linear-algebra

[charred stirrup]

this might sound silly, but is there a rule that stops the 3 from raising P and P^-1 by the 3rd power

[hallow bobcat]

The key is that you have P^(-1) and P next to each other when you write it

[dusky epoch]

matrix multiplication in general isn't commutative so (AB)^n is not the same as A^nB^n

[hallow bobcat]

^

[dusky epoch]

$PDP^{-1}PDP^{-1}PDP^{-1} = PD(P^{-1}P)D(P^{-1}P)DP^{-1}$

[stoic python]

Ann:

[charred stirrup]

holy shit

and you get the identity 1 thing

fucking brilliant

thank you

what property of upper triangular lets you skip the RREF and say that vectors are linearly independent?

to be more precise, the solution seems to be unfinished to me until you're able to put it into RREF

[slow scroll]

whenever the columns are linearly independent in REF. Basically, you can stop whenever its obvious that the columns are linearly dependent/independent.

(1,0,0) clearly can't be used to write (3, -2, 0) and same with the other vectors

[charred stirrup]

i know the dependent thing is we just have to make a row with all 0s

but is there something special about stopping at the upper triangular matrix?

[north sierra]

for A=PDP^inverse, when I find P, am i allowed to change one column of P to remove the fractions?

[charred stirrup]

@north sierra when you find your eigenbasis, i think you are allowed to scale any vector by any amount you want, because they will be linearly dependent

[north sierra]

in* but true

[charred stirrup]

that's also true lol i mean dependent as in scaling any vector doesnt change much (you can go back to the original vector with fractions without any problems)

[slow scroll]

well scaling a column or whatever will change the inverse matrix

[north sierra]

true

so my Pinverse will be off

[charred stirrup]

bruh how

that's a thing?

im so fucked

all these special properties i dont know about

[north sierra]

when you're working with a basis only

then i guess its okay to scale it

but not when you combine other basis together

?

is this true

[slow scroll]

wat? scaling a basis gives you a different basis

[north sierra]

yeah but you can still scale it

[charred stirrup]

no if u mean basis and not eigen basis

the different basis matters

it's a different transformation

[north sierra]

if a question said find the eigen basis, and i found it but it contained fractions couldn't i just scale it to remove it?

[charred stirrup]

you can for that

[north sierra]

and wouldn't it still be a basis

[charred stirrup]

basis and eigenbasis are different

basis is kind of like.. how u change i, j, k and stuff

[slow scroll]

okok. So scaling eigenvectors is fine, but you have to make sure you are inverting the corrected matrix with the scaled columns and not the original.

[charred stirrup]

brilliant

[terse mirage]

i,j,k

disgusting

[charred stirrup]

idk fam thats just how they taught it

mechanics teachers at my school are trash

[north sierra]

heres a good example

[slow scroll]

and just a hunch: any nice-ness you gain by scaling an eigenvector is probably lost when you invert the matrix for the sake of preserving determinant. i.e. if det(P^-1) = det(P)^-1

[north sierra]

yeah i would leave my basis alone if i had to do inverse tho

[terse mirage]

${\vec{e_1}, \vec{e_2} \cdots \vec{e_n} }$

[stoic python]

agentnola:

[charred stirrup]

yeah

i like that one

[terse mirage]

much better

[charred stirrup]

it has 0s and just 1 one

[slow scroll]

alright time to take diffeq final

[charred stirrup]

good luck

[north sierra]

gl

[charred stirrup]

thanks for helping out

[slow scroll]

thnx

npnp

[charred stirrup]

in a) how did they go from the identity matrix to saying showing its invertible, and in b) im confused about what's going on in general

oh

my god

they multiplied by A ^ -1

im such an idiot

[half ice]

If you have AB = I, then B is A's inverse by definition

[charred stirrup]

i kept looking at some of the invertible matrix theorem rules

[half ice]

I wouldn't try multiplying by A^(-1) if you're not sure that the inverse exists

[charred stirrup]

how does that "show" A is invertible?

Yeah that's why the questions seems more like "find an expression for A^-1"

than "show A is invertible"

[half ice]

I mean if you find a A^(-1) then A definitely has an inverse

[charred stirrup]

nice i see

what are they doing in part b) ?

[half ice]

But you shouldn't assume such a thing exists

[charred stirrup]

how can they say the RREF of 1/2 (A - 3I) returns the identity matrix?

oh

beacuse it is invertible

it must be able to turn into the identity matrix

[half ice]

👌

[charred stirrup]

How did they get rid of the 1/2 ?

[half ice]

But I don't see why that proves it has no eigens?

[charred stirrup]

yeah...

[half ice]

If 1/2 A is invertible, then A is too

[charred stirrup]

ooo

brilliant i've followed to the last line

[half ice]

OH I see

[charred stirrup]

how can they say no eigen basis

we need a null space to find an eigen basis?

ohh

[half ice]

What's det(A - 3I)?

[charred stirrup]

not 0

[half ice]

So 3 is not an eigenvalue

[charred stirrup]

what

oh

FUCK

3I is like.. eigenvalue 3

[half ice]

Clever question I liked it

[charred stirrup]

what does it mean by "complex conjugates" ?

ohh nvm i got it

they just made a typo in part of the answer

should be -1 - i

[dusky epoch]

if $A \in \bR^{n \times n}$ is a real matrix and $v \in \bC^n$ is an eigenvector of $A$ with complex eigenvalue $\lambda$ then $A\overline{v} = \overline{\lambda}\overline{v}$

[stoic python]

Ann:

[charred stirrup]

does the span of any vectors always include the 0 vector?

[empty copper]

Yes

[charred stirrup]

nice thanks

im sweating

too many properties to remember of stuff

[empty copper]

Stay hydrated

[half ice]

Because 0v + 0u + 0w is in Span(v,u,w)

[charred stirrup]

yeah I was just thinking maybe the constants not allowed to be 0 because you get a trivial solution

[half ice]

That has to do with linear independence lol but I know what you're talking about

[charred stirrup]

wont lie i miss 🅱️ aynex ur handle was straight fire

[half ice]

Is this alright? Fair trade-off?

[charred stirrup]

ur so awesome lol

[half ice]

Nah I'm on here teaching math. I clearly never got a life

[charred stirrup]

nah kids like me out here passing thanks to people like u

hopefully

how come the geometric multiplicity is in terms of the # of free variables, not the basic variables?

[vast torrent]

What's the defn of the geo. mult, there's more than one

[half ice]

Probability that the charpoly splits

[charred stirrup]

In this case you plug in the eigen values into the matrix and row reduce to find the eigen vector

[jagged saffron]

well the geometric mult is the number of linearly independent eigenvectors associated to an eigenvalue

[charred stirrup]

ohh

and we get the eigen vectors from the free variables

[jagged saffron]

yes

[half ice]

Oh ic

[jagged saffron]

now if you know how to construct a matrix from a basis, you'll understand that you need as many linearly independent eigenvectors as the geometric multiplicity

[charred stirrup]

Then what do the basic variables mean? in terms of the row reduced matrix

[jagged saffron]

what basic variables?

[charred stirrup]

since we associate eigen vectors with the free variables, what do we associate with the basic variables?

[jagged saffron]

huh?

I'm super confused, do you have an example

[charred stirrup]

yup sorry 1 sec im bad at explaining what I dont know

in the rref, what does "x1" mean conceptually?

x2 and x3 to me are the null space

[jagged saffron]

that means that x_1 has to satisfy the equation

[charred stirrup]

ye I get that part

Just like

[jagged saffron]

because we have x1 -3/2 x_2 + 1/3x_3 = 0

so we have two free variables right

x_2 and x_3

x_2 and x_3 completely determine x_1

[charred stirrup]

yeah we write our basic variable in terms of the free variables

but our eigen basis is the null of that RREF

[jagged saffron]

yes

[charred stirrup]

sorry I think i realized

I dont know how to say what I dont know

i feel like an idiot

[jagged saffron]

basically, the null space is determined by x_2 and x_3

which are free variables

so they can be whatever they want

[charred stirrup]

yes exactly

[jagged saffron]

but x_1 can be written in terms of x_2 and x_3

[charred stirrup]

what space is determined by x1?

the column space?

and the row space?

i think that's what Im not getting

[jagged saffron]

your column space of the matrix above is spanned by [1,0,0]

[charred stirrup]

why not [3,3,0]

[jagged saffron]

[3,0,0] is in the span of [1,0,0]

any scalar multiple of [1,0,0] is in the span

well I mean its impossible to have any non-0 variables in the second and third row

im bad at explaining

we have your matrix with 0's in all rows except the first row

[charred stirrup]

I thought only for row space we use the row in RREF

[jagged saffron]

but if we do Ax = b where A is that matrix, the first element of b is determined by your first row of A dot by the column of x

[charred stirrup]

for column space we use the column before we reduced to rref

[jagged saffron]

am confused

maybe im just wrong

[charred stirrup]

sry 1 sec

[jagged saffron]

[charred stirrup]

i am confused too

[jagged saffron]

Anyways i have class now

[charred stirrup]

i know one of the spaces we use the unreduced matrix row/col

[solar orbit]

I am taking linear algebra next semester and want to get a headstart over the break . I already am proficient in the basics(using augmented matrices to solve equations, etc), but have no real experience writing proofs which I know I will have to get good at. How do you learn to write a proof? I feel like I was never really taught how in previous math courses.

[snow oasis]

Is this room free to ask a question ?

I guess it is

Im doing geometry on algebra and got to find

Point D
So i tried finding point F

But when i check where it is on a 3D calculator

Im way off

Is my logic wrong here ? I made a point F thats on the BC line and the vector FA perpendicular to BC. Solved where it is, but its definitely not where its supposed to be

Any help will be super appreciated

Wait i messed up the input into geogebra, and some basic calculations ill be back in a sec to report

Ok its correct im just bad at inputting data lmao

[winter acorn]

So I was wondering if it’s possible to have a Q matrix which is orthogonal but not orthonormal (Q matrix from the QlambdaQ^T decomposition)

[north sierra]

how can i prove that this is not a vector space using scalar multiple

[gleaming knot]

Find a vector $v$ in there and a number $a\in\bR$ such that $av$ is not in there

[stoic python]

Icy001:

[north sierra]

could i use the -1 scalar?

[slow scroll]

sure

[north sierra]

i guess any <0 scalar would work right

[gleaming knot]

Literally any nonzero scalar works

Since the line $y=ax$ and $y=x^2$ intersect at $x=a$ and $x=0$ and can't have any more intersection points

[stoic python]

Icy001:

[odd yarrow]

what the heck is a vector space

[vast torrent]

it's a set of objects called vectors where the objects satisfy the vector space axioms

[odd yarrow]

what are axioms

[vast torrent]

rules

[odd yarrow]

will pay someone to tutor me in linear alg

if anyone is down

lol

[uneven peak]

For 1a, the solution Manuel is super confusing as to how to get the answer. Can someone help show me what I’m supposed to do?

[north sierra]

@odd yarrow let youtube be your tutor

lots of great videos on there

[silent dune]

Can someone explain how 11a is done

[winter acorn]

@silent dune you need to find an indépendant column vector which will not be a combination of the previous ones

In other words, I’d say complete the base

a cross product between the two should work

[native lodge]

oh boy...what is this person writing

[keen mirage]

let the third column of A be x,y,z. consider any of the cofactors of the matrix entries not on the same column or row as the 0 entry (so a_12 or a_32) to find y

then just consider any other cofactors and solve x and z simultaneously with y

(then consider cofactors of x,y,z for unknown cofactors in C)

[odd yarrow]

you can still find inverses of a 4x4 matrix with gauss jordan right?

no limit

[native lodge]

you can find nxn using gauss jordan

that's the beauty of linear algebra; things just generalize so easily and so naturally

[calm fulcrum]

Hello

so my professor wants me to take a huge amount of variables (36) in my dataset and turn it into linear regression?

I am so confused how would I do that? He did not give us an specific instructions

All i have done is this

[timber minnow]

@calm fulcrum looks awful lmfao

@calm fulcrum this should probably go in stat btw

and I think I can help with this

[calm fulcrum]

I know it looks terrible

My professor literally gave us no instructions on how to do any of this and was absent 2 weeks. @timber minnow any help would be amazing

[timber minnow]

let's talk in #probability-statistics

[calm fulcrum]

Okay going there now

[empty copper]

The easiest proof I can think of is by contradiction

the subspace can't possibly have a basis which contains more independent vectors than the dimension of the vector space itself

[lone quail]

What exactly is the eigenvalue of a conic or a quadric?

Geometrically

Like for linear functions we know that f(v) =lambda v, but for a conic?

[empty copper]

I don't see how the zero space relates to this

[dusky epoch]

"at most n" includes "less than n"

[lone quail]

Why does d hold true?

[dusky epoch]

what part are you confused about exactly

@lone quail

[lone quail]

Why is it if and only if b=0?

[dusky epoch]

is it "b=0 => set is a subspace" or "set is a subspace => b=0"

[lone quail]

Why not 1 or 2?

[dusky epoch]

which of these directions do you need clarified

"b=0 => set is a subspace" or "set is a subspace => b=0"

[lone quail]

Both

[dusky epoch]

ok well

alright

i think it might be helpful if we give our set a name

say W

so $W = { f : (0,3) \to \bR \mid f \text{ differentiable}; f'(2) = b }$

[stoic python]

Ann:

[dusky epoch]

@lone quail that ok?

[lone quail]

Yes

[dusky epoch]

ok so

you know the definition of a subspace right

[lone quail]

Yes

It must be closed under addition and scalar multiplication

[dusky epoch]

we need to check that for all f, g ∈ W and constants c ∈ R, that f+g ∈ W and c*f ∈ W

[lone quail]

As well as having 0

[dusky epoch]

so if b=0

and we have f, g ∈ W

f and g are both differentiable

and f'(2)=g'(2)=0

then f+g is also differentiable

and (f+g)'(2) = f'(2) + g'(2) = 0 + 0 = 0

[lone quail]

Ok so why f’(2) and not f’(1) for ex?

Why f’(2)?

[dusky epoch]

that's the definition of W that we were given

[lone quail]

Ok thank

I got it

[dusky epoch]

ok now

f ∈ W

and c ∈ R

(cf)'(2) = c * f'(2) = c * 0 = 0

so if b=0 then W is closed under addition and under scalar multiplication

and i will leave it to you to check that the zero function is a member of W

does this make sense so far

[lone quail]

Yes perfect

Thank you

[dusky epoch]

ok

so

does the other direction need to be clarified still

or in other words does it need to be clarified why W is NOT a subspace if b is NOT 0

[lone quail]

Hmm 🤔

Is it because it wouldnt be closed linearly otherwise?

Iike if it was 1, then a scalar times 1 might not be in the set (0,3)?

Please correct me if im wrong

@dusky epoch

[dusky epoch]

it's even simpler than that.

suppose b != 0

then W fails to even be closed under addition

(f+g)'(2) = 2b

and since b is not 0

2b is not equal to b

[lone quail]

Ahhh ok thanks

[meager tinsel]

Does anything good come from the fact that for some square matrices
AX = A = (X^T)A
?

[undone garnet]

any idea?

[stone valve]

Why do matrices work in electrical circuits? I've been using them for over 4 years now to solve circuits and recently introduced myself to using them in the complex field as well (2x2 matrix for every value in the equation) but is there some proof that explains why it works? I'm super interested

and cuz my matrix knowledge isn't superb I guess

the way I use it with kirchoff's law of voltages and law of currents

I might need to ask this in the other discord, but I was wondering if some multi-knowledge people could help me understand why it works to convert the Kirchoff equations to a matrix then RREF them to find a solution.

Afaik, RREF is just to solve equations in general by manipulating the rows and such

but the conversion from Kirchoff equations -> matrix baffles me for some reason lol

Could be ultra simple- but I'm just interested 🙂

[vast torrent]

What are you using them for

The simplest application of matrices is to solve systems of eqns linear in each variable

Is that what you're doing?

Do you have a link to someone explains kirchoff as matrices

If it's pde stuff i probably can't help 😓

But maybe i can

Ooh

The ones for summing current or voltage?

No calculus?

Yes okay

@stone valve do you have an example I can use to demonstrate

[stone valve]

Sec

I'll try to get a picture 🙂

might become a bit messy

just a few minutes

[vast torrent]

do you know what an isomorphism is

[stone valve]

no

[cursive narwhal]

When you're talking about Kirchoff's Laws, you're talking about forming a linear system of equations in terms of the currents and voltages?

[vast torrent]

do you know basic matrix arithmetic?

[stone valve]

yes

yes

[vast torrent]

you're asking about how to convert from complex equations to 4 real equations, yes?

[cursive narwhal]

'Why it works to convert he Kirchoff equations to a matrix then RREF them to find a solution'

[stone valve]

2x2 matrix per complex number I suppose as that's how I'm doing it now

[cursive narwhal]

^That is your main question?

[stone valve]

yes

why does it work

[vast torrent]

so

I'm going to define two objects

$\mathbf 1$ and $\mathbf i$

and I'm going to write a complex number a+bi as

$a \mathbf 1 + b \mathbf i$

let me bold it

[stoic python]

gfauxpas:

gfauxpas:

[vast torrent]

and I'm claiming that any complex number is going to satisfy

$(a \mathbf 1 + b \mathbf i) + (c \mathbf 1 + d \mathbf i) = $

[stoic python]

gfauxpas:

[vast torrent]

$(a+b) \mathbf 1 + (c+d) \mathbf i$

[stoic python]

gfauxpas:

[vast torrent]

and

$(a\mathbf 1 + b \mathbf i) \times (c \mathbf 1 + d \mathbf i)$

[stoic python]

gfauxpas:

[vast torrent]

as

uh let me make sure I dont make an arithmetic mistake here

$(ac-bd) \mathbf 1 + (ad+bc) \mathbf i$

[stoic python]

gfauxpas:

[vast torrent]

this is just multiplication of complex numbers

make sense? just a fancy new notation you dont see the point of yet

follow?

this you can check will have all the expected properties like

$\mathbf i \times \mathbf i = - \mathbf 1$

[stoic python]

gfauxpas:

[vast torrent]

$\mathbf 1 \times \mathbf i = \mathbf i$

[stoic python]

gfauxpas:

[vast torrent]

did I lose you or do you follow

[stone valve]

I'm following

[vast torrent]

oh I forgot to say a and b and c and d are real numbers

[stone valve]

doing 4 thigns at once atm - sorry for being inresponsive

[vast torrent]

okay, are you sitting in a chair with arms

because this might surprise you so much you'll fall over

if I wanted to

I can create 1 and i with matrices

and get all the same results as with regular complex numbers

$\mathbf 1 = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}, \mathbf i = \begin{bmatrix} 0 & -1 \ 1 & 0\end{bmatrix}$

[stoic python]

gfauxpas:

[stone valve]

yes I know this

[vast torrent]

so

[stone valve]

found it through youtube

[vast torrent]

2 equations in 2 unknowns

with complex coefficients

become 2 equations in 4 unknowns with real coefficients

err

no

4 equations

[stone valve]

yes

[vast torrent]

that's not enough to answer your question it sounds like

[stone valve]

not really

it's not the complex -> matrix conversion that I'd like to know

[vast torrent]

oh

[cursive narwhal]

'Why it works to convert he Kirchoff equations to a matrix then RREF them to find a solution'

[stone valve]

it's just the way you transform the circuit info to the matrix

[cursive narwhal]

^This was the original question that baffled him

[vast torrent]

oh oops

[cursive narwhal]

Okay, the idea here is this

[stone valve]

sec

I'll send a pic 🙂

messy handwriting inc

[vast torrent]

https://www.youtube.com/watch?v=_-GaXa8tSBE

[cursive narwhal]

You understand that Kirchoff's Laws create a linear system of equations that are written in terms of the currents and voltages in a circuit?

[stone valve]

yes lol

[cursive narwhal]

You introduce those currents and voltages as unknowns and then, you solve for them.

[stone valve]

this

[cursive narwhal]

Now, the idea here is that any linear system of equations can be transformed into a matrix by dumping their coefficients into a rectangular array of numbers. That rectangular array is exactly what we call a matrix.

Yes

[vast torrent]

there's another way that works, but it

's not immediately obvious why it works

is to take the real parts of each equation and set them equal to each other

etc

anyway

[cursive narwhal]

The process of reducing the matrix to RREF is, in effect, solving the equations to obtain each variable's values.

[vast torrent]

thats not your question nm

[cursive narwhal]

The idea behind all of this is to convert a physics/engineering problem into a mathematical one.

[stone valve]

just seemed weird that I can just slap this in a matrix and RREF this to get the actual current as I'm using the resistances and the voltage-

[cursive narwhal]

For example, if you consider the motion of a projectile, you will realize that it obeys a parabolic trajectory. The idea behind that is that we wish to reduce any kind of physical problem into a mathematical one. Once we can describe it using mathematical language, we can use mathematical solutions to describe something physical.

Well, RREF can be achieved through Gauss-Jordan Elimination and that's just a way to solve linear systems. For example, we would solve a linear system of 2 equations by eliminating one of the variables and solving for one of them first before solving for the other. It's the same idea here

[stone valve]

I just feel like I'm posing a question that is either simple to answer or I'm completely missinterpretting what I'm doing

xD

I just don't see clearly on how I can say that a matrix of only resistances with the last collumn being voltages produces a unit matrix except for the last collumn consisting of currents

wait

no

fuck

NEVERMIND

I understand now

ignore my retardation

Thank you @cursive narwhal @vast torrent ❤️

[vast torrent]

not understanding something isn't a bad thing

as my tutor tells me

the answer to a question being obvious after you know the answer

doesn't at all mean you knew the answer or you should have known the answer

[cursive narwhal]

Hmm i may have misinterpreted your question. Apologies for that. I tend to do that sometimes.

[stone valve]

Don't worry you both- did amazing to try to help

[meager tinsel]

oof, any suggestions?

about what I asked?

[cursive narwhal]

What did you ask?

[meager tinsel]

Does anything good come from the fact that for some square matrices
AX = A = (X^T)A
?

oh, and A here is symmetric

and everything is in R

[vast torrent]

for all X or some X

[meager tinsel]

for some

[vast torrent]

then nothing comes to mind

[cursive narwhal]

Ah, so we have $A = A^T$

$(AX)^T = X^T \cdot A^T = X^T \cdot A$

[vast torrent]

wait

for some X or some A

or for some X and all A

[meager tinsel]

for some both of them

[stoic python]

Abhijeet Vats:

[meager tinsel]

yeah

[vast torrent]

we dont know A = A^T, how do we know that if it's only true for some X

[cursive narwhal]

^That's pretty general for all matrices A & X that are multiplicatively conformable.

[vast torrent]

oooh A is symmetric

[meager tinsel]

because I said A is symmetric

[vast torrent]

missed that line

[cursive narwhal]

He said that A is symmetric

[meager tinsel]

yeah

[vast torrent]

:X

[meager tinsel]

I feel like it must be up to somthing

buuut

[cursive narwhal]

Okay, but you also said that $AX = A$, for some matrix X.

Thus, we have:

$(AX)^T = A^T = A =X^T \cdot A$

[stoic python]

Abhijeet Vats:

[cursive narwhal]

I think that's more of a specific thing rather than something general. Like, what I used to get to it is general because, for any two multiplicatively conformable matrices $A$ and $B$, we have:

$(AB)^T = B^T \cdot A^T$

[meager tinsel]

mind that A is not invertible
or at least doesn't have to be otherwise its boring

[cursive narwhal]

Unless someone wants to come in and talk about a more general application of the relationship that you have, I don't think it's anything interesting

Oh hmm

[stoic python]

Abhijeet Vats:

[meager tinsel]

well

this actually was my conclusion from something

an original exercises asked me to prove that
$\forall X: A^TAX = A^T$ X is pseudoinverse to A

[stoic python]

Dovahkiin:

[meager tinsel]

oh its called

Generalized inverse

[cursive narwhal]

Ah well, that doesn't seem like something i could assist with. I'm sorry 😦

[vast torrent]

So it's a statement for all X

[meager tinsel]

for all X

for which that thing works

what I previously posted had different X

and was somehow related to this

but maybe it wasn't helpful

I decided to show the original task

[odd yarrow]

anyone have a good tutorial on how to do svd

[median mantle]

How can I find the new coordinates of (0,1,-1) if it is rotated by an angle of pi/3 about the line (1,1,1)

[stone valve]

http://mathworld.wolfram.com/SingularValueDecomposition.html @odd yarrow

this?

[median mantle]

I have to find a matrix that rotates by pi/3 about the line 1,1,1

Now I found two vectors orthogonal to (1,1,1). I From this I calculated P. For the diagonal, by we have that the eigenvalue corresponding to (1,1,1) is 1. Setting the remaining eigenvalues as x and 1/x, I was able to calculate the matrix A = PDP-1 as a function of x. Now I just need the value of x.

[half ice]

You're best seeing what happens to the basis

[median mantle]

Yeah to see that is why I asked what happens to (0,1,-1) when rotated by pi/3 around (1,1,1)

I tried using the face The matrix A has trace of 1+2cos theta and det(A) = 1 implies det(D) = 1 but to no avail

[odd yarrow]

Can you make this into a matrix?

[lone quail]

@meager tinsel ok so i did pseudoinverses

I could try giving you a hand

[meager tinsel]

oof

cool this buut

I also did them

we can you know
compare our solutions or something)

[lone quail]

Yeah just let me look through my notes real fast

[meager tinsel]

Mine is VERY short

well

6 lines or smth

turned out this is kinda simple

[lone quail]

Post it

[meager tinsel]

$$A^TAX = A^T$$

$Im_A$ is orthogonal to $Ker_{A^T}$ -- since $\forall v, \forall w \in Ker_{A^T}, (Av,w) = (v, A^Tw) = 0$.

Another fact:

\begin{align*}
& (A^TA)(X) = A^T \
& A^T(AX - E) = 0
\end{align*}

meaning that we know $Im_{AX - E}$ is in $Ker_{A^T}$. Now, consider any vector $v$. We have:

$$u = (AXA - A)v = A(XA - E)v $$
and
$$u = (AXA - A)v = (AX - E)Av$$
Conclusion: -- $u$ is in $Im_{AX-E}$, but also $u$ is in $Im_A$, and they are orthogonal hence $u = 0 \Rightarrow AXA = A$.

[stoic python]

Dovahkiin:

[meager tinsel]

uu, smart bot

[violet field]

Can I get help with this?

Exam reviewing kinda lost

[lone quail]

Yeah pretty neat @meager tinsel

[meager tinsel]

Can't really do shit without geometry))

[violet field]

Part is a’s answer is 0.8

[meager tinsel]

Also, I proved it, and then I forgot what was given and couldn't explain it to a friend for like 10 minutes

almost gone crazy haha

[lone quail]

@meager tinsel another way to solve it

Wbich is do much simpler

You can just multiply by the inverse of the left side to both sides

(You can do that because At*A is a square matrix)

[meager tinsel]

its square

but

it doesn't have to be inversible

[lone quail]

You do it on the left side of both equation

It has

Theres a proof

[meager tinsel]

why

oh okay

[lone quail]

I will send it to you in a sec

[meager tinsel]

I want that proff other parts I understand

[lone quail]

But if you do it that way then you get the equation of the pseudo matrix on the right side

And thats finished

[meager tinsel]

that said, A is literally any matrix

sooo

I really doubt it

[lone quail]

Gimme a sec

Theres a proof

[meager tinsel]

I mean
jsut take a 2x3 matrix with 1 in the (0,3) and 0 everywhere else
and

[lone quail]

Yeah but theres a specific requirement for matrix A

Otherwise it doesnt even make any sense for it to have a pseusoinverse

[meager tinsel]

what is it?

[lone quail]

Nvm i did the pseudomatrices in a very specific context

[meager tinsel]

in my exercise I had only A is mxn matrix and thats it

[lone quail]

Yeah

It has to have max rank

Because generaly you use A for projections

And A is usually formed by the basis vectors of a space

So they have to be linearly independent

Which inplies that A has max rank

[meager tinsel]

oh

yeah in that case yeah

its trivial then though

buuuut

I just
didn't assume anything

[lone quail]

Yeah

@meager tinsel but how can you even say ImA when you just have a matrix

Dont you need a linear function

[meager tinsel]

.......

what of not

every matrix is a linear operator between some spaces

[lone quail]

Yeah but that matrix would be the product of At and A

to apply it to this cass

[meager tinsel]

no

A is a linear operator

any A

of any size

[lone quail]

Yeah but then the orthogonality between imA and ker At doesnt apply

[meager tinsel]

It does

[lone quail]

Hmm alright, we didnt dive too dip into this area in my course

@violet field this channel is for linear algebra, which is very different from the algebra done at high school

I suggest you post in #❓how-to-get-help

Never did it so i cant help you with that sadly

[odd yarrow]

how do you do projection with gram schmidt

[violet field]

@lone quail my fault it’s health econ, college Economics

Is there a section or similar discord you could push me to

[lone quail]

As long as its about maths you can post in #❓how-to-get-help

[violet field]

Oh okay

[lone quail]

@odd yarrow the theory behind gram shmidt

Is the following:

Lets say you are in R2

And have vectors, v1 and v2 which span the subspace of W

and what you want to do is find an orthogonal basis of W

then, by taking either v1 or v2 as a reference, you can remove the x or y components using gram shmidt

this is done by removing, for example, the projection of v1 on v2 from v1

so that the result is a vector having only the y-component (if v2 is parallel to the x axis)

[odd yarrow]

what did orthogonal mean again

[lone quail]

it means that their dot product is 0 (perpendicular)

so basically by removing the projection of v1 on v2, you are removing from it the component on the direction of v2

so, of course, they would become orthogonal to each other

[odd yarrow]

how did you calculate span again

cos something right?

linear algebra will b the death of me

[lone quail]

the span is the smallest set of vectors which fully describe a certain subspace

[vast torrent]

uh

[lone quail]

said simply

[vast torrent]

that's more a basis

[lone quail]

yeah the span is whats generated by the linear combination of said basis?

[vast torrent]

you can span things that arent bases

like span{[1,1],[2,2]}

[lone quail]

yeah but it wouldnt make any sense

[vast torrent]

of course it would

[lone quail]

why would it

[vast torrent]

because the span is the set of all linear combinations of the vectors

[lone quail]

yeah

[vast torrent]

nothing about the definition of span says you need to be spanning linearly independent vectors

[lone quail]

yeah

i was just trying to aid his understanding of gram shmidt

[odd yarrow]

still no clue

v confused

[vast torrent]

okay but don't misdefine terms to do so :/

[lone quail]

yeah, its just i have a but of trouble translating from time to time

we dont even have span here xd

[vast torrent]

ah

well, the span of a set of vectors is all possible linear combinations of the vectors

[lone quail]

+1

[vast torrent]

the span of a set of vectors is always a subspace

[lone quail]

@odd yarrow get it now?

[odd yarrow]

just gonna try to make everything into a system of equations and solve it

lol

partial credit hail mary

[vast torrent]

what was the original question

[odd yarrow]

how do you do projections with gram schmidt

[vast torrent]

okay. do you know what gram schmidt is

or what it's supposed to do

or when you use it

[odd yarrow]

make it orthonormal

[vast torrent]

make what orthonormal?

[odd yarrow]

or simplify vectors

vectors

[vast torrent]

to make a basis orthonormal

the formula for projection of one vector v on another vector u is

$\frac{\langle v, u \rangle}{\langle u, u \rangle} u$

[stoic python]

gfauxpas:

[vast torrent]

where <u,v> means the dot product

if u is a unit vector, the formula is simpler

$\langle v, \hat u \rangle \hat u$

[stoic python]

gfauxpas:

[vast torrent]

to project a vector v onto a unit vector u

[lone quail]

yep, then the reasoning is what i told you earlier

[vast torrent]

you dot product v with u, and multiply that number by u

this only works if u is a unit vector

this simplified formula

[median mantle]

Do we know how to calculate complex eigenvectors corresponding to a rotation

[odd yarrow]

any idea how to input lambda into a calculator

[vast torrent]

just use L

[odd yarrow]

uhh I dont think my cal has that

[vast torrent]

so use t or x or any other letter

[odd yarrow]

o gotchu

zzz didn't work gonna have to do svd by hand

rip

[median mantle]

Ended up doing this ffs.....

[north sandal]

Any help with this?

Find a basis of the subspace H of R3 spanned by the vectors x1, x2, x3 and use the basis to determine the dimension of H.

x1 = [-1 0 1], x2 = [3 12 7], x3 = [3 6 2]

I think I'm suppose to turn x1, x2, x3 into a matrix H, then find the subsp of H?

[vast torrent]

@north sandal you can do it that way but there's a shortcut

if all 3 vectors are linearly independent the dimension is 3. otherwise the dimension is lower than 3. can you tell if any of the vectors are linear combinations of the others? can you prove that or disprove that just by looking at them?

[north sierra]

how can i prove that this is linear

or show

or show that it isn't

without doing the zero vector trick

[quartz compass]

why

[lone quail]

@north sierra ok so, to show that it is linear

It means that there exists a and b such that f(av+bu)=af(v)+bf(u)

Where a and b are in the real numbers

And v and u are vectors

[north sierra]

yeah i know the axioms

just showing it

[half ice]

Go through the axioms lol

[north sierra]

i dont know how to start

😦

[quartz compass]

just have to try to show one axiom

[half ice]

Try this:
T(av + bu) | aT(v) + bT(u)
|
|
|
|

With a line down the middle

Simplify both sides, and show they come to the same thing

[lone quail]

Or that they dont

[half ice]

Exactly. Or, if it fails to, it should leave you with a nice counter example

[quartz compass]

you can even just do plain T(av)=aT(v) but I like how Kaynex is doing it too

[north sierra]

okay i will try this when im home

[lone quail]

@quartz compass no you need two vectors

[half ice]

Actually yeah that's not a bad idea split it into those two cases

[lone quail]

Because you have to prove the sun too

[quartz compass]

no lol @lone quail

[lone quail]

Sum

[quartz compass]

that is sufficient

[lone quail]

Nope

[quartz compass]

sorry you're wrong

[half ice]

Show both T(av) = aT(v) and T(u + v) = T(u) + T(v)

[lone quail]

In this case what you’re saying implies that this is linesr

But its not linear

[quartz compass]

it fails T(av)=aT(v)

that's enough

no need to keep going

[lone quail]

Yeah but it succeeds that

I think

[quartz compass]

no it doesn't, you should work it out yourself

[half ice]

Necessary but not sufficient

[lone quail]

Yeah

You need to prove the sum too

[half ice]

For T(av) = aT(v) to imply a vector space

[quartz compass]

it fails

so you can stop

[lone quail]

Yes if it fails you can stop

[half ice]

It will fail lol but Elite doesn't know that

[quartz compass]

I'm saying T(av) doesn't equal aT(v)

[lone quail]

But if it succeeds it doesnt tell you anything

[quartz compass]

which is why that's all you need to do

[lone quail]

You need to prove the sum too

[quartz compass]

obviously, but it doesn't succeed

[lone quail]

Yeah but sometimes its not obvious

You need to explain to him that he needs both conditions

Otherwise he will always do that and not know he needs the sum too

[quartz compass]

lol calm down already

[half ice]

We're all on the same page rofl

Nobody misunderstands the subspace test

[lone quail]

Yeah

The important thing is that @north sierra gets it

[north sierra]

wait is the sub space test and test to show a map is linear the same?

[lone quail]

For subspace you have to show that the result of the addition and scalar multiplication is in the same subspace

[north sierra]

i know

[lone quail]

To show that a map is linear you have to show that it respects the addition and scalar multiplication

[north sierra]

but are the same axioms?

[vast torrent]

The subspace test tests sets

[lone quail]

Yes

[north sierra]

oh okay

[lone quail]

Same axioms but the proofs are just slightly different

[vast torrent]

A neccesary cond for T to be linear, btw, is T[0]=0'

[north sierra]

ye

almost home

keen to try this out

[vast torrent]

T(x,y) fails that immediately

[north sierra]

in my question right?

[quartz compass]

they did ask for some alternative to that though, at least that's what I assumed he meant by "without the 0 trick"

[vast torrent]

Yes

[north sierra]

yeah i asked for alternative

[vast torrent]

Oh

[north sierra]

i wanna get good with the other axioms thats why

[vast torrent]

The properties of a linear map aren't axioms,i wouldn't think of it that way

[lone quail]

Yeah

Because they can be proved

[vast torrent]

No, i mean

It's just a property some functions have

If i want to test whether a function is odd

I don't consider the axioms of oddness

[lone quail]

Yeah but almost everything you do in linear algebra is based on the fact that said function is linear

So you always basically consider that fact even if you dont do it explicitly

[vast torrent]

But they're not called axioms, it's the wrong terminology

[lone quail]

Yep agreed

[vast torrent]

You don't say a continuous function satisfies the axioms of continuity

[north sierra]

lol true

[vast torrent]

It's just not the right wording

[north sierra]

i shouldnt use that

what should i use then?

[lone quail]

Anyways im not sure about english terminology so i leave that to you guys xd

[north sierra]

property?

[vast torrent]

Properties

[lone quail]

Yeah property sounds good

[quartz compass]

I'd use definition

[vast torrent]

Or definition of linearity yeah

That's better

I like that better

[north sierra]

okay

[lone quail]

Okay perfect 👌

[gleaming knot]

Group axioms...

[lone quail]

Who wants a super tough linear algebra question?

[quartz compass]

I don't see the difference between a definition or axiom personally, but I don't care enough to argue about it

[vast torrent]

Not me

[lone quail]

I know you want it 😉

Anyways for someone who feels confident about linear algebra this is the question

Determine the equation of the surface obtained by rotating the conic gamma in the plane xy, of equation x^2 -2xy+y^2-4x=0, around its principal axis. Classify the surface obtained and determine its geometric properties (axis of symmetry, planes of symmetry, centre and vertex) as well as writing its canonic form.

Tell me if anyone wants the worked out solution

[vast torrent]

It would be easier without the 4x but

There are tricks

[quartz compass]

diagonalize the symmetric matrix of coefficients to get the change of basis

[vast torrent]

Yeah but

[quartz compass]

although I think it's just a 45 degree rotation so I can do that regardless without doing that

[vast torrent]

What about the 4x

[quartz compass]

just let it chill outside

complete the square after rotation

or square(s) I should say

[lone quail]

@quartz compass you are getting close but its not 45 degrees

[quartz compass]

$$x=\frac{u+v}{\sqrt{2}}$$ $$y=\frac{u-v}{\sqrt{2}}$$

[lone quail]

You are right about diagonalosong tho

[stoic python]

Merosity:

[quartz compass]

that'd be my guess

of course I'm right

but I'm not gonna work it out cause I am busy, but that guess probably works to remove the xy cross term then you should be good

which is a 45 degree rotation

[vast torrent]

Why complete after rotation

[quartz compass]

I'd be surprised if it didn't

[vast torrent]

And not before

[quartz compass]

doesn't matter

[vast torrent]

Hm

[quartz compass]

do whatever works

[vast torrent]

I'd do that and use vectors [x,y,1] to catch the constant terms, right?

[quartz compass]

no, no homogeneous coordinates

just plain (x, y) vector

[vast torrent]

There won't be constant termsM

?

[quartz compass]

wait which way are you doing it, to try to catch the constant terms inside the square

yeah, this is why I wouldn't bother with completing the square first

[vast torrent]

Id get it in the form u²+cuv+v²+constant, idk how else to do it

[north sierra]

ok im gone

home*

[vast torrent]

And use homogenous coordinates

[north sierra]

T(av + bu)

[quartz compass]

nah

[north sierra]

what is the b vector?

[vast torrent]

There's an easier way?

[quartz compass]

diagonalize first

is easiest

[vast torrent]

How do i diagonalize with the 4x there

[quartz compass]

just leave it outside the matrix

[vast torrent]

Okay

[lone quail]

Hmm theres a specific matrix you have to use

[vast torrent]

So ill do that, then what

[quartz compass]

$f(x) = x^TAx + b^Tx +c$

[lone quail]

Let me send you the matrix

[stoic python]

Merosity:

[vast torrent]

Yes

[quartz compass]

diagonalizing gets you an orthogonal matrix so

$Px = u$

[stoic python]

Merosity:

[vast torrent]

So the radii stay the same size

[quartz compass]

that's your transformation

[vast torrent]

Of the diagonal matrix

[quartz compass]

$(Px)^T D(Px) + (Pb)^T Px +c$

[stoic python]

Merosity:

[vast torrent]

Ooh

[quartz compass]

$u^TDu + v^T u +c$

[vast torrent]

I like it

[stoic python]

Merosity:

[quartz compass]

then complete the squares

[lone quail]

Anyways the matrix you gotta use is the one at the bottom of the page

And to transform in canonic form you also gotta do a translation

[vast torrent]

Idk what canonic form is

[quartz compass]

lol

you don't have to translate

[lone quail]

You do

Because it has to have centre in the orogin

Or vertex or whatever this conic is

[quartz compass]

I just explained why you don't

[vast torrent]

Idk what canonic form is

[lone quail]

Believe me i studied conics a lot

You have to translate

[quartz compass]

it's cute but I'm not arguing with you over something stupid a second time

you don't understand what you're talking about I'm afrais

[lone quail]

Alright do what you like, do you want me to show you the textbook which explains you have to translate?

[vast torrent]

What is canonic form?

[north sierra]

can someone help me with the linear question i had earlier maybe in another channel 😢

[lone quail]

Its basically when the centre/vertex is in the origin

[quartz compass]

I am not saying your method doesn't work

I'm saying it's not the only way

[lone quail]

And you use canonic vectors

Im just saying you cant get to the right answer without a translation because thats not what canonic form is about

The canonic form of a conic is specific

For a conic to be in canonic form it has to have centre/vertex in the origin

@north sierra anyways go ahead

Whats the issue

[north sierra]

hey

gonna post again

so like i wanna use the addition property

and test it

not sure how to start

[lone quail]

Check this example

This is doing both at once but you should be able to extrapolate what to do

[north sierra]

the thing that confuses me is the T(x,y)

[lone quail]

Its the same as the example

Basically you have to show that

[north sierra]

idk i dont get that example

[lone quail]

F(v+u)=f(v)+f(u) for addition

So that means take a generic vector v

x,y

And a generic vector u

x’,y’

[north sierra]

so in my case what would the u vector be?

[lone quail]

x’,y’

[north sierra]

yeah but not sure how it would look like

like what values would it have

[lone quail]

Any values

[north sierra]

x`+5?

[lone quail]

Like this

As you can see you cant go forward

@north sierra

Anyways im getting some sleep now its past 1am here

[wintry steppe]

It feels very iffy because that's the second time I did the problem and my previous answer was different.

[north sierra]

can u send a better pic

@wintry steppe

np

[half osprey]

I think it's all good @wintry steppe

[wintry steppe]

Had to cut my name out

[north sierra]

lol

[uneven bloom]

Yes, this is correct

[wintry steppe]

Thanks @half osprey and @uneven bloom !

[half osprey]

np

[wintry steppe]

Is there a shorter way to get to the same (or even a slightly different) simplification?

[uneven bloom]

A(A^2(BA)^(-1))^T=
A(AB^(-1))^T=
A(AB^T)^T=
ABA^T=
-ABA

[wintry steppe]

Lol I made it so convoluted

[uneven bloom]

As long as it’s valid, it’s fine

[half ice]

@north sierra
Still looking for it?

[north sierra]

for what @half ice

[half ice]

T(x,y) = (x+1,y+1)

[north sierra]

yeah i need help

lo

l

for the addition property

[half ice]

Try this:
T(v + u) | T(v) + T(u)
|
|
|
|

[north sierra]

okay but what is v and what is u in this case?

i dont know what those vectors are

[half ice]

Each are vectors in your space. You're working in R² right now

[north sierra]

but what about T(x,y)?

[half ice]

T is a transformation that takes in a vector from R² and outputs a vector from R²

[north sierra]

yeah

[half ice]

You can call such a vector (x,y). I called it v instead.

[north sierra]

oh so (x,y) is v?

in your example

[half ice]

Well, any arbitrary vector can be represented as either (x,y) or v

or u

These are labels that are simply names for the object we're working with

Plus if I write them all as (x,y) it's gon get messy

Well, I suppose you have to.

[north sierra]

so for addition we need to do something like $T((x1,y1)+(x2,y2))?$

[stoic python]

The-Elite:

[half ice]

Yes exactly

[north sierra]

ohh so instead of writing $T((x1,y1)+(x2,y2))?$ you let v = (x1,y1) because itll be messy?

[stoic python]

The-Elite:

[half ice]

Can you write that without a T?

I write T(u + v) = T(u) + T(v) because it conveys the additive definition no matter what vector space we're actually working on

But fair, you need to put in (x1,y1) in order to simplify it

[north sierra]

ok

so like whats next?

[half ice]

T[(x1, y1) + (x2, y2)]
= T(x1 + x2, y1 + y2)
= (x1 + x2 + 1, y1 + y2 - 1)

That simplifies to that

Follow how I applied the transformation?

[north sierra]

one sec

how did you get the last step?

oh actually i see now

yeah i follow

[half ice]

Okay cool. Now, seperately, we simplify
T(u) + T(v)

That is,
T(x1, y1) + T(x2, y2)

[north sierra]

ok

[half ice]

Can you do that one? It's not bad

[north sierra]

yeah i think

T(x1,x2) = x + 1, y - 1
T(x2+x2) = x + 1, y - 1
?

[half ice]

Oh mb I thought it was (x + 1, y + 1)

Fixed my mistake above

[north sierra]

okay

should i include the x1, x2 in the transformed one ?

T(x1,x2) = x1 + 1, y1 - 1
T(x2+x2) = x2 + 1, y2 - 1

?

or remove the 1,2 thing

[half ice]

Yes we want to work with these where they are not the same vector

[north sierra]

oh ok

[gleaming topaz]

You should have T(x1,y1) and T(x2,y2) and yes include the 1 and 2

[north sierra]

aright

now i just add them togther?

(x1+1, y1-1) + (x2+1,y2-1)?

[half ice]

Yus, that's T(x1, y1) + T(x2, y2)

[gleaming topaz]

yeah and if it's a linear transformation that is supposed to equal T(x1+x2,y1+y2)

Well one of the steps atleast

But the other condition is alot easier to show

[north sierra]

im confused on adding these

[half ice]

Remember when adding two vectors in R² you just add their components

[north sierra]

$\begin{pmatrix}x1+1\ y1-1\end{pmatrix}+\begin{pmatrix}x2+1\ y2-1\end{pmatrix}$?

[stoic python]

The-Elite:

[half ice]

Can also write them like that yeah

[north sierra]

$\begin{pmatrix}x1+1\ y1-1\end{pmatrix}+\begin{pmatrix}x2+1\ y2-1\end{pmatrix}:=:\begin{pmatrix}x1+x2+2\ y1+y2-2\end{pmatrix}$

[stoic python]

The-Elite:

[half ice]

Exactly yes

That's T(u) + T(v) you just calculated

[north sierra]

earlier we had $\left(x1:+:x2:+:1,:y1:+:y2:-:1\right)$

[stoic python]

The-Elite:

[north sierra]

for T(u+v)

so they're not equal?

[half ice]

Now you're getting it

[north sierra]

lol

[half ice]

T(u + v) ≠ T(u) + T(v)

[north sierra]

ayy

okay thanks so much

omg

[half ice]

Now, we did things the hard way

Normally you just look for a counter example

[north sierra]

like?

[half ice]

However, what we did could have proven T is a linear transformation

And counterexamples can't do that

(But T isn't a linear transformation here so the result is good)

[north sierra]

so lets say the addition rule somehow proved that T is a linear transformation

we would still have to check the other properties in case right?

[half ice]

We could have proven the addition rule

But you need more to prove T is a linear transformation

Also can do
T(av) | aT(v)
|
|
|

And simplify each on their own side of the line