linear-algebra

[simple hornet]

though perhaps you could make the vector $\vec{v}$ be the values of x, and the vector $\vec{y}$ be the values of y you get by just substituting the values of x into y =mx + b

[stoic python]

mirzathecutiepie:

[mild igloo]

I dont have the linear equation yet

im supposed to be solving for the linear equation of the line of best fit using a system of linear equations

I just dont know where im supposed to be getting the equations for the system

[simple hornet]

or maybe $\vec{v} = (m + b, 2m + b,3m+b)$ and $\vec{y} = (2,3,5)$

[stoic python]

mirzathecutiepie:

[mild igloo]

I think I may have gotten it

[simple hornet]

ah do share

im curious too

[mild igloo]

so I made three vectors AB BC AC

used those as the equations and then the y values of the origional points as the y vector

not sure how to use the commands to show you the matrix but I think its right considering it should have no solution and I get a row of zeros which means infinite?

[simple hornet]

an example?

yeah a row of zeros means either no or infinite solutions, it depends on what you have

like if you have one row $[0~0~0 \mid 123]$ that's no solutions

[mild igloo]

right

[stoic python]

mirzathecutiepie:

[mild igloo]

I have row of zeros equal to zero

[simple hornet]

and if 123 was 0 you'd have infinite

yeah

[mild igloo]

so im not sure if thats correct then because the question says that row reducing should show no solution

however infinite solutions could be seen as no solution as there is no singular solution

[simple hornet]

actually i believe there's something called linear regression

[mild igloo]

yes

[simple hornet]

you might want to look into that

no solution means there is zero solutions

[mild igloo]

yea so what im supposed to be doing right now is called least squares regression which is the math behind a linear regression that gives line of best fit

[simple hornet]

infinite solutions means you have as many solutions as you want

ah i see

[mild igloo]

I just dont know whether my initial matrix is right

[simple hornet]

ive heard the name before lol

[mild igloo]

yea im sure you will come across it eventually 😄

[simple hornet]

hm lemme get out a pen and paper and muck around with this

we'll see, stats is not my thing

[mild igloo]

yea ive taken stats which is the only reason I know about it

that was actually the first thing I did when it said find line of best fit was solve it like stats

but that uses calculator and my prof wants this done by hand

[simple hornet]

ooo

did you get something to work?

[mild igloo]

So using the matrix I setup that im not sure is correct the rest of the math works

im gonna do everything out then compare my answer to my calculator answer

[simple hornet]

oh try that, do share once ur done tho

[mild igloo]

if you wanna follow along this is the full question

will send work when finished

[simple hornet]

ah i think i got a solution

consider this system of equations

$\begin{bmatrix} 1 && 1 \ 2 && 1 \ 3 && 1 \end{bmatrix} \begin{bmatrix} m \ b \end{bmatrix} = \begin{bmatrix} 2 \ 3 \ 5 \end{bmatrix}$

[stoic python]

mirzathecutiepie:

[simple hornet]

you can row reduce this

and the last row row reduces to $[0~0~\mid~-1]$ (if i did the row reduction correct lmao)

[mild igloo]

so where did you get the column 111 for that matrix

just cuz mine is almost identicle

[stoic python]

mirzathecutiepie:

[mild igloo]

just [1, 1, 2]

[simple hornet]

Okay so it says form an equation in m and b

[granite oriole]

could i get some assistance when one of you free up

[simple hornet]

so we have y = mx + b, and we want this to have it go through (1,2) (2,3) and (3,5) (or at least come close to) so therefore we'll just substitute in the values and we have 3 seperate equations for x = 1 we have y = m + b, for x = 2 we have y = 2m + b and for x = 3 we have y = 3m + b

sure :)

[mild igloo]

oooooh did you create a y=mx+b equation for each

[simple hornet]

yeah

[mild igloo]

then create vectors using the m and b

ooooh

[simple hornet]

and then just make it m and b

yeah

And then we just row reduce

[granite oriole]

its solving second order cauchy euler de

[simple hornet]

u should probably try row reducing it on ur own btw my row reduction is famously horrible

[mild igloo]

ofc

[simple hornet]

ooOP i know jack shit about diff eqs

[mild igloo]

how did you get the b row tho im curious

just solve out?

[granite oriole]

oh wrong section

my bad

[simple hornet]

b doesn't change at all r

*right

[mild igloo]

well for different lines it would

[simple hornet]

so for each equation we have (1)b

We're proving that we can't have any one line that goes through each point though

so we can fix a value of b i think

[mild igloo]

okay

Yea that makes sense

u genius

[simple hornet]

not if the row reduction turns out to be wrong lmao

okay one question

based on the one paragraph i've read on least squares approximation

is it basically just trying to minimize the variance (in a sense)

[mild igloo]

the row reduction by my math is correct

[0, 0 | 1] = no solutions

[simple hornet]

thank god lmao

[mild igloo]

so the new system would be
[1 1 1 | 10]
[1 2 3 | 23]

after transpose and multiply

or is that not how I would set it up to solve for m and b?

[simple hornet]

one sec im doing that step rn

[mild igloo]

v = A^Ty but

im not sure what that looks like

[obsidian bluff]

hey can i ask for advice here?

im trying to understand this algorithm/proof for diagonalising quadratic forms

[mild igloo]

hoooly shit thats linear?

[obsidian bluff]

here Beta are entries of the matrix of the quadratic form, but i dont understand how beta changes from the bullet points above where i have reordered the basis elements

yeah these are bilinear maps

[simple hornet]

okay i believe you'd have the equation

[14 6] [m] = [23]
[6 3] [b] = [10]

Was too lazy to do the whole bmatrix thing

ah im afraid i can't help with that, someone else will come along though and help surely

i have no idea about that lamo

[red prawn]

If beta_11 = 0 you've already switched it with something else nonzero

[mild igloo]

oooh I see what you did

[obsidian bluff]

im still not sure i understand apopheniac

eli5?

[red prawn]

It's the step in the second bullet point

[mild igloo]

for some reason I kept thinking A transpose would cancle a but its not inverse its transpose

[obsidian bluff]

okay so when i interchange b_1 and b_i, what even happens?

[red prawn]

Well then that b_i becomes b_1, which has nonzero entry on the diagonal

[simple hornet]

bahah yeah it happens sometimes to me too

[obsidian bluff]

so the matrix for this form also changes?

[red prawn]

Because that's why you chose b_i to begin with, to satisfy this criterion

[simple hornet]

I believe $(A^T A)^{-1}$ is known as the pseudoinverse

[stoic python]

mirzathecutiepie:

[red prawn]

Yes, we're rearranging it by changing the basis one step at a time

[simple hornet]

basically you force nonsquare matrices to have an inverse

[obsidian bluff]

how exactly does the matrix change?

[simple hornet]

okay i should probably stop interuppting

[obsidian bluff]

is it just that beta_11 and beta_kk swap?

[red prawn]

Whenever you change bases of a vector space, the matrix of a linear transformation is changed via conjugation

A -> UAU^-1

[mild igloo]

@simple hornet would the solution of A*A^T not be a 2x3 matrix?

3x3

[obsidian bluff]

i see

[simple hornet]

yes

[obsidian bluff]

so this Beta_11 in step 2 is completely different

would be clearer if it was called Gamma_11

?

[simple hornet]

so you can "force" an inverse for nonsquare matrices (if the resulting matrix itself is a singular matrix then it might be a problem but at least we tried)

[red prawn]

Yes it is something brought over from another row/column and now sits in the 11 spot of the matrix, after changing basis elements around

[mild igloo]

right but I dont see how A*A^T became a 2x2 matrix in the system you setup

[simple hornet]

we aren't doing A* A^T it isn't commutative

Av = y and we multiply by A^T

On the left

so we get A^T*Av = A^Ty

[mild igloo]

yea but I thought av was the matrix
[1 1]
[2 1]
[3 1]

[simple hornet]

Yeah ur right

we take the transpose

[1 2 3]
[1 1 1]

[mild igloo]

and mutliply them

[simple hornet]

and we then multiply that

[mild igloo]

and get a 3x3?

[simple hornet]

yeah

No we get a 2x2

Look at the transpose one

we have 2 x 3

and look at the normal one

we have 3 x 2

2 x 3 matrix times 3 x 2 matrix is a 2 x 2 matrix, the multiplication eats up the middle terms

[mild igloo]

I must be doing my multiplication wrong then

[simple hornet]

oof yeah

[mild igloo]

or do I just

oh

[simple hornet]

did you get it?

[mild igloo]

I just have them backwards

[simple hornet]

yeah that's what i was saying

[mild igloo]

oooh okay

[simple hornet]

the operation isn't commutative

meaning a * b is not the same as b * a

[mild igloo]

yea

my brain cant keep track of all the rules at once

I make mistakes like that often

[simple hornet]

bahaha don't worry

i do this so much it isn't even funny

we all do

[mild igloo]

so I got your matrix but flipped?

[6 3]

[14 6]

is my whole multiplication process straight up wrong?

im taking column 1 of Av multiplying it by row 1 of A^T

[simple hornet]

hm

are you sure you're doing

[1 2 3] [1 1]
[2 1]
[1 1 1] [3 1]

[mild igloo]

mine has 111 on top

thats wrong I assume

[simple hornet]

yeah

i was confused when taking the transpose too but you can remember it this way

[mild igloo]

1st column 1st row

[simple hornet]

$\begin{pmatrix}

a_{1,1} && a_{1,2} \
a_{2,1} && a_{2,2}

\end{pmatrix}

[stoic python]

mirzathecutiepie:
Compile Error! Click the reaction for details. (You may edit your message)

[simple hornet]

this is a generalized 2x2 matrix

you just switch the coordinates

so $a_{2,1}$ goes to where $a_{1,2}$ is

[stoic python]

mirzathecutiepie:

[simple hornet]

if that's confusing don't worry about it ur way of remembering it is good too

oepjif

[mild igloo]

so how am I solving for m and b here?

reduce matrix and x1 is m x2 is b?

[simple hornet]

Yeah just row reduce

and then back propogate

or do gauss jordan

[
\left
[\begin{array}{cc|c}
14 && 6 && 23 \
6 && 3 && 10
\end{array}
\right]
]

lmao

[mild igloo]

I think I did something wrong

got 43.6 for b

[simple hornet]

ooof

let me try row reducing it

[wintry sphinx]

[
\left[
\begin{matrix}
1 & 0 \
0 & 1
\end{matrix}
\right]
]

[stoic python]

Saccharine:

[simple hornet]

i see thank you

lmaooo

[mild igloo]

lol

even row reduce calc says the same

we either did something wrong

or this is the answer

[stoic python]

mirzathecutiepie:
Compile Error! Click the reaction for details. (You may edit your message)

[mild igloo]

and this is NOT what my calculator gave me

[wintry sphinx]

[
\left[\begin{array}{cc|c}
14 & 6 & 23 \
6 & 3 & 10
\end{array}\right]
]

[simple hornet]

yeah we probably did something wrong then

[mild igloo]

😦

man

[simple hornet]

lmao

[stoic python]

Saccharine:

[mild igloo]

I had 23 and 10 switched

[simple hornet]

wow

this is classic me stuff we're doing here

[mild igloo]

from doing bad translation earlier

😄

[simple hornet]

bahaha i was like how are we going to rework all of this

thank you saccharine

btw

[wintry sphinx]

that's why you use a computer algebra system that also prints out the latex for you

$\left[\begin{matrix}1 & 0 & \frac{3}{2}\0 & 1 & \frac{1}{3}\end{matrix}\right]$

[stoic python]

Saccharine:

[wintry sphinx]

ew

[simple hornet]

$
\left[
\begin{array}{cc|c}
14 & 6 & 23 \
6 & 3 & 10
\end{array}
\right]
$

[stoic python]

mirzathecutiepie:

[simple hornet]

bow before me

[mild igloo]

new answer match

[simple hornet]

btw sacchrine what CAS did you use?

[mild igloo]

LETS GO

[simple hornet]

bahahha

thats great

yea i graphed it it seems to work pretty nice

[wintry sphinx]

I just use sympy

[simple hornet]

sympy formats stuff too?

damn i didnt know

[wintry sphinx]

you can do latex(something) to get the latex of the thing

[simple hornet]

oh shit really

thats so cool

[floral thistle]

I just use sympy
@wintry sphinx In what scenarios do you use sympy?

[wintry sphinx]

basically anything that involves simplification of anything by hand

or differentiation or integration

[vocal prairie]

I'm guessing it should be, since it seems to satisfy all the vector space axioms. Am I correct?

[last gazelle]

can someone explain subspaces to me, mainly closed under addition/multiplication?

[raw sand]

do you understand the concept of closure

[last gazelle]

I havent heard that word before

[raw sand]

closure is just being closed

do you understand a space being closed under addition and multiplication

[last gazelle]

i think so

[raw sand]

like if you had r^3

[last gazelle]

it just means that the collection of vectors called V satisfy the parameters of V

im watching my lecture vids rn but they dont really help im not good at picturing concepts like this in my head

[raw sand]

i think the easiest way to explain closure is that if you perform the closed operation on two components of the space then you get a component of the space

[last gazelle]

so two vectors inside of the collection

could you hop in vc call and help me with an online question? i understand if u cant for any reason

[raw sand]

i cant go in VC rn

[last gazelle]

ok np

[raw sand]

its simple to extend the definition to a subspace

you know what a subspace is right?

[last gazelle]

its something that is basically flat in spaces R^3 and beyond

thats how my prof explained it not word for word though

[raw sand]

the simplest way to explain it is that a subspace is just a space within a larger space

[last gazelle]

and it satisfies 0 vector and is closed under addition and mutltiplication

[raw sand]

http://www.math.jhu.edu/~nitu/subspace.pdf

this paper is really good

to understand it

and its fairly simply put

it talks about closure and subspaces

exactly your questions

[last gazelle]

could i send you a question from my hw?

[raw sand]

ok

[raw sand]

the one you answered is correct

thats good

theres one other true one

[last gazelle]

ok the way i understand it is that

in a situation where a,b and c = 0

like a 0 vector

then it satisfies the condition

is that correct in some way?

[limber sierra]

yes, the vector $\begin{bmatrix}0\0\0\end{bmatrix}$ is in that set, since $0 \cdot 0 \cdot 0 = 0$.

[raw sand]

exactly

[stoic python]

Namington:

[last gazelle]

so for question 1

what im thinkin gis that it is false

because any one of a,b,c can = 0 which satisfy the condition

but they could also not = 0

[raw sand]

it is false

[last gazelle]

so its false

[limber sierra]

elaborate

[last gazelle]

like a b and c could be any value but 0

and therefore its false

[limber sierra]

could you give an example of two vectors from this subset, where their sum is not a member of this subset?

i think you have the right idea but its hard to parse

[last gazelle]

i dont udnerstand what that means

[limber sierra]

"closed under addition" means that, if v, w are two vectors in this subset, then v+w is in the subset as well

so to show that the subset is NOT closed under addition

[last gazelle]

like in some imaginary case a = 1, b = 2, and c = -3 so abc != 0

[limber sierra]

you need to find two vectors

v, w

...no

[raw sand]

that doesnt satisfy abc=0

[last gazelle]

yea

so its false

[limber sierra]

$\begin{pmatrix}1\2\-3\end{pmatrix}$ is not a member of that subset, so we dont care about it

[stoic python]

Namington:

[raw sand]

it is false but that is not the reason

[last gazelle]

basically my understand of this is that i just put imaginary situations in

which is wrong im now learning

[limber sierra]

your justification is invalid

let's go back to the definition of being "closed under addition"

[last gazelle]

ok why is the vector i gave not a member of the subset?

[limber sierra]

"closed under addition" means that, if v and w are two vectors from the subset

then v+w is also a member

the vector you gave is NOT a member of the subset

but that doesnt matter

[raw sand]

because it doesnt satisfy abc=0

[limber sierra]

that tells us nothing about whether its closed under +

in order to show that something is NOT closed under addition

you need to give an example of two things from that subset

such that, when we add them

[last gazelle]

oh

[limber sierra]

we no longer have things in that subset

[last gazelle]

so is a = 1, b= 2, c=0 a member of the subset

[limber sierra]

that is in the subset, yes

[last gazelle]

because at least one of them must make the multiplication = 0

so i have to try and find a situation in that subset

where its not true

?

[limber sierra]

maybe i should give an example

let's take, say, the even numbers

the even numbers are closed under addition

since when i add 2 even numbers

say 4+6 or 10+14 or whatever

i get another even number

[raw sand]

you have to find a situation where the sum of the two vectors in the subset arent in the subset

[limber sierra]

but the odd numbers are NOT closed under addition

since when i add odd numbers

say 3+5

it's possible that i get something that is NOT an odd number

in order to show that this subset is not closed under addition, you need to find two vectors $\begin{bmatrix}a_1\b_1\c_1\end{bmatrix}, \begin{bmatrix}a_2\b_2\c_2\end{bmatrix}$ that are members of this subset (so $a_1b_1c_1 = 0 = a_2b_2c_2$), but their sum is NOT in the subset

[stoic python]

Namington:

[last gazelle]

so i have to find a case where two vectors added in the subset dont fit the subset

ahhh

its false because

vector 1
a= 1, b=0, and c =2
+
vector 2
a=0, b= 2, c=3

no longer satisfies the condition?

[limber sierra]

there we go, yes

both of those vectors are in the subset

[last gazelle]

wow

[limber sierra]

since they satisfy abc = 0

but their sum does not

[last gazelle]

i finally understand

[limber sierra]

so the set is not closed under addition.

hence (1) is false, as you said

[last gazelle]

ok so what about closed under mutliplcation

must be true?

if i use the same vectors the values that = 0 after being multiplied by any numer that !=0 are still 0

[raw sand]

closure under scalar multiplication is true yes

[limber sierra]

right

you can think of this as

[last gazelle]

wow this is not as hard as i thought it was

[limber sierra]

we know $\begin{bmatrix}a\b\c\end{bmatrix}$ satisfies $abc = 0$

[stoic python]

Namington:

[limber sierra]

so if we consider $r \cdot \begin{bmatrix}a\b\c\end{bmatrix} = \begin{bmatrix}ra\rb\rc\end{bmatrix}$

[stoic python]

Namington:

[last gazelle]

so i basically pull any two random vectors that satisfy the parameters and see if they can make the parameters no longer true

[limber sierra]

we now care about whether $ra \cdot rb \cdot rc = 0$

[stoic python]

Namington:

[last gazelle]

by either multiplying or adding

[limber sierra]

but we know $abc = 0$, so we can rearrange $rarbrc = r^3abc = 0$

[stoic python]

Namington:

[last gazelle]

ah i see

[limber sierra]

(since anything times 0 is 0)

so indeed

[last gazelle]

im not good at conceptual math but this still helps a lot

[limber sierra]

when we multiply by a scalar r

[last gazelle]

thanks so much

[limber sierra]

we still get something

where the product of its entries is 0

anyway, one important aside:

[last gazelle]

ok

[limber sierra]

in order for something to be closed, it needs to be true for ALL vectors

whereas in order for something to NOT be closed, you just need to find ONE example where it fails

[last gazelle]

is there an easy way to test that

or do i just have to do trial and error a couple times until i get an answer

[limber sierra]

you can often make some observations that simplify the process

but in general, no

[last gazelle]

hm ok

[limber sierra]

theres no "formula" or "algorithm" you can apply [well, okay... technically you can see if you can determine a space it's isomorphic to, say by the first isomorphism theorem, but that's a far more technical topic]

[last gazelle]

ok so basic question but is the 0 vector in 3 space just [0, 0, 0]

[limber sierra]

yes

[last gazelle]

ok

haha

thank you

[last gazelle]

ok im back and confused again

how can i multiply two vectors that have all values >= 0 and get a negative value?

and same goes for adding two positive vectors

[limber sierra]

well, you cant multiply column vectors

(using linear algebra techniques)

i'm assuming it's asking you to show closure under scalar multiplication

in which case, the scalars can be any number

in particular, they could be, say, -1

[last gazelle]

ok since i cant multiply column vectors i have to assume its multiplication by a scalar

?

[limber sierra]

i mean, in the first image you posted, it explicitly said "closed under scalar multiplication"

[wintry steppe]

Hey, has anyone here taken an honors linear course ?

i don't know what "honors" means, but i did take the highest level LA my uni offered

if you have a question,

I did ask

what are you gonna do with the honors LA students

||or maybe ?||

[wispy swift]

@wintry steppe I figured out T = x +1 is injective but not a linear transformation

[wintry steppe]

very nice

[wispy swift]

I'm not a maths people but want to study about machine learning so getting to study maths

[simple hornet]

hii can someone explain if this is true?

or at least why this is true

to me this is like saying if you multiply $\frac{d}{dx}$ with a function you'll get its derivative

which i think is kind of weird

[stoic python]

mirzathecutiepie:

[dusky epoch]

it's just notation

don't read too much into it

the multiplication isnt really multiplication in the usual sense but rather it's application of the operator

[simple hornet]

yeahh but

this wouldn't be a matrix then right

this is just an arbitrary linear operator of the form $T(x)=b$

[stoic python]

mirzathecutiepie:

[dusky epoch]

who said it was a matrix

L is an operator on a function space

function spaces tend not to be findim aside from toy examples

[simple hornet]

Ah i see

i assume you can only have matrices in finite dimensional vector spaces?

[dusky epoch]

to express operators as matrices your space needs to be findim and you need to have a basis for it

[simple hornet]

aa so i can't really seem to wrap my head around this at all? this multiplication seems to make no sense

nvm i got it lol

[dusky epoch]

A is a matrix assembled out of the col vectors a_i

[crystal oracle]

@simple hornet Is that LADR? The two exercises 10 and 11 you posted earlier.

[simple hornet]

yes that is LADR

lol how'd you know

@crystal oracle

[waxen jacinth]

Anyone that can assist with this? Idk how to use the knowledge of geometric multiplicity to compute the two eigenvectors that correspond to eigenvalue -1. How do two different vectors spawn when the -1 is all there is?

[whole parcel]

You need a basis of Ker(A-(-1)I_3)

[crystal oracle]

@simple hornet I know because i am currently doing LADR, although a much later chapter

[waxen jacinth]

Yeah I tried it

I get this result when substituting the -1 eigenvalue in the aug. matrix

And idk wt that means lol. I see only one eigenvector whose entries are all 0

[whole parcel]

you need 2 non colinear (alpha, beta, gamma) that give you that

[waxen jacinth]

Well just pick a vector that spans an independent plane

I guess 1,1,0 and 1,0,1

Or u could include 0,1,1

I suppose..

[simple hornet]

@crystal oracle oh wow that's cool, what chapter are you on?

[whole parcel]

if you pick 1,1,0 you don't get (0,0,0)

[waxen jacinth]

Uh

Well they all have to be 0

Im confused

W/e

Ill fail

[whole parcel]

w8 I see your mistake

[waxen jacinth]

So theres hope

For my person

[whole parcel]

the last one is -7*

fck I can't count

w8

[waxen jacinth]

So why was I doing 4

[whole parcel]

that should work better

[waxen jacinth]

Is beyond me

Im so extremely idiotic

That I cant even

Either way

Lemme see this real quick

So this is it

Now what...

[whole parcel]

C1+C2 = 0

[waxen jacinth]

Oh is it because there are two free variables that I can compute two different eigenvectors?

[whole parcel]

you can compute 2 eigenvectors because the dim(eigenspace of -1) = 2

you have C1+C2 = 0 and C1+C3 = 0

[waxen jacinth]

Got it I think

[whole parcel]

so (1,1,0) and (1,0,1)

[waxen jacinth]

Simply speaking in noob terms, there are two free variables

Which is why I get two eigenvectors

So I have to find two separate "arbitrary" values for x2 and x3

From which I get two different e.vectors

[whole parcel]

well dim(2) <=> plane so you need 2 non colinear vector of your space to describe it

[waxen jacinth]

Dimension was what again xD

The number of free vars?

[whole parcel]

yes

[waxen jacinth]

Oke

Thanks!

[whole parcel]

np

[waxen jacinth]

Holy u were so helpful

I may need to come back later to annoy this channel some more w my ignorance though

[whole parcel]

don't worry everybody need help sometimes and so do I rn

I'm trying to implement the SVD for 2*2 real matrixes, so I need A=S U V with U and V rotation or reflection matrixes and U a scalar one

the issue is that I sometimes get -A

[waxen jacinth]

Cant help u there I cant understand half of wt ur saying

Sorry

Hope u find someone thats actually capable lol

[whole parcel]

so I choose S and V so that they match wikipedia svd page requirement (orthonormal eigenvectors of the right matrices) and fix their 'type' (rot or relf) so that the determinant is right

but am I missing something? I didn't get a proper class on the subject so I could be missing something simple

[jolly roost]

would anyone be able to help me with this

[simple hornet]

basically you want to find a linearly independent vector

A vector that is not a linear combination of the first two

Can you think of a vector that you can't get by adding or subtracting multiples of the two vectors?

Try a few permutations

[jolly roost]

so i just need to come up with a vector that you cant get by adding or subtracting u and v?

in that case couldnt i just choose literally any value?

like [100,100,100]?

[dusky epoch]

in that case couldnt i just choose literally any value?
not quite

[atomic flint]

http://prntscr.com/vjm2v3

I am struggling on part g

I have managed to get all the other parts but that last part

just don't understand where to start or what quite to do

[wintry steppe]

so in part (e) you found a basis of R^3 of eigenvectors for the matrix of f in the basis S

can you take that basis of R^3 and give me the corresponding elements of R_2[x]? (specifically, can you tell me what elements of R_2[x] must have those eigenvectors in R^3 as their coordinate representations in S?)

those corresponding elements will form the basis T that part (g) asks for

that sounds complicated but take a second to unpack it

[simple hornet]

@jolly roost you need to choose a value that you can't get by adding or subtracting MULTIPLES of u and v

so formally what you'd do is you'd add the two with generalized coefficients and form an equation of the form Ax = b where A is a matrix of coefficients and b is your proposed solution, and then you'd row reduce and show that this system of equations has no solutions

[simple hornet]

does anyone have a link to some proofs of the properties of determinants?

like the basic properties and also the proof that the determinant gives you the area of the parallelopiped in n-dimensions

[wintry sphinx]

most of the properties of the determinant can be derived from the procedure of calculating it via row reduction

the wikipedia page has a good overview of them

[simple hornet]

i see

how do you calculate it by row reduction? You mean the product of the diagonals thing?

[wintry sphinx]

basically, you know that the determinant of an upper triangular matrix is just the product of the diagonal elements

and the row operations affect the determinant differently:
scaling a row by c scales the determinant by c
interchanging rows scales the determinant by -1
adding a scalar multiple of a row to a different one does not affect the determinant

[simple hornet]

I see

But i thought the determinant was defined in terms of adjoints and laplacian expansions

[steady fiber]

that's not really a "good" definition of determinant

it's more of a consequence from how determinant should be defined

[simple hornet]

i see

i do agree, laplacian expansions are a pain

but how are they defined then?

or how is it defined*

[steady fiber]

the R^n x R^n -> R function that is multilinear and alternating in the rows, and that maps the identity to 1

it's defined by its properties

and you derive the rest from the properties

[simple hornet]

I see but the properties seem to be too good to be true in a sense, so I'd expect to see a proof for it

and it doesn't provide good intuition, does it? Maybe I haven't studied enough about this for it to be intuitive but still

[steady fiber]

I mean there's no proof for it lol

that's literally the definition

or at least, a definition

you can't prove the definition, it's just how it's defined

but, in certain lin alg courses, you would be able to see everything you'd normally think of determinants is implied by those two properties, along with the requirement that det(I) = 1

proofs of deriving stuff like the expansion formulae etc could probably also be found online

[smoky lily]

https://i.imgur.com/ZJ2viQn.png

why is my answer wrong?

[simple hornet]

im sorry im tired af so don't exactly know what im saying lol

idk quadratic forms, someone else will come along and help

[gray dust]

@simple hornet defines det as an oriented volume & derives a formula

[simple hornet]

oh wow that's good thank you

ill look into it

damn someone wrote all of this by hand

[sharp tusk]

holy

is this the sweaty side of the server

[simple hornet]

probably is a professors notes

[sharp tusk]

id hope so

[gray dust]

i slightly rewrote a teacher's old notes

[smoky lily]

can someone teach me how Diagonalizing a quadratic form works?

i understood my notes, but when i actually do it, i think i might be misunderstanding the question

https://i.imgur.com/ZJ2viQn.png

[wintry sphinx]

you basically write the quadratic form as x^T A x

then diagonalize A which gives you your change of basis matrix

[smoky lily]

Thank you!!

[nocturne jewel]

If A and B are symmetric square matricies and AB is diagonal, how do I show AB=BA ?

[wintry steppe]

hm

[clear epoch]

hi

i need help

i dont understand this 😦

[wintry steppe]

@nocturne jewel (AB)^T = B^TA^T

[nocturne jewel]

Omfg

Cause diagonal is just a specific symmetric

Right

[wintry steppe]

do you know how to show A and B commute now?

[nocturne jewel]

Yeah

The previous part of the question was if AB is symmetric and I did that fine lol @wintry steppe

[dreamy iron]

I just found this cool playlist of lectures based around Axler's LADR 3rd edition. I've seen 2nd edition lectures floating around, but this is the first time I've seen lectures for 3rd edition.

The channel is less than a year old. The videos are even younger. These come from a California State School, Cal Poly SLO.

I hope it may be of use to others going through LADR.

https://www.youtube.com/playlist?list=PLBUiHiRFQhsI--yc2PoCcK17fUR_mNJNH

[smoky lily]

why was the range =2 here? what is determining the range?

[warm niche]

Is there such things as eigenvalues/vectors for nonsquare matrices?

[wintry steppe]

good luck making sense of Av = lambda v when one side is a different size than the other

look into singular values, maybe

they aren't a generalization afaik, but they are very closely related to eigenvalues

[smoky lily]

https://i.imgur.com/s6afW0G.png

how can i find the kernel of a linear transfomration?

[wintry steppe]

write L as a matrix and do something with that

[smoky lily]

hm can i have another hint please

[wintry steppe]

you know how L acts on the vectors (1,0,0), (1,1,0), and (1,1,1), which constitute a basis of R^3. this allows you to write L as a matrix relative to this basis, and, say, the basis 1, x of P^1(x)

or you can just write L(x,y,z) = 0 and solve

[last gazelle]

can someone help me understand how to do 2, 3 and 4

[odd kite]

{a, b, c} is just a set of three vectors. You should probably know what span is at this point. But if you need to learn it, you could try this video. https://www.youtube.com/watch?v=k7RM-ot2NWY

[last gazelle]

Ive seen that video i love 3blue1brown

Im just not sure what method id use to find 2 3 or 4

1 i know its linear combinations so i did that easy

[odd kite]

no, I think you're pretty confused

[last gazelle]

ouch

is the answer for 2 3 vectors

and im just overcomplicating it

because its not span in that question

[odd kite]

let's go back to 1. when it asks you if w is in {v1, v2, v3}

[last gazelle]

yea

[odd kite]

it's asking if w is literally one of those three

has nothing to do with linear combinations

that comes in when span is involved

[last gazelle]

i thought it meant that the vectors v1, v2, and v3 are able to get vector W when multiplied by scalars and added together

[odd kite]

no that's asking if w is in span { v1 v2 v3}

[last gazelle]

so i did the RREF of v1,v2, v3

so 1 is no

becuase theres no span involved

[odd kite]

so the key thing here is to understand the difference between a list of vectors and the span of that list

[last gazelle]

my understanding of span is that its the space that a set of vectors can reach when multiplied by a scalar and added together

[odd kite]

yeah

[last gazelle]

ok so how do i determine the number of vectors in the span{v1, v2, v3}

is that where RREF comes in?

if the matrix has infinte solutions then the answer for 3 would be infinite? hypothetically

[odd kite]

let's take a simple example. Span{v1} includes for example 0, 0.0001v1, .00002v1, and so on and everything in between

it's the number of points on a line, which is

[last gazelle]

infinite

[odd kite]

yeah

same is true when you have multiple vectors

span of anything (besides the zero vector) is going to be infinite

[last gazelle]

ah that simplifies it a lot

thank you for your help

[odd kite]

yw

[junior panther]

Wait where’s the part of the server for people who have a hard time with division??

My duel digit iq can’t handle all this

[wintry steppe]

[dusky epoch]

@junior panther #prealg-and-algebra or any of the questions channels

do you still need help?

[keen flame]

What are the best sources to study Linear algebra? I was using khan academy but he doesn't have any practices beyond vectors

[wintry steppe]

a decent textbook should serve you well

[keen flame]

Recomendations?

NVM. found #books-old

[wintry steppe]

strang
roman
fucking kek

strang i don't know about, but roman is way too advanced for what you want, probably

i like the book by friedberg (also with two other guys, insel, spence), and the book by axler - these are theoretical LA books

[keen flame]

I've did linear algebra course for com-sci but failed miserably.

[wintry steppe]

Does anyone know how to do this? I've tried substituting some variables. I'm aware a span is regarding linear combinations and is so that all things are covered?

I've tried to do some substitutions but I feel like I'm not getting anywhere with it

[odd kite]

you subbed in z in the exp for w?

that should get you somewhere

[wintry steppe]

exp meaning?

[odd kite]

expression

failure at being lazy 😦

[wintry steppe]

you mean with w, I sub in 16x + y + 4z?

is that what you are saying?

[odd kite]

no you sub in z= .... into that

[wintry steppe]

ah right

z = -4x - 2y?

[odd kite]

yeah it says that

[wintry steppe]

I have tried that

and well, I just sort of like

I'm unsure what exactly I'm trying to pull out, if you get me?

[odd kite]

what'd you get

[wintry steppe]

So for me

for part A for example

I got:

lemme find it

ay + bw = cz

....

y(a-7b + 2c) + (4c)x = 0

am I going completely off the rails or am i on the right track?

[odd kite]

you aren't answering my question

if that's what you mean

[wintry steppe]

that's what I got

[odd kite]

$w= 16x + y + 4z = 16x + y + 4\underbrace{(-4x - 2y)}_z $

[stoic python]

Timon:

[odd kite]

now simplify that

[wintry steppe]

wait, I don't need to substitute in generic variables a, b, c, etc?

why is that?

[odd kite]

because it's what I told you to do

if you do it you should be able to see something

[wintry steppe]

I've noticed during my calculations, it cancels out the 16x

so you'll obtain

-7y

[odd kite]

yeah

so w is proportional to y. what does that mean about span(w) and span(y)

[wintry steppe]

...

I see.

so that therefore means their spans are the same

because you can obtain the "other" span by scalar multiplication

and it's closed under scalar multiplication

correct?

[odd kite]

well it's right that the spans are equal

[wintry steppe]

so C must be correct

[odd kite]

yes

[wintry steppe]

and it's the only one?

because I made a mistake in my working out

what I did for example with part A was:
ay + bw = cz

then I expanded it out and I got some letter mess

y(a-7b + 2c) + (4c)x = 0 this basically

but I noticed you didn't use generic letters, why is that?

[odd kite]

because I'm lazy and I didn't need to think about how to solve this problem in general. I could quickly work it out without resorting to such things, so why bother

[wintry steppe]

are you meant to use generic letters or?

[odd kite]

I assume you can use whatever method gets you the right answer

[wintry steppe]

and what would be the method that ensures you know a statement is true?

because that is a bit where I'm struggling. I don't know what exactly I'm looking for?

I can tell it's something to do with span, and linear combination and covering all possibilities

[odd kite]

well what do you think about A? since span(y) and span(w) are the same, span(y,w) = span(y) = span(w). Does that make sense to you?

[wintry steppe]

part A can't be true

span(y, w) = span(y) = span(w) as shown what you have told me and the reasoning you've used (implicitly)

wait

hold on

yeah

part A is false

or rather it's not necessarily true

z relies on x and y

[odd kite]

yeah, it could be true depending on what x is but it's not "must be true"

[wintry steppe]

precisely

because the LHS is essentially the RHS, but with co-efficient of x being zero

and hence, the LHS span is more restrictive than the RHS, so therefore the spans can't be equal

B therefore also can't be necessarily true as well

due to similar reason

oh wait, it's starting to click now

wait hold on

is part B necessarily true?

span (x, (-4x-2y) ) = span(-7y)

no wait, it's not

the RHS can't get any x, but the LHS can remove the x and enlarge the y as much as it wants

I'm gonna go with C and D being necessarily true?

awesome!

I got it right!

Thank you so much! It clicked at the end and it makes sense! I can treat x as being (1, 0) and y as (0, 1) and just see how if the spans of each other can cover one another

@odd kite

[odd kite]

ah, yw

[wintry steppe]

but I see the short cut you did

you are meant to have the generic constant variables

however, it's closer under multiplication and you can work without using it

I added them being silly, when really I should treat them as element entries

[tawdry yacht]

7. a carpenter is buying supplies for a job. the carpenter needs 4 sheets of oak paneling and 2 sheets of shower tile board. the carpenter pays 99.62 for these supplies. for the next job the carpenter buys 12 sheets of oak paneling and 6 sheets of shower tile board and pays 298.86. he also spends 139.69 on 1 sheet of shower tile board and 8 sheets of oak paneling. how much does each item cost individually?

help ^ pls

[wintry steppe]

I think this is a #precalculus question @tawdry yacht

#prealg-and-algebra seems ideal

you want me to help out?

@tawdry yacht has been answered, no need to answer

[wintry steppe]

@barren plank Yo, gotta borrow you

[barren plank]

what's the issue

[wintry steppe]

I'm sort of not getting how to calculate it

got any pointers?

[barren plank]

literally find a vector that's in both U and W

[wintry steppe]

I can tell it wants me to find the intersection, and find some vector that when you multiple it, it includes all the stuff of the intersection of U and W

Yeah, I tried 0 0 0

and it was like "nah"

[barren plank]

a nontrivial vector

[wintry steppe]

right

[barren plank]

naturally

[wintry steppe]

gotcha

and that makes sense

you find one

[barren plank]

do you know how to solve general linear systems?

[wintry steppe]

and the span will be all multiple stuff

I do, but I wanna see how it manifests itself

[barren plank]

well this is a linear system

[wintry steppe]

a(-3, 3, -4) + b(4, 8, -1) = c(5, -1, 4) + d(25, 1, 5)

correct?

[barren plank]

yes

[wintry steppe]

then what?

is there an easy thing I'm missing out?

[barren plank]

this is a linear system

[wintry steppe]

I get you gotta solve it and find specific values of a, b, c and d

[barren plank]

solve it

yes

[wintry steppe]

long way?

no matrix magic?

[barren plank]

well

[wintry steppe]

wait, I can see

yeah

I can cancel some stuff out

[barren plank]

the columns of your matrix are column vectors of U and W

[wintry steppe]

go on?

[barren plank]

it's got 3 rows and 4 cols, so the kernel is at least 1-dimensional

find the fundamental solution

idk, I don't see any conceptual issues in this problem

it's just a shut up and calculate type thing

[wintry steppe]

I don't get the 3 rows and 4 columns?

oh wait

I get it

wait

yeah

wait, no I don't

:/

oh wait

I think I do get it

@barren plank wait I'm still stuck

:/

[barren plank]

with?

[wintry steppe]

like this thing you were talking about

3 rows and 4 cols

[barren plank]

yes?

[wintry steppe]

I can't see it :s

[barren plank]

$\begin{pmatrix} -3 & 4 & 5 & 25 \ 3 & 8 & -1 & 1 \ -4 & -10 & 4 & 5\end{pmatrix}$

[stoic python]

mniip:

[wintry steppe]

ah right

I was working on that

and manipulating it

oh right, yeah

I put it into RREF

[barren plank]

that gives you the solution

[wintry steppe]

so do I put it into REF?

[barren plank]

yes?

do you know how to solve general linear systems?

[wintry steppe]

I do but I'm getting cut for time with a homework assignment

hence why my head is going fuzzy

that and I am all nighting

I got into REF

now what?

oh wait

god I have 6 minutes

and I have no clue

@barren plank

[barren plank]

is this a test

[wintry steppe]

homework assignment

not a test

and it's timed

[barren plank]

literally just solve the linear system

I don't understand what you're having problems with

[wintry steppe]

final question and I've gone blank

ahhhhhhh

[bold python]

I have a question, if I were to show that a vector $\bf p = (1, -2,-3) \in\text{col} A$ where $A$ is a $3\times 3$ matrix I have to check if $A\bf x = \bf p$ but if the system has infinitely many solutions can I conclude that $p\in \col A$

[stoic python]

homomorphism:
Compile Error! Click the reaction for details. (You may edit your message)

[dusky epoch]

bad tex

but anyway, p ∈ col(A) iff Ax = p is consistent ie has at least one solution

uniqueness does not matter here

[bold python]

alrighty, thanks Ann

[wintry steppe]

just a question, how can I check if my answer is correct?

nevermind I got it wrong

and it was (5, 1, -1)?

how is that obtained?

[obsidian bluff]

does this just mean all evalues are Real?

im confused about how this prop is worded

[half ice]

@obsidian bluff
All eigenvalues of A are real, and there's at least one.

[obsidian bluff]

thanks

[prime patrol]

Okay so

If I have a set of matrices, particularly these ones

And the span of this set is equal to the range of another matrix, how would one find an example of this other matrix?

[half ice]

The matrix would be
[0 1 1]
[-1 1 0]
[1 -1 2]

Note that the range refers to all outputs of
[0 1 1] [a]
[-1 1 0] [b]
[1 -1 2] [c]

Which, if you matrix multiply that out, is just another way of writing
[0] [1 ] [1]
a[-1] + b[1 ] + c[0]
[1 ] [-1] [2]
Which is the span you want

@prime patrol

Swapping the columns around doesn't change the range so you can make more matricies this way.

I'm sure there's other matrices you could make too, that cover the same space.

[royal ore]

Hi. How can I tweak this to find the parabola of best fit through the points

[last gazelle]

if i have a matrix where the bottom row is all 0s and im asked for the span

is that considered the 0 matrix? and therefore the span is not infinite

[half ice]

I assume you are actually being asked for the span of the columns. In that case, the span is still infinite

[last gazelle]

ah ok

[gritty kelp]

Can someone explain how these multipliers are found?

I dont ever remember hearing about them before, so I dont know what their relation to the pivots are

I multiplied the first row by 1/2 to get the second pivot, and I multiplied the second row by 2/3 to get the third pivot, so I can sort of see where these numbers are coming from. What I am not sure about is how we determine the order of the multipliers, and why is 0 a multiplier?

I need to have the order because I need to use them in this equation, so there are pivots that correspond to certain multipliers that have a certain place in the equation.

[red prawn]

the multipliers are negative of the numbers you've found, because they want to write an expression like x-my, as subtraction.

[quiet heron]

i have am really lost on my whole math homework today can someone help me it is only 6 qoustions?

?

<@&286206848099549185>

[quiet heron]

is this supposed to help me?

[half storm]

@royal ore how do you normally show that map a map between vector spaces is linear?

Think about that first and if you've still got problems come back.

[wintry steppe]

@quiet heron just post your questions instead of asking if you can be helped

[quiet heron]

ok

[wintry steppe]

either just solve the linear system and pick the right solution, or just check each one and pick the one that solves it

you know, take each pair (x, y) that you're given and see if it satisfies the two equations

[quiet heron]

ok so thats where i am confused i put the problem in y=mx+b but right after that i am lost

[wintry steppe]

you don't need to do thay

you can literally just put each pair into the linear system and see if the equations are satisfied

like for (-2, 0), just plug in x = -2 and y = 0 and see if the equations hold

repeat until you found the right solution

the smarter way to do this would be to solve the system, but plug and chug works too

[quiet heron]

ohh wow thanks

[wintry steppe]

(note that there's only one solution, so you don't have to worry about multiple)

[quiet heron]

ok

is it the second answer

?

[unreal goblet]

This may be out of the blue, but do you know how to solve this without the multiple choice?

[quiet heron]

no if i am being farley honest

[unreal goblet]

Okay, I'll walk you through it then

Sometimes you don't always have the multiple choice, so you can't always rely on it

[quiet heron]

ok

[wintry steppe]

this is what i meant by just solving the system btw

[unreal goblet]

Do you know the substitution method? Have you learned it yet?

[quiet heron]

no

[unreal goblet]

Take a look at this

I used your question but I did it step by step

So read it and then ask me any question or anything you're confused about

shoot

7+4=11

but its zero either way

Sorry about that the 7x+4x= 11x actually

[quiet heron]

can yo u explain the second step further

i dont get it

[unreal goblet]

okay so since we isolated the y

you have to insert that into the 7x-4y=8

you have to put the (-x-2) in the y spot of that equation ^

to solve for x

after you insert (-x-2) you do the foil method

so it would be you basically expanding the equation

(-4)(-x)=4x and (-4)(-2)=8

so now you have 7x+4x+8=8

7x+4x=11x and you move the 8 over to the other side so it cancels out and becomes a zero

11x=0

0/11=0

so then x=0

[quiet heron]

i am sorry is that a problem

if it is i will stop

[unreal goblet]

no no

text me privately if you'd like

i can try to help you

[quiet heron]

ok

[unreal goblet]

i think this isn't the right category for it though, pre-algebra would be more like it

[rigid reef]

holy crap i suck at linear algebra mcq

does anyone know of any practice for that

[arctic hazel]

hi, i have a question about eigenvalues and diagonalization

i have to find all eigenvalues of the transformation T: P_2(R)->P_2(R), where T(f(x))=xf(1)+(x+1)f'(x)

i could type that out in latex but i think i did all the math about it anyway here:

a) is what i wrote right? and b) doesn't this mean the set of eigenvectors is linearly dependent, meaning it cannot be diagonalized?

[half ice]

The logic is good. However, 0 is never an eigenvalue by definition.

[arctic hazel]

oh whoops, that might be a mistype

i think the eigenvalue was supposed to be 1 there

[half ice]

Because T is a finite dimensional linear operator, you can write it as a matrix. Consider f(x) = ax² + bx + c. Then:
Tf = x(a + b + c) + (x + 1)(2ax + b)
Tf = ax + bx + cx + 2ax² + bx + 2ax + b
Tf = 2ax² + (3a + 2b + c)x + b
So the coefficients are given by a matrix multiplication:
[2 0 0]
[3 2 1]
[0 1 0]
Describing T

,w {{2,0,0},{3,2,1},{0,1,0}}

[arctic hazel]

yeah this format makes a lot more sense

ok i should have been doing it this way from the start

[half moss]

I'm trying to learn 3D rotations. I feel like I understand these rotation matrices, are they not just like Given's rotations?

Anyways, I'm wondering why the negative sign is moved across the diagonal in the second rotation matrix, the Y axis rotation.

Does this have something to do with the system being left-handed?

if that sign wasn't moved then I feel like everything would make sense

[red prawn]

In the first one, $R_z$ the positive sin term is "to the right" of z, ie, applied to the x column.

[stoic python]

Apopheniac:

[red prawn]

So for the matrix $R_y$, the column to the right of y gets the positive sin term.

[stoic python]

Apopheniac:

[half moss]

Thank you for the response, but I don't understand that explanation 🤔

[round coral]

if you construct each matrix yourself visualising the rotation you will understand what is happening. Of course you only need to do this once and you will get it

a very brute way, but this is how I first understood it. There may be better ways

This is good to read too https://en.wikipedia.org/wiki/Rotation_matrix

[arctic hazel]

hi, it's me again

i need to come up with a proof for $c_A(x) = x^n - \tr(A) x^{n - 1} + \ldots + (-1)^n \det(A)$

[stoic python]

Snodlop:

[arctic hazel]

where c_A(x) is the characteristic polynomial

i have an example for 2x2 matrices, but i'm not sure how to expand it to nxn

ok thanks, i'll let you know if there are any problems with that

[ivory knot]

i'm supposed to make it so that both sides are the same
(2k+1)(k-1)+4(k+1)-3 = (2(k+1)+1)((k+1)-1)

[dusky epoch]

@ivory knot wrong channel, see pins

[ivory knot]

aight

thanks for the heads up

[arctic hazel]

thanks viburnum, your tip helped a ton

[patent cloak]

I don't know where to ask this, but, if you take a square matrix as an oriented and ordered collection of vectors of some kind (like some kind of collection that corresponds to a volume form,) is there any geometric identification for the way the two "forms"/collections of vectors are being combined?

[odd kite]

I don't understand what you are asking

nvm brain failed

2-vectors correspond to anti-symmetric matrices

this is the combination of two vectors by the exterior product

sorry I'm tired

the combination of N vectors of N dimensions is a generalized volume

[patent cloak]

My question was incompetent and probably for the same reason

I'm thinking of a context where each matrix is just treated as an ordered set of vectors regarded at the same time as a geometric entity

What it geometrically means to multiply the matrices - what we can say in intuitive terms about what happens to the ordered sets of vectors

@odd kite

[odd kite]

has no meaning afaik

well

other than it applies a linear transformation to each of the vectors

so essentially skews,scales, rotates, or flips whatever the matrix represents

I guess

sorry I'm really not 100% right now due to tiredness

[viscid kernel]

When u apply kirchoffs current law and get a 7x7 matrix in the form of Ax = (2,1,5,3,3,4,1) and you end up having a free variable after 30 minutes of row operation 🔪💀🔪💀🔪💀

[half ice]

Seems like a bad way to do the question

[rough olive]

Is the gradient the same thing as the slope? I'm reading up on material on this introductory course on machine learning and my book states the following, which i have trouble understanding:

It's about linear regression and the formula for a linear equation is y = mx + b, no?

where m is the slope, right? So is w in this case the slope or gradient as the author calls it?

He goes on to mention this:

To me it seems like $w_0 = b \quad w_1 = m$. I just want someone to confirm that I have understood this correctly.

As b is the y-intercept and m is the slope in the classic linear equation mentioned above

[stoic python]

Resident Tracksuit Advisor:

[rough olive]

I don't understand this. It just seems like another way of writing y = mx+b

why is this considered a better model than the classic linear equation written above?

[dusky epoch]

I don't understand this. It just seems like another way of writing y = mx+b
it is

the author should go on to extend this principle to higher dimensions tho

[rough olive]

But why not just write m_0, m_1, m_2 then

Maybe I'm asking too many questions. I should read further, he'll probably answer it.

[dusky epoch]

you're not "asking too many questions" you're just trying to get too far ahead rn

the author starts with simple stuff as a base to progress on to more complicated models

[tawdry palm]

Guys if all leading principal minors of a symmetric matrix are 0, what will be the definiteness?