linear-algebra

[spare widget]

the above does this for a linear map

ah ok, guess I misunderstood what you were trying to do

[zinc copper]

.

We saw bases for multilinear forms first semester so the proof is quick

But it doesn’t really tell me anything

[zinc timber]

@zinc copper general bilinear forms are not really an endomorphism

[zinc copper]

But both are identified with nxn matrice so there must be a connection somehow

I’ve learnt to take nothing as a coincidence in math anymore

[zinc timber]

bilinear could be from VxW

[zinc copper]

Si right now I mainly mean bilinear forms on V

[zinc timber]

the nxn matrix only comes when both the spaces are the same

so you mean from VxV ?

[zinc copper]

So elements of V^* \otimes V^*

[zinc copper]

Yeah

[zinc timber]

[zinc copper]

On VxW you’re right it wouldn’t be nxn

[zinc timber]

idk what kind of connection you are looking for

[zinc copper]

Me neither lol

[zinc timber]

if the bilinear form is +ve definite then it can be thought of as an inner prod

(also has to be symm)

[zinc copper]

I mean it’s not so much thought of as defined there is it 🤔

But yeah I’m looking for something else

[zinc timber]

I kinda understand what you want but also kinda don't

lol

[zinc copper]

Cause you get an associated endomorphism which you can describe explicitly by just looking at the matrix but I can’t assign meaning to this endomorphism relative to the bilinear form

[zinc copper]

Is there even a name for this map/matrix

[zinc timber]

ig you can think of a bilinear form as an endomorphism of V\otimes V?

hmm..

[zinc copper]

Yeah but then you don’t have nxn matrices

So different kind of identification

[zinc timber]

ye

[zinc copper]

I think this was somewhat on track I’ll see again on paper when I’m done eating

[zinc timber]

it's like you are applying the transformation before dot'ing them

like x.y but with x.By

[spare widget]

looks like changing the basis for y

and then inner product

[zinc timber]

"basis" change then you also need to change the basis of x

[spare widget]

you can assume x is already there

[zinc timber]

probably not the right interpretation

[spare widget]

it's like pairing x from that new basis

with an y from the old

so you need to transform y to the new one

[zinc timber]

you are literally changing x then

[spare widget]

you're just treating it as defined wrt some other basis

not changing it

[zinc timber]

like 'same coordinates' but in 'different basis' is longer x

[spare widget]

$([x]_B)^T[y]_B = ([x]_B)^TB[y]_C$

[stoic python]

criver

[spare widget]

that's how I would interpret this

i.e. you're given a covector with respect to some basis, and a contravariant vector with respect to another, so you have to transform the contravariant vector to the same basis in order to be able to feed it to the covector

[zinc timber]

what if B is singular

you interpretation falls apart

[spare widget]

then it goes to some subspace

[zinc timber]

transformation is much more accurate

[zinc copper]

Yeah my earlier interpretation is right I think: consider the unique endomorphism $\psi: V^* \to V^$ such that $\psi(e_i^) = \left<e_i, \bullet\right>$. Then its dual has matrix A in the basis $\mathscr{B} = {e_1,…,e_n}$

Not particularly enlightening but I’ve never found anything to be enlightening with dual morphisms anyways

[stoic python]

𝓛ittle ℕarwhal ✓

[zinc copper]

Hmm I guess in a sense a dual morphism can be extended to a bilinear form on V* x V

[spare widget]

That's what I mean by paiting the covector and contravariant vector

Typically that's considered a measurement of some sort in physics

The contravariant vector being the thing to be measured and the covector being your measuring device/sensor

[short stirrup]

what do you mean? like a software to put the matrix into?

[spare widget]

https://en.wikipedia.org/wiki/3D_modeling

[zinc copper]

right that makes sense

[spare widget]

then I interpret the matrix just as a change of basis or transformations of the contravariant vector

[short stirrup]

im not trying to create a complex model of the skateboard, just a representation of the length and width, maybe just a 3d rectangle the same length and width of the skateboard.

[spare widget]

then apply the rotation matrix to the vertices of the rectangle

if you want to rotate it around its center, then you'll have to translate it to the origin -> rotate -> translate back to its original location

[short stirrup]

what software can i use to visualize that? i downloaded the free trial of matlab but im not sure if i can use that for this

[spare widget]

https://www.anim8or.com/

https://www.blender.org/

[spare widget]

MATLAB should be able to do it too

[short stirrup]

alright, thank you.

[wintry steppe]

Quick question regarding matrices: If I factor a matrix to yield 1's along the diagonal and zeroes elsewhere is it still considered an identity matrix?

[tranquil steeple]

What do you mean?

[spare widget]

if you mean eigendecomposition, then QIQ^T = I, so you're technically factoring the identity matrix still

if you mean A = A * I = I * A, then this is clearly doing nothing

[vivid flume]

As long as there are 1s along the diagonal and the other elements are 0s its an identity matrix

[bold ermine]

help #help-22

[spare widget]

what have you tried?

[bold ermine]

should i send ss or something?

[spare widget]

nvm, someone's already helping you

[lavish jay]

I'm probably constructing this matrix wrong because myanswer differs from the book one, my matrix is $\begin{bmatrix} 1/2 & 1/2 \ 1/3 & 2/3 \end{bmatrix}$

[stoic python]

Guilhotina ᓇᘏᗢ

[lavish jay]

and also i don't think i understand leontief model correctly, the rows are the producers of goods and collumns the consumer of goods ?

[primal fable]

is there a good visual interpretation to what's happening on in a linear transformation when the algebraic and geometric multiplicities are different?
my reason for asking this is an attempt to understand why you can't have a jordan canonical form even if the field is algebraically closed

[gleaming knot]

Can't have a jordan canonical form even if the field is algebraically closed? What?

[primal fable]

you're only given the consumption, nothing about generation. you know that v (your economy state) must be an eigenvector of eigenvalue 1

so Av = v and you now need to solve for A

[gleaming knot]

As for visual interpretation, I'm taking a liking to comparing the generalized eigenspace with k[t]/(t^n)

[primal fable]

if I'm wrong here my bad, my interpretation was that you don't always have an eigenbasis for your domain but I might be mistaken and thinking that's equivalent to not having a JCF.

[gleaming knot]

So, having an eigenbasis is equivalent to saying the Jordan canonical form is a diagonal matrix

The cases where there are 1's on the off-diagonal is precisely the case of not having a basis of eigenvectors

[primal fable]

oo gotcha

[primal fable]

think you could expand on this? I'm assuming this is quotienting by an ideal but I'm lost as to why that's connected

[gleaming knot]

k[t]/(t^2) for example is the k[t] module with basis 1 and t, where t times t is 0

This corresponds to any Jordan block of the form $\begin{psmallmatrix}\lambda&1\0&\lambda\end{psmallmatrix}$

[stoic python]

Icy001

[gleaming knot]

Where the k[t] module corresponding to $(k^2,T=\begin{psmallmatrix}\lambda&1\0&\lambda\end{psmallmatrix})$ (here $k$ is the ground field) is isomorphic to $k[t]/(t^2)$ (note that $(T-\lambda)^2=0$

[stoic python]

Icy001

[gleaming knot]

Geometrically the Spec of $k[t]/(t^2)$ is a "thickened" point

[stoic python]

Icy001

[primal fable]

hmmm, hopefully I'll get it after looking over some notes, gimme a hot second but regardless thanks!

[lavish jay]

How can I do it? The problem says : Show that for this matrix the Leontief equation** x - Cx = d** has a unique equation for every vector d if $c_{21}c_{22} < 1 - c_{11}$

[stoic python]

Guilhotina ᓇᘏᗢ

[lavish jay]

i found the inverse of it

$\begin{bmatrix} 0 & \frac{1}{-c_{21}} \ 1 & \frac{-(1-c_{11})}{c_{21}c_{22}} \end{bmatrix} $

[stoic python]

Guilhotina ᓇᘏᗢ

[gray dust]

@brittle ingot #prealg-and-algebra

[brittle ingot]

thank you

[mossy coral]

Can someone explain why is matrix multiplication is not commutative but associative?

I understand why it is not commutative, but I dont understand how something can be associative but not commutative

[nocturne jewel]

cause commutativity and associativity are different properties

with no actual connection b/w them

[distant schooner]

seems to me like the step above (4) should be =, not <= right?

[hoary void]

no its pretty much just saying that something <= |something| (the middle term is what makes it need to do this)

u dot v doesn't have to be positive

[distant schooner]

oop didnt catch that

thanks

i thought it said 2(u.v)

didnt see that norm :((

[neon lake]

I'm having trouble with this exercise. For anyone wondering it is not an exam I'm taking, but one from around 2011, so do not worry. This is its translation:

a) Calculate the values ​​of parameter a so that the system is not independent (r(A)=r(A|B)= number of variables).

b) Is there a value of a for which x = 1, y = –3, z = –1 is the only solution in the system?

[torn stag]

@neon lake One approach: Put the system into a matrix. Row reduce the matrix. Find values of a for which the reduced system has infinitely many solutions (is that what "not independent" means?).

[neon lake]

Yes, I tried to do that, the solutions of the triangular matrix I got were a=2, but solutions said a=2 and a=9 (there's a a-9 in the 2nd row of the same matrix and I dont know what it has to do...)

Here, look

[timid sage]

If you aren't sure of how to solve a problem, I often find the best first step is to graph it. In this case, what we're looking for is two values of a such that at least two of the three resulting planes are parallel.

If we graph this in 2d (leaving out the first column), we can see that for a=9, the lines that represent the second and third rows are parallel, which is a fast way to confirm its a valid answer

[neon lake]

I suppose it is either parallel or one on top of each other

BTW thanks

[karmic dagger]

This is from discreet math, but it's linear manipulation

can anyone explain to me how the second line is achieved?

[stoic python]

Arkid

[karmic dagger]

and the next step after that

[neon lake]

Using determinants I got it right

[neon lake]

Thanks

[neon lake]

Sorry for interrupting you

[karmic dagger]

no worries

please ping me when done so I can rebring my question

[neon lake]

I think Im done

[karmic dagger]

if anyone can help clarifying this it'd be nice

[lavish jewel]

it's because of the determinant

if you have a matrix M with determinant |M|, then scaling the matrix by c (i.e. you now have cM) yields |cM| = c^n |M|

where the matrix M is of size nxn

the vertical bars around the matrix denote the determinant

[karmic dagger]

so the straight lines surrounding the matrix mean det

[lavish jewel]

still i don't quite get what they did

[teal grotto]

what about the first row of the matrix? i’m not following that part

[lavish jewel]

me neither

[teal grotto]

now it’s even more confusing haha

[lavish jewel]

i have no idea what they did

they subtract the first row from all the others, sure

[karmic dagger]

that makes sense yea

[lavish jewel]

but idk where they get that (t+1)^3 from

that doesn't look right to me

[karmic dagger]

I'll ask him then, the explanation you gave makes sense

and now I see too why it's confusing

[lavish jewel]

i know they want to boil it down to a case where taking the minors makes it easy

so they want to do it along the first col

[karmic dagger]

what about from to the second step of the second line

where he gathers the first column in the first row

[lavish jewel]

i don't understand your question

[karmic dagger]

assume LHS is correct

how is RHS achieved?

[lavish jewel]

idk, that looks wrong to me too

[karmic dagger]

it's okay, still thanks a lot

[lavish jewel]

i get the impression they knew the result ahead of time and then did random nonsense to reach it

[karmic dagger]

lol maybe

I'll ask them directly nonetheless

[tranquil steeple]

A trick is to see that it is a circulant matrix, and eigenvalues are given by $f(t_j)=\cos(t_j)+\cos(2t_j)+\cos(3t_j)$ for $t_j=(j-1)2\pi/4$ and $j=1,2,3,4$

[karmic dagger]

What's the intention behind it?

[tranquil steeple]

You have the spectrum and can construct the charpoly from that 🙂

[karmic dagger]

Ah I see, that's a nice formula

[stoic python]

Sven-Erik

[tranquil steeple]

(changed n to 4)

[fringe sierra]

intuitively, associativity lets you move the parenthesis around without affecting the result, so you can do both $(a \circ b) \circ c$ and $a \circ (b \circ c)$. however you can't swap operands around, that's what commutativity lets you do.

matrix multiplication and function composition are two standard examples that are associative but in most cases not commutative. go ahead and try it out for yourself and it'll become more convincing

[stoic python]

crossbeam

[normal void]

To show that the transformation $T_\omega : \mathbb{R}^{MN} \rightarrow \mathbb{R}^{MN}$ given by the below equation is linear, I start off by showing that $T(u+v) = T(u) + T(v)$

[stoic python]

FantaSkink

[normal void]

But I'm not sure whether this is adequate

[stoic python]

FantaSkink

[normal void]

Am I on the right track?

[stoic python]

FantaSkink

[mystic dagger]

anyone knows a software to check if a basis is orthonormal?

[wintry steppe]

could anyone link me to some resource that explains the notaiton used here ?

[spare widget]

Einstein notation

$v^i = \sum_j A^i_j u^j$

[stoic python]

criver

[spare widget]

the v on the rhs should read u

someone has a typo

[lime zinc]

Let V be a vector space over C and T: V to V is a linear operator such that for v in V there exists n in N(naturals) satisfying T^n (v) = v. Show that T is diagonizable.

I tried to solve this like : As every transformation has matrix representaion so we can say A^n x= x or
(A^n-I)x=0 implies A^n=I , which tells us that all the eigenvalues of A are just nth roots of unity , which are distincts, so that means A is diagonizable implies T is

Is my solution ok?

[scenic fulcrum]

No because n depends on x

[lime zinc]

but for every n this still the roots of unity

[scenic fulcrum]

Ok and if I take

1 1
0 1

The eigenvalues are 1 which is also a root of unity

And in your exercise you have NOT A^n = I

What is n ??? It is not even defined

[bold sun]

hey i need some help on this question part c

im so stuck like how would i show its surjective??

[spare widget]

Pick an arbitrary vector and find its corresponding polynomial

Surjective means that for any element of the codomain you can find its preimage

here it's fairly trivial

[bold sun]

hmm can i just use the 1 they gave like a0, a1,...an?

[teal grotto]

if you have a complex vector (a0, a1, ..., an), then how can you get a complex polynomial of degree at most n?

[bold sun]

noo im doing something worng

oh no it would just be the column vector 1,x,x^2,....xn right?

[halcyon spindle]

I feel like your thinking too hard about this.

[spare widget]

try forming this

$\begin{bmatrix} 1 & x & \ldots & x^n \end{bmatrix} \cdot \begin{bmatrix} a_0 & a_1 & \ldots & a_n \end{bmatrix} = \sum_{i=0}^na_ix^i$

[stoic python]

criver

[spare widget]

this is just your coefficients multiplied by the basis functions

you know how in R^n you can write a vector as

$\vec{v} = \sum_i a_i \vec{e}_i$

[stoic python]

criver

[spare widget]

well here it is the same, except e_i are the functions 1, x, x^2, ..., x^n (your monomial basis)

you can in fact write the map S

as

$\frac{1}{k!}\frac{d^k}{dx^k}\biggr|_{x=0}$

[stoic python]

criver

[spare widget]

for each of the components

the above will extract those

[elfin hawk]

Nvm I didn’t read it properly

[halcyon spindle]

How is f(x+y) = f(x) + (y)?

How is f(cx) = cf(x)?

No it’s not right.

that not what you had before.

[wintry steppe]

lmao

[halcyon spindle]

Lol now everything is gone.

[grave garden]

What does this mean guys

[zinc timber]

minimal polynomial of A

that means the monic polynomial of least degree s.t. m(A) == 0

alternatively, it's the monic poly generating the ann(A) in F_A[x]

[cold temple]

Can anyone help me to show this?
I checked in the Wikipedia, they used the young’s inequality
But I didn’t learn that in my courses
Is there any ways to start the proof?
Originally I want to try the mathematical induction
Can I just use basic knowledge of vectors and Cauchy Schwartz inequality to prove it?

[bold sun]

only saw this now i but ok that kind of makes sense-i haven't really learnt it as like writing it in terms of derivatives and factorials tho yet- Thank you though so much!!

[zinc timber]

@cold temple there's a typo

it should be $\sum |x_i y_i| \leq \norm{x}_p \norm{y}_q$

[zinc timber]

and also, why can't you use young's ineq?

[cold temple]

Oh yes I can use this as well

[zinc timber]

then?

[cold temple]

Then $\frac{1}{p}+\frac{1}{q}=1$

[stoic python]

Trenton

[cold temple]

The holder’s conjugate

Then perhaps I can use the natural log

So after taking ln, the power will be bring to the front

[cold temple]

I guess I can get something like this

[zinc timber]

assume $X_n = x_n/\norm{x}_p$ and $Y_n = y_n/\norm{y}_q$

[zinc timber]

then apply young

[cold temple]

Hopefully I didn’t get your message wrong

I’ve got a careless mistake

[zinc timber]

upto this is correct

[cold temple]

Ok

[zinc timber]

notice that $\sum |x_i|^p = \norm{x}_p^p$

[zinc timber]

so they will cancel each other out, that's the reason we normalized them

I hope rest is clear?

also, use _p or _q to enforce which norm you are using, otherwise it'll get confusing as there are multiple norms

[zinc timber]

here

[cold temple]

Noice

[zinc timber]

[cold temple]

Thank you so much!!!

[zinc timber]

[snow jetty]

real and symmetric matrix = self - adjoint matrix or is there any difference? lol

[lavish jewel]

self adjoint is more general

but indeed, real symmetric matrices satisfy the definition

[zinc timber]

yeah but not the other way around in case of complex

nvm you said real

[left mural]

Hi guys,
Disclaimer: This is a homework question.
Question: True or False. the nullspace of a 3X4 matrix cannot consist of only the zero vector.
I know of this theorem, except I don't know how to apply it :

[limber sierra]

That theorem is not helpful for the question.

[left mural]

What would you suggest ?

[limber sierra]

do you think the statement is true or false?

[left mural]

I think that the statement is false because the nullspace of A(mXn) is closed under linear combinations. Although, I am not sure if this includes the zero vector.

oh wait

it's true isn't it ?

[lavish jewel]

there are... several problems with what you had said

it is indeed true

[left mural]

cannot consist of only the zero vector

that's probably the statement I missed

[lavish jewel]

do you understand why, though?

[limber sierra]

so you missed the entire question?

[left mural]

yes happens 😄 😅

[left mural]

yea. The nullspace is any sort of vector when multiplied by the matrix A equals zero

Ax=0

[lavish jewel]

all right

[left mural]

the x can be anything in R^n and a zero vector too

[limber sierra]

so how do you know that there's a nonzero solution to Ax = 0?

[lavish jewel]

now think about the maximum rank a matrix can have, and the relationship between the row and column rank

[limber sierra]

by "nonzero solution" i mean

[lavish jewel]

and also about the rank-nullity theorem

[limber sierra]

no matter what matrix A you pick, there exists an x such that Ax = 0 and x is not the zero vector

[left mural]

This thing ?

[lavish jewel]

yes

[limber sierra]

if you haven't covered rank-nullity, instead imagine what your matrix must do to a basis (say the standard basis of ℝ⁴)

and then use linearity, i.e. aMx + bMy = M(ax+by)

using this, see if you can find a way to construct a vector that isn't 0, but when you multiply M by it, gives 0

[left mural]

hold on, let me start with this: the max rank a matrix can have is one less than the total number of columns ?

[limber sierra]

the key step is noticing that ||your images must be in ℝ³, but you have 4 of them, so they must be linearly dependent||

||why does this let you conclude the statement is true?||

[left mural]

ok let me figure this out with the info given

Thanks guys I'll let you know if I need more help

[wintry steppe]

guys A is a matrix nxn in K field and P polynomial, S an invertible matrix and ofc S^-1AS. Is it true if i say that P(S^-1AS)=S^-1P(A)S ?

[dim ivy]

Resolving the components of the vectors in the sense finding the vectors right?

[void cloud]

anyone know how to approach this?

for (ii), i'm assuming the answer would be 6.099? since it's the entry with the largest magnitude in the first column

[halcyon spindle]

Yes.

[void cloud]

i'm not sure about (i) and (iii)

if there's no pivoting, then would the first pivot just be 0.2115 ?

[halcyon spindle]

These terms are new to me, so hold on a sec while I look it up.

[void cloud]

[halcyon spindle]

Take a look at this to help you. https://courses.engr.illinois.edu/cs357/su2013/lectures/lecture07.pdf. Sorry 😢 I am lazy to read it right now.

But it does look interesting.

[torn stag]

@tired lance Yes please. Write my dissertation on proving the Collatz conjecture please.

[stoic python]

arcuzie

[final lance]

nvm

is there a way to simplify A*A^T

[fringe sierra]

is A a matrix and ^T transposition?

[final lance]

Yes @fringe sierra

[dreamy depot]

Dumb question does $(\mathbb{N}, \mathbb{R})$ from a vector space ?

[stoic python]

Zophike1

[wintry steppe]

depending if the natural set contains 0 or not

[dreamy depot]

yes we including zero

[wintry steppe]

yes i believe it is right

there are like axioms to know if something is a vector space or not

[dreamy depot]

Ye I know I was just double checking

[wintry steppe]

zero vector is a necessity

[dreamy depot]

yeah I figured 🙂

[peak lodge]

how do i go about converting this into general form?

i know general form involves an orthogonal vector but i dont know how to get that from this

[nocturne jewel]

what's general form in this context?

[peak lodge]

ax+by+cz = d although where [a,b,c] is orthogonal

[nocturne jewel]

that's a plane though

[peak lodge]

but i think this is a line and not a plane so idk how to

yeah

[nocturne jewel]

there's no such thing for R^3 lines

[peak lodge]

oh damn, did not know that

hmm

well this is the actual question. and im thinking it involves projection but having trouble with that line equation

[hoary void]

~~2 vectors:

1. the vector orthogonal to the line's direction
2. a vector through a point on the line and Q
   What can you say about the actual length of the first vector mentioned? how might you find it using the second vector i mentoined?~~

[peak lodge]

1 is where im having trouble, i dont know how to get a vector orthogonal to x

[peak lodge]

like in general lets say i have a vector [5,8,7,9,0,1,2,3] how do i go about finding a vector orthogonal to this?

[hoary void]

yeah that's actually my bad. there are infinitely many vectors orthogonal to a vector in R^3 (and above). so im gonna do some more looking around and if i find a sol ill post it here

i was thinking of shortest distance between point and a plane in R^3

[hoary void]

https://www.youtube.com/watch?v=5F4rsBMtFxM

[wintry steppe]

I have A is a matrix nxn in K field and P polynomial, S an invertible matrix and we know SAS^-1 (similarity transformation). I found out P($SAS^{-1})=SP(A)S^{-1}$ is that true? Because my inital guess would be just P(A). Please someone help so I can prove it by induction

[stoic python]

dim99
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[dusky epoch]

but yes, $P(S^{-1}AS) = S^{-1}P(A)S$

[stoic python]

Ann

[wintry steppe]

thanks @dusky epoch

[turbid hedge]

got part i (i think), but idk how to start part ii

[lavish jewel]

write down how the polynomial would look if the coefficients satisfy the conditions on V_a

that T is an isomorphism should tell you how many basis elements you need

[turbid hedge]

uhhh... lol x_X

[lavish jewel]

what did you get for part 1, then?

[wintry steppe]

whats the difference between a square and symmetric matrix?

[lavish jewel]

a symmetric matrix is a special kind of square matrix

the matrix $\begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix}$ is square and not symmetric

[stoic python]

Edd

[wintry steppe]

ah ok

square is when its n x n

and symmetric is when u = u^T

[lavish jewel]

mhm

[wintry steppe]

cool ty

[bold ermine]

can someone help me with this? I already asked this but I am still confused

for dimension isn't the formula n(n-1)/2 so the answer should be 10? cuz (5)(4)/2
but its 7? idk
and idk how you get the basis, im confused

[lavish jewel]

what formula is n(n-1)/2? i don't see how that comes up here

first, they tell you A is skew symmetric

the property you mentioned would be true if that were all the info, but you have extra conditions

after that, they tell you that the sum of the components of the first 2 rows of A above the main diagonal is 0

this means you have a sum of 7 elements = 0

one way to interpret this is as a hyperplane with normal [1,1,1,1,1,1,1]

this is 6 dimensional

then you have a similar condition on row 3, so now you have a sum of 2 elements = 0

again a hyperplane, this time with normal [1,1]; this is 1 dimensional

[radiant ridge]

If i have a diagonal matrix that where the diagnoal elements are all distinct, is then the rational canonical form just the same matrix?

[wintry steppe]

when is (ww^T)^T = ww^T

is it just hwen its symmetric

and square

[lavish jewel]

always

[radiant ridge]

Is there a difference between rational canonical and rational normal form?

[bold ermine]

oh so you subtract the ones which are making 0?

but why?

[lavish jewel]

those are already fixed, you can't choose those

[bold ermine]

but the sum is 0

not the elements

[lavish jewel]

the elements too

the main diag has to be 0

and so does a_45

[bold ermine]

yea so 6 elements are 0 right?

[lavish jewel]

mhm

[bold ermine]

wait but what will be the dimension for the given matrix

[lavish jewel]

what?

[bold ermine]

I mean subspace

mb

[bold ermine]

also i got this formula cuz in skew symmetric matrix aij=-aji so we have n-1 entries to consider we add up the number of entries, that is
1+2+⋯+(n-1)

[lavish jewel]

yes but that's not the case here

you have fewer degrees of freedom because of the extra constraints

you're ignoring the constraints

the dimension would AT MOST be the one you wrote just now

but as i showed you above, it's smaller

[bold ermine]

i didn’t understand this

[lavish jewel]

for example

you are told that x + y + z = 0

this is the same as 1x + 1y + 1z = 0

which is the same as [x,y,z] dot [1,1,1] = 0

this last one is the equation of a hyperplane

[bold ermine]

but what does that have to do with this

[lavish jewel]

this

hyperplanes are n-1 dimensional

this exactly tells you the dimension of what you're looking for

[bold ermine]

i’m so confused

[gray dust]

@bold ermine apply edds logic to the conditions given by sums

heres 1 https://i.gyazo.com/b96f4ac497fa568077218dbe4d7331b3.png

the 7 entries above sum to 0. we can write it as the eqn of a hyperplane with normal [1,1,1,1,1,1,1], which is 6 dimensional

[blissful vault]

I really don't see it (highlighted part). Why is this useful? We can now write any vector in terms of the orthonormal basis, but I don't see why that is important. The next page is Gram-Schmidt Process. Linear Algebra Done Right btw.

[nocturne jewel]

It gives an easier way to compute coordinate vectors

cause the ith component is just $\langle v,e_i\rangle$

[stoic python]

Mosh

[nocturne jewel]

so $v=[\langle v,e_n\rangle,...,\langle v,e_n\rangle]_B$ where $B:={e_1,...,e_n}$

[stoic python]

Mosh

[nocturne jewel]

whereas to write v wrt a non-orthonormal basis, it'd be harder to have a systematic way of computing the components

[blissful vault]

I had to chew on that. I think that makes sense. With a non-orthonormal basis, we don't have an easy way to write some vector in terms of it. With an orthonormal basis, we have a standard procedure.

[nocturne jewel]

yes

and that standard procedure is do a shit ton of inner products

which is usually actually easy to do

[blissful vault]

Got it. Thanks for the help.

[grim leaf]

basically in the "vector space" sense, a "vector" is not necessarily a euclidian vector?

[nocturne jewel]

yes

Vectors period are things that can be added together and scaled, and obey the axioms of vector spaces

to which R^n spaces are vector spaces

(assuming standard + and scaling)

[grim leaf]

i see

interesting

[proud veldt]

by decompose a positive definite Hermitian form into real and imaginary parts. How to show any two forms are equal if they have same imaginary part?

[languid sphinx]

Is a rank 0 matrix considered linearly independent?

I.e. 'nothing'

[lavish jewel]

a rank 0 matrix is a matrix of zeros

so it's lin dep

[languid sphinx]

What if there's no matrix

0 x n

or m x 0

[lavish jewel]

you can use an empty set construction to show every element of it is linearly independent from all other matrices

i think that works, yeah

[zinc timber]

0xn matrix

waah

philosophers are like

[lavish jewel]

https://tenor.com/view/obito-uchiha-tobi-naruto-peace-out-bye-gif-15428029

T:V->

[zinc timber]

lol

[zinc timber]

is it even possible to have 0xn matrix

[lavish jewel]

it's called empty matrix

off the top of my head, i mostly see utility in coding

[zinc timber]

like if you think in terms of T:R^n->

like you can't map something to an empty set

[lavish jewel]

wikipedia has a very short snippet

An empty matrix is a matrix in which the number of rows or columns (or both) is zero.[73][74] Empty matrices help dealing with maps involving the zero vector space. For example, if A is a 3-by-0 matrix and B is a 0-by-3 matrix, then AB is the 3-by-3 zero matrix corresponding to the null map from a 3-dimensional space V to itself, while BA is a 0-by-0 matrix. There is no common notation for empty matrices, but most computer algebra systems allow creating and computing with them. The determinant of the 0-by-0 matrix is 1 as follows regarding the empty product occurring in the Leibniz formula for the determinant as 1. This value is also consistent with the fact that the identity map from any finite-dimensional space to itself has determinant 1, a fact that is often used as a part of the characterization of determinants.

[zinc timber]

holy

det = 1 lmao

[lavish jewel]

empty product 😌

[zinc timber]

[lavish jewel]

kinda like the 0 vector space being spanned by the empty set

[zinc timber]

ye

[lavish jewel]

i have never seen it used for interesting stuff though

[zinc timber]

interesting to philosophers

[languid sphinx]

Not really, just some extreme cases which either a) easy base case for induction or b) just for consideration of 'non-negative integers' vs 'strictly positive integers'

n=0 base case are even easier than n=1 base cases, if there is nothing to compute :D

I'm just asking to see because my primitives of the proof require 'linearly independent matrices' so I was wondering if empty matrix is part of it, if so it can be an example for the proof

[zinc timber]

[lavish jewel]

i don't think that makes much sense

the empty cases are vacuously true

even if the nonempty cases aren't

[zinc timber]

yeah that's what i was thinking

like empty set as a base case doesn't feel right

[lavish jewel]

we prove the matrix of all zeros is linearly independent (given some set, etc)

the base case of the empty matrix is obviously true

[zinc timber]

although idk if it affects the induction argument

[lavish jewel]

now let...

[frigid kettle]

How come the 2nd question got an answer of 77mm. Can someone mind interpreting it?

[teal grotto]

the n = 0 case doesn’t imply the n = 1 case here

[frigid kettle]

The question is,
A scene points at coordinates (400, 600, 1200) is respectively projected into an image at coordinates (24, 36) and the camera's principal point is at coordinates (0,0,f). Assuming the aspect ratio of the pixels in the camera is 1. What is the focal length of the camera?

[frigid kettle]

How come the 2nd question got an answer of 77mm. Can someone mind interpreting it?

[frigid kettle]

Someone plz reply I'm desperately in need

[lime zinc]

How can i find the dimension of the space of linear operators on R^n , such that it maps v to w , where v and w are non zero vectors. Any hint? (By using operator form not matrix form)

[lavish jewel]

given nothing other than L:V->V, where V is R^n?

[lime zinc]

just one thing more that linear operator maps v to w where v and w are non zero vectors

[dusky epoch]

{A in L(V,V) | Av = w}

this?

and wym by using "operator form and not matrix form"?

[lime zinc]

yes in mathematical form.

[lavish jewel]

so what i was gonna suggest was to pick a basis and then use matrices

but if i understood your question correctly, you want to talk of invertible matrices, which don't form a vector space

[lime zinc]

but why?

[dusky epoch]

can i have an answer to my question

[lime zinc]

n^2-n

[dusky epoch]

no

[dusky epoch]

.

[lime zinc]

like writing it into matrix form

[dusky epoch]

so you reject any solution method that involves matrices or bases in any fashion?

[lime zinc]

let me check question one more time.

[dusky epoch]

why not send the question here and let us see it for ourselves

[vocal isle]

hey folks, I have 3 vectors in R3: n1, n2, n3. The values that make up n3 are given. I need to find n1 and n2. I know the following:

n1^T n3 = 0 (n1 is orthogonal to n3)
n2^T n3 = 0 (n2 is orthogonal to n3)
n1^T n2 = 0 (n1 is orthogonal to n2)
n1^T n1 = 1 (n1 is a unit vector)
n2^T n2 = 1 (n2 is a unit vector)

Now we have 5 equations and 6 unknowns which makes sense because there are an infinite number of basis vectors that are perpendicular to n3. However, I want to find just one solution by putting these equations in matrix [A]x = b form and solving using the pseudo inverse. In this case x would be in R6 and would be x = [n1;n2]

Can someone help me find the matrix [A] (5x6) and vector b (5x1) that does this?

[spare widget]

do you specifically need the pseudo-inverse solution or would any solution do?

because if any solution is fine then pick any vector that is non linearly dependent with n3

then orthonormalize it

and finally produce the other as the cross product of the two known ones

[vocal isle]

yes i need pseudo inverse sadly (long story)

[spare widget]

are numerical solutions ok?

if not then just do the SVD manually, or maybe using a symbolic solver

[vocal isle]

numerical solution is fine

(infact its preferred)

i'm intrigued by your idea of using the SVD. How do you expect that to work?

[lavish jewel]

this would be a lot easier if you just consider a basis for the plane orthogonal to n3

and then do gram schmidt

[vocal isle]

interesting!

I've never heard of gram schmidt, but it looks like it needs a matrix as an input. WHat matrix would that be?

[lavish jewel]

a matrix whose columns you want to make an orthonormal basis out of

the first column is kept as is, so you'd put n3 there

[vocal isle]

the only input info i have is n3 (3x1 vector)

[lavish jewel]

how do you know what pseudo inverses do if you don't know G-S

[vocal isle]

matlab has a pinv() function 😄

[lavish jewel]

if you know n3, you can make n2 by inspection

and n1 by taking their dot product

then all you need is to normalize n2 and n1

[vocal isle]

found this online

[vocal isle]

i can do this by hand for some simple n3 vectors no problem

[lavish jewel]

and for complicated ones as well

[vocal isle]

but letting n1(3) = 1 for example, but i was hoping to code up a way to do this fo rany n3

[lavish jewel]

i think you have... bigger problems

[vocal isle]

if the determinat of the matrix is 0 it's not invertable

[vocal isle]

?

[amber osprey]

Exercise 7.3. Let 𝑉 be the vector space, which consists of all differentiable functions 𝑓: [−1, 1] → ℝ. Which of the following quantities are subspaces of 𝑉?

(a) {𝑓 ∈ 𝑉 | 𝑓(0) = 0},
(b) {𝑓 ∈ 𝑉 | 𝑓(1) = 0},
(c) {𝑓 ∈ 𝑉 | 𝑓(−1) = 1},
(d) {𝑓 ∈ 𝑉 | 𝑓′(0) = 0}.

can someone throw me a bone

[zinc timber]

🦴

[lusty pumice]

What does it mean for a set of vectors in a vector space to be closed ?

[subtle walrus]

do you have more context?

[lusty pumice]

"The space of any finite set of vectors in a vector space is closed under addition and scalar multiplication"

But I don't know what it's referring to when it says closed

[subtle walrus]

a set is closed under addition/multiplication when those operations on the members of the set dont allow you to "leave" the set

so this is not true?

[lusty pumice]

The answer was true

but I have no idea what it was referring to

[gray dust]

is this the full question?

[spare widget]

https://jcgt.org/published/0006/01/01/

It gives you one possible solution

[tough veldt]

An operation is 'closed' within a set if that operation takes parameter(s) from the set as input and it output something that's also in that set.

You also say 'the set is closed under the operation' to mean the same thing

[lusty pumice]

Yeah

[gray dust]

do u have a pic

[royal tangle]

Ty

[amber osprey]

very funny

[icy blade]

to do this, all i have to do is make a 3x3 matrix with them and if the det is not 0 they are linearly independent?

[nocturne jewel]

That is sufficient yes

Since they are both TFAE statements that the matrix is invertible

[weary flare]

https://tenor.com/view/cat-chips-hello-chat-cat-chips-hello-gif-21385431

just started learning lin alg on my own time and

im kinda confused on matrix multiplication

[spare widget]

$C_{ij} = \sum_{k=1}^{K}A_{ik}B_{kj}$

[stoic python]

criver

[spare widget]

What are you confused about?

[weary flare]

oh wait i think i figured it out

textbook that i was recommended was kinda confusing

[spare widget]

The definition or how one comes up with said definition

You can always try various textbooks

See which you like best for a specific topic

[weary flare]

eh yeah

ive been told that it's a good idea to also study lin alg alongside calc

[spare widget]

idk about calc, but alongside analytic geometry it makes sense because you gain geometric intuition

As far as calc goes, probably makes sense studying it alongside mechanics

[amber osprey]

how do I take the gradient of this

[spare widget]

not #linear-algebra

but use the chain and product rules

[amber osprey]

it is literally an exercise in my linear alg course tho :/

[spare widget]

it's technically calculus

eitherway product and chain rule

Differentiate wrt a, then wrt b, then wrt c

[amber osprey]

the entire thing

\partialf(a,b,c)/partial a?

and so on?

[spare widget]

what is the definition of gradient?

[amber osprey]

i dont know

I just know you get the slope and shit

[spare widget]

well look it up

[amber osprey]

you get the vector field

w/e that means

[spare widget]

https://en.m.wikipedia.org/wiki/Gradient#Relationship_with_derivative

[amber osprey]

yea

so

am I right or wrong?

something like that?

[spare widget]

no

[amber osprey]

with the partial in b and c

[spare widget]

The gradient is a vector

what you wrote is a scalar

even provided you fix the db, dc

Look up the section I linked

[amber osprey]

man im garbanso at math

I not getting anything from what you linked

[spare widget]

$\nabla f (p) = \begin{bmatrix} \frac{\partial f}{\partial x_1}(p)\ \ldots \ \frac{\partial f}{\partial x_n} (p) \end{bmatrix}$

[stoic python]

criver

[spare widget]

on your case p = (a,b,c), x_1 = a, x_2 = b, x_3 = c

[amber osprey]

yea

isn't that what i did

whats wrong with that

[spare widget]

What you wrote is

$\sum_i\frac{\partial f}{\partial x_i}$

[stoic python]

criver

[wintry steppe]

Why are you doing these additions tho?

[amber osprey]

so it is correct

you just want me to multiply?

[spare widget]

no

The gradient is a vector

[amber osprey]

yea that doesnt help me tho

[spare widget]

$\nabla f (p) = \begin{bmatrix} \frac{\partial f}{\partial x_1}(p)\ \ldots \ \frac{\partial f}{\partial x_n} (p) \end{bmatrix}$

[wintry steppe]

Yeah, you contrust a 3d vector

[stoic python]

criver

[wintry steppe]

Not do a summation

[amber osprey]

please remember I am not good at math telling me it is a vector does not help me

[spare widget]

Do you know how a R^3 vector looks like

[amber osprey]

yea 3d

(a,b,c)

[spare widget]

Ok, now instead of a,b,c put the partial derivatives there

[wintry steppe]

It's literally that

[amber osprey]

huh

okay gimme a sec

?

without the big ass d

ofc

[wintry steppe]

yeah

[amber osprey]

okay so far so good

but how do I compute it

;D

[spare widget]

lmao no

pdes means partial differetial equations

You need the product and chain rules

[amber osprey]

can you give me an example?

maybe for the first one

this is the chainrule right?

[halcyon spindle]

Yes.

[spare widget]

https://tutorial.math.lamar.edu/Classes/CalcI/ChainRule.aspx

[amber osprey]

y is the function f(a,b,c)?

[spare widget]

You should read a book or something

[amber osprey]

Yea

I have read a book or something

I forgot

[spare widget]

It will be more productive

I don't think most people here will be willing to solve it for you, for didactic reasons

[amber osprey]

I dont want anyone to solve it

I just want to know what is u and y here

[wintry steppe]

If you are already familiar with it, and you read it again, it will be easier to remember

[halcyon spindle]

Functions.

[amber osprey]

y is f(a,b,c)?

[spare widget]

ok, you want someone to spoon feed you the chain and product rule

[amber osprey]

telling me what y and u is, is spoon feeding?

[wintry steppe]

You can literally read the article that criver gave you in like, 5 minutes or so

[spare widget]

yes

[amber osprey]

ok

[wintry steppe]

Read the article and if you have specific doubts we can help you to solve'em

[tired fossil]

hey guys, does deta not equaling zero have something to do with this?

The determinant of A not being equal to zero

[restive raft]

the fact the inverse exists is the important part

[tired fossil]

yes

if A is invertible then detA /= 0 corrrect?

[icy blade]

how is this vector space not closed under addition

[frozen vigil]

suppose y=(x-1)^3

in this case

u=x-1

y=u^3

dy/dx=dy/du du/dx=d(u^3)/du d(x-1)/dx

[hoary void]

[0, 2] and [-1, -1]

for example

[frozen vigil]

Something like this, maybe it help?

[oblique prairie]

@amber osprey just read this, it won’t take long and it will help

as long as you have some knowledge on derivatives that is

[hoary void]

derivatives would also be more appropriate for #calculus

[oblique prairie]

also yeah people say you don’t need calculus for LA

but i sure do see a lot of calculus in my LA book

[hoary void]

what book

[oblique prairie]

linear algebra done wrong

[hoary void]

why wrong instead of right

[oblique prairie]

parody in a way

says in the first few pages

anyways this is kind of a discussion so

#serious-discussion if you want

[icy blade]

how is this not closed under vector addition and scalar mulitplication

[native rampart]

Is t supposed to be real?

[icy blade]

thats all the info i have sadly

[native rampart]

Consider
$(a+t_0^2)-(a+t_1^2)=t_0^2-t_1^2=a+(t_0^2-t_1^2-a)$

Weird

[stoic python]

Drink Drake

[native rampart]

So pick $t_0, t_1$ such that
$t_0^2-t_1^2-a$ is negative

[stoic python]

Drink Drake

[icy blade]

why did you do (a+t)-(a+t)

[native rampart]

because of the form cx+y

Combining both scalar multiplication and addition

[icy blade]

ok but what if i wanted to prove them seperately how would i proceed

[native rampart]

If you are asking for motivation,we want to get a case where t^2 is negative

[icy blade]

i just dont understand for what values of t its not closed under vector addition

because a e R numbers so like how can it not be closed

[native rampart]

For all values of t,it will be in the set

But if t^2 were negative,there would be no such t

[icy blade]

so zero vector, va, sm

i dont understand how va, sm fails

[native rampart]

Oh wait t is a fixed constant,mb

[icy blade]

oh wait i think im just dumb

what is R

isint that like only positive

[native rampart]

Set of reals

[icy blade]

ok nvm

so its still gonna be part of real numbers then

even if i mulitply by -1 or whatever

[native rampart]

$ca+ct^2=(ca+(c-1)t^2) + t^2$

[stoic python]

Drink Drake

[native rampart]

So it will never fail?

Unless c can be over complex nonreal nunbers

[icy blade]

i dont understand your formula on top

what exactly does it show

[nocturne jewel]

@native rampart what are you even saying tbh

[nocturne jewel]

🤨

[native rampart]

Scalar multiplication will land you inside the set

[nocturne jewel]

No it wont

[native rampart]

Always

[nocturne jewel]

c is not always 1

so ct^2 != t^2 always.

[native rampart]

ca+(c-1)t^2 is a real number

If c is real

[nocturne jewel]

no

it's a polynomial

that's not in the space.

[icy blade]

so mosh what did you mean by If you add 2 elements, you dont get an element in V

[native rampart]

Ok,I am dumb

[icy blade]

me too bruh

[native rampart]

I was thinking t is a real variable

[nocturne jewel]

$a+t^2+b+t^2=a+b+2t^2$

[stoic python]

Mosh

[nocturne jewel]

the RHS isn't in V

but the 2 polynomials a+t^2 and b+t^2 are

[icy blade]

what does RHS mean

[nocturne jewel]

right hand side

and

left hand side

[icy blade]

why is it not part of V, how does what you wrote violated a e R

[nocturne jewel]

2 != 1

[nocturne jewel]

it.... doesn't?

a and b are real numbers

I just didn't feel the need to say that explicitly cause I'm implicitly taking polys from the set

[icy blade]

ok so what we have 2t instead of 1t what exactly does that change

[nocturne jewel]

It's no longer in the set

cause... 2 isnt 1

so the coefficient of t^2 is wrong

[icy blade]

but why does it have to be 1 im so confused

[nocturne jewel]

cause it's t^2, not ct^2

or some variable coefficient

[icy blade]

, shouldnt it be written on the right hand side then, it only tells us a e R

[nocturne jewel]

what

Ok

From the top:

[icy blade]

i thought the right hand side was supposed to be like the rules for the set or something

[nocturne jewel]

if $a,b\in\mathbb{R}$, then $a+t^2\in V$ and $b+t^2\in V$

[stoic python]

Mosh

[nocturne jewel]

yes?

[icy blade]

yes

[nocturne jewel]

now $a+t^2+b+t^2=(a+b)+2t^2$

[stoic python]

Mosh

[nocturne jewel]

yes?

[icy blade]

yes

[nocturne jewel]

Now, does this match the form exactly as stuff in V?

Well, we check both terms

is a+b a real number?

[icy blade]

yes

[nocturne jewel]

great

is the coefficient of t^2 1?

[icy blade]

ok so basically if i got some stuff like this, after the va, sm it has to match the coefficients exactly

[nocturne jewel]

it has to match the form of the elements

and any rules that dictate their form

[icy blade]

ok but we dont have to match a,b because its said that it has to be real set

[nocturne jewel]

yes

the only requirement on the constant terms is that they remain real

[icy blade]

alright then, thx my g

actual legend on here ngl

[nocturne jewel]

Now apply a similar process for scaling

[icy blade]

so like 2(a+t^2) = 2a which is part of a e R but 2t^2 does not match so its not closed under SM

[nocturne jewel]

🤨

2a+2t^2, yes

$c(a+t^2)=ca+ct^2$, where $c\in\mathbb{R}$

[stoic python]

Mosh

[nocturne jewel]

however c is not always 1

thus the coefficient of t^2 isnt guaranteed to match

[icy blade]

alright then, big brain stuff right here

[nocturne jewel]

$U:={a+ct^2|a,c\in\mathbb{R}}$ however, is a subspace of $\mathbb{R}[t]_{\leq 2}$

[stoic python]

Mosh

[icy blade]

still havent gotten the hang of subspaces yet but i will save that for later

[nocturne jewel]

It's just testing the 3 requirements

[stoic python]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by Hack With Techno Boy#6902

[dusky epoch]

wolfram alpha is not a do-your-homework tool

also your question isn't linear algebra

[timid hornet]

hmm yea

[timid hornet]

:/ it is not my h.w.

it was a sample paper for iit enterance

[spare widget]

Does anyone have a suggestion for a numerical iterative method for quickly computing an underestimate of the smallest eigenvalue of a SPD matrix?

(I am aware of the power method, I would like something better)

[sudden stratus]

Hi, I'm currently self-learning MIT's OCW linear algebra course and am pretty lost with one of their problem sets (here: https://ocw.mit.edu/courses/mathematics/18-06sc-linear-algebra-fall-2011/ax-b-and-the-four-subspaces/matrix-spaces-rank-1-small-world-graphs/MIT18_06SCF11_Ses1.11sol.pdf)

I was already having trouble understanding Problem 11.2b, but for part c) I completely don't understand how the dimension of the "column space" is 6. Could someone help derive this for me?

More importantly, I also don't get why the dimensions of the pseudo-"null space" and "column space" should add up to the dimension of the space of inputs to M. I suspect I'm lacking in some form of conceptual understanding...

[lavish jewel]

you need to take a step back

do you know what the dimension of a vector space is?

and have you seen the rank nullity theorem, or alternatively what strang calls the "fundamental theorem of linear algebra"

[sudden stratus]

I think I do. It's the number of elements in the basis?

[sudden stratus]

the latter sounds familiar, but I can't recall it... I'll look it up!

[lavish jewel]

well, the column space is the vector space spanned by the columns of a matrix

so you can find its dimension by finding out how many lin indep columns a matrix has

[sudden stratus]

Right. The "column space" (in quotes, because even the question acknowledges it's not a real column space I think) that was derived in the earlier part of the question consisted of a matrix populated by variables, so I don't think I can do elimination to find the lin indep columns and thus the basis.

Besides, the matrix for the "column space" has 3 columns - so how can the dimension be 6?

Apologies for the long-windedness, and thanks for your input!!

[lavish jewel]

so

the formulation is a little tricky

you should instead think of it as some linear transformation T:V->V

where V is the vector space of 3x3 matrices

V has dimension 9

the transformation T:V->V is doing something to the matrices in V, and outputting only a subset of the matrices in V

presumably, this subset of V will be a subspace

if it helps you visualize it, you can treat the 3x3 matrices as vectors in R^9

then the transformation T:V->V in some basis is given by a 9x9 matrix

and the question involves the columns of this 9x9 matrix

not the columns of A

the reason they write "column space" is that they didn't give you any matrix

so they actually mean the image of the transformation $T: X \mapsto AX, ,, X \in \mathbb{R}^{3 \times 3}, ,, A = [...]$

the image and the kernel of T

[stoic python]

Edd

[sudden stratus]

Thank you so much for your response @lavish jewel! I took a while to digest it and had to do some googling here and there... but ultimately I think I get what you were trying to convey.

I really appreciate your help!!

[spare widget]

didn't we discuss that already

you can pick any orthonormal system with one of the vectors being n3

there are infinitely many such solutions

here's one possible solution with a numerically stable algorithm for computing it: https://jcgt.org/published/0006/01/01/

[vocal isle]

Yep we did! And yes, I can absolutely do it this way. But I really want to find a linear algebra solution to this by solving for [Bt; n3t] [B n3 ] = I

[spare widget]

all of the suggested solutions are "linear algebra solutions"

[vocal isle]

and yes, you're right there's an infinite number of solutions (because there are an infinite number of paris n1 and n2 that are all orthogonal to each other and orthogonal to n3). But I only seek 1 (without preference). I'm really looking for a way to do this using linear algebra. I've already looked into Gram–Schmidt and that works as well once i define a set of linearly independant vectors n1 and n2

[spare widget]

But I only seek 1 (without preference).
did you read the paper I linked?

[vocal isle]

yes i have

[spare widget]

then you should be aware it gives you one solution

[vocal isle]

it does

but i car eabout the method

[spare widget]

Gram-Schmidt is linear algebra btw

[vocal isle]

and i was hoping specifically there would be a way of inverting B^T or something along those lines (using matrix algebra)

[spare widget]

I am unclear what linear algebra means for you, or what kind of solution you expect to get

[spare widget]

there are infinitely many pseudo-inverses of it

that's the whole point of having infinitely many solutions

you can pick any solution (e.g. the one from the paper)

and produce any other by a rotation and potentially a reflection

[vocal isle]

ah yep, once you get n2 then you could easily multiply it by the rotation matrix about n3

and that would get you n1

[spare widget]

rather once you have n1 and n2

you can rotate bouth around n3

to get any other solution

[vocal isle]

yes

i agree

[spare widget]

for solutions with different handedness you need a reflection too

[vocal isle]

let's say that the RHS of this equation wasn't the identity (but it was symmetric & positive definite). How could you solve for a solution of B?

in other words, i want to turn this isn't a complete abstract matrix algebra problem that has no connection to orthogonality

i'm looking for a proof something along hte lines of this (although it's problem statement is slightly different)
https://math.stackexchange.com/questions/2858299/solve-xtx-a-for-x

[lavish jewel]

real symmetric matrices are diagonalizable

that's wall you need there

[spare widget]

you can do Cholesky for X^TX provided A is spd

[lavish jewel]

if A is symmetric, A = QDQ^T

then take the sqrt of each element in D

A = Q D^1/2 D^1/2 Q^T

then X^T = Q D^1/2

[spare widget]

won't this result in complex D^{1/2} in the general case?

[vocal isle]

and in the specific case where A = I then Q = I and D = I, right? So wont that mean that X = I ?

[lavish jewel]

in general yeah, criver

if A = I, all orthogonal matrices work

by definition

[vocal isle]

so is there a way i can solve for B using a similar method to the one you just used?

[lavish jewel]

oh, you know n3?

[spare widget]

it just shifts the problem to choosing a basis for the 0 eigenvalues

[lavish jewel]

you can make infinitely many solutions by picking n1 and n2 as an orthonormal basis for the plane orthogonal to n3

[spare widget]

you will have to make that choice regardless how you rewrite the problem

[lavish jewel]

yeah there is no unique way to do this

[spare widget]

if you add some additional constraint that is linearly independent then you can get a unique solution

it's what they implicitly do here: https://jcgt.org/published/0006/01/01/

it's also what you would do if you were to pick a vector linearly independent from n3 and applied Gram-Schmidt

then the remaining vector is defined up to sign

[wintry steppe]

what's the difference in "LA done right" "s.friedberg linear algebra" on the approach to the subject?

[halcyon spindle]

I see friedberg as a more complete linear algebra book. Theory and computation where axler is just theory .

Also determinant is not done properly in axler from what I have heard.

Friedberg would be a better introduction to the subject than axler is what I am saying.

[normal void]

How do I show that $T(u+v)= T(u) + T(v)$ for an image convolution with a kernel?

[stoic python]

FantaSkink

[normal void]

Our entire year has been stuck on this one for weeks :/

[lavish jewel]

simple enough

you have your function defined for some input O

now substitute O = (U + V)

then distribute the omega term to both and split the sum

if you want something more esoteric, you can show that, after vectorizing the image O, the 2D convolution shown here is equivalent to multiplying a matrix with a 2 level block toeplitz structure

and matrix multiplication is linear

it might help to do it with a 1D case first, where the convolution is then represented by a toeplitz matrix without a block structure

[warm grotto]

[normal void]

I'll try that, thanks man

[fallen karma]

$||AB|| \leq ||A|| ||B||,$ where A and B are n-by-n matrices.

[stoic python]

seth.delacroix

[fallen karma]

Is there a way to do it in general, or can I pick a norm and then argue from equivalence of norms?

[lavish jewel]

these are operator norms?

induced by vector p norms, or?

because it's not true for some special norms like the max norm

[fallen karma]

Oh bet

Ok that answers my question

[lavish jewel]

i think it's true for the norms induced by the p norms though

if you construct the operator norm as max_x norm(Ax) with norm(x) = 1, you can apply the definition twice

i.e. ABx, subs Bx = y and then apply the def to Ay, and then to Bx

[wind shore]

so why is the answer E?

since i dont find any other vector that would also work, and i assume the 0 vector is also a scalar multiple of that vector

[nocturne jewel]

0 is a scalar multiple of any vector

cause 0v=0

[wind shore]

i agree. so why is E actually false?

[nocturne jewel]

Solve the system and find a basis for the kernel?

[novel marsh]

How difficult is linear algebra to self teach with a prerequisite of calc 3?

[hoary void]

@oblique prairie

[grim leaf]

about as hard as any other math class tbh

if you've already learned calc 3 you've hopefully already touched on some linear algebra

like matrices and vector spaces and whatnot

[novel marsh]

How comprehensible are the concepts relative to calculus?

[wind shore]

right which i think (-5,1,1,0) is the right vector, so E should not be false, so E is not the answer, but E is (nvm there is another vector that is part of the kernel that i overlooked, solved*)

[meager harness]

<@&286206848099549185>

what i dont get is that there is supposed to be a matrix A that gets multiplied to f(t) and yields T(f(t)) but wouldnt there be t variables in there instead just a vector of constants f(a1) ... f(an)

[teal grotto]

no t is just a point in the domain of f. t is actually not necessary for this problem and only adds confusion.

[gray dust]

@meager harness we multiply the coordinate vector of f wrt the basis by A, not f itself

[wintry steppe]

For any transformation to be a projection, it just has to be endomorphic and idempotent right?

[gray dust]

@wintry steppe yes

[wintry steppe]

Thank you

[tranquil steeple]

[-8,2,0,1]

[novel needle]

Hi

what is the significance of k=/= i,j

wouldn't it be the same if it hadn't written k=/=i,j

[stoic python]

Drink Drake

[novel needle]

its the levi civita symbol

[native rampart]

You are correct

It shouldn't make a difference

[novel needle]

just to make sure, this sum should give me = eij1eij1+eij2eij2+eij3eij3

right

[native rampart]

Yea

[novel needle]

but then it says k=/= i,j doesn't that mean my i and j for each term has to be different ?

for example

[native rampart]

No constraint on that

Only that k≠i and k≠j

[novel needle]

if i = 3 and j = 2

i would just get

e321e321+e322+e322+e323e323

here k = j

doesnt this violate k =/= i,j

[native rampart]

Yea,you skip cases of k where k is I or j

Something seems wrong with the equation you wrote

If i=j it's clearly wrong

And k!=i,j is not needed, because the term will reduce to 0 in those cases

$\sum_{k} \varepsilon_{ijk}^2 = 1, i≠j$

[novel needle]

this i understand, i was just wondering what the significance of writing k=/= i, j is

[stoic python]

Drink Drake

[native rampart]

There's none

[novel needle]

mhmm i see thank you

[native rampart]

Where did you find this equation exactly

[native rampart]

This one seems like a cleaner version of your version

[novel needle]

i got this from a classical mech textbook

and physics textbooks usually make minor mistakes with math maybe

[native rampart]

Yea,or maybe the author didn't want students to expand terms that have j=k or i=k

Because they will reduce to 0 anyway

[novel needle]

yeah

[native rampart]

If you want a meaning for this(and hence your statement) this basically says if i and j are fixed different numbers, there's exactly one permutation as k varies over everything else

For example say i=2,j=3