linear-algebra

[lavish jewel]

and ofc, the row space is the ortho complement of the null space

null space is another name for Ker A

[keen coyote]

but since it's least square doesn't b not equal Ax

so then you can't write b as something in the range of A with svd

[strong bison]

Is there a fast way to solve this big matrix?

[keen coyote]

nvm nvm

[wary lily]

yes, using CAS

[lavish jewel]

yeah, you only get the component of b that was in the column space of A

the other one gets mapped to the 0 vector

you need to specify b = bs + bn and x = xs + xn

some of those get mapped to 0, others not

[dire thunder]

wym by "solve matrix"

[bitter shuttle]

What is the projection of some vector v on a unit vector u

How come it’s not just they inner product multiply by u

[lavish jewel]

wdym?

sounds to me like it should be

[wary lily]

|u| cos(theta) = (u . v)/|v|

that's the "shadow" of u on v

the magnitude

that's multiplied with v since v is unit already

in case it isn't unit, you divide it by |v|

which result in this formula:

[lavish jewel]

that's the same as they wrote up there, just with u swapped with v

apparently they ran into some case in which (v dot u) u with unit u didn't work?

[strong bison]

what's CAS?

[strong bison]

Gaus Jordan Elimination to get S_0, S_1, etc expressed in variables

[lavish jewel]

RREF it using your favorite software

otherwise, the rows actually look kinda friendly

you can start from the second to last equation parametrizing stuff and back-substituting

[strong bison]

RREF online calculator shows this

as final answer

but doesn't make sense does it?

[lavish jewel]

did you augment the matrix?

[wary lily]

nah, it's square

[lavish jewel]

otherwise, at least you know for sure the problem has a solution

[wary lily]

yeah, unique

but, there may be some catches

[lavish jewel]

you have 8 rows and 7 cols, is that already augmented?

or did i read somth wrong

[wary lily]

first row appears to be headers

[lavish jewel]

oh oops

so 7 eqs 6 vars

[strong bison]

yes sir

[lavish jewel]

this kinda seems to say there is no solution, then?

[strong bison]

it does

[lavish jewel]

i'm hypoglycemic at the moment, so if anyone has more insight than i do, feel free to correct me

[wary lily]

no, the rref has a 0... 1 row

that's definitely inconsistent

as you said

[lavish jewel]

that's what it looks like to me lol

[strong bison]

hmmm might have to take a step back and check the equations i made the matrix with

[wary lily]

I was wondering tho, when a system has n variables and n+1 equations, for it to have a solution, at least one row needs to be a multiple of another, right?

then that row doesn't apply any constraints and we get down to n euqations.

[lavish jewel]

either a multiple or in general a linear comb., yes

[wary lily]

Ahh, ok

thanks

[lavish jewel]

that's pretty much what the definition of "consistent" means

[dire thunder]

just think about linear eqs in terms of dimensions of domain and codomain of linear map

[waxen flume]

https://i.imgur.com/lvqTpc4.png

For part b, if the set of 3 vectors are linearly independent then they form a basis

does that make it sometimes or always?

[lavish jewel]

are vectors always lin indep?

[waxen flume]

no.

[lavish jewel]

sometimes?

[waxen flume]

yeah

[lavish jewel]

so, sometimes 😛

[waxen flume]

thanks @lavish jewel

[lavish jewel]

no prob

[solid flower]

is $n=4$ in both $z^4=(re^{i\theta})^4$ and $z^{\frac{1}{4}}=(re^{i\theta})^{\frac{1}{4}}$?

[dusky epoch]

why are you raising theta to the 4th power

[solid flower]

sorry im trying to raise all of re^itheta to the 4th

im just new to latex lol

[dusky epoch]

$(re^{i\theta})^4$

[stoic python]

Ann

[dusky epoch]

also to fix the error you need to write \& and not just &

[solid flower]

ah

[lavish jewel]

what is n supposed to be?

[solid flower]

the number of roots

[stoic python]

ahmad_11

[solid flower]

there we go , thx

any ideas?

maybe im just thinking of it the wrong way

basically the question is if you have something like find all the roots of z^4=16, then yea n=4 and theres 4 roots but if its something like (1+sqrt(3)i)^1/2 why are there 2 roots to that

[dusky epoch]

every complex number other than 0 has two square roots

it's an immediate consequence of fta

[solid flower]

right

[lavish jewel]

maybe a substitution would help make it more clear?

say you have z^4 = w

the solutions of that are the z = 4throot(w)

[solid flower]

right

that makes sense

and if you have z^1/4=w then the solutions to that are z=w^4

[lavish jewel]

i guess doing it backwards requires you to be careful

if you are already given one of the roots for that, there is only one way to raise that to the 4th power

but you have to be aware there are 3 other roots that will give the same result

since the two problems are linked

so it depends on whether you start with z or with w how you go about it

[solid flower]

hmm yea

i dont like this stuff

[lavish jewel]

why haha

[solid flower]

idk probably because i have a terrible foundation in radicals

from h.s

[lavish jewel]

this is your chance to make up for it

[solid flower]

you're right

the more i look at it

i start seeing different things about it

[lavish jewel]

that's the spirit

my personal experience has always been pretty similar. find a cool new topic, realize i don't know anything or i had a ton of misconceptions, cry myself to sleep, start understanding the stuff, enjoy the profits

i guess sometimes that's just how it goes

[solid flower]

lol yup

are you studying or have studied math in school too

[lavish jewel]

i'm a lowly engineer

doing a phd, which probably puts my math level somewhere around like 1st or 2nd year undergrad mathematician at best

[solid flower]

oh wow whats ur phd in?

im in first year electrical eng

[lavish jewel]

signal processing

[solid flower]

xD nice

[wheat prairie]

how did the multiplication got split into a summation of two products here?

[mortal juniper]

is this true or false

[lavish jewel]

@wheat prairie which part do you mean?

[wheat prairie]

i just don't know how the first simplification was done

[lavish jewel]

$\hat{\Phi} w = \begin{bmatrix} \Phi w \\ \sqrt{\lambda}I_m w \end{bmatrix}$, yeah?

[stoic python]

Edd

[lavish jewel]

if you subtract $\hat{y}$ from this you end up with $\begin{bmatrix} \Phi w - y\\ \sqrt{\lambda}I_m w \end{bmatrix}$

[stoic python]

Edd

[lavish jewel]

now you multiply this from the left by its transpose and you're done

ah, with a factor of 1/2, apparently

@mortal juniper i cannot give you the answer, but i can help you think of a special case. what happens if the rank of the matrix A is r < n < m?

[mortal juniper]

in that case we cannot cross multiply the matrices right?

[lavish jewel]

wdym cross multiply

[mortal juniper]

dot product my bad

it would be trivial right

[lavish jewel]

dot product?

[mortal juniper]

yes

At . Ax

?

[lavish jewel]

that's not a dot product tho, if A is a matrix

[dusky epoch]

@brazen venture we don't speak mandarin here, sorry

[brazen venture]

sry i was explaining

[solid flower]

mud?

[dusky epoch]

you said something that began with cao ni ma which to my knowledge is the chinese equivalent of "fuck your mom"

[lavish jewel]

maybe they were explaining how to do that

[mortal juniper]

oh i c thats a trivial solution

becos it can contain a 0 vector

[lavish jewel]

i don't understand what you mean

[mortal juniper]

sry if im dumb 😦

zero matrix i mean

[little crater]

Not sure if this counts as linear algebra but if I have 3 vectors and I know they share some plane, how does one find that plane?

[tulip glacier]

is null space equivalent to trivial solution?

[mortal juniper]

you can use the Gram-Schmidt process

[lavish jewel]

@tulip glacier nope

that's kinda cursed, you don't need to go that far

bringing a shotgun to a knife fight

[tulip glacier]

huh cursed?

[lavish jewel]

i was answering poropichu and brandon

[little crater]

what is Gram-Schmidt process? sorry this was just something from a calc course so it might be out of my league

[lavish jewel]

just make or take 2 lin indep vectors

and take their cross product

this gives you the plane's normal

[tulip glacier]

lol

[little crater]

I was told to find the determinant of <1,2,3>,<3,4,5>,<6,7,9>

[tulip glacier]

isnt both null space and trivial solution solving for Ax=0?

[mortal juniper]

just wondering if after gram-schmidt process, can {v 1, v2, v3, v4}

​ be normalized to become an orthonormal set?

[native rampart]

Yes

[mortal juniper]

how do i show the proof for it

[little crater]

You all are speaking in tongue to me ;V

is this something you learn in linear algebra though?

[wintry steppe]

What do you wanna prove?

[mortal juniper]

becos i cant really show my prof how to prove this statement

[lavish jewel]

the trivial solution always exists for that. the null space always contains the 0 vector. it can contain other stuff too, though, depending on the matrix

[wheat prairie]

but then if we transposed the column vector where does the transpose come from. I mean it is imp for dimensions to match but I don't follow the math

[little crater]

the whole gram-schmidt process

[mortal juniper]

its on google

[little crater]

yeah ik but i was just curious if you learn that in linear algebra

[tulip glacier]

lol my school didnt even bother to differentiate them for me...

[lavish jewel]

oh, you meant you didn't know where any of it came from

[wheat prairie]

yeah sorry xD

[lavish jewel]

the 2 norm squared of a vector can be expressed as x^T x

you can see that carrying that product out will give you the sum of the squared elements of x

so if x = phi w - y, it gives you the sum of squared differences

if you minimize that, that is the "least squares solution"

[wheat prairie]

oh no i know this part...

[lavish jewel]

the solution that minimizes the sum of squares of the differences

so which transpose did you mean?

[wheat prairie]

like i know how the summation of square differences is transformed to matrix

i just don't understand how to deal with vector of matrices like here

[lavish jewel]

it's the same thing

phi w, even tho phi is a matrix, yields a vector

since it's multiplied by w

[wheat prairie]

you mentioned multiplying this by its transpose, the simplificatoin of this confuses me

sorry if my question was unclear

[lavish jewel]

if you pay close attention, that vector has another vector as the first element, and yet another vector as the second element

so writing it like that really means it's a really long vector

or a "block vector", as some call it

alternatively, if you treat it as if it were a vector containing two elements u and v, then you can do x^T x = u^Tu + v^T v

where u and v are the stuff up there

you know, phi w and sqrt(lambda) I w

[wheat prairie]

so sort of instead of multiplying the whole thing we split it up

[lavish jewel]

i mean, the whole thing IS multiplied

just like you would normally multiply any x^T x

[wheat prairie]

yeah i just mean since this is basically like the dot product you can split it up

[lavish jewel]

first element squared plus second element squared, etc

[wheat prairie]

yeah exactly and here in the simplifcation it is two elements squared, namely phi*w -y and sqrt(lambda)I * w

right?

i think i got it now, thanks!

[lavish jewel]

aight

[ember shard]

What does onto and one to one mean?

I kinda of know some of their characteristics, but I'm not entirely sure what it means for a matrix to be either

[hazy sparrow]

3blue1brown spoiled me

linear algebra is just linear transformations of space

all the visual intuitions make so much sense

but now every other video that i watch lacks the visuals and just show the steps

[wintry steppe]

how can i proof something is a dot product?

[lavish jewel]

onto or surjective would mean the columns of the matrix span all of the output space R^m

[dusky epoch]

the verb is "to prove", and you prove that it satisfies the definition of an inner product

[lavish jewel]

one to one would imply the matrix is square and full rank

[dusky epoch]

namely, you need to show that your "something" is a function which takes two vectors as input and returns a number as output and satisfies the following:

1. linearity in the first argument: <ax + by, z> = a<x,z> + b<y,z>
2. symmetry: <x,y> = <y,x>
3. positive-definiteness: <x,x> ≥ 0 for all x, & is 0 only when x itself is 0

[wintry steppe]

ye, prove. But i mean, i have here the conditions it must satisfy, but idk how to start

https://gyazo.com/7be81be087e45ffcc2f5f93e3a9a873a

[ember shard]

Is there a way to visualize this?

[wintry steppe]

first condition is < af + bg, z > = a < f, z > + b < g, z >

[dusky epoch]

yeah that's linearity

also uh... if you're writing in latex please use \langle and \rangle

@wintry steppe linearity follows directly from the fact that the derivative and integral are both linear

[lavish jewel]

i have no good way of showing this, because a matrix maps (some part of) R^n to R^m

[wintry steppe]

ah because integral (a * something) = a * integral (something)?

[dusky epoch]

that's what i said yes...

[wintry steppe]

can i do the same even if i have a sum on the first part of the dot product?

[hazy sparrow]

can someone help me understand the idea behind the inverse matrix? i understand it sort of "undos" a matrix, but i don't know how it ties in with the idea of a determinant and span

i tried watching videos but they just recite formulas and don't show what is happening in linear space

[dusky epoch]

im not sure what you mean

are you under the impression that you're not allowed to use the linearity of the integral when there are other things in the formula even if those things don't affect it?

[wintry steppe]

no. i mean, i know the constant can exit the integral. from on my dot product <x,y>, for the linearity, x become a*f + b*g

do i just need to substitute?

[lavish jewel]

pretty much

you substitute stuff in and check that the properties hold

[wintry steppe]

okey. now, for the conjugated one

if a is a constant

what is a conjugated?

or am i missing something?

https://gyazo.com/76bd034f6234b4e4fa8ebd91007c8741

[lavish jewel]

are you working with complex numbers?

[wintry steppe]

no

son on R is the same, right?

[lavish jewel]

then the complex conjugate of a real number is the same real number

[wintry steppe]

ok so this is extended for C?

[lavish jewel]

mhm

[wintry steppe]

forget it, i understand myself xd

like, it is written with the conjugate cuz it isnt restricted to R

so can i just say this will be satisfied as well? like the left part?

assuming we are working on R

[dusky epoch]

i mean, you could prove your thing is symmetric

or you could go through the same legwork twice for linearity in both arguments

if you wanted

[wintry steppe]

that is because multiplication is (?) like a * b = b * a

[dusky epoch]

commutative

[wintry steppe]

so to prove the left side, i did (a * f + b * g) * z * t^2 bla bla bla

[dusky epoch]

.-.

[wintry steppe]

here i will have the same but swaped

sorry idk how to write with the bot xD

[dusky epoch]

that isnt the issue

the issue is that you're signing yourself up to do all the same work all over again

[wintry steppe]

🤷‍♀️

[lavish jewel]

for reference, their username is "i don't know what's going on" 😛

[wintry steppe]

and that is relevant because..........? xd

what ever. The next thing to prove is... idk the name in english. "hermiticity"? i dont know. <x, y> = conjugate <y, x>

and again, if i am not working with complex numbers, it should be the same, right?

[lavish jewel]

that's what ann said earlier about symmetry

[wintry steppe]

yes, but i dont know how to prove it is symetric. The reason it is is because the product is conmutative, isnt it?

[lavish jewel]

it is called symmetry for the reals, conjugate symmetry/hermitian symmetry for the complex

over the reals, yes, it has to do with multiplication being commutative

[wintry steppe]

okey okey

and the last thing is to prove it is positive defined

but idk. First i though about the sign of the functions. t^2 and t^4 are always positive, but f and g could be positive and negative, so the product remains negative

can u give me a hint?

[lavish jewel]

read carefully. positive definiteness is for g = f

[wintry steppe]

ah lol

true

[dusky epoch]

this really feels like making a mountain out of a molehill

all these properties are like... obvious for the most part

[wintry steppe]

just saying. maybe things u dont know are obvious for others. Stop saying these trash comment that aport nothing to a help channel

[lavish jewel]

easy there, ann did give you all the tools and definitions all the way up there. i have only been re-reading that to you

just keep working on your stuff at your pace

[wintry steppe]

he/she didnt gave me anything ^^'

[dusky epoch]

if you dont know someone's pronouns you can say "they", just saying

though i am a she

and yeah i gave you the definitions (and even the names!) of the 3 properties you need to prove

[wintry steppe]

i dont care what u are tbh

[dusky epoch]

😒

mkay.......

i mean look

if you look at your formula

then swapping f and g around will not change anything

hows this not obvious

[wintry steppe]

this is the only name u gave me. And if u are so clever, u have may notice eng is not my main language, so i dont study math on english, so i dont know the names

[dusky epoch]

you ignored this then?

[wintry steppe]

actually yes, i missed that comment, sorry

[dusky epoch]

and hows it not obvious that symmetry + linearity in 1st argument automatically implies linearity in 2nd argument too

[lavish jewel]

let's just drop it here, all right?

everyone have a nice day

[dusky epoch]

meh

[nocturne oracle]

dont tell me what to do @lavish jewel

[lavish jewel]

ok, have a bad day

my bad

[glossy parcel]

🤬

i was having a good day until now

[north steeple]

this would be the bassi right
{[1 0 1], [1 1 1]}

[native rampart]

That's a basis

Yes

[lavish jewel]

looks ok, yea

[north steeple]

ok just wanted to make sure

[native rampart]

How did you get that basis?

[lavish jewel]

i'm also curious, because (1,0,1), (0,1,0) seems a lot easier to find 😛

[hazy sparrow]

found this interesting proof of inverse matrices

[native rampart]

That's kinda the normal proof

[dusky epoch]

a proof by explicit computation

[hazy sparrow]

what is going on here?

[storm totem]

need some help can anyone do

[lavish jewel]

multiply left and right by one of the values a,b,c, or d @hazy sparrow

[vocal spade]

I hope this is the correct channel for this
could I get some help with this one? https://i.imgur.com/wo0wBjy.png

in a) I'm thinking of just reducing it to 3D rotations. So since phi1 and phi2 are not on the same line, we can write v1=phi2-<phi1,phi2>*phi1 and use that to form a basis for H by phi1,v1,v2,...,v_n (where all v_i for i>1 are orthogonal)
Then let the unitary operators V_i be rotations of vectors in the 3D subspace spanned by phi1,v_{i-2},v_{i-1} to the 2D subspace spanned by phi1,v_{i-2} st. taking V_1*V_2*...*V_{n-2}=V \in G_1 would rotate the vector to the 2D subspace spanned by phi1, v1. Then have some operator in G_2 which does the last rotation to phi1. I think that should all work, but I'm not sure about 3b) and 4)

wait fuck, I'll try to fix the discord formating

in b) I'm thinking I could use that the 3D rotation group is isomorphic to SU(2), but I'm not sure how

[sonic osprey]

Aren't there usually going to be more than two such G_i?

[vocal spade]

Well I can just multiply the unitary operators to get new unitary operators, so I should be able to reduce it to two G_i s, I think

[hazy sparrow]

so i get up to this part

but the line of reasoning fails for me after this

[true kraken]

Hey i have a question, how would i solve this using geometric description?

[lost ermine]

anyone know how to do dis?

[wintry steppe]

For question (iv), why would we do what the hint says and take Av dot Aw?

Why not just see if v dot w = 0?

[sonic osprey]

how are you going to calculate v dot w?

[wintry steppe]

Oh my bad..

[sonic osprey]

The problem is that you don't really know much about v or w

the only thing you know is that they're eigenvectors for some eigenvalues

which is why you have to bring A into the picture, because you don't know much about v,w themselves

[wintry steppe]

I'm not too sure how to solve it then

Would it still be in the form Av dot Aw = 0? Or do we have to do something strange with a transpose?

[stable kindle]

stupid question, can you not find the eigenvectors and eigenvalues explicitly

[sonic osprey]

It tells you to look at Av dot w

Something you might have covered is that Av dot w is equal to v dot A^Tw

but you know that A = A^T

[wintry steppe]

I didn't remember this

[sonic osprey]

idk check your notes

this is what they want you to use I'm like 90% sure

[stable kindle]

no but actually can you not just find the eigenvectors explicitly

[sonic osprey]

I mean yeah you could

[wintry steppe]

The book I am using is Linear Algebra and It's Applications

I just tried looking through the paragraphs I had to cover and I couldn't find Av dot w = v dot A^Tw

[sonic osprey]

did you define the dot product as

v dot w = v^T w

[wintry steppe]

That's how it's written in the book, yeah

[sonic osprey]

Then yeah

Just use that

to see that Av dot w = v dot A^Tw

[wintry steppe]

I don't really know how that justifies this

It has nothing to do with matrices does it?

[sonic osprey]

Not really

just apply the definition

Av dot w = (Av)^Tw = v^T A^Tw = A^Tw dot v

[wintry steppe]

Jesus

[sonic osprey]

?

[wintry steppe]

I just realised that I would not be able to answer this kind of question

A bit worrying

[sonic osprey]

I mean, you just follow your nose

They tell you to look at the inner product of Av and w

so you write out the definition

that's (Av)^Tw

Then you apply your transpose rules to expand this out

v^T A^Tw

[wintry steppe]

I didn't even process Av as a vector to be honest, rather as two distinct entities

[sonic osprey]

Uh

That's what matrices do?

They're a function that takes in vectors and outputs vectors

[wintry steppe]

Yeah, I understand

Not questioning it

Just, didn't come up in my mind when looking at this equation

[marble lance]

Well, you can only apply the dot product on vectors. So if you didn't realize that, that's a lil worrying. But nbd if you just started working with inner products.

[sonic osprey]

I mean, you can only take the dot product of vectors of the same length, so Av dot w can only make sense if Av is a vector

[marble lance]

Sniped

[stable kindle]

um

[nocturne jewel]

<u,u> only makes sense \s

[solid flower]

are these are equal $$ -2c_1e^{-x}+c_2e^{5x} = c_1e^{5x}-2c_2e^{-x}$$ , i got a different answer using a different arrangement of $\lambda$ and $\vec{x}$ in my D and P

[stoic python]

ahmad_11

[solid flower]

would they be equal only if c_1 and c_2 are the same?

[wintry steppe]

does this look correct?

[stable kindle]

WLOG they're equal

[wintry steppe]

mn -xy = 0 therefore parallel

[stable kindle]

you can just rename c1 as c2 and c2 as c1

[solid flower]

WLOG?

[solid flower]

ah

[stable kindle]

without loss of generality

ie. 'it doesn't matter which way round they are'

[solid flower]

ah

[wintry steppe]

W-what does that mean @wintry steppe

Please don't bully me...

[wintry steppe]

Is the range of a matrix A an orthogonal basis, by definition?
I am asking this because I am not sure if we are always allowed to use the "Orthogonal Decomposition Theorem" find the projection of a vectory y onto R(A)

Does it expect me to first find a basis for R(A), check if the basis is an orthogonal basis, and then apply the "Orthogonal Decomposition Theorem"? Or is R(A) by definition an orthogonal basis, allowing me to skip that check?

[wintry steppe]

Sorry but we skipped over the vector space chapter :(

So I'm not sure what that entails. Could you elaborate?

[lavish jewel]

i somehow doubt that is true

why would you be looking at bases, then?

[wintry steppe]

We are following Linear Algebra and its Applications by Lay et al.

And Chapter 4 is about vector spaces

3 is about determinants

As you can see on the schedule, we skipped both of those

[lavish jewel]

what is your definition for a basis?

[wintry steppe]

I only know how to get determinants under certain conditions, like with 2x2 matrices with the method in eigenvalues

I only know a basis from the subspaces paragraph

Where it is a linearly independent set that spans a certain subspace

Could you please help me with the interpretation of that question?

[lavish jewel]

i mean... what they told you is exactly it

it is a vector space

in general, a subspace of a vector space

[wintry steppe]

Do my questions not make sense?

[lavish jewel]

they do, it's just that

i have no idea how else to put it haha

you kinda missed what linear algebra is all about

what doesn't make sense is your course

[wintry steppe]

This is the first year this lecturer is giving the course

And it's really confusing to be honest, he even apologised about it

So I am somewhat desperate for assistance elsewhere

[lavish jewel]

i would recommend you go to the chapter where they explain what a vector space is, or ask the guy to explain it

[wintry steppe]

His lectures are also very difficult to understand, come across as trains of thoughts rather than presentations to other people

There is no real time anymore for that, sadly

[lavish jewel]

because every other definition from that point on will include vector spaces and subspaces in some way or another

the range of a matrix is the subspace spanned by its columns

the set of all the linear combinations of its columns

there isn't much more to say

[wintry steppe]

Right, and the range is also the column space

But I am just wondering

Can I use the "Orthogonal Decomposition Theorem" on a R(A)?

Because the orthogonal decomposition theorem talks about an orthogonal basis

[lavish jewel]

you would need to find an orthogonal basis for the subspace first

[wintry steppe]

I'm not sure if the basis for the range of a matrix is always orthogonal

[lavish jewel]

there is no "the" basis

there are infinitely many

[wintry steppe]

Okay, I am not sure if any basis of a range of a matrix is always orthogonal

[lavish jewel]

no

[wintry steppe]

Or if I need to specifically look for one

[lavish jewel]

any set with enough linearly independent vectors is a basis of some subspace

you can take the columns and gram-schmidt them

[wintry steppe]

Right, I would have to apply the gramm-schmidt method to be certain that it is an orthogonal basis

So you think that is a necessary check for a question like this one?

https://cdn.discordapp.com/attachments/540211747613704221/823668292333994034/unknown.png

[lavish jewel]

it is

[wintry steppe]

Thank you for helping me by the way, it's such a relief...

Would you be willing to look at the questions I have posted in #help-9 ? Because this is a combination with two other questions

[lavish jewel]

i don't have the time rn, i gotta go sleep

[wintry steppe]

Maybe the context will make clear to you what is expected, since I am a bit confused still about the other ones

Oh, okay.

Sleep well, and thank you for the help

[wintry steppe]

whats wrong here?

[hallow cliff]

2 looks true to me

and idk about 3

looks false

[wintry steppe]

could u pls explain why 2 is true?

[hallow cliff]

i may have foiled wrong but

[wintry steppe]

no you were right, ty

[hallow cliff]

you get $I_nI_n + I_nA^{-1} + AI_n + AA^{-1} = I_n + A^{-1} + A + I_n = 2I_n + A + A^{-1}$

[stoic python]

magnusChadson

[wintry steppe]

So anything multiplied by $I_n$ is just itself?

[stoic python]

request new nickname

[hallow cliff]

yes

[wintry steppe]

i see, ty!

[hallow cliff]

np

[lost ermine]

really need help with this

anyone plz

[dusky epoch]

what's Gamma?

the field over which everything happens?

@lost ermine do you still need help or did you figure it out

[wary lily]

Let $A$ be an $n \times n$ matrix. Is there any relation between $det(A)$ and $det(A^{-1}$?

[stoic python]

az

[dusky epoch]

yeah theyre reciprocals

assuming A is invertible

[wary lily]

like $\det(A) = \frac{1}{\det(A^{-1})}$?

[dusky epoch]

yes

btw \det exists

[wary lily]

O, thanks

[stoic python]

az

[wary lily]

[wary lily]

I can deduce that U is in triangular form.

[dusky epoch]

no you can't

nobody said U must be triangular

[wary lily]

then it's transform flips the upper to lower and vice versa?

[dusky epoch]

transpose

[wary lily]

transpose

[dusky epoch]

but you don't know that U is triangular

[wary lily]

i meant

so, U^t must be equal to U^(-1)

[dusky epoch]

yeah sure

U^T = U^-1

[wary lily]

O

[dusky epoch]

what

i just parroted what you said back at you

[wary lily]

I think I have the idea

I know, you confirmed

but I have an idea now

let me formulate it

[stoic python]

az

[wary lily]

@dusky epoch

[dusky epoch]

verbose but correct

[wary lily]

thanks

[wintry steppe]

Could someone tell me what this is based on?

[lavish jewel]

what about it troubles you?

[wintry steppe]

I'm just not sure which rule allows for this

[dusky epoch]

$y^TAx = (A^Ty)^Tx$

[lavish jewel]

if that's a matrix and vectors over R^n and R^m, the argument for why that holds is pretty easy. the transpose of a scalar is the same scalar

[stoic python]

Ann

[wintry steppe]

the transpose of a scalar is the same scalar(edited)
Sure, so then we can say
<Ax , y> = <A^T x , y>

[lavish jewel]

no, that's cursed

you don't even know if you can multiply A^T by x

[wintry steppe]

The transpose of a scalar is the same scalar?

[lavish jewel]

yes, but A is a matrix

[wintry steppe]

Okay let me give some context

[lavish jewel]

look at what Ann wrote instead, since my argument seems to be confusing for you

[wintry steppe]

In the question they state
A = A^T

[lavish jewel]

ok, but you hand't told us that haha

[wintry steppe]

And that we have to use Av = w to show that v and w are orthogonal

Yeah sorry

[lavish jewel]

you didn't mention any of this before

but still, look at what ann wrote up there

[wintry steppe]

But every time I give long contexts I tend to not get responses, so I try to narrow it down

Which is my fault, sorry

[wary lily]

maybe find a middle ground for context

A = A^T is profound

[lavish jewel]

so your A is a symmetric matrix, meaning it is diagonalizable. is A also full rank?

(invertible)

[wintry steppe]

With full rank you mean that there is a pivot in every column (that the dimension of R(A) = n?)

Then yes

I think so

This is the context

[lavish jewel]

aha

[wintry steppe]

(iii) would be -1

Then we need to carry the fact that A = A^T to (iv)

We start out with Av dot w

I saw this explanation. I get all steps except for the middle one
<Ax, y> = <x, A^T y>

[lavish jewel]

that was crucial because i was planning on directly using that result, but i see now they have asked you to prove it

[wintry steppe]

Sorry for not giving full context, hopefully it didn't throw you off

[lavish jewel]

here is how it goes

we will use exactly what ann wrote up there

gimme one sec to sketch it cuz i'm too lazy to latex right now

[wintry steppe]

Thank you :>

[lavish jewel]

oof office decided to start updating, i guess i'll do it in paint

something like that

oh oops, typo

[wintry steppe]

I can follow most of this, but I just don't know why
<Av, w> = <v, A^Tw>
Is this a transpose property?
Is this a inner product property?
I can't find it

[lavish jewel]

that's kinda straightforward

it's what ann wrote

it's a consequence of (A^T)^T = A

[wintry steppe]

I don't know how what Ann wrote explains it, she starts with a transpose of a vector

[wintry steppe]

OHH

Thank you so much

[open pivot]

Hey I don't know if this is the right channel but I have been stuck on this for a bit I'm not sure how to get to the right answer

This is my working so far

I think there has to be something wrong with my method because d isnt supposed to be a vector obviously

I think I understand how the method works but I was just using the formulas, deriving A and b from q(x) to get p and then d

[hazy sparrow]

does anyone have any good resources for self-teaching linear algebra?

[novel hamlet]

If you look for starting id suggest youtube 3brown1blue he explain basics well

[tame mural]

IMO you need a book, and 3b1b says so too

His videos are just for intuition

In sum his videos might be maybe 2 hours, maybe even less

No video course will be that successful in cramming a linear algebra course into 2 hours

[hazy sparrow]

yeah i watched the entire 3blue1brown series 😄

[novel hamlet]

Yeah, but he explain the intuitition on the topic quite nicely

[hazy sparrow]

i feel like he gave a good intuition to help dive deeper into it

[novel hamlet]

Tbh i still dont know what I do in real life with how to calculate null and column space for a matrix

[hazy sparrow]

but yeah i felt like i shoved a semesters worth of linear algebra in 2 hours

[hazy sparrow]

which book would you recommend?

[zealous junco]

im confused of the last 2 lines, i thought the proof was over when they proved R = I which is when they said "last row of R cannot be 0". So i don't see the point of the text after that.

so as A~I then A must be invertible

[nocturne jewel]

The line above the E= is just them rephrasing/simplifying what they have to prove

not the proof itself

"We wish to show that R=I. That amounts..."

[zealous junco]

yea but once u showed R = I, then since R = PA then A~I and we are done right

[slate siren]

how would you begin to do this problem

I thought something along the lines finding the span of both linear combinations in terms of x,y,z

and these spans are the same equations for x,y,z but im not sure if thats enough and I don't understand why it works which means i cant justify

[nocturne jewel]

if (1,4,-2) is on the first line, then you're done

since the direction vectors are the same upto a scalar

[slate siren]

ah I see

so they are the same line if 1,4,-2 lies on the first

I understand

yeah I noticed that they had the same direction vector but couldn't figure out how to use that

seems so obvious now

[nocturne jewel]

they're the same line if they have the same direction vector and start at points on the same line

[tame mural]

I'd try Gilbert Strang's book because he has a lecture series and you can pair it together. I'd also recommend any book which is popular enough to have lots of online discussion you can Google.

[hazy sparrow]

thanks 🙂

[novel hamlet]

I know the intuititiioin on this one but i dont know how to prove it

A and B are 2x2 block matrices

and multiplication is defined for all A_jk and B_ki

[dire thunder]

@novel hamlet you can just do from definition

[novel hamlet]

i need to prove it,

[dire thunder]

yes you prove this from definition

i mean expand A and B's as usual matrices

[novel hamlet]

i know how matrix multiplication is defined and i tried it but it became quite fast hell of down notifications

[dire thunder]

@novel hamlet look

[novel hamlet]

i ended up starting this way

[dire thunder]

i will try myself a bit later

need to do tea

but you may start from A, B as usual matrices

and then collapse to 2x2

or vice versa

[dire thunder]

show that this can be transformed to AB from usual matrices

hi drake

how are you doing

or even simpler approach

just look at values of c_ij where C=AB

[novel hamlet]

quick scetch tells me the B needs to be as vide as A is tall

and then same applies to the inner partitions

wich would lead them being square matrices

[native rampart]

I guess I am not having a bad day

[dire thunder]

your matrices looks too big

you should have two 4x4 matrcies

[native rampart]

Yea,Just find the entry in ith row and jth row in AB

[novel hamlet]

there is 2 4x4 block matrixes

[dire thunder]

[native rampart]

Compare it with the i,j th entry you get from that block matrix product

Should be same

[novel hamlet]

ah yes, had typo theere

[dire thunder]

anyway, do you see intuitively why it is true?

[novel hamlet]

yes i see it, i just dont know how to write it mathematically

it is quite logical its just good old substitution trick

[dire thunder]

if you want you can just write down all explicitly and do cumbersome computation

but what i would suggest you is just to argue about values of c_ij

[novel hamlet]

im not sure if it is sufficient to define A_iJ and B_ij have multiplication defined therefore i can substitute matrix A_ij with X and B_ij for Y and do standard matrix multiplication

[little crater]

can someone clear up what this means. |A| != 0 where A is a matrix

[dire thunder]

determinant ig

[little crater]

|A| means what again?

[dire thunder]

wym

[dire thunder]

determinant

[little crater]

okay

okay so what is the difference between cross products and determinants exactly

or is it that you just use determinants to solve cross products

[dire thunder]

[novel hamlet]

lets say i have these matrixes and i want to multiplicate them

[dire thunder]

sorry wrong emoji

man

[dire thunder]

i am suggesting you here

[novel hamlet]

i decide that A upper left corner can be substituted as X and B upper left corner Y and so on and get then 2 4x4 matrixes with substituted values

[dire thunder]

write explicitly the value of (i.j) entry

in terms of entries of A and B

[tulip glacier]

um guys i managed to solve the first part of this question but im curious why when S becomes the solution space here, i need to transpose it?

[plain saffron]

From

#❓how-to-get-help

Rule 3

Stick to one channel and don't post the same question in multiple channels. Please don't ask for help in other channels if no one is responding in the one you have posted your question in.

[wary lily]

are there any benefits to using Cramer's rule?

[native rampart]

No

It's less efficient than Gaussian elimination

and is more resource heavy

[wary lily]

why is it taught then?

useful later or in some other context?

[native rampart]

idk

I guess useful in some very obscure cases

[wary lily]

your honesty and straightforwardness is a public service

[limber sierra]

its useful in some proofs

not really computationally

the wikipedia article for cramer's rule provides some example proofs where its handy and simplifying

but it also mentions that its less efficient for computations, confirming what i said:

Cramer's rule implemented in a naïve way is computationally inefficient for systems of more than two or three equations.[7] In the case of n equations in n unknowns, it requires computation of n + 1 determinants, while Gaussian elimination produces the result with the same computational complexity as the computation of a single determinant.[8][9][verification needed] Cramer's rule can also be numerically unstable even for 2×2 systems.[10] However, it has recently been shown that Cramer's rule can be implemented in O(n3) time,[11] which is comparable to more common methods of solving systems of linear equations, such as Gaussian elimination (consistently requiring 2.5 times as many arithmetic operations for all matrix sizes), while exhibiting comparable numeric stability in most cases.

it cites http://web.eecs.utk.edu/~ielhanan/Papers/JDA2011.pdf, which states that cramer's rule is comparable to LU decomposition in terms of numerical stability and asymptotic speed

but still slower in practice

(up to a constant multiple)

As can be seen in Table 4, the algorithm runs approximately 2.5 times slower than the execution time of Matlab, independent of matrix size, which closely corresponds to the theoretical complexity analysis presented above

[wary lily]

thank you

[wintry steppe]

cramer's rule is that useful handwaving tool that i use when i want to prove something about invertibility of matrices and smoothness

[lost ermine]

Sorry for not replying, Gamma is just a field of characteristic 0, you can just think of it as R.

[wintry steppe]

e.g., GL(n, R) is a lie group, proving inversion is smooth uses cramer

[lost ermine]

thats the problem if u wanna see it again]

[dusky epoch]

mhm

okay

if im reading this right it looks like we must have $P(t) = \alpha_1 t$ with all other coefficients zero

[verbal shard]

its more fun tbh

[stoic python]

Ann

[lost ermine]

thats what i thought but it seems to trivial dont u think?

but yea it looks like thats the only way

[verbal shard]

i was wondering if there is some way to represent a trilinear form as the product of vectors and a 3D matrix?
or
generally as a matrix vektor product

[dusky epoch]

well it feels like any other coefficient being nonzero would break linearity

so

[lost ermine]

yea u r right

[wintry steppe]

What does it mean for Ax=b to have a solution for every b?
Is this another phrasing of 'is there at least one free variable'?

[nocturne jewel]

It means what it sounds like

any system which can be expressed as Ax=b will have a solution

[wintry steppe]

But is this related to the idea of "no solution, a unique solution or infinitely many solutions"

Or is it related to mapping/onto?

[nocturne jewel]

"have a solution"

it refers to the existance of solution(s)

since it says solution

[wintry steppe]

Sorry, it just threw me off that it says 'for every b'.
Couldn't it be the case that there is a consistency condition which would allow for infinitely many solutions, but not for a specific b?

[nocturne jewel]

if there's infinitely many solutions, then there's a solution

[wintry steppe]

So it's just asking if there are free variables?

[nocturne jewel]

FTIM says that Ax=b has a unique solution for all b, if A is invertible if that's what you're referring to

[wintry steppe]

I'm not sure what FTIM is, sorry

[nocturne jewel]

fundamental theorem of invertible matrices

[wintry steppe]

Wait, so that would mean that there should be no free variables, right?

If a matrix is invertible, it has n pivot positions

[nocturne jewel]

yes

since rank(A)=n

[lavish jewel]

this also works for nonsquare matrices

no need to look at invertible ones

only invertible from one side

[wintry steppe]

injective matrix moment

[lavish jewel]

it can have infinitely many solutions and still work

wouldn't it be surjective?

[wintry steppe]

in this case sure

i just think it's funny to call matrices injective/surjective

[nocturne jewel]

you're an injective matrix

[lavish jewel]

😛

anyway, you should rather talk about the rank, number of rows, and number of columns

[wintry steppe]

I was watching videos on this

And it just confuses me because here he says there are infinitely many solutions if there are not n pivot positions

And that n pivot positions would mean that there is only a unique solution

But maybe he means for a specific b?

That if there are n pivot positions, there is a unique solution for every b?

[lavish jewel]

it means the matrix has full row rank

and that n >= m

for A of size m x n

[wintry steppe]

But was what I just said correct?

I'm sorry Edd, I'm pretty dumb

[wary lily]

imagine a system of n equation in n variables

[lavish jewel]

the case you mentioned with n pivot positions is a full rank (invertible) matrix

there is exactly one solution for each b

[wintry steppe]

Okay thank you, and when there is a free variable, does that mean that there are infinitely many solutions for each b?

[lavish jewel]

but your matrix can be rectangular and still have solutions for every b

yeah

[wintry steppe]

i agree, there are very few solutions to my problems

[wary lily]

what may help you to better visualize/digest this is to write your system in three forms:
1- As a matrix equation
2- As a combination of a vector equation
3- As an augmented matrix

you can also write it as a system of equations

which is pretty much the same

[wintry steppe]

Would you be able to tell me what you would do when faced with a question like
"Does the vector equation Ax=b have a solution for every b?"

It would help out just knowing how someone would go about doing so

Sorry for asking so much

[wary lily]

there are other equivalent statements

but that's as far as I got

look at this and tell me what you don't understand

[stable kindle]

i mean matrices are just linear transformations right

i just visualise various violations committed upon the cartesian plane

now considering that there are some matrices which absolutely crush the skeleton down to like a single line, or even the zero matrix which causes an implosion to the origin

[wintry steppe]

I have no clue what a cartesian plane is

[wary lily]

also, the theorem I posted answers the question: "Does the vector equation Ax=b have a unique solution for every b?" Which is a bit different with your original question.

[stable kindle]

well, for those matrices you're not gonna get all the things

wait what

[wary lily]

maybe you call it smth else in your lang, Bes

[wintry steppe]

Yeah, I'm not sure if it's semantics or if there is an actual difference

That is what I am worried about

[wary lily]

there is an actual difference

unique solution and infinitely many solution, that is

[wintry steppe]

'a solution', do they mean singular, or can there be multiple? stuff like that

[stable kindle]

i mean

there can be multiple solutions

but not for every b

[wary lily]

then when they say every b, they mean unique?

[wintry steppe]

I wrote the entire Lay's Invertible Matrix Theorem a couple of times, it just doesn't stick sadly..

[stable kindle]

not necessarily

[wary lily]

same book moment

[stable kindle]

but if it does have a solution for every b, then it's unique anyway

i think

[wary lily]

that's what I'm saying

[stable kindle]

no but that's not what they're saying

in the proposition it doesn't have to be one and only one solution, as i understand it, just >= 1

and then it turns out that it's only 1 anyway, i think

[wary lily]

when they say: "a solution for every b", it can only be true when every b has solution, in which case solutions will be unique.

anyway, the theorem I'm working with is about an n * n matrix

[fervent gulch]

Give a scalar equation for a plane at distance 9 from the plane P with equation −2x−2y−z = 4.

[wary lily]

@wintry steppe is what you dealing with n * n matrices?

[wintry steppe]

Yeah this is a square matrix

[wary lily]

OK, then the theorem I posted applies

[wintry steppe]

What do I check for if it's not a square matrix, if I may ask?

[wary lily]

so, you do know how to solve this, right?

when it's square

[fervent gulch]

anyone

[wintry steppe]

Just check if there are n pivots, no?

[wary lily]

yes, n pivots, or if the det is not zero, which means it is invertible, or there isn't any zero row/cols in reduced row ech form, or all the vectors/columns are linearly independent.

any of these, means every of these

we clear about this?

[wintry steppe]

Yes, thank you

[wary lily]

OK, now when the matrix is not square

the rows represent equations/possible constraints

the columns are the number of variables

you need to row reduce this to echelon form, the augmented matrix

first answer the consistency question:

is there any row of the form [0, 0,... x], where x is none-zero

[wintry steppe]

In the augmented matrix, right?

[wary lily]

this would equal an equation of the form 0x_1 + 0x_2... = x, where x is not zero

which can never be true

if you find such a row, there is no solution at all

system is inconsistent

right?

[wintry steppe]

Right

[wary lily]

now, if it's consistent, we need to answer the uniqueness question

are there more unknowns than equations?

[fervent gulch]

anyoneee

[wary lily]

then there are infinite many solutions

if not, the solution is unique

which, coincidentally happens when we dealing with an invertible square matrix

[wintry steppe]

With unknowns and equations you mean columns and rows, right?

[wary lily]

yes

but, say rows with pivots for constraints

a full zero row, doesn't count as an equation/constraint

makes sense?

it's always true

[wintry steppe]

Yeah it does

I'll keep that in mind

Thank you for taking the time to help out az, you're a blessing

[wary lily]

OK, now a point that may help

you have a non square matrix and are asked if it has a solution for every b

if you end up with a zero row

the system will not have solution where the b has a non zero entry in the corresponding row

bc that will be a [0 0 0 0 x] where x is not zero

that means, it has infinitely many solutions for some bs

but there are specific bs where it has no solution

if you get that too, you have the essence of this subject AFAIK it

[fervent gulch]

y'all left me on read

[wintry steppe]

That makes sense, thank you

[reef sleet]

What are some engineering applications of linear algebra?

[lavish jewel]

computer graphics, machine learning, physics simulations

[reef sleet]

I'm in class right now so I can't take a good look, could you explain or provide some examples please?

[wintry steppe]

just ignore your class

[hazy sparrow]

computer graphics: projection from 3d objects into 2d space (your screen)
machine learning: you can represent your inputs and weights calculations as dot product multiplications
physics simulations: vector fields come up a lot and can be used to determine where a particle can end up

just google applications of linear algebra lel

[reef sleet]

oKK got omegasullied my bad 😔

I will Google after ty

Thanks for the examples Charles!

[wintry steppe]

m x n
rows x columns
The columns need to span the dimensions of rows
So n >= m for it to be onto
Is this correct reasoning?

[slow scroll]

yep

@wintry steppe

[wintry steppe]

Thank you :)

yw

[dull pewter]

why does this always result in the same matrix

[tame mural]

I get different values from u

I wonder if u just have a numerical error

[turbid temple]

well Im completely lost here

I need to factor that but cant find a way to

[desert light]

ummm

i’m pretty sure now you’re at x^3 -8x^2-19x-12

[turbid temple]

yeah

Now im trying to do it with 1 instead of 2

[desert light]

that 38 is supposed to be -38

and the -12 should be -88

[solid flower]

not sure about c) here

is norm only defined for vectors?

so i cant have norm(scalar) ?

[wintry steppe]

well

the norm of a scalar absolutely makes sense

as in, it's just its absolute value

[solid flower]

right

[wintry steppe]

but you should check back to your book's definition of norm

because

who knows

maybe it only defines norm for vectors?

[solid flower]

yea the back of the book says the expression doesnt make sense

[wintry steppe]

norm(scalar) makes sense

and indeed

[solid flower]

"In mathematics, a norm is a function from a real or complex vector space"
$\mathbb{R}=\mathbb{R}^n\big|_{n=1}$

[stoic python]

ahmad_11

[wintry steppe]

if your vectors have one entry

i.e., are scalars

[solid flower]

righttttt

[wintry steppe]

then the usual norm is just absolute value!

[solid flower]

cool!

[wintry steppe]

so i'm tempted to answer "yes" to part c

it's not a great question imo

[solid flower]

true

[ebon veldt]

Let $P_3(\mathbb{C})$ denote the complex vector space of polynomial of degree 2 or less. Let $\alpha,\beta \in \mathbb{C}, \alpha\neq \beta$. Consider the mapping $L: P_3(\mathbb{C}) \rightarrow \mathbb{C}^2$ given by $$L(p)=\begin{pmatrix} p(\alpha) \ p(\beta) \end{pmatrix}, \ \text{for} \ p \in P_3(\mathbb{C})$$ \

Consider the basis $V=(1,X,X^2)$ for $P_3(\mathbb{C})$ and the standardbasis $E=(e_1,e_2)$ for $\mathbb{C}^2$. Consider the matrix presentation $_E[L]_V$ for $L$ with respect to $V$ and $E$

[stoic python]

ScapeProf

[ebon veldt]

How do I do this? Im just confused because V is dim 3 and E have dim 2, so how does that work when finding the matrix presentation?

[quartz compass]

shouldn't change anything

isolate where each of the basis vectors from P_3(C) get mapped to

so what are your basis vectors of P_3(C) I'll help you out @ebon veldt

[ebon veldt]

(1,00),(0,1,0),(0,0,1) right?

[quartz compass]

I was gonna say 1, X and X^2

[ebon veldt]

oh ye

[quartz compass]

so where does the polynomial p(x)=1 get mapped to

[ebon veldt]

to (1,00)?

[quartz compass]

where's this come from

that's not the mapping, go back up there what do they say L(p) = ?

it should be a 2 dimensional vector

[ebon veldt]

p(\alpha),p(\beta)?

[quartz compass]

yeah good

so now specifically our basis vector, p(x)=1

what does this get mapped to

[ebon veldt]

oh 1(\alpha),1(\beta)?

[quartz compass]

what's 1(\alpha)

is that 1*alpha?

[ebon veldt]

yeah

[quartz compass]

nope

p(x)=1

p(alpha)=1

you're just plugging into a polynomial here nothing spooky 😛

[ebon veldt]

ah that makes sense

[quartz compass]

this determines the first column of our matrix

$[L] = \begin{pmatrix} 1 & ? & ? \ 1 & ? & ? \end{pmatrix}$

[stoic python]

Merosity

[quartz compass]

see how if you were to multiply by a column vector on the right, let's say [3,0,0]^T

that'd correspond to taking the polynomial p(x)=3

and getting out the column vector [p(alpha), p(beta)]^T = [3,3]^T

so let's do the next basis vector now

p(x)=x

what's L(p)?

[ebon veldt]

p(\alpha)=X, p(\beta)=X

[quartz compass]

😬

p(x)=x

plug in x=alpha

let's try something easier

p(x)=x

plug in x=7

what is p(7)

[ebon veldt]

So thats [7,7]?

[quartz compass]

no

for the question I just asked it's just a polynomial

[ebon veldt]

p(\alpha)=7?

[quartz compass]

p(7)=7

stop

we're just talking about a polynomial here

like you learned in middle school

p(x)=x

means whatever you plug in for x is what you get out

so if I plug in x=3, then p(3)=3

what L(p) does is it takes the polynomial and plugs in alpha and beta to p

@ebon veldt you with me so far?

we're working towards what L(p) is when p(x)=x

[ebon veldt]

Think so just confused cos doesn't a polynomial have 3 unkowns, \alpha+\beta X+\gamma X^2?

[quartz compass]

we're just transforming the basis vectors

the columns of a matrix are exactly where the basis vectors are sent to

[ebon veldt]

Alright

[quartz compass]

so doing that is all we need yup

so recap, we took p(x)=1 our first basis vector and found L(p)=[1,1]^T

[ebon veldt]

so L(p) when p(x)=x is [p(x),p(x)]^T?

[quartz compass]

now we need to take p(x)=x and see what L(p)=? here

not quite

it's [p(alpha), p(beta)]^T

what's p(alpha)=?

[ebon veldt]

ah so just [x,x]^T

[quartz compass]

wrong

[quartz compass]

answer this question

[ebon veldt]

p(alpha)=x

[quartz compass]

...

p(x)=x

p(alpha)=x

[wintry steppe]

This is contradictory isn't it?

[quartz compass]

this channel is occupied right now @wintry steppe

[quartz compass]

this is not how you plug into a polynomial!!! @ebon veldt

[wintry steppe]

Take your time, just curious

[quartz compass]

p(x)=x

what is p(11)=?

@ebon veldt

[ebon veldt]

oh okay so [alpha,beta]^T and for x^2 its [alpha^2,beta^2]^T

[quartz compass]

you sure?

[ebon veldt]

uh

[ebon veldt]

well p(11)=11

[quartz compass]

ah just so for when I help future students, could you explain why you thought to write this

and what made you realize what p(alpha) should be

[ebon veldt]

hmm not sure tbh but it makes sense that p(alpha)=alpha I guess

But thanks @quartz compass

[quartz compass]

lol you're welcome

[glad hamlet]

anyone mind explaining how to go from -2/(1-i) to -1-i?

i cant wrap my head around it

[quartz compass]

multiply by the conjugate divided by itself, (1+i)/(1+i)

[glad hamlet]

ohhh

[unborn bear]

Hey im trying to recreate affine transforms in desmos: https://www.desmos.com/calculator/u8kgcldzqd. But i have problem that draggable points doesnt end on up on circle.

[quartz compass]

how's it supposed to behave?

[unborn bear]

Theese points should lie in circle as they define my transformation

[dusky epoch]

wait... so what are the requirements for your circle?

it needs to pass through your two points?

or what

[unborn bear]

So point [1,0] is transformed to [a, c] by affine transformation so is [0,1]->[b, d]

[quartz compass]

what's your affine transformation, how does it have anything to do with a circle?

[unborn bear]

Ok i know the problem, I have wrongly transformed the circle.

Its not affine actually, it's just 2x2 matrix mutltiplication

of 2d points/

[mortal juniper]

can we take the difference between these 2 eigenvectors which is (1,-1,1) and say it is associated to the eigenvalue 6?

[quartz compass]

try to show you can take any linear combination of eigenvectors with the same eigenvalue and get a new eigenvector with the same eigenvalue

[dusky epoch]

if Av = 6v and Aw = 6w then is it true that A(v-w) = 6(v-w)?

[mortal juniper]

yes

[dusky epoch]

well then

is the difference of two 6-eigenvectors also a 6-eigenvector?

[mortal juniper]

yes