linear-algebra

[lavish jewel]

symmetric matrices can be diagonalized, yes @trail dirge

what do PSD stand for here? @fallow jolt

[fallow jolt]

I got it

Positive semidefinite

Also

[lavish jewel]

aight

[fallow jolt]

It should be positive definite

[lavish jewel]

yes

[fallow jolt]

HW had a typo

[lavish jewel]

otherwise no inverse

[fallow jolt]

But yeah we good

[lavish jewel]

yep

kk coo

[fallow jolt]

Actually @lavish jewel, I proved M^-1/2 exists, but how do I prove its symmetric?

Is the inverse of a symmetric matrix also symmetric?

[lavish jewel]

a symmetric matrix can be written as Q D Q^T

with Q being unitary, i.e. QQ^T = I = Q^TQ

you can invert that simply as Q^T inverse D inverse Q inverse

[fallow jolt]

Gotcha!

One other thing, I'm not quite sure how to prove this:

[lavish jewel]

which is of the same form

[fallow jolt]

Multiplying a positive definite matrix on both sides with a symmetric matrix produces another positive definite matrix.

I need it to finish my proof, but I'm having a bit of trouble proving it

[lavish jewel]

it doesnt look true to me but i might be wrong

is the symmetric matrix invertible?

[fallow jolt]

Yes, I actually am multiplying a PD matrix on both sides by the M^-1/2 we talked about above

[lavish jewel]

ah, then yes

[fallow jolt]

How could I prove it?

[lavish jewel]

gimme a few min to think about it

what is the matrix in the middle

also symm?

[fallow jolt]

I want to do

M^-1/2(M - N)M^-1/2

Where M and N are PD

@lavish jewel

[lavish jewel]

mhm

let me see

[trail dirge]

I meant orthoginally diagonalized

all with same element

for example 2 by 2 matrix with all elements 1

[lavish jewel]

still yes

[trail dirge]

yeah but how

I can't seem to orthogonally diagonlize it

[lavish jewel]

do the evd

here you go @momo

a bit messy, but i think that is rigurous

thw first line is a definition of PD matrices, oops

not just symm, cuz the factors might not exist then

everything was PD here tho

[fallow jolt]

Wow

This is amazing

Thank you!!

[lavish jewel]

what level are you at? i hope you're at least in uni cuz otherwise that won't make much sense lol

i forgot to ask before going HAM on decompositions

[fallow jolt]

Yeah I'm currently taking an ML course at a university

So we are going over linear algebra pre-requisites and stuff

[lavish jewel]

oh, then this is about right

[fallow jolt]

Yep! I can follow it 🙂

[lavish jewel]

EVD, SVD, etc

aight, sweet

[fallow jolt]

Ty :)))

[lavish jewel]

i only called it D there, but that's the diagonal matrix of the reciprocals of the square roots of the eigenvalues. should be ok

[zealous junco]

its insane how i forgot everything about lin alg i learned half a year ago 😦

now suddenly I need it for numerical methods and struggling again lol

[lucid cedar]

I was wondering if anyone here knew anything about the equation z = ax + (1-a)y 0<a<1?
Im trying to understand why this plots a line segment between x and y.

[lavish jewel]

ah, the definition of convexity

[lucid cedar]

Yes im in an optimization class doing LP

[lavish jewel]

the name of that is the convex combination of the points x and y

[lucid cedar]

oh it has a name

i bet i can google a proof of it easy

[lavish jewel]

i can hopefully convince you with 3 examples

[lucid cedar]

even better, id love to see your examples

[lavish jewel]

try a = 0 and you get y. x = 1 and you get x. a = 0.5 and you get the average of x and y

which is between the two

[fallow jolt]

@lavish jewel Sorry, I see how you proved its symmetric, but how do we know its PD?

[lavish jewel]

that's what i meant. the first line was actually PD -> A = B^T B

not just symm. otherwise the matrix B doesn't exist

regarding the convex combination, you can set x and y as columns of a matrix, and a and 1-a as elements of a vector that multiplies the matrix

[fallow jolt]

ohhhh

I see

[lucid cedar]

its intuitive to see that a the convex combination generates vectors that essentially make a small inner triangle to the original vectors IE:

It just seems pretty incredible that these fractional vectors managed to land on that line

Your examples help though. The average case for sure.

[lavish jewel]

that's why i also suggested to put x and y as columns in a matrix

[lucid cedar]

I dont quite follow that line

[lavish jewel]

they become a sort of basis and span a space that is a straight line

[lucid cedar]

oh i see what ur saying

[fallow jolt]

Could I get a hint as to how to begin this problem?

[native rampart]

Hint: if x is an eigenvector what will x^T (Ax) be

[lavish jewel]

ah, yes, rayleigh quotients

https://en.wikipedia.org/wiki/Rayleigh_quotient

[fallow jolt]

Excellent job from the discord preview bot

[lavish jewel]

ye boi

[fallow jolt]

Which section should I look at?

[lavish jewel]

anyway, follow the hint you were given. the link i sent follows up with complex matrices

try the suggestion they gave you, this one is pretty straightforward. you can read the link after

[fallow jolt]

I'm not sure what x^TAX is

[native rampart]

Are you familiar with dot products?

[fallow jolt]

Yes

[native rampart]

If X is an eigenvector AX=cX for some eigenvalue c,now X^T AX=X^T(AX)=c X^T X=c ||X||^2

[fallow jolt]

Ah yeah, that makes sense

Why would we want the norm to equal 1?

[native rampart]

Because if norm was 1 this is just c

The eigenvalue of X

[fallow jolt]

Is there a range for possible values of a norm?

[native rampart]

Should be >=0

||x||=0 iff x=0

[fallow jolt]

Oh, so we just set the norm equal to 1 so that lambda is isolated?

[native rampart]

Yes

[fallow jolt]

Gotcha, so we're done then?

[lavish jewel]

you have to show what the vector x should be

[fallow jolt]

Why?

[lavish jewel]

follow the logic to find that if x is the eigenvector corresponding to the largest eigenvalue of A, then the expression is equal to the largest eigenvalue

if A is symmetric PD, it has full rank and you can express any vector x as a sum of the eigenvectors of A

arrange the eigenvalues and eigenvectors in decreasing order and go from there

[fallow jolt]

Can we not just say let x be the eigenvector corresponding to the largest eigenvalue of A, followed by this ^, and be done?

[native rampart]

Yes

Well,If you take max of all eigen values, you get the max eigenvalue

[fallow jolt]

Hm, that is true...

[zealous junco]

but that's not done right?

[lavish jewel]

i'd do it in two steps

i'd say x is an eigenvector

show what happens

and then say it has to be the eigvect of the largest eigval

[zealous junco]

you also need to show that for any vector in V, it's x^TAx is less than if x was the maximum eigenvector

or why not?

[native rampart]

If x is not an eigenvector,That will still be lesser than the max eigenvalue

[lavish jewel]

what i said about A being a full rank matrix with orthogonal eigenvectors is an extra argument, because you end up with a convex combination of all the eigenvectors

with max when all the weight is at the largest eigenvalue

that would be the most general way of showing it

[fallow jolt]

My logic is this line showed how x^TAx = lambda, and why norm needs to equal 1

Therefore, we're left with the expression that we need to prove

[zealous junco]

yea^ because all symmetric matrix have eigenspace = V

[fallow jolt]

So I'm not sure why we need to do any more

[zealous junco]

because you also need to show given any vector v in V, it's smaller than that eigenvalue i think

[lavish jewel]

yeah

that's what's missing

my suggestion is to expand A

x^T Q D^(1/2) D^(1/2) Q^T x

[fallow jolt]

What's V, and how can a vector be smaller than a scalar?

I'm not quite following

[lavish jewel]

from there it is evident that x^T A x is the norm of the vector D^(1/2) Q^T x

[zealous junco]

capital V i meant is the vector space

[lavish jewel]

i'll let you peeps handle it, having 3 explanations isn't gonna help 😛

[fallow jolt]

@zealous junco How can a vector be smaller than a scalar?

[zealous junco]

tbh im also thinking about it and not 100% sure

let me think about it

[fallow jolt]

Ah ok

I'm probably going to sleep, just dm me if you figure it out

It's nearly 4am here LOL

Thank you all

[keen flame]

What would be the determinant of a matrix A size nxn
which the mid row is 0 and the rest are 1's?
(0,...,1)
(...,...,...)
(1,...,0)

[native rampart]

0

[keen flame]

@prime vapor how come?, sorry

@native rampart how come?

[native rampart]

Revise your notes

[keen flame]

there's a section about identity matrix, but not on this kind of matrix

in identity matrix by defintion it's going to be det(I) = 1, while trance(I)=n

[native rampart]

Do you know how to compute a determinant?

[keen flame]

manually, yes

[native rampart]

Just do it along the row filled with zeroes

[keen flame]

It'll be if n=3 than det(A) = 0
if n =2 than det(A) = -1

[zealous junco]

just want to confirm

if A is diagonalizable

then PAP^-1 = D, then P^-1 stores then eigenvectors right

because P^-1 is the change of basis matrix s.t. [a]stdbasis = P^-1 [a]eigenbasis

[woeful plank]

ooooo this channel is perfect for helping me with my homework

[south sedge]

is there a way to find out there is no solution to a system of equations without gauss-elimination?

[lavish jewel]

that's really the easiest way

try to solve it and fail

there are other ways, but they are sophisticated

EVD, SVD, and other matrix factorizations

[south sedge]

alright

another question, what is the reason for having a double absolute value of a vector?

isnt one enough?

[dusky epoch]

double absolute value?

you mean the double bar notation for the norm of a vector?

$\nrm{x}$ like this

[stoic python]

Ann

[south sedge]

yes

[dusky epoch]

to distinguish it from absolute value (or modulus if youre working in C)

[south sedge]

why would we wanna do that? aren't both the length of a vector?

[lavish jewel]

there's different ways of measuring length

what you're doing there is actually the 2-norm of the vector

let's see what wolfram says

,w l-p norm

[stoic python]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Edd#5182

[lavish jewel]

not very useful

[grave plank]

shilov uses the term "eigenrays"

this feels very non-conventional

[digital bough]

I have shown that (v1, v2) are linearly independent of each other in R^2. How do i prove that that they are a basis in R^2 or well how do I say that the span of the vectors is equal to R^2?

”Let V be a finitely generated vector space. The vectors v1,v2,...v_n in V is a basis in V if :
Span{v1,v2,...v_n} = V
and
The vectors are linearly independent.”

Or well, they never said “and”actually, they just listed

Is a basis if:
a)
b)

Not sure if I can interpret as in it is a basis in R^2 if one of the conditions are met or if both have to be true. I assume the latter

[nocturne jewel]

both have to be true

cause for example [0,1];[1,0];[1,1] span R^2, but they arent indep. so it's not a basis

[hoary osprey]

they're 2 linearly independent vectors in R^2

they have to be a basis

[digital bough]

Well lets say the general case

I prove an arranged order of vectors are linear independent in R^n how do I claim they also span R^n?

[hoary osprey]

u gotta show that any vector in R^n can be written as a linear combination of these vectors

[marble lance]

You can even try to prove the generalized version that if V is an n dimensional vector space then any linearly independent set of n vectors is a basis

[digital bough]

Why is a) listed then if linear independency implies it is a basis?

[lavish jewel]

that's not true

[marble lance]

Linear independence is not enough to show it's a basis

Notice in my theorem I said V has to have dimension n. So you must already know that it has a basis of n vectors. Then you can prove any other linearly independent set with n vectors is a basis.

If you don't know if it's finite dimensional or what its dimension is, then you still need to check that it spans the space.

[lavish jewel]

right. it's not enough that you have linearly independent vectors. you need enough of them, too.

a way of proving that it is a basis is converting the vectors in your set into the canonical basis, and then substituting the coefficients

[somber bronze]

hello, I was trying to reduce the rows of my matrix using the Gauss-Jordan algorithm, but I can't understand how in the notes of the teacher it makes the division by 8 of the row when it makes R4 - 18 / 8R3 because if I only do the difference without dividing I still get all 0's in the last row, could someone help me? Thanks in advance!

[lavish jewel]

the thing is that the goal was to make the 18 into a 0

nothing else

18 - 8*18/8 is 0

if you do r4 - 18r3, you get

18 - 18*8

that's not 0?

at any rate, the teacher did the operation in one go

you did it in two steps

that's the only difference

you took more steps

they directly followed the definition of gauss-jordan

same result, you're both correct. do whatever is easier for you as long as the operations are valid

[somber bronze]

what you write is correct and I understood what the goal but did my teacher make R4 - 18/8 * R3 on the intermediate matrix I wrote or did he apply the algorithm directly on the original?

[lavish jewel]

directly to the original

[somber bronze]

where is R3 still 8 16 or has it become 1 2?

oh sorry, got it

[lavish jewel]

the original

[digital bough]

Ok, thanks Edd and Luna!
R^n is a finely generated vector space and the vector space R^n for any n will always have a basis of dimension n. That means that in R^n I just need to show that the vector is linearly independent and has the same dimension as the standard basis, am i wrong?

[lavish jewel]

i see what you meant now, yeah, that procedure was kinda obscure

they should've stated the steps more clearly

[somber bronze]

you clarified a trivial doubt but very important to me, thank you very much

[lavish jewel]

no prob

what vector are you talking about?

[burnt prawn]

guys do u know that linear algebra comes form the arabs ^^

[digital bough]

Algebra does not linear

[burnt prawn]

heh...

heh sry im rlly not good at math...

im new here

[nocturne jewel]

I mean, not surprising since Arab mathematicians did alot, same with Babylonians

[digital bough]

I mean the word algebra is Arabic and the method of solving equations by adding or diving both sides is arabic but I wouldnt really say linear algebra is an Arabic invention though

[native rampart]

I mean,Who cares if "arabs used linear algebra first"

[burnt prawn]

Can someone explain whats intergration. Im a beginner.

[quasi vale]

we know ur a troll why are u bothering

[wicked axle]

Hello people

I have this graph

and I have been asked to "visually interpret" to determine what the frequency might be

how do i go about visually interpreting it?

[lavish jewel]

you guys are all in the wrong channel

[wicked axle]

suggestions on channel?

I thoguth my question was quite basic 😮

[lavish jewel]

@wicked axle try one of the questions channels

@burnt prawn try the calculus or a questions channel

[wicked axle]

Thanks Edd

[burnt prawn]

okity

[willow patio]

i dont have a question currently but i just wanted to say that linear algebra is thus far THE BEST MATH, i'm just learning what a linear transformation is now

and seeing how matrices transform inputs into outputs is very interesting

[native rampart]

That's algebra in general

[willow patio]

yeah i wish i had taken this in college

is analysis an extension of algebra or is a separate aspect?

i.e. real analysis

[native rampart]

Separate

[willow patio]

cool thanks, as an economist i have to find a way to take classes on that too

[digital bough]

Linear algebra so far is fun but I hate the computation part

[lavish jewel]

you usually don't do computations of this stuff by hand

suffer not

[tame mural]

Is there, um, a place where cheapo students go to peruse available current textbooks

in PDF form

[digital bough]

Am I allowed to talk about pirating it?

[tame mural]

Or just any forum where people talk about this stuff

[tame mural]

Is there an operation that allows you to take the entry-wise product of two n-length vectors?

[lavish jewel]

hadamard product

[tame mural]

thx

[lucid cedar]

hello LA chat:

So ive done my time in my senior algebra course. I know what all these terms mean

but the fact that the set is of linear equations is throwing me off

idk if im over thinking it or if theres just something im missing

[toxic karma]

can someone help me with some questions on my algebra home work

[tawny tulip]

@lucid cedar the terms stay the same, even if its a set of equations.

@toxic karma you can send here a question and people will answer, that's how this server works. (obviously let us know what you tried)

[toxic karma]

oh ok sounds good

[lucid cedar]

just to clarify an equation like y = 2x + 6 wouldnt be considered here because it does not pass through the origin correct?

[lavish jewel]

that's a linear equation

$[-2 ,,, 1]
\begin{bmatrix}
x \
y
\end{bmatrix} = 6$

[stoic python]

Edd

[lavish jewel]

you can see how this generalizes to Ax = b when you add more equations of x and y, yeah?

[lucid cedar]

right

okay

but if im testing this for linear dependence I would have.... Ax + Bx = 0? that does not seem right

a1Ax + a2Bx = 0 but that still looks definitely wrong

[lavish jewel]

both are wrong

[lucid cedar]

atleast i was right about that

[lavish jewel]

make up another equation like the last one

[lucid cedar]

y = 2x + 6 and y = 5x - 1

[lavish jewel]

$\begin{bmatrix}
-2 & 1 \
-5 & 1 \
\end{bmatrix}
\begin{bmatrix}
x \
y
\end{bmatrix} =
\begin{bmatrix}
6 \
-1
\end{bmatrix} $

[stoic python]

Edd

[lavish jewel]

how do you test for linear independence?

[lucid cedar]

I only really understand this term as linear combinations of vectors. if a set of vectors span another one then its dependent

[lavish jewel]

exactly

i just rewrote your system as vectors and matrices

do yo thang

[lucid cedar]

I dont want to be annoying and stupid, but.....

[lavish jewel]

build the extended matrix

(is that what it's called in english?)

ah, augmented

$\begin{bmatrix}
-2 & 1 & 6\
-5 & 1 & -1\
\end{bmatrix}

[stoic python]

Edd
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[lavish jewel]

look at the rows of the augmented matrix

are they linearly independent?

[lucid cedar]

mmmmmmmmmmmmmmm

yes

[lavish jewel]

ok

then the original equations were linearly independent

[lucid cedar]

how do i conceptually define this, the are linearly dependent when the coefficeints of all the terms are linearly dependent?

[lavish jewel]

when you cannot get one equation from the other

look at this

x + y = 1
x + 5y = 3
2x + 6y = 4

you see how the third equation is the sum of the top 2?

[lucid cedar]

yes

[lavish jewel]

$\begin{bmatrix}
1 & 1 & 1\
1 & 5 & 3\
2 & 6 & 4\
\end{bmatrix}

[stoic python]

Edd
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[lavish jewel]

do you still see it?

[lucid cedar]

yes

[lavish jewel]

it's literally the same thing

they're isomorphic

[lucid cedar]

mmmm okay that makes sense

alright im seeing it, sorry lol I didnt mean for this to be difficult

[lavish jewel]

it's aight

[lucid cedar]

its amazing how much time I can spend learning soemthing and then still feel like i know nothing about it

[lavish jewel]

the thing with linear algebra is that they tend to teach it in one of two ways

all with vectors and matrices

or based on the definitions of what vector spaces are

with vectors and matrices, people get lulled into thinking that is all linalg is

with the latter, holy crap,

[lucid cedar]

I took an intro class that was packaged with diffyQ. and then I took a follow up class with axlers book

The follow up class was a challanging semester as many of the helpers here can probably attest lol. but it was alot of symbol manipulation. I felt like sitting to ponder the material was too slow

too slow when you have looming test deadlines i should add

[reef sleet]

yeah I wish I could learn it based on what vector spaces are

I think my class focused on the matrices & vector part though

[granite tiger]

@rose grotto help

[wintry sphinx]

this is not linear algebra; next time ask in #prealg-and-algebra . You should review https://www.wikihow.com/Find-the-Equation-of-a-Line and see if you still don't know how to do it.

[toxic karma]

ok thank you for the help

[reef sleet]

What is this first theorem?

Is it not obvious??

The inverse of an inverse is the original thing

Just like the transpose of a transpose is the original thing

[wintry sphinx]

I don't think that follows straight from the definition

but it's a really quick proof

actually, it almost immediately follows from the definition lol

[reef sleet]

I would assume so lol

I can see why

[wintry sphinx]

but it's something that would require proof anyway

[reef sleet]

How am I supposed to interpret this?

Like it's not just the inverse of A

And it's not 1/A^2

Oh it's A^2 inverse

Weird

Earlier the textbook said that AC = BC doesn't always imply that A = B; so can you only cancel when the matrices A and B are invertible?

[nocturne jewel]

If C is invertible, then C^-1 exists

so then you can right and left multiply by C^-1

[reef sleet]

Which leaves you with A = B

Interesting

[humble oak]

could i get some help understanding how they get to the last line with the three equations?

what does it mean to equate coefficient of powers

ah nvm i got it

[zealous junco]

if A diagonalizable

then A = U D U^T can expressed?

U orthonormal

[native rampart]

If AA^T=A^TA yes(assuming we are talking over real matrices)

See Spectral Theorem

[zealous junco]

oh so ok

A has to be symmetric thx

[native rampart]

I mean A A^T =A^T A is a weaker condition

[zealous junco]

yea ic

[native rampart]

If we are talking complex matrices, this is no longer true

For that you would need AA*=A* A,A* is the conjugate transpose

[vocal prairie]

I need to compute the dimension of the vector space n x n matrices with trace 0. Would like some hints for the same.

I tried working with nxn matrices to check how many matrices I need to span the space. For the case where trace 0 is not the restriction, it has dimension n^2(I guess?). I don't know how things work when we have the trace 0 restriction. I'm assuming it should free me of the burden of the diagonal(and hence should have a dimension n^2-n), but not entirely sure.

[zealous junco]

yea n^2 if trae nonzero

trace(A) = 0 i haven't figured out a basis but im guessing n^2-1

because a11+...+ann = 0 is the only restriction

[vocal prairie]

Aaahhhh

That makes sense

[zealous junco]

actually yea

you could construct a standard basis B

Aij in B where Aij = ith row jth column 1, else 0

and so you see that once you constructed everything else Ann is a linearly dependent vector

[vocal prairie]

Thanks!

[zealous junco]

np

So if I have an induced norms on linear transformation T:V->V

then if A is the unique matrix of T,

is the basis the infinity norm works on the standard basis for A?, if say it induces infinity norm from V

nvm im being dumb

[floral thistle]

Could someone help me with this

?

I'm still cracking at it

[hallow cliff]

find the projection of x onto theta and theta_0

then subtract that projection from x

you get the component of x that is not in the hyperplane, and you can find its norm to get the distance

[floral thistle]

@hallow cliff You still there?

I'm trying to make sense of what is theta_0

geometrically

[lavish jewel]

by the description

theta is the normal of the plane

theta0 is a poimt on the plane

[floral thistle]

An offset is a point, @lavish jewel?

[lavish jewel]

sure

think about it. if you have a point, you can describe it as a vector pointing from the origin to that point

[floral thistle]

Yes

That's true

[lavish jewel]

but if you want to offset the origin to a particular point

then you get a new vector

say the original was x, w.r.t origin

[floral thistle]

Oh damn....

[lavish jewel]

the new one is x - x0

[floral thistle]

This is gonna be messy to explain in words

[lavish jewel]

just do head - tail

[floral thistle]

But I'm imagining the normal as a vector with its tail in the origin

And the offset is a displacement vector of that tail

[lavish jewel]

yes

no

not of the normal

the normal only gives direction

[floral thistle]

Then I still can't figure out the offset

[lavish jewel]

the offset tells you how vectors are related to theta0

and the normal theta tells you if the orientation is correct

do you know the definition of a plane?

or understand it geometrically? a drawing really helps here

[floral thistle]

I don't remember it

I know that in a plane

ax + by + cz = d

(a,b,c) is the normal vector to the plane

But I've always struggled with understanding the role of d geometrically

[lavish jewel]

how do you know if a vector is in a plane?

[floral thistle]

$\vec{n} \cdot \vec{x} = d$

[stoic python]

Max Hetfield

[lavish jewel]

what does that mean?

consider this

the normal is magnitude 1

[floral thistle]

Aha

[floral thistle]

Only this?

[lavish jewel]

whats another way of writing the dot product?

[floral thistle]

The square of the norm of a vector

[lavish jewel]

not quite

[floral thistle]

No?

I gotta go Edd, thanks for the help

I'm gonna try to keep thinking about what you said

[zealous junco]

I think I solved this but wanted to check

since I'm quite new to concept of norms & on lin operators

So to show this, is to show $||D||_\infty = \underset{||v|| = 1}{\sup}||Dv||$

[stoic python]

Anticipation

[zealous junco]

is equal to max of the Dii

[lavish jewel]

yes

infinity norms on the right, too, yeah?

[zealous junco]

yea,

where $\underset{v = 1}{\sup}||Dv|| = \underset{|x_i| ≤ 1}{\max} {|D_{11}x_1|, ...,|D_{nn}x_n|}$

[lavish jewel]

what's v on the right sight

[zealous junco]

same v right

like in infinite norm

[lavish jewel]

it's not there anymore tho

[zealous junco]

oh i meant v = [x1,...,xn]

since v is in R^n in the above question

[lavish jewel]

just go ahead and replace that by abs(x_i) <= 1

[zealous junco]

ah

so writing this is ok and equivalent?

just confirming

[lavish jewel]

yes

that's the definition of the infinity norm

anyway there is no v on the RHS so it makes it confusing 😛

[zealous junco]

does this make it better in any way

[stoic python]

Anticipation

[zealous junco]

And then I have to claim that if wlog I let |Dnn| to be the largest diagonal entry

then it is an upper bound and least of the upper bound?

to finish the proof

[lavish jewel]

almost

just address the signs, i guess

[zealous junco]

wait wdym

I thought ther r all positive value

[lavish jewel]

well D could have all negative values

but i guess in the case of a diagonal matrix it's ok

[zealous junco]

|D11x1| = |D11||x1| right

ah

ok

[lavish jewel]

since you're not adding up terms per entry

[zealous junco]

ic

so |Dnn| is an upper bound is just obvious

but it's a least because suppose I have a smaller one K, then I can design v = (0,...,1)

right

[lavish jewel]

yeah, you can first say that the expression is trivially maximized by letting all x_i = 1

[zealous junco]

for contradiction

[lavish jewel]

and then address Dnn

[zealous junco]

ah

ok thanks yea

[lavish jewel]

which smaller K do you mean

[zealous junco]

Like suppose K is a smaller upper bound K < |Dnn|, then what i meant was

I can just say let v = (0,....,1) then |Dv| = |Dnn| contradiction so there is no smaller UB

in infinity norm

i guess ur way is much better, like let all the i's be 1

so it naturally becomes max{|Dii| i = 1,...,n}

[lavish jewel]

i just don't see how considering K shows it's a least upper bound

we already said Dnn had the maximum magnitude

[zealous junco]

oh I meant to consider K after showing |Dnn| is an upper bound

[lavish jewel]

an upper bound to what

[zealous junco]

to the |Dv|, |v| = 1's

hol up i think im being dumb

[lavish jewel]

you owuld have to construct what the upper bounds look like first

the you mention has nothing to do with it

the K

[zealous junco]

I'm wondering if this is jutified

wait i'll write my full proof down

yea idk how to justify the last step?

i know it makes perfect sense

[lavish jewel]

i'm a little rusty at this but

i think you can just swap out supremum by maximum in this case

someone else will have to help you with the last part cuz i haven't done this, specifically, in a long time

[native rampart]

Kind of

[zealous junco]

ok thanks i mean everything before last line seems well justified right

[native rampart]

How are you defining arithmetic operations?

What do arithmetic operations mean?

On an arbitrary set

[zealous junco]

[native rampart]

Subtraction and divison aren't exactly operators

You can use addition and multiplication to define those 2

Addition with an additive inverse

And multiplication with a multiplicative inverse

That's a very specific thing in N

[zealous junco]

yea

[native rampart]

Yes

[zealous junco]

given an ordered field you can define up to Q I believe

rationals

just using 0 and 1

[native rampart]

I don't think you need an order for that

Every field has Q or Z/pZ as a subfield

[limber sierra]

how do you define multiplication as repeated addition in R?

[hint: you cant]

what would you be adding

??

what addition formula does pi * pi represent?

(aside: as you progress in linear algebra, you may learn that the dimension of R over Q is infinite, and in fact the basis is unconstructible)

(which implies that there's no way to define multiplication as repeated addition in R)

even R is a bit overengineered though

consider the finite field of size 2^2 = 4

let's label the elements, 0, 1, a, b

okay so consider the field of size 4

this has 4 elements, 0, 1, a, b

0 is the + identity, 1 the * identity

its operations are given as such

(taken from google images, sorry theyre messy)

as you can see, a*b = 1; but how do you represent this as repeated addition?

well a + a = 0

and b + b = 0

so if you "repeat addition" by a or by b

a bunch

you'll get either 0 or a/b

theres no way to get 1 from that

so theres no way to represent a*b = 1 as repeated addition.

it's a totally separate operation

now, it's related to addition by distributivity

but it's separate from it

[lucid cedar]

Thats wild

[limber sierra]

not really, {0, 1} forms a subfield of the field of 4 elements, and it's of prime order so it behaves like the integers

for fields that behave like the integers, it does make sense to connect multiplication to repeated addition

that's how multiplication in the integers (and hence in modulo fields) is defined, in fact

but i wouldnt worry too much about this

they are not.

[native rampart]

Elements

[limber sierra]

ab = 1

which is outside of the subset {a, b}

so it cant possibly form a field

even ignoring that issue, there's no additive identity

[lucid cedar]

They also require the identities too

[limber sierra]

right

if you just look at this part of the tables

you get a field

the unique field of order 2, in fact

but you cant do the same if a and/or b are involved

(unless it's the ENTIRE field)

one fact that's worth knowing is that all finite fields have order p^k for some prime number p and some natural number k

and moreover, two finite fields of the same order are isomorphic

this fact may not be discussed in a linear algebra class but it's good to have for a "gut check"

graph isomorphisms are one type of isomorphisms.

field isomorphisms are another.

it's just saying that two things are "the same", just with elements "relabelled"

[native rampart]

As sets yes

[zealous junco]

f(a*b) = f(a)f(b), f(a+b) = f(a)+f(b)

and f bijective

[split wedge]

Any guesses?

[limber sierra]

my guess is that this looks suspiciously like a test

[vocal prairie]

How does linear independence over spaces of function work? If I use coefficients x_1 and x_2, set them up as x_1(1)+x_2(t)=0, I get x_1+tx_2=0. Is this sufficient to conclude x_1=x_2=0? If t itself happens to be 0, x_1 would be forced into being 0, but x_2 could assume any value, right?

(I'm responding to part (a))

[limber sierra]

the zero vector in a space of functions is the zero function

ie the function that maps t |-> 0

anyway

from x_1 + tx_2 = 0 you can conclude tx_2 = -x_1

which would imply that tx_2 is a constant function (unless x_1 = x_2 = 0)

[vocal prairie]

Yes

[limber sierra]

but that obviously aint true since t varies

[vocal prairie]

Owwww

Okay, that makes sense.

[limber sierra]

if youd like, formally you can rearrange this to t = -x_1/x_2

which is constant

which is clearly a problem

but i dont think that last step is necessary

it should be fairly obvious that tx_2 cant be the constant function

(unless x_2 = 0, in which case x_1 = 0 as well)

[vocal prairie]

Makes sense. I was kinda getting confused between the variables and constants here; somehow I kept thinking x_1 and x_2 could keep changing as t changed so that the equality makes sense.

[zealous junco]

iw as thinking

if T^(n-1) ≠ 0

T^n = 0

does T have n eigenvectors

and ifso why

cuz T^n has n eigenvectors right

[lavish jewel]

you mean regular matrix exponentiation?

is T of any special structure?

[zealous junco]

yea nvm its jut general

i realized the D operator doesn't

all good

[lavish jewel]

anyway your wording was off

it does have n eigenvectors, the problem is whether the corresponding eigenvalues are 0 or not

[zealous junco]

wait D does?

[lavish jewel]

does what?

[zealous junco]

have n eigenvectors

[lavish jewel]

what is D

[zealous junco]

Differentiation on P_n -> P_n

[lavish jewel]

ah

[zealous junco]

so I was thinking D^n has n eigenvectors

but D doesn't

[lavish jewel]

D has infinitely many eigenvectors if it's a linear differential operator

as far as i recall

[zealous junco]

oh

i dind't calrify

P_n is vector space of ≤n degree polynomial

[lavish jewel]

oh, that's a lot more specific

just by looking at your question though, it has to be singular

[zealous junco]

Yea

[lavish jewel]

so at least 1 eigenvector as a 0 eigenvalue

mind you it still has n eigenvectors

just that past a certain point, the eigenvalues are 0

with multiplicity of the size of the null space

[zealous junco]

hold o

N(D-lambda*I) = [1, 0,...,0] isn't it, if I choose the basis {1,x,x^2,...,x^n}

[lavish jewel]

like, tell me the eigenvectors of $\begin{bmatrix}
1 & 0 & 0 \
0 & 1 & 0 \
0 & 0 & 0
\end{bmatrix}$

[stoic python]

Edd

[lavish jewel]

how many eigenvectors are there?

[zealous junco]

3

cuz if lambda = 1

there's 2

right?

[lavish jewel]

what is lambda

[zealous junco]

the eigenvalue

[lavish jewel]

i think i'm not understanding your english haha

[zealous junco]

there's 2 eigenvalues for that matrix

0, 1

and for 0, I think the eigenvector is [0,0,1]

for eigenvalue 1, the eigenvectors are [1,0,0], [0,1,0]

[lavish jewel]

yeah

[zealous junco]

my point is D is kind of different though

[lavish jewel]

what are the dimensions of D for a given order n of polynomials?

[zealous junco]

$\begin{bmatrix}
0 & 1 & 0
0 & 0 & 2
0 & 0 & 0
\end{bmatrix}$

[stoic python]

Anticipation

[zealous junco]

Dimension? I mean if polynomial order n then

dimP_n = n+1

[lavish jewel]

so n+1 by n+1

then it has n + 1 eigenvectors

[zealous junco]

not necessarily right

some eigenvalues have multiplicity ≥ 1 and null space of that is small

for instance i mean

[0, 1, 0;

0, 0, 2;

[lavish jewel]

so you mean UNIQUE eigenvectors that are also linearly independent?

[zealous junco]

oh yea

[lavish jewel]

...

[zealous junco]

?

i thought when

you talk about eigenvector you mean it's linearly indep

[lavish jewel]

no

only for special matrices

[zealous junco]

yea from what i seen

that' be normal matrix or something

isn't it normal iff diagonalizable

[lavish jewel]

no

normal iff it commutes with its hermitian transpose

and this implies it is diagonalizable

but not backwards

[zealous junco]

oh ok

also last thing

similar matrices preserve eigenvalues?

[lavish jewel]

you are assuming your matrix is diagonalizable, i guess. idk if its, mind you. a lot of differentiable operators are hermitian though (or so-called "self adjoint"), so it should be ok

[zealous junco]

ok

thanks for the help 😅

[lavish jewel]

and yes, similarity transformations are "eigenvalue-preserving transformations"

(the EVD is precisely this)

[novel hamlet]

can someone help me to get to start with this problem: let f : R^3->R^3 be linear systems (x, y, z) → (3y−2z, 4x+2z, y−x). find Transformation matrix for system f from standard base (e1,e2,e3) to standard base (e1,e2,e3)

[gray dust]

find each f(e_i)

[novel hamlet]

yeah i get the idea of substityting x,y,z with e1,,e2,e3 but im not sure what happens when i need to "switch" base

[gray dust]

just do it. f(e_1)=?

[novel hamlet]

then i would get (3e_2+2e_3, 4e_1+2e_3, e_1-e_2)?

but im not sure how to get the transformation matrix out of this

[gray dust]

you wrote nonsense. that's not what e1,e2,e3 mean

the standard basis of R^3 is e1=(1,0,0), e2=(0,1,0), e3=(0,0,1)

[novel hamlet]

so i should be getting matrix like this out when i input the standard ones in

and do the math

[gray dust]

find each f(e_i) and those will be the columns of the matrix

[novel hamlet]

thats what i got out from calculating that 1st colum is 3e2-2e3 (got it wrong there the -2 should be bottommost)

[gray dust]

again that's not what the standard basis means

for example e1=(1,0,0) so f(e1)=f(1,0,0)

[novel hamlet]

now im kinda confused what im supposed to do on this then

[gray dust]

for example e1=(1,0,0) so f(e1)=f(1,0,0)
we find f(1,0,0). plug x=1, y=0, z=0

[gray dust]

you keep thinking e1, e2, e3 are components of a vector. they are not. they are vectors themselves defined here

the standard basis of R^3 is e1=(1,0,0), e2=(0,1,0), e3=(0,0,1)

[novel hamlet]

so when i plyg them in i get e1= 0,4,-1, e2=3,0,1 and e3=-2,2.0?

[gray dust]

you mean f(e1)=(0,4,-1), etc

[novel hamlet]

yes

[gray dust]

then those vectors make up the columns of the matrix we want

[novel hamlet]

then what about the part 2 i need to change them from standard base to standard base

that should be the transformation matrix?

[gray dust]

each f(e_i) is ALREADY its own coordinates in the standard basis. no more work to do

[novel hamlet]

yeah that is the part that got me confused on this one as well. why i eed to change them to standard basis if they are already in standard basis

[gray dust]

in GENERAL we must obtain the coordinates of each output with respect to whatever basis of the output space we have. however vectors in R^3 are already their own coordinates with respect to the standard basis of R^3

[novel hamlet]

yeah, i think i get this idea now

Let V be a K-Vectorspace and f:V→V and ker(f)∩im(f)={0}. can i assume that f∘f=f.

[gray dust]

that's not enough. we must in fact have $\ker(f)\oplus\im(f)=V$

[stoic python]

RokabeJintarou

[novel hamlet]

yeah that is the part im trying to prove, i found nice math stackexchange but it reguires f∘f=f.

[tame mural]

Just curious, when would that not be the case?

for infinite-dimensional spaces?

Because I thought ker(f) ∩ im(f)={0} meant you have a direct sum

if not, then I better update my notes

[gray dust]

@tame mural we must have ker(f)+im(f)=V too in order for ker(f) oplus im(f)=V

i imagine what you're proving is $\ker(f)\oplus\im(f)=V$ iff $f^2=f$

[stoic python]

RokabeJintarou

[gray dust]

@novel hamlet in which case you must prove an implication both ways

[novel hamlet]

basically i got same problem as https://math.stackexchange.com/questions/12122/show-that-v-mboxkerf-oplus-mboximf-for-a-linear-map-with-f-circ

but that i know ker(f)∩im(f)={0} .and not that f∘f=f .

[gray dust]

then you don't really have the same problem

[novel hamlet]

back to google it is then

[novel hamlet]

back to transformation matrices, i have

i ahve done (e11,e12,e21,e22)

but im not sure how to get 4x4 matrix out of them

[lavish jewel]

you'Re not listening to what they're telling you

[novel hamlet]

wait do i just put them in the 4x4 matrix like

[gray dust]

not they, me

[lavish jewel]

this problem is multilinear algebra btw. choose your poison of vectorization, unfoldings, or contractions

[novel hamlet]

do i just blap them here like this

[gray dust]

the issue is you don't know what basis means

[novel hamlet]

i know the 2x2 basis vecotrs are, and i know how to run them trough the function, the problem is i dont understand how i am supposed to get 4x4 matrix out of 2x2 matrices

[lavish jewel]

so many problems in one sentence

i forfeit my life

[novel hamlet]

sorry english is not native to me

[lavish jewel]

the math is wrong, not the english

[gray dust]

do you know what e11 is?

[novel hamlet]

they look like that

[gray dust]

ok E, not e

F(E^11)=?

[gray dust]

F(E^11) is wrong

[novel hamlet]

ah yes im stupid

it should copy the row

[gray dust]

no

F(X)=MX, not XM

[lavish jewel]

thou shallt not commute the matrices

[novel hamlet]

yeah i think where i got the problem

[gray dust]

what are the coordinates of F(E^11) in the standard basis of R(2x2)?

[novel hamlet]

isnt it that {[a,0},[c,0]}?

wen i run it trought that MX

[gray dust]

tell me the definition of coordinates wrt a basis

[novel hamlet]

basis is any amount of vectors that span base ie. all other vectors can be expressed as their linear combinations.

[lavish jewel]

so... what linear combination do you need here

[gray dust]

i asked for the definition of coordinates in a basis, not a basis itself

also that's not a complete definition of basis

[novel hamlet]

coordinates are some linear combination of basis

[gray dust]

no

[novel hamlet]

eg. 2d space point (3,4) is same as 3(e1)+4(e2)

[gray dust]

in a general vector space, how are coordinates in a basis defined?

[novel hamlet]

doesnt it work the same way, coordinate in space are linear combination of basis of the space?

[gray dust]

@red ether channel occupied

coordinates are NOT the linear combos themselves

the coordinates of a vector in a basis is a tuple made of the unique coefficients that make the vector a linear combo of the basis vectors

[novel hamlet]

isnt that same as coordinates are some linear combination of basis (there is only 1 unique way of doing so)

[gray dust]

the vector $$F(E^{11})=\m{a&0\c&0}$$ is written as a linear combo of $(E^{11},E^{12},E^{21},E^{22})$, the standard basis of $\bR^{2\times 2}$, as $$F(E^{11})=aE^{11}+0E^{12}+cE^{21}+0E^{22}$$ so the coordinates of $F(E^{11})$ in the standard basis is $(a,0,c,0)$

[novel hamlet]

https://cdn.discordapp.com/attachments/540211747613704221/807265747881295912/unknown.png

so where is that m here?

since that A looks like M*E11

[stoic python]

RokabeJintarou

[novel hamlet]

isnt that the thing i have been doing?

[gray dust]

not at all. look at the coordinates of F(E^11)

[novel hamlet]

i think i realize that now

wait a sec ill do the math

[gray dust]

note especially that the coordinates of F(E^11) is a tuple with 4 entries

[novel hamlet]

so end result looks like that?

i just slap them as colums to 4x4 matrix

like in the previous example

[gray dust]

note especially that the coordinates of F(E^11) is a tuple with 4 entries
THAT'S why we expect the standard matrix of F to be a 4 by 4 matrix

[novel hamlet]

yeah now i understand this

and how this works

[gray dust]

no really, make sure you get the definition of coordinates in a basis

the coordinates of a vector in a basis is a tuple made of the unique coefficients that make the vector a linear combo of the basis vectors

[lavish jewel]

may i ask why you preferred doing everything symbolically instead of explaining with an example?

[novel hamlet]

easier to use later on + all textbook examples are symbolically

[lavish jewel]

i meant the other person, but sure

[gray dust]

how is the latex not an example

[lavish jewel]

i just meant a simple b = Ax was easier since it's already in the form they need

ofc isomorphic and whatever, but maybe clearer to see

[gray dust]

what

[lavish jewel]

using a coordinate vector x and putting the linear combination directly as a matrix-vector product

doesn't matter, don't sweat it

[gray dust]

there's 0 need to rewrite that way

[lavish jewel]

sure

there's 0 need to write it as you did too

the original problem is a 3-mode tensor product with a vector

but some formats are easier to grasp

it's fine, don'T mind me. i'm nitpicking

[gray dust]

there's no nitpicking for you to do. but i'll nitpick you instead

i directly used a definition of coordinates. write each f(..) as a linear combo of the basis vectors. the coeffs go into a tuple. that's it

[lavish jewel]

i understand and i agree

[gray dust]

that's an example of finding coords. don't say it's not

[lavish jewel]

my wording was poor indeed

you didn't seem to get your point across tho, so a different way of doing the same thing, which was already correct and properly exemplary, might have been helpful. that is what i meant

it is literally of no import now that it's solved, i'm just being annoying

[novel hamlet]

Im confused about the notation on my assingment, how do you read/understand A=[e1....ek 0....0] ∈ R^nxm

[lavish jewel]

see...

[gray dust]

show a picture

[novel hamlet]

there is nothing beween ek and 0

[gray dust]

each e_i is likely an m-tuple and 0 is really a column of 0s

[novel hamlet]

question is: let f:v-w be finite vecotr spaces show that bases for V and W exist that linear map f transformation matrix is A

[lavish jewel]

see, the reason i asked is that i tried and they deleted my message

so that is why i was asking about all this

[novel hamlet]

basically i need to show that (v1...vn) and (w1...wn) exist with that kind of A

but the prroblem is i dont know how does that matrix A look

[gray dust]

each e_i is likely an m-tuple and 0 is really a column of 0s
each entry you see really represents a column

[novel hamlet]

soo basically i have K colums

and under them just colums of 0?

[gray dust]

k columns e_1,e_2...,e_k then n-k columns made of 0s

[novel hamlet]

this is one of these math courses that want me to just go and hit my head on wall

[digital bough]

I think i am fucked. I am halfway through this book and kinda rushed the chapter about determinants.

I got three weeks left to learn all stuff about linear maps required for spectral theory and applied specteal theory. I just need to be able to solve some differential equations and some difference equations (recurrence relations) to pass.

Is this even doable? Planning to just focus on the other courses instead and do Linalg some other time.

[lavish jewel]

determinants aren't very important

the eigenvalues are the key there

[native rampart]

Hello,axler

[lavish jewel]

the determinant is a proxy for degeneracy

who is axler?

[wintry steppe]

[lavish jewel]

hmm?

maybe it has a different name in english

rank deficiency?

no replies 3;

[hollow holly]

idk if this is the right channel but I need help with question b

[lavish jewel]

for a), what have you tried

[hollow holly]

for a) I proved there exists an inverse of the function

[lavish jewel]

aight

how about c

[hollow holly]

I proved that the function preserves distance

[lavish jewel]

sounds about right

[hollow holly]

I've been using this to get started

I have a point p(s, t) and I need to show that the point exists on m, but I'm lost

[lavish jewel]

idk, not gonna lie :x a lot of people have been lurking around. maybe they can help

[hollow holly]

lol nw

<@&286206848099549185>

[somber lintel]

Does the geometric formula (|a||b|sin(theta)n) for the cross product work in all situations?

say I have two vectors: <0, 0, 5>, and <6, 3>. I don't think that this formula would work, correct?

because there is no unit vector (n) perpindicular to these two vectors

[lavish jewel]

that operation is not defined, unfortunately

[somber lintel]

sorry, what does it mean for an opperation to be undefined?

[lavish jewel]

it does not exist

you cannot even calculate the cross product between those two vectors

[nocturne jewel]

Cross product only works if the 2 vectors are in R^3

[somber lintel]

o dam

so they both have to be on a plane?

[lavish jewel]

no?

[somber lintel]

oh wait no

[lavish jewel]

well

[somber lintel]

they have to be on the plane which contains ihat and jhat, or khat

[half storm]

All vectors are in some plane so yes but no

all you need really is that they are vectors in r^3

[lavish jewel]

they have to exist in 3d space

[somber lintel]

they have to be on a plane with at least two unit vectors

[lavish jewel]

that's all

[nocturne jewel]

R^3 vectors are cartesian vectors with an x,y,z

[lavish jewel]

all planes have at least 2 basis vectors that describe them

i think you have more than one problem here :x

[nocturne jewel]

you can make the R^2 vector into and R^3 vector though

by recognizing the xy plane is located at z=0 in a 3D space

[lavish jewel]

that's kinda... wishy washy 😛

[somber lintel]

huh, sorry I am little bit confused. So in what situations are you able to calculate the cross products and which ones are you not?

[lavish jewel]

if you have 2 vectors in 3d space (i.e. with 3 elements)

you can do it

[somber lintel]

so if you have the vectors <0, 0, 5> and <6, 3, 0> can you calculate the cross product?

[lavish jewel]

yep

[nocturne jewel]

yeye but the point is made lol

[somber lintel]

but the geometric formula (|a||b|sin(theta)n) would not work? You would have to use the "elemental" formula which subtracts the elements and stuff?

cx = aybz − azby
cy = azbx − axbz
cz = axby − aybx

this one

[wintry sphinx]

both work

not sure what you're asking here

[somber lintel]

i am confusion

[wintry sphinx]

the formula |a||b|sin theta only gives you the magnitude of the cross product, anyway

[somber lintel]

here lemme make my question more clear

[wintry sphinx]

the cross product can be calculated for any pair of vectors in R^3

[nocturne jewel]

you should usually use the non-geometric formula, cause how else do you find the n vector for the geometric formula?

[wintry sphinx]

orthogonal complement

[somber lintel]

ik, i just want to make sure i understand it

[nocturne jewel]

is that what the normal vector is called?

[wintry sphinx]

basically; the orthogonal complement is a subspace

and in the case of a plane in R^3, it's just the span of the normal vector

[nocturne jewel]

so yes...?

[lavish jewel]

ask away

[wintry sphinx]

yeah basically

[gray dust]

@wintry sphinx don't confirm if you know he's wrong

[somber lintel]

so |a||b|sin(theta) gives you the scalar which represents the magnitude of the vector returned by the cross product. Then n is the unit vector that is at right angles to both vectors. But for the vectors <6, 3, 0> and <0, 5, 0>, there is no unit vector which is perpindicular to both of these vectors. So, to my understanding, this formula: |a||b|sin(theta) is just scaling a unit vector. But in this case there is no unit vector perpindicular to both vectors.

shit

how do I get rid of the bubble

[nocturne jewel]

put a \ in front of one i believe

[wintry sphinx]

your assertion that for those two vectors that there is no vector orthogonal to both of them is just wrong

[lavish jewel]

isn't (0,0,1) perpendicular to both?

bruh

[wintry sphinx]

two linearly independent vectors always span a plane

[lavish jewel]

i can give you an alternate way to think of it

[wintry sphinx]

and the normal vector to that plane is orthogonal to both of them

[lavish jewel]

you can always find a plane that passes through 3 points

and planes are defined with help of a normal vector that is perpendicular to the plane

yeah?

[somber lintel]

yes

[lavish jewel]

ok

you know you can get a vector by subtracting two points in space, yeah?

and that vector points from one of the two points to the other

so far so good?

[wintry sphinx]

basically you can always find a plane between two vectors

[somber lintel]

I see, and n is the vector perpindicular to that plane

[lavish jewel]

(it depends on how you interpret the vectors, calm down a sec)

yeah

you grab the origin, and the points the vectors point to

that's 3 points

[somber lintel]

so, is n always a unit vector

[lavish jewel]

for a plane, yes. in a cross product, no

[wintry sphinx]

the vectors perpendicular to the plane are not necessarily unit vectors, but you can always normalize it to make it a unit vector

[somber lintel]

alright, lemme open like a 3-d gepahing calculator to see this stuff . My brain is overloading sorry

[nocturne jewel]

Why did the co-ordinate vectors of the functions go into the matrix as rows and not columns if they're transposed?

[wintry sphinx]

I don't think it matters; since you're trying to figure out whether it's a basis or not, the matrix is always square

and then whether the matrix is invertible or not (i.e. the basis vectors are lin independent) can be ascertained by either row or column operations

[somber lintel]

alright, so say I had these two vectors

and I wanted to calculate the cross product of it. How would the |a||b|sin(theta)n formula work?

Im probably mistaken, but the vector <0, 0, 1> doesn't seem to be perp to both?

[wintry sphinx]

but you can clearly see that there's a plane between them

[lavish jewel]

rref works on rows, so it shows that rows are linearly independent. so you put the vectors you wanna test as rows. as stated though, the row rank and column rank of a matrix are equivalent

[wintry sphinx]

and that there is a vector that's perpendicular to both

[somber lintel]

yes, I see that plane

[wintry sphinx]

just imagine the normal vector to that plane

that's clearly perpendicular to both of those vectors

[somber lintel]

an I understand that there is a vector perpindicular to both.

[wintry sphinx]

that is basically in the same direction as the cross product

[somber lintel]

I just don't understand how you can calculate the cross product for these two kinds of vectors using the |a||b|sin(theta)n

[lavish jewel]

what vectors did you plot

[wintry sphinx]

(or opposite, but nitpicking)

you can't really calculate the cross product with that formula

you can only calculate how long it is

well, there's the whole solve a system of equations thing to figure out the direction, but the point is that formula alone won't get you the cross product

[somber lintel]

so, just to make sure that I am understanding correctly: This formula |a||b|sin(theta)n will only work if you have two basis vectors in that same plane (or you change the location of the basis vectors to the plane that passes through the two vectors a and b, but that is probably not practical)

[nocturne jewel]

Ok cause the next example had them inserted as columns so I was confused lol

[wintry sphinx]

no, I think you're thinking of it in a very improper way; even if what you say is technically correct, it's missing the point

I suggest writing the formula like

$||a \times b|| = ||a||||b||\sin \theta$

[stoic python]

Saccharine

[wintry sphinx]

that formula only gives you how long the cross product is

i.e. the magnitude

not the direction

[somber lintel]

https://en.wikipedia.org/wiki/Cross_product

[wintry sphinx]

the two vectors that you're crossing are always in the same plane; this is just a fact of vectors

[somber lintel]

your leaving out the 'n' part of the formula

[wintry sphinx]

I'm not

The way I wrote it says exactly the same thing

the reason why I wrote it that way is because you don't seem to understand what the n means, and it's not particularly relevant

and it's pretty abstract

[somber lintel]

the equation you wrote is correct, and I understand that, it is jut a part of the defintion for the cross produt

[wintry sphinx]

n is literally defined as "the unit vector in the direction of the cross product"

the point is that the formula gives you the magnitude