linear-algebra

[rain echo]

yeah

do you like hockey?

I don't like any sports

but hockey is the one I am the most ok with

oh sorry

this is an on topic channel

[mild igloo]

anyone around to help me with a least squares line of fit for polynomial?

[rain echo]

sure

[mild igloo]

so this is what im given

the part I usually have trouble with is creating the matrix

[rain echo]

alright you think about what the equations say

you have ax²+bx+c=y for each pair of x and y

this is a linear combination of a, b, c

so we are finding a matrix which acts on (a,b,c) and hope the output is the vector with all the y values

so the first row is the first equation is month x=1 so x²=x=1

so the coefficients multiplying a,b, and c are 1,1,1

[mild igloo]

right

[rain echo]

then for x=2 the coefficients are 4,2,1

[mild igloo]

second column is the x values? from the ordered pairs?

[rain echo]

second row is the equation that gives you a(2)²+b(2)+c=4.4

so the y vector has the second entry as 4.4

and the entries along row 2 are 4,2,1

then for row i you get a(i)²+b(i)+c=y(i)

so the ith row is i²,i,1

make sense?

[mild igloo]

I think so but only cuz im looking at the book rn

the second matrix in m is what ur talking about right

first column of 1s then column of xs then colum of i^2?

[rain echo]

yeah, the book is doing a different order of columns than I am

[mild igloo]

im sure im misunderstanding somewhere

[rain echo]

but that is still right

[mild igloo]

okay lemme make the matrix then if you dont mind could u check it?

[rain echo]

sure

[mild igloo]

thank you

?

OH I see how you were explaining it now

[rain echo]

yes!

[mild igloo]

WOO

[rain echo]

in general, you are solving for the coefficient vector

[mild igloo]

omg I actually understand you magician

[rain echo]

usually polynomials are thought of as a linear combination of powers of x

like take 3 x² and 5x and so on

so this way of thinking of it is counter intuitive a little

cause youre solving for the coefficient vector

so you have the think of having x² a's instead of a x²'s

I'm glad that helped!

[mild igloo]

yea so ive done a simular problem for a line before and I didnt quite understand how it applied to a quadratic

but the way you explained it made it all click

now I just make that matrix equal to the vector of y variables and solve yes?

[rain echo]

yes

solve with all the A^TA stuff

[mild igloo]

oh yea forgot about that

multiply both sides

then solve

[rain echo]

yes solve ATAx=ATy

[mild igloo]

reactions are hard

thank you

[ocean sequoia]

can someone help me walk through this proof why do we need to show B = XY^t

where did the V come from

and why is Y^t w = 0 come from

[warm niche]

oh shit I'm doing SVD too

but idfk y it works

[upbeat mauve]

I'm confused on how this 2nd matrix is obtained

I figured it would be row reduction from the c matrixies on the left side and [T]b on the right side but that wasn't it

[wintry steppe]

the columns of the matrix are just the representations in the basis C_1, ..., C_4 of T's action on the basis E_1, ..., E_4

so for example, if you wanna compute the second column

[stoic python]

TTerra:

[wintry steppe]

the (unique!!) scalars in this linear combination will be the second column of your matrix

[stoic python]

TTerra:

TTerra:

TTerra:

TTerra:

[wintry steppe]

and that's the second column

the computations for the other three columns are similar

@upbeat mauve

no row reduction

[upbeat mauve]

Oh yeah I understand now thanks

[wintry steppe]

Why is there such scarcity regarding material on Kronecker Capelli's theorem? I have it for a test, anyone know of a nice article/video explaining it?

OH Rouche Capelli's theorem is the same???

ok epic

[rain echo]

don't know any videos or articles about it

but I could explain it

[odd kite]

It's covered in pretty much every linear algebra course (?)

[rain echo]

it doesn't often get named I guess

but I've seen it all ove

over

[odd kite]

yes it's not always named

[acoustic zodiac]

Given the set $P_n[x]$, subspace of P[x]$, why is its dimension $dim P_n[x] = n + 1$ if its basis is all powers of x from 0 to n?

[rain echo]

cause there are n+1 powers of x

the first is x^0

the second is x^1

and so on

[acoustic zodiac]

Oh damn yeah, you start counting the "ns" at 1, not at 0

thanks

[rain echo]

np

[wise flare]

im confuse

is there a way to anilitaccly find if a function is diagonaliable

[limber sierra]

analytically?

[wise flare]

yes

im not given the matrix. just the character polynomial

so i have found the eigen values but cannot find the eigen vectors ofc

it's 12x12.

[limber sierra]

can you factor the char poly?

[wise flare]

yep. i have found the eigen values

and some have multiplicity

[limber sierra]

are you familiar with the concept of a minimal polynomial?

a linear map is diagonizable if and only if every factor of the minimal polynomial is linear

so if you know what a minimal polynomial is and how to find them, thats probably the most direct approach

[wise flare]

is that my only option ?

im not very familiar with that

[limber sierra]

well, if you could determine the dimensions of the eigenspaces, you could use that

[wise flare]

i read something similar to that but have no idea what that means

i assumed i needed to have the matrix on hand (which i do not)

[limber sierra]

well, its hard to give general answers

might be worth skimming the first couple pages of http://math.mit.edu/~dyatlov/54summer10/ws18.pdf, it has some practice problems (with solutions) of this nature

but it depends heavily on how the char poly looks

are all its factors linear?

[wise flare]

no

one complex value

[magic timber]

quick question when solving a linear equation and the answer you get is something like 1 = 1, 2 = 2, 3 = 3, etc what does that mean? That it's proportional?

[ashen bear]

hmm what do you mean?

if you end up with a bunch of tautologies that often means you have infinite solutions to your linear system

[rain echo]

just forgot a definition in order for the direct sum of 2 vector spaces to be defined (assuming they are subspaces of the same vector space) must the 2 original spaces have {0} as their entire intersection?

or is direct sum more of a union thing, like the smallest subspace containing both

[wintry steppe]

generally in linear algebra "direct sum" has two meanings

if you're referring to two subspaces of a given vector space, then direct sum means what's called the internal direct sum

and it refers to, well, their sum, so long as they intersect only at {0}

regardless of how two subspaces V, W of some larger vector space intersect, their sum V + W is always the smallest subspace containing both (since you're taking group theory, compare with generated subgroup stuff)

so it's kinda like the union in that regard, yes

just the union will never be a subspace unless one subspace in the union contains the other (classic LA exercise)

the other meaning of direct sum is basically just the product lol (it's called the external direct sum sometimes)

@rain echo

[rain echo]

alright thanks for clarifying

I'm going through an old linear algebra book ("old"=1970) to see how much the subject has changed, and it is honestly so different

it's like giving a new perspective

as well as teaching me some history I guess

[mild igloo]

if A =4x6 matrix

it can have rank 1 2 3 or 4?

anything else?

[rain echo]

0

if A is all zeros

[mild igloo]

ah'

would it change if the matrix was reverse?

no, right?

[rain echo]

by reverse do you mean transpose?

no, cause A and A transpose always have the same rank

[mild igloo]

yes

thank u

[winged belfry]

So if a vector or scalar is = 0, does that mean linearly independent or dependent

[rain echo]

any set with the zero vector is linearly dependent

[winged belfry]

So if a function has one value = 0, it's LD

like if x1 = 0

oops, i mean v1

[rain echo]

v1 cannot be zero

cause that implies x1=-x2

[winged belfry]

I just mean, if any of them equaled zero, would it be LD or LI

[rain echo]

LD

[winged belfry]

Ty

wonder why this dude was lying https://www.youtube.com/watch?v=1I8nGrJKv-g

[rain echo]

ohhh

that's different

ok

so if a vector is zero

then any set containing that vector is dependent

but

a test for linear independence

is to take a linear combination of vectors

which equals zer9

so like if x,y,z are vectors

[winged belfry]

O

[rain echo]

you look at ax+by+cz=0

And if you can prove that the only solution to this is a=b=c=0, then x,y,z are linearly independent

[wintry steppe]

ethan you're so good at linear algebra

[thin bloom]

Does hilbert space discussion come here?

[native rampart]

Guess so

[raw sand]

or analysis

depends how deep you wanna go

[rain echo]

analysis is probably a better place

[clever pumice]

Can you only have as many linearly independent eigenvectors as you have dimensions?

Like say a linear transformation in R3

Can only have 3 LI eigenvectors, maximum?

[odd kite]

you can only have 3 linearly independent vectors in R3, period.

[clever pumice]

lol, when you put it that way, right

I was thinking more like, vectors that aren't just scalar multiples of another

I know linear independence is more than that though

So say in R2, you can't have 3 eigenvectors that don't end up being scalar multiples of another right?

[odd kite]

@clever pumice Each eigenvalue λ has a subspace associated with it and Lv =λv for all vectors in that subspace, called an eigenspace. If the dimension of the eigenspace is 2 or more, there are an infinite number of eigenvectors in that space which are not scalar multiples of each other. In R2, the identity matrix has an eigenspace of all R2 and has an infinite number of eigenvectors that aren't scalar multiples of each other.
It's common to choose a set of linearly independent vectors that represent each eigenspace. (They form a basis of each space). Only when the eigenspace is 1D are all vectors in that subspace scalar multiples of each other (all vectors on are on one line).
From my previous comment it should be obvious that there are never more than N linearally independent eigenvectors, since there are at most N linearally independent vectors of any kind.
It's also true there can be up to N eigenvectors.

[thin bloom]

Does series x^n converge if |x|<1 on a hilbert space(or whatever)?

[round coral]

@clever pumice Yeah, you can have many eigenvectors but the length of the basis of them if it exists must be smaller than or equal to the dim of the vector space. Am I right Timon?

[odd kite]

yes

[round coral]

Have you heard of Pumba

[odd kite]

no

[clever pumice]

Thank you!

[round coral]

@odd kite Haven't you watched Lion King? Pumba was a pig and Timon was his friend.

[thin bloom]

Are you answering me

[odd kite]

ah yes

it's actually a reference

[thin bloom]

I guess no?

So let me reiterate

Does series x^n converge if |x|<1 on a hilbert space(or whatever)?

[odd kite]

@thin bloom sorry I don't know enough about functional analysis to answer

[thin bloom]

It's okay

Anyone confident in this field? I think it should be pretty basic

[round coral]

@thin bloom I think you must go to one of the advanced mathematics sections maybe?

[thin bloom]

? I mean, this is too basic for analysis I think

[round coral]

Sorry, not for me, I haven't studied it yet.

[odd kite]

yeah Timon, it's a reference to stories about fleeing into the woods.

[thin bloom]

Hmmm

[round coral]

@thin bloom Although I don't know about Hilbert space, but for real numbers it is convergent. Quite useless, you would already know it I think

I think normally also, it is oscillating, I am not sure.

[thin bloom]

I forgot about how complex numbers behave

[unique sluice]

Can someone give me an intuitive explanation of Hermitian matrix ? I know the definition and the process , but I can't interpret its geometrical meaning . It would be helpful if someone can at least direct me to some resources

[odd kite]

There's a lot to say about Hermitian matrices. I assume you are familiar with some properties, like symmetry of the conjugate transpose and complete basis of eigenvectors.

Some of the significance is plainly on the algebraic side of things

It's true for example that every matrix can be written as a sum of Hermitian and anti-Hermtian part, which is often a useful expression

[wraith dune]

Sorry if I am disturbing anybody, but could someone help me prove this?

[odd kite]

From the context of classical Lie groups, Hermitian matrices are related to the tangent space of unitary groups. A unitary matrix is generated by matrix exponention of generators which are in the tangent space of the group at the identity. If H is a Hermitian matrix then exp(iH) is a unitary matrix.

You may be able to have some geometrical sense of what a Hermitian matrix does to its eigenvectors

@wraith dune you can consider each point A, B, etc as a vector and $\vec{AB} = \vec{B}-\vec{A}$, and also $(\vec{F}+ \vec{G}).\vec{H} = \vec{F}.\vec{H}+ \vec{G}.\vec{H}$

[stoic python]

Timon:

[wraith dune]

Yeah, I end up getting BD - AD - BC - AC

Am I somehow supposed to group the vectors now?

@odd kite

[unique sluice]

@odd kite I am sorry , but my knowledge in Linear algebra is very minimal, I don't know most of terms you just mentioned . I know only basic Linear algebra and some intermediate topics like diagonalization ,SVD etc . I was interested in hermitian matrix in order to understand pseudo inverse of a matrix

[odd kite]

like was trying to say they come up in a lot of different contexts and there's a lot to say

[unique sluice]

Ok , I understand .Why hermitian matrix comes into picture for this particular case ? What does it have to do anything with inverse of matrix ?

[hollow cobalt]

find the solution if it is unique; otherwise, describe the infinite solution set in terms of an arbitrary parameter k.
4x-2y+6z=0, 5x-y-z=0 and 2x-y+3z=0

can anyone solve this for me?

[round coral]

@hollow cobalt if you want a solution without solving use symbolab or wolfram. Computer is better for this if you are lazy

[odd kite]

@unique sluice I'm not especially familiar with pseudo inverses but, starting with the basics if we have Ax = y and A is a singular matrix or rectangular, then we can have multiple solutions for x. So there isn't just one unique G where x=Gy and x solves Ax=y. So it appears they add various 'nice' conditions on G, which have some motivation behind them. When all four Moore–Penrose inverse conditions are satisfied G becomes unique. It happens that those properties also cause the relation you posted to be true when $H^H H$ is invertible. I'm not able to say much more than this on the subject at the moment.

Hermitian conjugation and Hermitian operators have an important role in linear algebra which is probably has something to do with why these particular set of conditions where chosen for the Moore–Penrose inverse

but I have very little experience using them

[stoic python]

Timon:

[wraith dune]

Could someone help me prove this?

It just says that in a foursided shape ABCD: ... AD is perpendicular with BC

Then prove that ..

[round coral]

what a weird labelling of quadrilateral, in normal cyclic labellings AB and CD are never adjacent

[jaunty wing]

Can anyone help me with diagonalizable matrixes?

[dusky epoch]

post your question

[jaunty wing]

I'm going to try to translate it at it's best

[dusky epoch]

what language

[jaunty wing]

Catalan

[dusky epoch]

oof ok

[jaunty wing]

Its about eigenvectors and eigenvalues

Knowing that A is a diagonalizable matrix that admits the eigenvectors (-1,2,2), (2,2,-1), (2,-1,2) and that A*(5,2,5)^T = (0,0,7)^T, find the eigenvalues and the matrix A

This is an optional that I have for tomorrow, I will know how to do it after class, but I like doing them before

My doubt is

How do I find a matrix out of eigenvectors without knowing the eigenvalues

[round coral]

in diagonalization the matrix that is diagonal one is actually the one which has eigenvalues on the diagonal. So I think find out the eigenvalues from eigenvectors

[jaunty wing]

And how would I find the eigenvalues of the eigenvectors without knowing the matrix?

do I just factor them?

[round coral]

what is this I can't understand it A*(5,2,5)^T = (0,0,7)^T ?

[jaunty wing]

I will send a pic for reference

[round coral]

I suggest use this A(5,2,5) = (0,0,7)

there is a purpose it is given

[jaunty wing]

Yes, but I have been very stuck here

I thought I should use a power iteration

I also know that 7 could be an eigenvalue of (0,0,1)

[round coral]

you already have the eigenvectors make the matrix of them then use A = PDP^-1

write a,b,c in the diagonal of D

and find the inverse of the eigenvector matrix

[jaunty wing]

So I have to find D first

[round coral]

yes D is what you have to find, then you will know the eigenvalues

you have P, you can calculate P inverse

take D with a,b,c constants on diagonal and 0s everywhere and multiply these to get A and solve the equation

[jaunty wing]

Can I use 7 and it's eigenvector(0,0,1) for this?

[round coral]

Please read what I have said, I have explained almost everything.

[jaunty wing]

Yes 🙂

[jaunty wing]

I assume this is what I will use when I have the matrix A to see if it gives me the same result no?

[wintry steppe]

Sorry for being stupid but I'm having a hard time understanding this geometrically:
"If we have a dot C(xC, yC, zC) on the like AB, and AC/CB = m, then:
xC = (xA + m*xB) / (1 + m), ...".
I just don't understand why the formula works.

[round coral]

@jaunty wing yeah that is what you have to use, find A in terms of constant eigenvalues say a,b, c then use the equation

[jaunty wing]

In other words once I have the matrix, I will multiply it by (5,2,5) and if it gives me (0,0,7) it means I did it right

[round coral]

you have to equate it with (0,0,7) to find the eigenvalues

[jaunty wing]

Alright I'm lost again

[round coral]

if you get (0,0,7) directly then you already know the eigenvalues, then what's the use

Look over the diagonalization of matrix again. And see my messages then

[jaunty wing]

Alright

[silver tree]

If I'm given a subspace in R^4, and I know the subspace U= span{v1, v2} (where I know the vectors of v1 and v2).

How am I supposed to show that the dimension of U is 2? At first, I thought that a simple explanation would be enough (since it has 2 different vectors in span), but then they wouldn't write "show." Am I overthinking it?

[dusky epoch]

you need to verify {v1, v2} is linearly independent

[silver tree]

They are linearly independent.

I was thinking if that was enough, when the assignment says "show."

[dusky epoch]

yes it is

[silver tree]

I see, thank you a lot.

[round coral]

I had a question, that if I take two operators over R^infinity , say T and S where T is right shift operator and S is left shift operator like S(z_1, z_2,.......) = (z_2, z_3, .........) and T (z_1, z_2,.......) = (0,z_1, z_2, .........) . When we try to find the eigenvalues of ST, we do S (T(z_1, z_2,.......) ) = K(z_1, z_2,.......) upon simplifying it we get K(z_1, z_2,.......)= (z_1, z_2,.......) and so we get out eigenvalue K=1 but if we do T(S(z_1, z_2,.......)) = K(z_1, z_2,.......), we get no eigenvalue at all as T(S(z_1, z_2,.......)) = ( 0,z_2, z_3,.....) and if we equate it with K(z_1, z_2,......) there seem to be no eigenvalue K which satisfies the equation
Why is this so? For finite dimensional vector space, ST and TS have the same eigenvalues but not when they are operators over infinite dimensional vector spaces? am I correct? But I don't understand why it is so that they don't have the same eigenvalues

[rain echo]

(0,anything.....) is an eigenvector of TS with eigenvalue 1

and also (1,0,0,0,0...) is an eigenvector with value 0

the last part is the real mystery

[round coral]

eigenvalue cannot be 1 for TS as as 1. z_1 is not 0

it can only be 0 when z_1 is 0

[rain echo]

yes

so the vector(0,z2,z3...) is an eigenvector

[rain echo]

maybe I'm dumb but that looks like it should be impossible

my brain rejects that as a possibility

[grave halo]

I have no clue man

Maybe that's why none of my peers have managed to solve it xd

[rain echo]

if hf=cf doesn't that mean h=c

[grave halo]

as long as f is nonzero

[rain echo]

forget about the self adjoint part I feel like in order for L to have any eigenvalues at all then h must be constant

[grave halo]

Yeah we've thought so too

But how do we get two distinct ones then

[rain echo]

idk I would give up

[grave halo]

Can it be periodic perhaps?

[rain echo]

then what

[grave halo]

Well ok so

nah nothing

this shit is impossible

we need h(t) to spit out two values at the same time

it's not even a function

[rain echo]

yeah it almost looks probably impossible

like is there a hole in the logic to saying that if hf=cf and f isn't zero then h=c

[grave halo]

maybe we have to put some limit to the t

like for certain values of t only

[rain echo]

or maybe

h is like a step function

[grave halo]

but it has to be continuous xD

h in V

[rain echo]

oh you're right

k even better

let me construct a function

it's like flat at -1 for a bit

then there's a line segment that goes up

then it's flat again at 1

let's just say

[grave halo]

Yeah so a piecewise

[rain echo]

yeah

still continuous

[grave halo]

yup

[rain echo]

then if f is zero where h is not -1

hf=-f

but if g is zero where h isn't 1

then hg=g

[grave halo]

Yeah but legit the ugliest fucking answer known to mankind

[rain echo]

hahahaha yes

but it's still an answer!!

and actually

I think it's semi pretty

[grave halo]

I like that term

semi-pretty

[rain echo]

yeah hahaha

it isn't a clean answer

[grave halo]

Ok so

It's Swedish but you get it

Now I construct the shitty function

[rain echo]

oh are you in sweden?

[grave halo]

Yeah

[rain echo]

are you getting a pure math degree?

[grave halo]

Nop! Engineering Physics at KTH

[rain echo]

oh I know KTH!!

[grave halo]

Lessgo

[rain echo]

ok can I dm you please

[grave halo]

Yeah sure lol

[jaunty wing]

I got a 3 x 1 matrix with a, b and c after multiplying PDP-1 * (0,0,7) and I assume these are the eigenvalues you told me about @round coral

But now I'm stuck at finding a b and c

I think I know what to do now, (5,2,5) are abc

[honest imp]

Would x1 be considered a free variable? o.o the whole column for x1 are 0’s

[jaunty wing]

Hmm, it's clear that x1 = x2 and x2 is 0

x2 is free

so x1 = x2

[honest imp]

oh wut i thought x2 was 0

[jaunty wing]

🙂

[round coral]

that's good

I am not sure that's correct though

[jaunty wing]

I guess 5,2,5 are not a, b, c

it gives me a decimal number and I assume it shouldn't, by how these exercises are prepared

[silver tree]

@honest imp I don't get the same reduced matrix as yours - I get this.

[half ice]

Note there's an elementary step you can take to get Suu's, so they are equivalent. @silver tree

This is basically just considering the equations in a different order.

[silver tree]

Oh yeah, I'm completely blind

I apologize.

[half ice]

No no you're good lol

I don't even know what the original question was

[wise flare]

can a matrix be diagonalizable if it is linearly dependent ?

[silver tree]

As far as I know, a Matrix can only be diagonalized, if a n x n matrix has n linearly independent eigenvectors.

[wise flare]

a matrix with a full row of zeroes is linearly dependent by defaulkt right ?

because of the existence of a free varibale

[silver tree]

Not necessarily (as far as I know)

[trail mulch]

what about $\begin{bmatrix}1 & 0 \ 0 & 0\end{bmatrix}$ ?

[stoic python]

js:

[wise flare]

well i mean something like

a 3x3 an the last row is all 0s

that matrix above is also linearly depenent btw

[trail mulch]

yeah

but it's diagonzable

whatever that thing is

[silver tree]

Well, how many eigen-values have you found?

[trail mulch]

2

λ = 1, 0

[silver tree]

Yeah, then you'll need to find a eigenvector for each eigen-value - if they are linearly independent. Then that matrix can be diagonalized.

[trail mulch]

there are two independent eigenvectors though

(1,0) and (0,1)

[silver tree]

Then it can be diagonalized.

[trail mulch]

but it's linearly dependent

that was the question, right?

[silver tree]

Since it's a 2x2 matrix, with 2 linearly independent eigenvectors.

Oh.

I thought we were talking about Eigenvalues and Eigenvectors.

[wise flare]

so how can you tell if a matrix is truly diagonaliable

[trail mulch]

n linearly independent eigenvectors

[silver tree]

^

[wise flare]

im confuse

wym linearly inepenent vectors

after you fin the eigeb vlaues an eigen vectors

you also have to fin if those vectors are linearly inepenent ?

[trail mulch]

have you encountered diagonalizing a matrix A like A = V D V^{-1} yet?

[silver tree]

I actually have a question:

If i have 2 2D subspaces, U1 and U2 and I know their span vectors and that they are orthogonal on each other.

How am I supposed to make an orthonormal basis (q1, q2, q3, q4) where (q1, q2) $\epsilon$ U1 and (q3,q4) $\epsilon$ U2?
I though of just normalizing it, but the basis vectors in U1 aren't orthogonal on each other.
if needed, I can show the actual vectors.

[stoic python]

Philoshōten:

[ocean sequoia]

Could you do a change of basis for U? Such that q1,q2 are orthogonal assuming they span the space

[silver tree]

Well, then I should simply take one of the basis and make a ortoghonal vector on that vector?

[ocean sequoia]

I mean I think that would be ok as long as it’s still a basis

Tbh I’m taking an educated guess I could be wrong

[silver tree]

I'll try it out later then - thanks for the help!

[rain echo]

use gram schmidt it's a process to turn a set of LI vectors into an orthonormal basis for the space they span

[wise flare]

how is this correct

multiplicity of 2 means there are eigenvalues

0 5 5

[round coral]

@wise flare lambda =5 with multiplicity 2 means the eigenvalue 5 has 2 linearly independent eigenvectors associated with it

[wise flare]

so everytime a matrix has multiplicity it isnt iagonilable ?

[round coral]

The reason is given above in the image you posted

and what you said is not true, if the eigenvalue 5 had a multiplicity 3 then the matrix would have been diagonizable whatever

a diagonal matrix could be made

[wise flare]

im confuse

[round coral]

Try finding out eigenvectors and eigenvalues yourself, diagonalize the matrix on your own. Only then will you understand what is happening. Calculator can only give you answer

[wise flare]

i have foun all that

Wdym ?

It can anything

Depending on the values you choose

-2
-2.5
1

That was my vector

[round coral]

wrong question I asked. How many distinct eigenvalues are there?

[wise flare]

2 (but really 3)

2

[round coral]

only the distinct count, there are only 2 distinct so you can't make a 3X3 diagonal matrix

[wise flare]

But in that case

[wise flare]

how is this true

thsts still 2 dinstinct

[round coral]

Sorry I take back my words, I said wrong. Sleepy!!

[silver tree]

On which vector though?

[rain echo]

you have 2 2d subspaces

[round coral]

The no of distinct eigenvalues must equal to the dim of the vector space your operator is on

[wise flare]

So basically a matrix with multiplicity by default is not diagonlzable

[rain echo]

in each, pick 2 vectors which are LI, then use gram schmidt on that pair

[wise flare]

Is that true

[silver tree]

Oh, so for instance pick q1 and q3, and use gram schmidt to find q2 and q4?

[rain echo]

yes

[wise flare]

Because there is like geometric and algebraic multiplicity or something t

I’m confused

[round coral]

I don't know that. But the theorem I have proved, is that if you have an n X n matrix then for it to be diagonlizable you need n distinct eigenvalues

and here there are only 2 distinct eigenvalues

[silver tree]

But how are you sure that q1, q2, q3 and q4 are in their respective subspace?

[wise flare]

Right but the contrapositive of your statement makes mine true

[rain echo]

pick q1 and q2 in one subspace

[wise flare]

That’s what I’m confused

[silver tree]

Since q3 and q4 are already orthogonal, then I don't need use gram schmidt on that, right? I simply need to normalize it. but for q1 and q2, I need to do it.

[rain echo]

yes

[wise flare]

@round coral I agree with you

I’m asking if the opposite of that Stevens is true

So a matrix has an eigenvalue with mutlitplicity. Is it immediately not diagonlixable

[rain echo]

no

[silver tree]

Alright, thank you very much - I'm reading about this in advance, and thus not that knowledgeable at the moment.

[wise flare]

The premise of that book statement alludes to it tho

[round coral]

@wise flare Actually I studied a bit and found out that only when geometric multiplicity is equal to algebraic then only the matrix is diagoni...

algebraic multiplicity you may know when we find Null(A - lambda I) , then we get it.

the polynomial whose roots are the eigenvalues, the multiplicity of those roots is the algebraic multiplicity

[wise flare]

Right but this leads to a bigger question.

To find the geometric multiplicity. You need to matrix A

[round coral]

the geometric multiplicity didn't you find just now, it is 2

[wise flare]

That’s the algebraic

If the geometric was 2 then the matrix would be diago

And it is not

[round coral]

I checked it I am right. See the polynomial you got with which you found the eigenvalues

[wise flare]

The issue is that say you have some arbitrary matrix A that is like a 15x15 and you are only given the character polynomial. How would you know if it’s diago. You cannot do the geometric check (A-lambda I)

That’s the flow in the theory that I am failing to understand

[round coral]

I don't know that, I only deal with the theory aspect and am not much considered with the computation

[wise flare]

That’s still the theory

You cannot truly find if a matrix is diag or not just based on the eigen values

[round coral]

and checking geometric is not so tough, after all you just have to find the eigenspace

[wise flare]

You need a matrix for that

[wintry steppe]

ah yes, define a matrix to be diagonalizable if its algebraic and geometric multiplicities coincide

[round coral]

@wise flare Now you believe me, Tterra is an OG . He has already studied all this stuff

[wise flare]

I never disagreed with you

I disagreed that you cannot find if a matrix is diago or not with just its character poly

You need the matrix itself

[wintry steppe]

an nxn matrix with n distinct eigenvalues is always diagonalizable

[wise flare]

That’s true

[wintry steppe]

but yeahthe characteristic polynomial alone tells you nothing

well

not much

given any polynomial you can find a matrix that has it as its characteristic polynomial

[round coral]

@wintry steppe Sempai 👍

[wintry steppe]

so just the char poly doesn't tell you a whole lot

[wise flare]

But then that means that if a matrix had multiplicity. It shouldn’t be diagonalizable

#linear-algebra message

This is literally what that picture is saying

[wintry steppe]

it probably means algebraic multiplicity

and in that case the geometric multiplicity of 5 as an eigenvalue is strictly less than the algebraic multiplicity, 2

so it cannot be diagonalizable

since diagonalizability, as kanishk mentioned, is equivalent to alg mult = geo mult for every eigenvalue

[wise flare]

I don’t understand

[wintry steppe]

ok so the characteristic polynomial of this thing is

(also i mistyped, pls reread)

ugh discord is being annoying one second

ok so like this thing has characteristic polynomial x(x - 5)^2 up to a sign, so the eigenvalues are 0 and 5

the algebraic multiplicity of an eigenvalue is defined to be its multiplicity as a root of this

so 0 has algebraic multiplicity 1, and 5 has algebraic multiplicity 2

the geometric multiplicity is defined to be the dimension of its corresponding eigenspace

and i'm assuming once you work out the dimensions of the eigenspaces (as symbolab did i guess), you find that 0 has geometric multiplicity 1 and 5 has geometric multiplicity 2

diagonalizability is equivalent to the algebraic and geometric multiplicities agreeing for each eigenvalue

in this case, that does not happen

so the matrix isn't diagonalizable

if there is more confusion @ me

wow many typos

pre-coffee math isn't a good thing

ok chum it shouldn't have typos anymore

[honest imp]

How would I find out c after doing what I’ve done thus far? I thought about saying that each eigenvalue corresponds to a eigenvector which all of them would be considered linearly independent, and since there’s three linearly independent vectors they can span the space R3

[wintry steppe]

there’s three linearly independent vectors they can span the space R3
yes

[honest imp]

[wintry steppe]

eigenvectors corresponding to distinct eigenvalues are linearly independent

and you have three distinct eigenvalues

so you're correct, i'd just be a bit careful with the wording though

[honest imp]

Ah okay, yea sorry I’m not great at explaining mathematical ideas/concepts

[wintry steppe]

it's a skill you develop over time

frankly

it's probably the only reason i help out in these channels

[wise flare]

this

how is this fiun @wintry steppe

foun

[wintry steppe]

just look at the characteristic polynomial

[wise flare]

?

not that

[wise flare]

i meant this

sorry

[wintry steppe]

are you asking how you find the dimension of the eigenspace?

[wise flare]

the geometric multiplicity is defined to be the dimension of its corresponding eigenspace
and i'm assuming once you work out the dimensions of the eigenspaces (as symbolab did i guess), you find that 0 has geometric multiplicity 1 and 5 has geometric multiplicity 2
diagonalizability is equivalent to the algebraic and geometric multiplicities agreeing for each eigenvalue

that is the same multiplicity as the arithmetic multiplicity so if they agree

how is it not iagnonliable

[wintry steppe]

sweet fucking christ i made so many typos

sorry

don't worry i have my coffee now

oh jeez

yeah i've probably just confused you further, i think i misread things

sorry

give me a few minutes to collect my thoughts and i'll write something that isn't blatantly wrong

i basically owe it to you after dragging you through all this

matrix has characteristic polynomial x(x-5)^2, up to a sign
so we have eigenvalues 0 of alg. mult. 1 and 5 of alg. mult. 2
to determine diagonalizability we need to find the corresponding geo. mults.
the matrix will be diagonalizable if and only if the geo. mult. of 0 is 1, and the geo. mult. of 5 is 2
when you compute the dimensions of the eigenspaces, they're both 1
so the geo. mult. of 0 is 1, and the geo. mult. of 5 is 1
these do not match up with the algebraic multiplicities, so the matrix is not diagonalizable

if this has a typo i will leave the server

[wise flare]

right

but is this process possible if only given the caracter polynomial

[wintry steppe]

if given only the characteristic polynomial? i'd say no, not really, since finding the geometric multiplicities depends on the matrix itself

[gray dust]

what's a caracter polynomial

[wintry steppe]

open up the representation theory book and figure out

[rain echo]

det(xI-A) is the polynomial

[gray dust]

ew i don't like picking up extra factors of -1

[vocal isle]

Hey folks I have a system of linear 2nd order differential equations [M] and [K] are 3x3 matrices, F(t) is a 3x1 time dependent vector, and X is a time dependent vector I'm trying to numerically solve. Without using 'modal analysis' does anyone know how I can solve the following:
$$
[M]\ddot{X}+ [K]X = F(t)
$$

[stoic python]

The Smelly Cherry:

[simple hornet]

I think the diff eq channel would be better suited to that oof

also one question of my own

im having doubts about the way i proved the forward implication

I think the union of the two subspaces $U_1 \cup U_2$ is going to contain zero and it will be closed under scalar multiplication. We need to search for problems in additive closure.

[stoic python]

mirzathecutiepie:

[wintry sphinx]

you are correct

[simple hornet]

I said that $U_1 \cup U_2$ had to be closed under addition (since it is a subspace as we assumed). This is the same thing as saying $u,v \in U_1 \cup U_2 \implies u + v \in U_1 \cup U_2$. So we must have either $v,u \in U_1$ or $v,u \in U_2$. If $v,u \in U_1$ and $v,u \in U_2$ then $U_1 = U_2$. (This is the part im iffy about) If $v,u \in U_1$ and $v,u \not\in U_2$ then $U_2 \subset U_1$

[stoic python]

mirzathecutiepie:

[simple hornet]

i think the last part is wrong

i did a similar argument for when $v,u \not\in U_1$ and $v,u \in U_2$

[stoic python]

mirzathecutiepie:

[simple hornet]

i think my problem was assuming that set inclusion obeys trichotomy

[wintry sphinx]

mmm yeah it's sketchy, since there are $u, v \in U_1 \cup U_2$ such that $u + v \in U_1 \cup U_2$; it's just not all of them

[stoic python]

Saccharine:

[simple hornet]

Wait wdym

oh yeah

i get it

I should probably try another method then, right

[wintry sphinx]

I think it's probably easier to consider disjoint subspaces U and V; it's not hard to prove that u + v is not a member of either subspace

[simple hornet]

tho i believe this statement is sound, right $v,u \in U_1$ and $v,u \in U_2 \implies U_1 = U_2$?

yes the answer key does it by contradiction

ill try that then

[wintry sphinx]

I think that statement is sound, if you consider all v, u in the union, but I don't see how it gets you anywhere

[simple hornet]

yeah

[wintry sphinx]

you're basically just saying that the union is a subset of each space

which is way too specific

[simple hornet]

yeah i guess

i think my approach is a little convoluted

[wintry sphinx]

you could go about this proof in this way:
If one subspace is contained in the other, then their union is a subspace (trivial)
If one subspace is not contained in the other, i.e. both differences (U1 - U2) and (U2 - U1) are nontrivial, then they are not closed under addition, and you can exhibit a vector like this

(U1 - U2) and (U2 - U1) are disjoint, so it reduces to the disjoint proof, and you can find a vector u in (U1 - U2) and a vector v in (U2 - U1) such that u+v is not in (U1 - U2) union (U2 - U1); further, u + v is not in the intersection of U1 and U2, so it can't be in (U1 - U2) union (U2 - U1) union (U1 intersect U2) = U1 union U2

this is the proof that makes intuitive sense to me, but it's a bit convoluted

[smoky lily]

can a basis of 0 , 0, 0 be a kernal of a linear trans?

[ocean sequoia]

no a basis needs to be linearly independent

and the zero vector isnt

[slow scroll]

what do you even mean by 0, 0, 0?

a basis for a vector space is not the kernel of any linear transformation

at least in finite dimensions

i don't think that can be true in infinite dimensions either actually
besides the point. mostly likely a mis-woreded question

what is PCA?

[ocean sequoia]

principle component anaylsis

dimnesionality reduction

after choosing the k eigenvectors from the eigendecomposition of the covariance matrix we project our original data onto the column space of those k eigenvectors correct?

[smoky lily]

https://i.imgur.com/G3le6ph.png

my work is on the right and the correct ans is on the left (in red)

what steps am i missing here?

only thing i realized is that 1.769 is equal to 23/19 seems like im missing a lot of steps tho..

[cunning arch]

I'm kinda blanking out, but: "Use this to write the image Nx as a linear combination of b_1 and b_2."
So if my Nx is [1&0 \ 0&1/4] [x_1 \ x_2] and b_1 is [2 \ 1] and b_2 is [1 \ -1], I'm a bit confused on how to write it as a linear combination
Edit: rip formatting

Wait am I supposed to do N(b_1) and N(b_2)?

[slow scroll]

Nx is a linear combination of (1,0) and (0,1/4) so you need (1,0) and (0,1/4) as linear combinations of (2,1) and (1,-1)

[cunning arch]

So for [1,0] it would be b_1 + b_2?

Oh wait that doesn't work

[smoky lily]

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

is the basis of this kernel equal to -10, 15?

[slow scroll]

are u familiar with change of basis strawberry?

[cunning arch]

Kinda, it's what I did in the first part of the problem (not shown in the screenshot)

[slow scroll]

do u have a screenshot? I want to make sure I'm not about to overcomplicate this

[cunning arch]

And A = [3/4 & 1/2 \\ 1/4 & 1/2], given in the problem

[slow scroll]

is the E basis the standard basis?

[cunning arch]

Yes

Wait I think I just set up a system of equations from it?

Am I overthinking this or underthinking it lmfao

[slow scroll]

I'm asking myself the same question atm

[cunning arch]

Lolol I think I might just ask them @ office hours tomorrow. Before I go, do you have an idea how to find N^{1000}x? There has to be some shortcut out there, right?

[slow scroll]

yeah, N is a diagonal matrix, so you can take corresponding powers of the diagonal entries

[cunning arch]

Ah I think I've seen that around somewhere. Thank you so much for helping!

[slow scroll]

np

[hollow finch]

how do you prove that a linear transformation is surjective?

proving it isnt is pretty easy bc you just need a counterexample

[odd kite]

look at the span of the columns

[rain echo]

find rank

compare it to dimension of the codomain

[hollow finch]

the student im trying to help hasnt learned how to make a matrix out of a general linear transformation yet

[rain echo]

oh

[hollow finch]

okay yeah rank makes sense

but the lack of a matrix is still a problem

[rain echo]

hmmmm

you should be able to get around it

find a basis for the domain

and determine whether the space spanned by their images in the codomain is the entire space

possibly

but yeah tough without a matrix

[round coral]

@hollow finch there are many ways, one is the way to find rank of the matrix of linear map, other is to find the null space, then use rank nullity theorem,

there is one also that every surjective linear map has a right inverse

[hollow finch]

Aha rank nullity theorem thats smart

[rain echo]

yeah that's a smart idea

[hollow finch]

If rank(T) is equal to the dimension of the codomain then it is indeed surjective right?

[rain echo]

yes

[round coral]

if it is an operator the job becomes really easy, surjective and injective mean the same thing

[hollow finch]

true

[round coral]

an important point to note also that the dim of the vector spaces say if the linear map is from U to V then dim U must be > = dim V for the map to be surjective but if dimU < dim V then there is no chance for that at all

[half ice]

What's the example? If you can't make a matrix out of it, your best bet is really just to reason that you can hit every point

[shell pond]

if a divide c and b divide c and gcd (a,b)= d then ab divide cd . how do I proceed to proof this

[arctic hornet]

HI quickie question, does homogenous system mean systems with trivial solutions?

[odd kite]

homogenous system means Ax = 0

[arctic hornet]

Oh I get it

dis is dumb but what about Ax=b?

non-homo?

[odd kite]

that's called inhomogeneous

[arctic hornet]

I see

[odd kite]

but yeah right idea

[arctic hornet]

thank u

Um so that means homo systems only has 0 as solution sets?

[odd kite]

no, if A is singular

then it can have others

if A is invertible then yes, only 0

[arctic hornet]

Oh okay, I have to get pass to invrses then to fully understand

i just assumed cuz of the 0 at the end

[odd kite]

we call the solutions of Ax=0 the kernel or nullspace, you'll probably get to that

eventually

[hollow cobalt]

For the following system of homogeneous linear equations, find the solution if it is unique; otherwise, describe the infinite solution set in terms of an arbitrary parameter k.

4x-2y+6z=0, 5x-y-z=0 and 2x-y+3z=0

Can anyone solve this for me?

[austere cedar]

I can't solve it for you. I can give a hint though: Write the question using a matrix representation.

[dusky epoch]

@hollow cobalt no

we do not give out answers here

[hollow cobalt]

@stabulo should I solve it by determinant method?

[austere cedar]

I'm not sure what the determinant method is.

Do you mean using the determinant to find the inverse matrix to then apply it appropriately to the matrix representation?

Or do you mean Cramer's rule?

[arctic hornet]

Hey I managed to pass the inverse topic -- so if its inversible, that means it has one solution?

nvm

[round coral]

@hollow cobalt if you want solutions write the matrix in symbolab or wolfram or any other good site

[arctic hornet]

just realized -- linear algebra is teaching me how to find X in different ways?

now im learning LU Factorsz

[round coral]

Well linear algebra is not only finding X, there is way much to it

[austere cedar]

If the determinant is non-zero, there will be a inverse matrix, so a unique solution will exist. The proof is simple.
If Ax=0 is the system. If A is invertible (detA is non zero) then A^(-1) exists
A^(-1)Ax = A^(-1)0
Matrix multiplication is commutative, meaning you can put the brackets in any way:
( A^(-1)A )x = 0
Ix = 0
x = 0.
This also shows, if A has an inverse matrix, the only solution to the "homogenous" system is (0,0,0).
Assuming I've not made any mistakes, only just starting to learn Linear Algebra.
@arctic hornet

Wrong tag! sorry! I would delete but then it would just be a invisible ping, I argue that's worse.

@hollow cobalt

[hollow cobalt]

@austere cedar Thanks a lot... I took your hint

And I guess I can solve now

[austere cedar]

You can also view it geometrically as the intersection of the 3 planes.

If you plot them you will see they intersect at the origin.

I guess you asked a related question, so the ping wasn't too bad actually.

[arctic hornet]

just want to know -- if matrix is invertible, then it has 1 solution?

[marble lance]

What is the solution of a matrix?

The system of equations represented by the matrix has 1 solution

[dusky epoch]

matrices don't have solutions

it doesnt make sense to speak of a matrix having one solution or not

[arctic hornet]

something like X= A-1*B

cuz its invertible

[dusky epoch]

if you want to talk abt a system of equations then you talk about a system of equations

[arctic hornet]

so like the solution for this system of equation?

[dusky epoch]

yes, if a matrix A is invertible then the system Ax = b has one and only one solution

[arctic hornet]

nice

thanks for that

[austere cedar]

Notice the lower cases used on x and b, they are matrices but it's usually written in lower case, since they are vectors, to indicate this you should use bold or arrows. You can modify a proof by having b instead of 0.

You had the write idea I imagine, just need to be careful with language.

[arctic hornet]

I understand, thank you for that

[arctic karma]

am i allowed to ask for help with homework (i've already finished it)

or does it fall under "Requesting help during an exam is a bannable offense."

[austere cedar]

If people got banned for asking for help with homework, there wouldn't be a server left.

[arctic hornet]

hey -- about linear transformation, is there another term for it? Its really confusing

theres so many kinds O.O

[dusky epoch]

wym kinds

another term for what exactly

[arctic hornet]

like

I dont get it

[dusky epoch]

don't get what

[arctic hornet]

R2 and R3

[dusky epoch]

?

R^2 and R^3 are vector spaces

[arctic hornet]

Rotation in R2, Reflection

its like scalars?

[dusky epoch]

wh

[arctic hornet]

oh man

this is bad

Im transforming the system linearly?

[dusky epoch]

some particular classes of linear transformations get special names because they have special properties that make them st-

what

oh god

[arctic hornet]

Idk after my determinant topic--it jumped to linear transforms

suddenly I see more than 5 letters

[acoustic path]

hey guys is this defitition correct?

the determinant of a matrix is the scalar for the new area/volume after a transformation

[austere cedar]

Obviously a geometric pictures only works for 2D and 3D, in the 2D case it's the scale factor for which areas are scaled by and in 3D it's the scale factor for which volumes are scaled, the sign indicates orientation. If you go around each point in the same order you will have a direction (clockwise or counterclockwise) if it's negative after transformation, going in the same order of points will have the opposite direction (2D case), I'm not sure on a "nice" description for the 3D case though.

[viscid kernel]

@austere cedar it also works for 1D, s scalair is just a vector in 1D, the determinant ( trivial ) gives the length of that vector.

[austere cedar]

Good point.

[rough olive]

I have a course right now regarding analysis of data and we have gone through different types of algorithms (regression, clustering, classification etc.), loss functions (MAE, MSE, RMSE), and matrix notation. Anyway, one of the exercise tasks is the following:

I'm not entirely sure how to start though. In our book, $\bar{x}$ is notation for the average or the sum of all $x_0 \ldots x_n$

[stoic python]

Resident Tracksuit Advisor:

[rough olive]

So as far as I can see, the above is equivalent to a row vector that looks like this $[x_0,:\ldots:, x_n]$ that is multiplied to a column vector with the same values

[dusky epoch]

i would assume indexing starts with 1 and not with 0

[stoic python]

Resident Tracksuit Advisor:

[rough olive]

Oh yeah you're right

I study CS so I keep thinking everything is zero-indexed lol..

anyway

I am unsure of how to differentiate this.

Any tips or hints would be greatly appreciated.

Usually, with a function like $f(x,y)=4x^2+y^2$ the gradient would be $\nabla f = <8x, 2y>$. But for some reason, when it's written in vector/matrix notation my mind is all blank.

[stoic python]

Resident Tracksuit Advisor:

[dusky epoch]

\langle and \rangle, or \left< \right>

anyway

you could rewrite this function coordinate-wise

$f(\bar{x}) = \sum_{k=1}^n x_k^2 + c$

[stoic python]

Ann:

[dusky epoch]

x here is a col vector by the way

[rough olive]

Oh yeah that's true. I just assumed they wanted us to do it with matrix/vector notation, but they're not strictly forbidding us from doing this. They're the same thing anyway, just a different notation.

Thank you so much, this makes it a lot easier!

How would I go about rewriting this?
$$
f(\bar{x})=\bar{x}^{T} A \bar{x}+\bar{b}^{T} \bar{x}+c
$$

It's mostly the $A\bar{x}$ that's causing confusion

[stoic python]

Resident Tracksuit Advisor:

[dusky epoch]

$\bar{x}^T A\bar{x} = \sum_{i,j=1}^n a_{ij} x_i x_j$

[stoic python]

Ann:

[dusky epoch]

$\sum_{j=1}^n a_{ij}x_j$ is the $i$'th component of $A\bar{x}$

[stoic python]

Ann:

[rough olive]

Okay yeah makes sense. How would I go about differentiating something like this? Is the derivative simply $a_{ij}\cdot 1\cdot 1 = a_{ij}$ when you're differentiating with respect to x?

[stoic python]

Resident Tracksuit Advisor:

[rough olive]

With a sum like this:
$$
f(x)=\sum_{i=0}^{n} a_{i} x^{i}
$$
The derivative is:
$$
f^{\prime}(x)=\sum_{i=1}^{n} i a_{i} x^{i-1}
$$
But in my case there's both i and j. So what would the derivative of $a_{ij}$ be?

[stoic python]

Resident Tracksuit Advisor:

[marble lance]

What do you mean by the derivative of a_(ij)? Isn't a_(ij) just a constant entry of a matrix?

[rough olive]

I suppose it is yes

Would the derivative of the above then be
$$
f^{\prime}(x)=\sum_{i, j=1}^{n} ija_{i j} x^{i-1} x^{j-1}
$$

[icy merlin]

im sry for accidentally jumping in

[stoic python]

Resident Tracksuit Advisor:

[rough olive]

It's fine no worries

[marble lance]

Are you confusing exponents with subscripts?

x_i is the i-th component of the vector x

It is not x to the power of i

[rough olive]

Oh fuck

Lol

What would the derivative then be? if they're all constants and we're differentiating with respect to x then surely $a_{ij}$ would be 0 whilst the two x variables will be 1, which means the derivative would be 0*1*1 = 0? Or am I still not understanding this correctly

[stoic python]

Resident Tracksuit Advisor:

[red prawn]

It's like if you have for example $F(x,y) = axy$ . The derivative wrt x is $F_x(x,y) = ay$, with respect to y you get $F_y(x,y) = ax$
The a's are coefficients as before, focus on which variable you're differentiating w.r.t.

[stoic python]

Apopheniac:

[rough olive]

Well, I'm asked to find the gradient $\nabla f(\bar{x})$ of:
$$
f(\bar{x})=\bar{x}^{T} A \bar{x}+\bar{b}^{T} \bar{x}+c
$$
But usually when finding gradients, you have one or more variables. In this case, it's simply saying to find the gradient of $f(\bar{x})$, which as far as I understand is the same thing as saying find the gradient of:
$$
f\left(\sum_{k=1}^{n} x_{k}\right)
$$

[stoic python]

Resident Tracksuit Advisor:

[half storm]

The gradient is just a vector whose components are the partial derivatives of the function

So just take the partial derivatives of the thing on the right-hand side

[rough olive]

Yes but how can you find partial derivatives when there's only one variable indicated on the left side? Wouldn't it just be the derivative of the whole thing in respect to x?

I thought partial derivative implied at least 2 variables

[half storm]

$\bar{x}$ is avector

It's a function of several variables

[stoic python]

TheDon:

[rough olive]

yes it's $[x_1, \ldots, x_k]$

[half storm]

$f(\vec{x}) = f(x_1,x_2,\dots , x_n)$

[stoic python]

Resident Tracksuit Advisor:

[rough olive]

But how would I go about differentiating with this notation? It confuses the hell out of me

[half storm]

differentiate with respect to each of the variables

just like how you would normally find the gradient.

[stoic python]

TheDon:

[rough olive]

I'm sorry I just don't understand it. Would you mind walking me through it in an ELI5 fashion?:/

[red prawn]

Maybe try it for n=1, n=2

[red prawn]

$\frac{\partial}{\partial x_k}\left( f(x_1,x_2,...,x_n) \right)$ will be the kth coordinate of the gradient

[stoic python]

Apopheniac:

[marble lance]

$\pdv{x_k}$

[red prawn]

omg ty so much you have no idea

lol

[stoic python]

Lunasong:

[marble lance]

Lol

[winged belfry]

one sec

So I was doing #1 C and was wondering if what I was doing was correct.

1. I setup the equations as follows:

eq 1: -2a + b + 2c
eq 2: a + 3b + 4c

2. scaled equation 2 by "2"

(2a + 6b + 8c) + (-2a + b + 2c) = 7b + 10c = b = -10c/7

3. plug b into equation 1.

(-2a -10c/7 + 14c/7 = 0) --> -2a = -4c/7 ---> a = 4c/14

before i go any further, (-4c/7) / -2 is 4c/14 right?

[dusky epoch]

is there an image that's meant to be here

also -2a + b + 2c is an expression not an equation

[dusky epoch]

so you're trying to explicitly establish whether or not a set of 3 vectors in R^2 is linearly independent?

[winged belfry]

yes

[dusky epoch]

you're kinda wasting your time there

a set of 3 vectors in R^2 will always be not LI

nmw

[winged belfry]

but i got them all = 0

[round coral]

it's impossible

check it again

[dusky epoch]

you did?

(a,b,c) = (2, -10, 7) would like a word with you

[winged belfry]

I got b = -10c/7 and a = 4c/14, plugged those into the 2nd expression to get the expression:

4c/14 -60c/14 + 56c/14 which comes out to c = 0

or did it just come out to 0

and not c = 0

[dusky epoch]

you should have had the equations: -2a + b + 2c = 0 and a + 3b + 4c = 0

[winged belfry]

I did

[dusky epoch]

there is also an issue with your work but if properly formatted it will become correct so whatever.

[winged belfry]

i had to scale the first expression by 3

[dusky epoch]

I got b = -10c/7 and a = 4c/14, plugged those into the 2nd expression to get the expression:

you would have gotten 4c/14 -60c/14 + 56c/14 = 0

or just

0 = 0

which tells you nothing

[winged belfry]

doesn't 0 =0 tell me that C doesn't equal 0 so it is linearly dependent

[dusky epoch]

it tells you nothing about c

c can be whatever

therefore the set of vectors is LD

[winged belfry]

ok ty

[royal ore]

What do I use to check if a list is orthonormal?

[round coral]

A list of vectors is called orthonormal if the vectors in it are pairwise orthogonal and each vector has norm 1. In other words, a list {e1,..., em} of vectors in V is orthonormal if 〈ej, ek〉 equals 0 when j = k and equals 1 when j = k, for j, k = 1,...,m. For example, the standard basis in Rn is orthonormal.

[dusky epoch]

@royal ore the definition of orthonormal

[nocturne jewel]

Are cofactors numbers or matrices?

[obtuse mulch]

Anyone here experienced with sparse matrices?

[gray dust]

@nocturne jewel cofactors are numbers. the cofactor matrix of A is the matrix made of A's cofactors

[nocturne jewel]

$C_{i,j} = (-1)^{i+j} \cdot |A_{i,j}|$ right?

[stoic python]

moshill1:

[gray dust]

|A_ij| is odd notation

rather we write M_ij, the (i,j) minor of A

[gray dust]

a hint is to not even think of C's entries. think of the size of I so that the sizes of CI & C match. then think of the simplest such I where CI=C

[royal ore]

so it's not a 3x4?

[nocturne jewel]

My prof said either det(A_ij) or |A_ij| meant the same thing

[acoustic zodiac]

i've used both

|A_ij| seems more convenient to me

[gray dust]

A is the matrix. A_ij is the (i,j) entry of A. neither det(A_ij) or |A_ij| make sense to say

if you mean to say C_ij=..det(A) that's wrong

the formula is C_ij=(-1)^(i+j)M_ij where M_ij is the (i,j) minor of A

[royal ore]

how would I do the C*I = C question

[nocturne jewel]

@gray dust a_ij is the entry, A_ij is the matrix with the ith row and jth column removed

[gray dust]

@nocturne jewel then that's ok

[hollow finch]

It can't be a 3x4 or the multiplication wouldn't be defined

C is 3x4 and we want CI to be a 3x4 as well (the matrix C specifically)

So we need (3x4)*(mxn)->(3x4)

We can find m and n by

1. Making sure the multiplication is defined
2. Making sure the result has the correct dimensions

[jaunty fossil]

so what is the difference between Hadamard product and the matrix product?

[wintry steppe]

one you dot product rows and columns and it works for matrices with the same "inner dimension"

that's the matrix product

one you just multiply corresponding entries for matrices of the exact same size

that's the hadamard product

a big difference is the matrix product is rarely commutative while the hadamard product is

[jaunty fossil]

so i'd have A(i,j) and B (i, j),both of the same dimension, and I'd do A(i) * B(j)?

@wintry steppe

[wintry steppe]

explain your notation pls

https://en.wikipedia.org/wiki/Hadamard_product_(matrices)

[rain echo]

oh yeah ok thats different from how i have seen it, i usually just write A(i,j) to mean the determinant itself so with abs value signs, which also can mean determinant, I thought you were taking the abs value of the determinant

[jaunty fossil]

If A of shape m * n (column, row) and B is of Shape n * p, then C (the matrix product) is of shape C m * p

@wintry steppe

[wintry steppe]

the matrix product yes

[jaunty fossil]

Then what I did there was just times the elements of each row by each column of each matrix?

[wintry steppe]

i don't understand what you mean

[jaunty fossil]

Sorry, I just started learning Linear Algebra

I just can't understand what the matrix product is going to do with those matrices

I'm going through the deep learning linear algebra section of MIT

[wintry steppe]

if $A$ is $m \times n$ and $B$ is $n \times p$ then the matrix product $AB$ is the $m \times p$ matrix whose $i$th row and $j$th column entry is $$ (AB){i,j} = \sum{k=1}^n A_{i,k}B_{k,j}, $$ which is the dot product of the $i$th row of $A$ with the $j$th column of $B$

bot slow today

[stoic python]

TTerra:

[wintry steppe]

on the other hand

if $A$ and $B$ are both $m \times n$, then the Hadamard product $A \circ B$ is the $m \times n$ matrix whose $i$th row and $j$th column entry is $$ (A \circ B){i,j} = A{i,j}B_{i,j} $$

[stoic python]

TTerra:

[wintry steppe]

key differences

[rain echo]

ewwwww

i have never seen that product

what on earth is that

[wintry steppe]

1. matrix product is defined for matrices of different sizes, hadamard only for matrices of the same size
2. matrix product is rarely commutative, hadamard product is always commutative
3. the matrix product says something about linear transformations, the hadamard product on the other hand idk

[hollow finch]

its the way that every beginning linear algebra student wants to multiply matrices

[wintry steppe]

pretty much lmao

the "identity with respect to the hadamard product" is just the matrix with all 1s

[rain echo]

hahaa that is pointless literally what can you do with that product

[jaunty fossil]

lol yeah when they said matrix product i was like: wow finally something easy to do

[wintry steppe]

idk

[prisma pier]

could the relationship b/w the hadamard product and the trace of a product give any intuition for what the hadamard product means?

or is there not really intuition

[wintry steppe]

the hadamard product gives you a commutative unital ring

which might be nice

you can call it a cring

[rain echo]

yes

[wintry steppe]

cringe

[hollow finch]

other than being easy to compute im curious to know what it could possibly tell us

[rain echo]

its like arithmetic but multiple number at a time

[prisma pier]

I looked up unital ring and saw anime 👀

[wintry steppe]

yeah it's the name of one of the SAO arcs i think

[prisma pier]

[wintry steppe]

to wikipedia one goes https://en.wikipedia.org/wiki/Hadamard_product_(matrices)

seems like you can say a few nontrivial things

[prisma pier]

there's also hadamard division apparently 👀

[hollow finch]

hah i was just looking at the wikipedia page

https://puu.sh/GQEMs/09012f0345.png

not a lot of applications

i guess its nice to have a commutative ring of matrices

[odd kite]

well I think it actually gets used a fair bit in engineering, especially with vectors rather than matrices

in numerical stuff

but it often goes by a different name

the makers of MATLAB apparently decided to call it 'array multiplication'

but I have used that operation quite a bit

.*

[rain echo]

so its just when you want to do like lots of calculations and keep them in a box or what

[odd kite]

seems like I used to with things like time series

sampled data points

[cunning arch]

Just take a look at the hint and you can set up a system from that. An arbitrary 3x3 matrix so:
$\begin{bmatrix} a&b&c\d&e&f\g&h&i\end{bmatrix}$
And multiply that by each of the e basis vectors = each of the corresponding new basis vectors

[stoic python]

strawberrypocky:

[royal ore]

i see what's going

on

but how can i explain it

sorry just bad at explaining

[round coral]

you have to make a matrix of a linear map. Do you know how to do that

you are given where the basis vectors are sent to

[royal ore]

not really sure how to

been learning so much topics at once

[round coral]

see this https://www.youtube.com/watch?v=LlbzPlS3ZQk

don't know if you have the book so

[hollow finch]

If e_i is the ith standard basis vector (zeros in all entries except for a 1 in the ith entry)
then notice that Ae_i gives you the ith column of your matrix. Ae_i is also where e_i gets mapped to, so they are one in the same.

for example
$$A\vec{e}_1=\begin{bmatrix}1\0\1\end{bmatrix}$$

[stoic python]

nix:

[hollow finch]

but Ae_1 is the first column of your matrix

so... thats the first column of your matrix

[royal ore]

the third column is being multiplied by two

[hollow finch]

you can get more general than this by using coordinate vectors but thats the basic idea

[royal ore]

oh wait

am i getting this confused

[round coral]

@royal ore watch the video I gave. Study first the book then ask here.

[hollow finch]

probably. this is a question which you could do without scratch work

its easier than it looks but requires conceptual understanding

[round coral]

@royal ore it your concepts aren't clear no matter how much people say here, you will get more and more confused

so go back read again

any problem then come back

[royal ore]

wait

which is the main basis

cause if i'm using a 3x3 matrix example

how does that get applied

[round coral]

there is no "main " basis

[royal ore]

so what does it mean when it maps the one basis to the other basis

like how does that get applied to the 3x3 matrix

[cunning arch]

i think they mean the standard basis? it’s just <1,0> <0,1> in R^2, same pattern for R3. 3Blue1Brown has a great conceptual video for it

[hollow finch]

when you construct a matrix for a linear transformation you have to do it for specific bases in the domain and codomain

because otherwise how do you know you did it correctly if the vectors we write don't mean the same thing?

imo the hint for this problem is a big waste of time and scratch paper :/
column perspective is all you need in this case

[cunning arch]

i mean change of basis, the long way, is still kinda pretty important to understand imo, but i could be totally wrong. if they don’t understand the concept rn, i think it’s worth it to use the scratch paper & work through it, rather than going straight to the shortcut. kinda like jumping to the power rule in calc 1 before you understand limits

lol i’ll shut up bc i’m still in linalg but yeah

[hollow finch]

nono i see where youre coming from. i just think that column perspective is more fundamental/important than change of basis so it should go first, especially since this problem doesn't actually require a change of basis at all. however yes i agree conceptual understanding of change of basis is incredibly important

since change of basis is such a tricky topic with really confusing formulas (at least imo)

[ionic meteor]

A least squares question. I want to fit a circle to 4 points. (0,2),(0,3),(2,0),(3,1). Ive found the following matrix equation to find A B and C. How do i insert my x's y's and n into this matrix?

[round coral]

what a weird circle how can it touch both (0,2) and (0,3)

[dusky epoch]

there is nothing weird about a circle intersecting the y-axis twice

[round coral]

oh yes I remembered

thank you for pointing that out

but this question can be done more simply using just the general equation of the circle

[ionic meteor]

It's for an assignment and i am supposed to show how i find the equation of the circle

And i'm just really stuck in what x_i actually means?

As i see it i have 4 x values and i don't know which to use. Do i take the average of them? 😄

[native rampart]

The points are (x_1,y_1),(x_2,y_2)...

For example: sigma(x_i^2) is just x_1^2+x_2^2+x_3^2+x_4^2

[ionic meteor]

Oh i see

Hold on, let me try it

[pine scaffold]

if x^2 + y^2 = r^2 what is x+y in terms of r?
Can anyone help?

[dusky epoch]

impossible to determine uniquely