]

kxrider:

[slow scroll]

you are proving that these are vector spaces, not just subspaces, so you may have to show that sums of polynomials are commutative and associative and stuff like that.

[sand oak]

only one example suffices? we dont need a total 3?

[slow scroll]

sorry wdym?

ah okay good. Thats what i thought they might be looking for

so prove that the set is closed under scalar multiplication, vector addition, and contains the zero vector. Those are the three properties

same thing for 2. $\text{span} \right{ \begin{bmatrix}2\3\7 \end{bmatrix}, \begin{bmatrix}1\0\1 \end{bmatrix} \left} \subset \mathbb R^3$

[stoic python]

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

[slow scroll]

or however tf u latex that

[sand oak]

ty. i might have a few more problems

[slow scroll]

alright

[sand oak]

prove that the following are not vector spaces

[dusky epoch]

show that these don't satisfy the vector space axioms

find, for each one of these, an axiom that it fails

what's holding you up?

[slow scroll]

for 7, create two vectors (x, y, x^2 + y^2) and (u, v, u^2 + v^2) and show that these two vectors added together (x + u, y + v, x^2 + y^2 + u^2 + v^2) doesn't satisfy 1st coordinate squared plus second coordinate squared equals third coordinate squared

[dusky epoch]

that's showing lack of closure under addition ^

[digital yew]

can someone help me prove that

if $$[\Omega, \Theta] = 0$$, then $$e^{\Omega}e^{\Theta} = e^{\Omega + \Theta}$$

[stoic python]

habajiga:

[digital yew]

how would i prove this?

not using lie algebras and stuff, if u can put it in dum dum terms it would help :c

[broken hawk]

are omega and theta matrices and [,] is the commutstor?

or is this more general

if it's matrices or anything else where the argument makes sense, I would show that for all n, the partial sums of the power series with the first n terms are the same

and then let n to infinity

$$e^{A+B} = \lim_{N \to \infty}\sum_{n=0}^N\frac{ (A+B)^n}{n!} = \lim_{N \to \infty}\sum_{n=0}^N \frac{\sum_{k=0}^n \binom{n}{k} A^k B^{n-k}}{n!}$$ where I use commutativity to apply binomial theorem

[stoic python]

Sascha Baer:

[broken hawk]

And then that is expressable as a product of the two power series

[charred stirrup]

I am not sure of the definition of "row-space"

Does that mean the length of a row in the matrix?

so for a 5x4 matrix, the rank will be '5' ?

given that we're talking about rows x columns

[empty copper]

The row space is the subspace spanned by the row vectors... Like it's written there

[digital yew]

hey @broken hawk thanks you, yes they are matrices

to me i was thinking of using zassenhaus formula

but if im being honest i was hoping to get a proof from scratch, like starting with a taylor expansion or something, but i was unable to complete it

otherwise can just use this

[dusky epoch]

[digital yew]

did i do something wrong

https://en.wikipedia.org/wiki/Baker–Campbell–Hausdorff_formula#The_Zassenhaus_formula

heres where i got that btw

[dull hemlock]

Why is linear algebra notation so clunky and inelegant? I feel as if there is a better way to do linear algeba.

[dusky epoch]

can you show an example of linear-algebraic notation which you find clunky?

[dull hemlock]

[bunch of shit]

[half ice]

Oof quite the opposite, I would say linear algebra takes what should be a very complicated concept and uses beautiful notation to make it easy

[dull hemlock]

Why can't we just write lines as "ax" instead of this vector nonsense?

[half ice]

Vectors are above lines. You use vectors when cartesian equations don't work

As well, vectors can represent anything with a certain set of properties. You've probably seen linear algebra treat polynomials as vectors

🅱ectors

I vote new name.

[dull hemlock]

We take the average of 2 lines (ac+bc)x/2 to aquire a sum of vectors or lines. And if we want a certain length of the vector, we can restrict the domain.

where c is the largest domain of ax

We can go on like this to recreate all of linear algebra.

But I suppose we'll have to put up with this abuse.

[half ice]

You have to understand math is about oppression that's just what we do best

[quaint heart]

You can remake linear algebra using lines with projective geometry

This is pretty well known

[dusky epoch]

@dull hemlock are you being serious rn

[broken hawk]

I find it incoherent, personally

[dusky epoch]

as do i

they failed to answer my question, which would have been straightforward to answer had they not been full of shit

[dull hemlock]

@dusky epoch Hey that was very rude of you. I am full of dirt, not shit. Have some class.

Also I identify as a book. Use my pronoun, nazi.

[dusky epoch]

??? wow, name-calling?

[rigid cypress]

[dusky epoch]

@dull hemlock i saw that.

[dull hemlock]

no you didnt

[dusky epoch]

yes i did. right before you deleted it.

[dull hemlock]

No.

shut your up

[rigid cypress]

what'd they post

[dreamy fiber]

👀

[dusky epoch]

okay, you cannot be reasoned with.

@rigid cypress they posted an image corresponding to the phrase "the joke went right over your head".

[jagged pendant]

talk less shit, more linear algebra

[dusky epoch]

hoooo

bold move

[half ice]

🤔

[jagged pendant]

ok, your choice lmfao

[rigid cypress]

woog your roles are back!

[rare barn]

question, if a matrix A has a characteristic polynomial of 7x^3+4x^2-3x-6 , is A^T 's characteristic polynomail -7x^3-4x^2+3x+6?

x is lambda

[dusky epoch]

the char polys of A and A^T match

and if you're allowing them to differ by an overall constant multiplier, you might as well identify those two you've written

[rare barn]

got it, ty

If A is a square matrix with only nonzero eigenvalues, then A is invertible. (true/false?) I know this is a condition for the invertible matrix theorem, but does the determinent of A ALSO need to =/ 0 for a matrix to be invertible, or does this one condition work out

[jagged pendant]

?

if it has no zero eigenvalues then that's sufficient

it has no kernel

so it must be injective and hence bijective

[broken hawk]

right

I overthought it

explains why I couldn’t think of counterexamples :P

[wicked trellis]

Can anyone help me with this

[broken hawk]

hint: ||a diagonal matrix is diagonalizable||

[wicked trellis]

So it’s always invertsble?

[broken hawk]

nope

the zero matrix is diagonal

since every non-diagonal entry is 0 (it doesn’t matter what’s on the diagonal)

[wicked trellis]

Oooo

[broken hawk]

of course I’ve now solved the exercise for you

so sorr about that

[wicked trellis]

😂

It’s a very small part at least

[charred stirrup]

I need some help classifying definitions

Is it correct to say that all subspaces are subsets, but not all subsets are subspaces?

[half ice]

Yes

[charred stirrup]

@half ice yoo thanks again fam

i appreciate you

[half ice]

Np. Is that all you wanted to know about it?

[charred stirrup]

ill probably be back with more questions soon lol, thanks alot

[half ice]

Np good luck

[charred stirrup]

when i'm trying to reduce an augmented matrix

why am I not allowed to add any number x to a row?

for example a row is [1 5 9 | 7 ]

[half ice]

Cuz you can't! That's not an elementary move

[charred stirrup]

oh wait

im just dumb lmao

sorry yeah I get it

[half ice]

Imagine solving
x + y = 1
x - y = 2

And I "added 5 to the first row":
6x + 6y = 6
x - y = 2

You'd probably look at me funny

[charred stirrup]

yup I see

uhh

[half ice]

Actually that's the same as multiplying by 6 in this case, but it didn't have to be

[charred stirrup]

for the elementary moves

[half ice]

If I chose better numbers

[charred stirrup]

can I do something like row 1 - (c)(row 1) ?

c is some scalar?

[slow scroll]

yea. Thats the same thing as doing (1-c)row1.

[charred stirrup]

ahhh I see

thank you so much

[slow scroll]

np

[autumn anvil]

actually for x + y = 1 adding the same number is the same as multiplying the same number regardless of number choice, because the first row is all the same number

[slow scroll]

im pretty sure he meant number choice as in different coefficients for x, y, 1

[autumn anvil]

but for like row 2 you can see by +3, x - y = 2 would become 4x - 4y = 5 which isn't the same

ah then yeah. but that's interesting anyway

[charred stirrup]

the nullity of a matrix A can be found by taking the # of columns and subtracting the rank

but what exactly is the definition of nullity? what is significant about it that I should know?

[slow scroll]

err wait

dim(Null())

dimension of the null space/kernel

[charred stirrup]

can you define dimension and null space for me? if we have a (m X n ) matrix then the "dimension" is defined as m X n ?

" The null space of a matrix is any set of vectors that you can multiply a matrix by which gives you 0 as a result. " correct?

[slow scroll]

The "dimension" of a vector space refers to the number of vectors in a basis for the space. For example, the rank of a matrix refers to the number of linearly independent columns of a matrix, and the nullity of a matrix refers to the number of free variables in the solutions to Ax=0

Yes to your definition of null space.

[charred stirrup]

god your explanation is really straightforward and i feel retarded for not getting it

I feel like I am missing some fundamental understanding that's gonna help me understand it

[slow scroll]

rank + nullity = columns comes from the fact that free variables correspond to the lack of column pivots and rank refers to the number of column pivots.
existing column pivots + missing column pivots = number of columns

hmm what part is confusing?

[charred stirrup]

the words "spanning" , "pivot", "dimension", "basis"

I understand that a "span" is a SET of all possible linear combinations for any number of vectors

but what is "spanning"

and I think "pivot" is also very easy to explain, but I am just having trouble understanding my notes alone

ok wait I think I get your explanation for dimension

[slow scroll]

"spanning" usually refers to when the span of a set of vectors represents every vector of a particular vector space.

Pivot rows/columns are important because they have certain qualities like being linearly independent from other pivot rows/columns

[charred stirrup]

if I am in R^3, have a vector (v1, v2, v3), then the dimension is 3 because the BASIS of R^3 needs 3 vectors

[slow scroll]

right but the dim(span{(1,0,1), (0,1,0)}) is a 2D subspace of R^3 for example

[charred stirrup]

is it cool with you if I try breaking that down?

[slow scroll]

sure

[charred stirrup]

dim(span{(1,0,1), (0,1,0)}) = dim ( some linear combination)

[slow scroll]

right, and the span of any set of vectors is a subspace

[charred stirrup]

dim = # of vectors that form a basis of a space

[slow scroll]

right, and notice the two vectors I included would form a basis for that subspace

[charred stirrup]

for the two vectors you gave v1 and v2, if I have v1 + 0v2, then the dim( v1+ 0v2) is going to be 3 or 2?

[slow scroll]

"v1 + 0v2" is not a space, so dimension doesn't apply there

[charred stirrup]

why is v1 + 0v2 not a space? that is not a valid linear combination of the two?

[dusky epoch]

v1 + 0v2 is a vector.

[charred stirrup]

correct, which happens to be a linear combination where the scalar applied on v1 is 1, and the scalar applied on v2 is 0 ?

[dusky epoch]

...okay, wait

what was the original question

[slow scroll]

right, so its just an element of the 2D space, span{(1,0,1), (0,1,0)}

[charred stirrup]

OHHHH

i have one more question that's probably gonna make my understanding complete

what if I have v1 + v2 ? so we get a vector (1, 1, 1 ) ?

that is no longer an element of the 2D space ?

[dusky epoch]

no, (1,1,1) is in span {(1,0,1), (0,1,0)}

[slow scroll]

no it still is lol. Think of it this way, all linear combinations of (1,0,1) and (0,1,0) lie on a plane in 3 space. Yes, they are all vectors in R^3, but they also form a 2d subspace (like a plane) within R^3

[dusky epoch]

and in presenting the linear combination of v1 and v2 (specifically, 1v1 + 1v2) that equals it, you proved that (1,1,1) ∈ span {(1,0,1), (0,1,0)}

[charred stirrup]

please tell me im understanding that correctly

[slow scroll]

yea but it could have any arbitrary orientation

[charred stirrup]

thank god

[dusky epoch]

okay first off it's $\bbR^3$, not $R_3$

[charred stirrup]

yeah i get what you're saying

[stoic python]

Ann:

[dusky epoch]

and second, a subspace doesn't look like what you drew. it's an infinite plane, not a square

[charred stirrup]

I drew a portion of the plane

which i should have said

is that fair?

let's say I made a restriction on the span that would give me a set of linear combinations that just so happens to make that square

please tell me im getting this

[dusky epoch]

your picture is kinda hard to judge either way tbh

like i can't say you're completely clueless but i can't say you're really getting it either, however vague that notion may be

[slow scroll]

I just want to add one more thing about spanning systems that may be useful in the context you heard it. "Spanning," "generating," or a "complete" set of vectors refers to the fact that any vector in the space can be represented as a linear combination of the vectors in the set (presumed to be spanning).

In the context of linear transformations, the span of the columns of a matrix give you the "range" or "image" of a matrix. Suppose we have a linear mapping from R^n to R^m. That means i plug in vectors from R^n and i get out vectors in R^m.

Suppose the span of the columns of the matrix representation of this transformation mapped to some proper subspace of R^m. I.e. the range of the transformation is not all of R^m. That means the columns are not spanning, and that there exists some b in Ax = b that does not have a solution x. This is called "inconsistency."

Just had to get that off my chest. Anyway i gtg to bed

[charred stirrup]

@slow scroll thank you, really appreciate it

[slow scroll]

np np c:

[charred stirrup]

I can successfully identify whether two or more vectors are linearly dependent but I dont know what the significance of linear dependence/independence is

[dusky epoch]

have you watched 3b1b's essence of linear algebra series

bc i think it gives a good intuition for those

[charred stirrup]

thanks for resource

gonna check it out asap!

[broken hawk]

(start at the beginning)

[winter reef]

if an endomorphism has dimension n and it doesn't have n lin independent eigenvectors then its not diagonalizable, right?

[dusky epoch]

wdym by an endomorphism having dimension n

[winter reef]

a linear transformation from Rn to Rn for example I mean

sry poor wording

[brittle juniper]

I'd say that's true

as the contraposition is true

[sage mantle]

hey , i got asked to compute the characteristic polynomial of a matrix only composed of one number (a0) , i know how to compute the characteristic polynomial of 2 by 2 and 3 by 3 matrix but how am i supposed to do this?

[sand hedge]

isn't it just a0?

[brittle juniper]

or X-a0?

[sand hedge]

o

a-lambda

[dusky epoch]

a 1×1 matrix?

[sand hedge]

kek

[sage mantle]

so i don't have to compute a determinant?

or anything of that sort

[brittle juniper]

you could do the usual determinant thing

[sand hedge]

I mean det of a 1x1 matrix is just the entry

so a - 1x1 identity matrix * lambda

det(a-lambda)

= a-lambda

no?

[brittle juniper]

aren't characteristic polynomials supposed to have 1 as the dominant coefficient?

[sand hedge]

hmmmmmmmmm

not sure

[sage mantle]

i'd say that the most logical answer would be a0-lambda

[sand hedge]

its what i'd go with unless someone has an alternative

[brittle juniper]

they do!!

[sand hedge]

[brittle juniper]

so X-a0 instead of a0-X

[sand hedge]

idk symbo agrees with me

[brittle juniper]

awwh that's a different def from Wikipedia

[sand hedge]

[sage mantle]

what do you mean by 1 as the dominant coefficient x.x

[sand hedge]

wikipedia fake news tbh

[brittle juniper]

Well then follow the choice that was made by your teacher

[sand hedge]

dumb question tbh

[sage mantle]

i never went to class

but thanks for the help

[sand hedge]

[dusky epoch]

@brittle juniper le coefficient dominant s'appelle "leading coefficient" en anglais

😛

[brittle juniper]

xièxie ( ^_^)

[dusky epoch]

c'est quoi ça

c'est "merci" en chinois ?

[brittle juniper]

Yup

[dusky epoch]

[wintry steppe]

Hi I just have a quick question, can vectors in R3 span R2? My calculus teacher said it’s possible, because R2 is a subset of R3 but my linear algebra teacher said that they’re 2 different dimensions so it’s not applicable

[dusky epoch]

R^2 is not a subset of R^3

(x,y,0) and (x,y) are different things

[wintry steppe]

Subspace*

Sorry

[dusky epoch]

subspace implies subset

[wintry steppe]

Ok

Thanks

I felt like an idiot disagreeing with her in front of the class REEE

[half ice]

3D vectors can span a 2D space. Calling that space "R^2" is a weird choice.

Unless I'm not thinking about any specific cases, there's a bijection between that space, and R^2? This might be enough for your calc teacher to consider them equivalent

[wintry steppe]

so 3D vectors can span 2D spaces, but not R2

is that what you're refering to?

[dusky epoch]

"3D vectors"

[wintry steppe]

well

vectors in R3

[dusky epoch]

i think "3D vectors" is a horrible term and shouldn't be used

[wintry steppe]

ok lemme rephrase

[dusky epoch]

a set of vectors in R^3 can span a two-dimensional subspace of R^3

but that two-dimensional subspace is not to be identified with R^2 itself

[wintry steppe]

so vectors in R3 can span 2 dimensional spaces, but not R2

ok

thanks \o/

[oblique crag]

Can anyone help with #5 anx 4a

<@&286206848099549185>

[slow scroll]

what is Proj?

does it just mean $$ \begin{pmatrix} 0 & 0 & 0 \ 0 & 1 & 0 \ 0&0&1 \end{pmatrix} + \begin{pmatrix} 1 & 0 & 0 \ 0 & 0 & 0 \ 0&0&0 \end{pmatrix} $$ ?

[stoic python]

kxrider:

[slow scroll]

i would assume proj_yz means to project onto the xy plane, but idk what proj_x is supposed to mean

[oblique crag]

Proj is projection

[slow scroll]

also, that first matrix would be Proj_xy

yea but what does "Proj_x" mean?

[oblique crag]

No idea

😅

[slow scroll]

: D

[oblique crag]

I think it has something to do with projv+proj(orthog v)=I

If that helps

[slow scroll]

whats wrong with what i have above? the first matrix is Proj_yz and the second (would have to be) is Proj_x

[oblique crag]

Wait does it even matter

Ur adding them

Same thing no?

[slow scroll]

no it doesn't matter, I was just distinguishing between them, since Proj_x is still ambiguous here haha

[oblique crag]

Im guess

If u have the yz plane

The x plane is the orthogonal piece

Which is what the formula says projv+proj(orthog)=i

[slow scroll]

I guess?? but at the same time, wetf is "x plane" supposed to mean? It sounds like a top secret jet fighter to me.

[oblique crag]

Found this

So <x,0,0> and <0,y,z>

[slow scroll]

oh ok. that makes sense i guess

[oblique crag]

So x plane and yx plane which are orthog to each other

But idk how to prove that it =I3

I sub 3

[slow scroll]

I just did it
https://cdn.discordapp.com/attachments/540211747613704221/563052251950743729/195237017049759744.png

[oblique crag]

O

[slow scroll]

thats Proj_yz + proj_x

[oblique crag]

What does I sub 3 mean though

[slow scroll]

the 3x3 identity matrix

[oblique crag]

Oh...

So I stands for identity matrix?

[slow scroll]

yep

[oblique crag]

And the sub number is just the square

So I sub 5 is 5x5?

[slow scroll]

yea i guess. I've never actually seen that notation before but im pretty sure that is what it means

[oblique crag]

Ok cool

And uhh do u know how to do 4a? 😅 I was able to do 4b by doing proj L and just switching the signs to get proj orthog L

I know u get the unit vector from@the span

But then is that orthog L or do i still need to do@somethin to it

[slow scroll]

span{1,1,1}? Those are all the same vector

[rare barn]

I have no idea what to do here...

[dusky epoch]

do you know what "matrix for T relative to B" means

[winter reef]

does relative mean similiar/inverse?

[dusky epoch]

no

[winter reef]

ok, thought its some kind of synonym

[rare barn]

is it not this

[dusky epoch]

i guess that's the more general case

but point is

calculate where T sends each vector in B

[rare barn]

yes

[dusky epoch]

then write the results in B

i.e. as linear combinations of vectors in B

[rare barn]

i dont get why

[dusky epoch]

why what

that's what "matrix of T relative to B" means

[rare barn]

I still cxannkt figure out how this works

i dont see how they are getting this

[dusky epoch]

the hard way would be to write out a general linear combination of B and set it equal to each of the T(b_i) in turn

[rare barn]

but what are they doing here

i have no idea how they are gfetting those matrices

[dusky epoch]

those what?

[2, -1, 3, 1]^T, [0,0,3,1]^T and so on?

[rare barn]

yes

[dusky epoch]

that's just a matter of calculating the values of T at the basis vectors lol

[rare barn]

thats not what im getting though

[dusky epoch]

what are you getting

and might you be fucking up your arithmetic somewhere

[wintry steppe]

http://mathb.in/32578 can someone help on how to start ?

[earnest vessel]

What is L(E)? @wintry steppe

[slender yarrow]

set of endomorphisms E->E @earnest vessel

i'd answer the q if i didn't have to go to sleep

[wintry steppe]

Yes

[earnest vessel]

I would suggest looking at ker u

[wintry steppe]

@earnest vessel whan can I do with ker u there ?

[earnest vessel]

There is a useful criterion for linear maps that relates injectivity and the kernel

[wintry steppe]

ik

But I do not see how I can extract information abouter ker u in this case

[earnest vessel]

Okay, then let me phrase it differently

Suppose u were bijective

And take a nonzero element x in E

What can you say about u(x)?

[wintry steppe]

Ker(u) = {0}

Im(u) = E

[earnest vessel]

I mean about the element u(x)

[wintry steppe]

inversible ?

[earnest vessel]

What does invertible even mean for an element of a vector space?

No, you said ker u = 0

That is right

That means precisely that the only element that goes to 0 is 0 itself

In other words, the image of a nonzero element is a nonzero element

Is that clear to you?

[wintry steppe]

y

So if u^p is bijective E should be {0}

And I know that dime E > 0

So it's not possible to have u bijective

right ?

[earnest vessel]

Not really

What I meant is that u(u(x)) would also be nonzero

And so on

[wintry steppe]

true thx

[slow scroll]

Prove that if $A$ and $B$ are similar matrices, then $\det A = \det B$.

[stoic python]

kxrider:

[slow scroll]

im at my wits end on this one

[half ice]

Two matricies are similar if
PAP¯¹ = B

[slow scroll]

ahh wait, $\det B = \det (Q) \det(Q^{-1}) \det A$

[stoic python]

kxrider:

[half ice]

Yes yes

[slow scroll]

and $\det(I) = \det(QQ^{-1}) = 1$

[stoic python]

kxrider:

[slow scroll]

duh 🤦

[half ice]

Sure, also det(Q¯¹) = det(Q)¯¹

[slow scroll]

yes that would make sense

[half ice]

In general, det(Aⁿ) = det(A)ⁿ

[quaint heart]

Shitty alternative proof- similar matrices have the same eigenvalues

And determinant is the product of the eigenvalues

[slow scroll]

hmm haven't learned eigenvalues so can't use that anyway.

[quaint heart]

Ah okay

But yeah, determinant of product is product of determinant works

[rare barn]

here, im confused by Re z and Re w, isnt z and w nonreal ?

[wintry steppe]

a non real number refers to a number a + bi where b is not zero. a is not necessarily 0

[rare barn]

oh yeah 🤦

ty

i dont think that statement makes sense though, arent there possibilites where that matrix has real eigenvalues?

@wintry steppe

[winter reef]

in affine transformations, f(p)= f(p0) + f'(p0p) (f' linear part) is itr true for every vector that f' (a) - f'(b) = f'(a-b) ?

that doesnt work for just f, right?

[wintry steppe]

yeah idk zo, z = 1 + i, w = i are clearly nonreal, but clearly the matrix then has an eigenvalue 1 which is real.

besides, almost any real matrix satisfies this property!

[rare barn]

exactly

[dusky epoch]

why not take z arbitrary nonzero and w = iz, and see what comes of it

[wintry steppe]

Im not sure I understand what you are asking, dog, but if f'(x) is a linear function, then by definition f'(a)-f'(b) = f'(a-b)

oh yeah ann then u got a rotation matrix thats cool

I still dont really understand what they were trying to go for though

[quaint heart]

More generally if re(w) = im(z), then the matrix is symmetric and so has real eigenvalues

[slow scroll]

anyone understand why a basis is obtainable by finitely many column interchanges of the identity matrix? Obviously this works for a standard basis in R^n or something, but I am having trouble seeing this for an arbitrary basis.

[slow scroll]

Also part c of this question

I was stuck on the possibility that $P^N$ could just loop through a few permutations, never returning to the identity, but I came up with the idea that if we let $N>k$ then we can write $P^N = P^k$ to represent the situation where some finite composition of $P$ goes back to an earlier permutation, then I multiplied both sides by $P^{-k}$ to get
$P^{N-k} = I$.

[stoic python]

kxrider:

[slow scroll]

is this a valid proof? It seems a bit convoluted so I thought I'd ask if i forgot something. 🤷

[half ice]

@slow scroll
That is indeed a valid proof.

[slow scroll]

i have a new question lol....

[half ice]

Are you sure you didn't look it up? That mirrors the proof that all members of a group will return to identity after being multiplied by itself some number of times

Which is exactly what's happening with these permutation matricies

[slow scroll]

nope i have no clue what ur talking about

xd

[half ice]

Fair enough lel

[slow scroll]

completely stumped here. It has an even number of permutations because n! where n >= 2 always has 2 as a factor.

Odd/even permutations refer to the number of times two numbers are interchanged. I have no clue why half of the permutations would be odd.
https://cdn.discordapp.com/attachments/550859457308000276/563570955251744779/unknown.png

[half ice]

Hmm. Well, an even permutation has a determinant of 1, an odd permutation has a det of -1

[slow scroll]

ye

[half ice]

That's how you use the hint. How to solve that exactly half are even/odd...?

[slow scroll]

if A is a 9x9 matrix describing a permutation of (1,2, ...., 9) then det(A) = 1 if A represents an even permutation and det(A) = -1 if A represents an odd permutation. Thats as far as I've gotten lol

[half ice]

There's an even number of odd permutations

Is immediate

I'm not sure that's useful lel

[slow scroll]

how do you know that?

[half ice]

Two odd permutations, when multiplied, gives an even permutation. Multiply them all together, the result should be determinant = 1

There needs to be an even number of odd permutations, so (-1)ⁿ = 1

[slow scroll]

wait a second, det(A^n) where A is an odd permutation and n varies from 0 to Perm(9), alternates between 1 and -1. If it does that an even number of times, then half will be 1, and the other half will be -1. -1 corresponds to odd permutations so half of the permutations are odd

[full palm]

is this free?

[slow scroll]

is what free?

[full palm]

the chat

[slow scroll]

sure i guess. I've never paid a dime to use this chat

oh wait

🤦

Yea its free

[full palm]

lul.... maybe u could help me then. The set => {x( 1, 0 , 1) + x(1, 0, 0) + y(0, 1, 0) + z(0, 0, 1)} can be considered as R^3?

[slow scroll]

are x, y, and z just scalars or something?

[full palm]

they're equivalent to ( x, y, z)

[slow scroll]

not sure what you mean by that, but span{( 1, 0 , 1), (1, 0, 0), (0, 1, 0), (0, 0, 1)} = R^3

[full palm]

i'll post the whole question

were given two subspaces, W and U. Where U belongs to R^3 and is formed by (x,y,z) where z=x

and W same but with the rule x+y+z = 0

i have to prove that U+W = R^3

[slow scroll]

oh okay. Then yea, its just a matter of checking whether the span{( 1, 0 , 1), (1, 0, 0), (0, 1, 0), (0, 0, 1)} = R^3
which it does.

[full palm]

but how do i properly check it?

i know i'll have something like (2x, y, z+x)

which is a bit different than (x, y, z)

[slow scroll]

wait x + y + z = 0 and x=z for W?

[full palm]

no.... x+y+z= 0 for the W subspace

and x=z for the U subspace

[slow scroll]

(1, 0, 0), (0, 1, 0), and (0, 0, -2) form a basis for R^3

you can represent any vector in R^3 as a linear combination of those three vectors

[full palm]

thanks a lot for effort!

i just got a bit confused with that -2 and the - sign

[slow scroll]

i let x = 1, and y = 1, for my basis vectors, so x+y = 2. To satisfy x+y+z = 0, z = -(x+y), which is -2

[full palm]

got it. but that sounds different from the basis from R^3

and where went the (1, 0 ,1) vector?

[slow scroll]

you don't need it. its redundant. There are many bases for R^3. The conditions are that the vectors in the basis must span all of the space, and that none of the vectors are linear combinations of each other. I chose those 3 specific vectors for my explanation because it is easiest to see that those conditions are satisfied.

[full palm]

got it! that vector would be equivalent of having ( 1, 1, 0) right?

so if put that "2" on my test, teacher should still give me correct, right?

[slow scroll]

OH WAIT A SECOND big mistake

im a freaking idiot lmao one sec

[full palm]

no worries

[slow scroll]

This is our basis for W

hopefully you can see why,
x(1, 0, -1) + y(0, 1, -1) = (x, y, -(x+y)) for all x, y in R

We have 4 basis vectors to choose from now to make our new basis

[full palm]

got it

actually is a bit confusing

but i guess i'm just not that used to see things like this

[slow scroll]

we need to show that we can represent any vector in R^3 with these four vectors (or fewer)

@full palm Do you have a way of doing this? xd

[full palm]

i'm not sure

[slow scroll]

do you know about matrices?

[full palm]

i think so

could i break the two vectors on the W basis, to more vectors?

[slow scroll]

no. one sec

[full palm]

like for (0,1,-1) having: (0,1,0) and (0,0,-1)

[slow scroll]

okay so our goal is to show that the system
$$ \begin{bmatrix} 1\0\1 \end{bmatrix} x_1 + \begin{bmatrix} 0\1\0 \end{bmatrix} x_2 + \begin{bmatrix} 1\0\-1 \end{bmatrix} x_3 + \begin{bmatrix} 0\1\-1 \end{bmatrix} x_4 = \begin{bmatrix} a\b\c \end{bmatrix} $$ has solutions for all $a,b,c \in \bbR$.

[stoic python]

kxrider:

[slow scroll]

this is represented by the matrix $$ \begin{pmatrix} 1&0&1&0\0&1&0&1\1&0&-1&-1 \end{pmatrix} \begin{pmatrix} x_1\x_2\x_3\x_4 \end{pmatrix} = \begin{pmatrix} a\b\c \end{pmatrix}$$

[stoic python]

kxrider:

[slow scroll]

equation*

if we can show that the row echelon form of the matrix is consistent, (pivot in every row), then we will have proved that there will always be solutions, aka there is always a linear combination of columns that represents an arbitrary (a,b,c).

thats honestly the best I can come up with rn, other than saying that "these vectors are obviously not linear combinations of each other!"

[full palm]

and sir, you did a excellent job!

i tottaly got it

now

[slow scroll]

oh okay cool!

and if the vectors were linear combinations of each other, that would show up as a reduction in column pivots. Since there are 4 vectors here, at least one of the vectors is guaranteed to be a linear combination of the others, so there should not be a pivot in the last column in theory

[full palm]

got it!

[slow scroll]

sick

[full palm]

again, thank you so much for your effort!!! it meant a lot to me and helped to save a lot of time!

[slow scroll]

np glad i could help!

[full palm]

have a great night, thank you very much!

[slow scroll]

np u too

[sage mantle]

hello , do you perhaps know any good linear algebra 1 online courses that you'd recommend? i have a linear algebra 2 midterm next week but my basics are way too weak for me to understand linear algebra 2 correctly and my teacher didn't put any notes during the first semester

[winter reef]

MathTheBeautiful Channel on yt has a nice course

[sage mantle]

i'd prefer courses as in written since i don't have much time sadly

[broken hawk]

you're gonna have to tell us what Linear Algebra I entails

curricula are not standardized globally

[plain fjord]

Hmm, basis in R^3 can't have four vectors, it must be exactly 3 to span whole R^3. If there are 4, one is a linear combination of one of the others

Oh sorry, looks like someone already said that above.

Ah maybe the question was already answered and dealt with.

I just finished 2 linear algebra courses at my university, one started at Fall and one at January

[snow snow]

So I've got a parabola (y=2x^2)
I move it to the coordinates (2,-4)
How do I find the equation of the new parabola?

I'm unsure of how to find the equation of a parabola in general

[charred stirrup]

is nullity the dimension of null space?

[slow scroll]

yea

[charred stirrup]

ayee you always helping me

thank you

[slow scroll]

np np :p

[charred stirrup]

the nullity is basically the number of vectors that DONT form the basis?

[slow scroll]

... for the column space, yes

[charred stirrup]

column space = span of column vectors

is there such thing as a nullity for the row space?

[slow scroll]

its called the left null space

there is no secret for figuring it out like the row space tho. You just have to solve A^Tx=0

[charred stirrup]

I thought row space is the span of the row vectors?

[slow scroll]

it is, but Col(A) = Col(A^T).
I was just saying that there is nothing analogous for null spaces

[charred stirrup]

Ahh, I see

thank you

[slow scroll]

np

[charred stirrup]

for column space, say I have a 3x5 matrix, what exactly is an example of the column space? - I get that it's a set of linear combination, but how would one calculate one possible linear combination of that set?

[slow scroll]

The column space is the set of all values that the variable of b can take on in Ax=b

If you wanted one possible linear combination, just plug in a vector for x i guess

[charred stirrup]

could you do a practical example? what is one possible linear combination for this matrix?

i'll probably get it then