linear-algebra

[radiant yarrow]

more like (0,a2*(-a2))

[lavish jewel]

idk what you'Re doing

did you mean 0 element for multiplication or for addition?

[radiant yarrow]

Addition

is defined like that right

[lavish jewel]

for addition it's the one i wrote

you get a zero vector if you add (-a1, 0)

[radiant yarrow]

Oh okay

got it

[lavish jewel]

wait

the 0 vector is one such that v + 0 = v so

so you'd need (0,1) as the zero vector

[radiant yarrow]

For vector sum to be 0, you can define (-a_1, 0)?

[lavish jewel]

that's the one i meant earlier, yeah

that's the additive inverse

[radiant yarrow]

got it

[lavish jewel]

"-v"

and the zero vector would be (0,1)

gotta get a bit creative

we're busy working on a problem rn

[radiant yarrow]

@lavish jewel If additive identity is (0,1) and the sum of additive inverse is (0,0), is it fine for the elements to be different?

[lavish jewel]

hmm no

i guess we have to take another look at the additive inverse then

how about (-a1, 1/a2)?

[radiant yarrow]

In the axioms, it's shown that there exists element denoted by 0 such that x+0 = x

and also there exists an element such that x+y= 0

[lavish jewel]

yeah, it means the same 0 vector for both

[radiant yarrow]

Yes. But the zero vector isn't the same in this vector space

Is it a contradiction?

[lavish jewel]

that's why i said we had to take another look, try (-a1, 1/a2)

this should yield (0,1), which does work

[radiant yarrow]

Okay

[lavish jewel]

this is also a good point for you to look back at the previous question with that v = {0} thing

maybe now you see what i meant by the 0 vector not necessarily being the 0 you were thinking of

or + not being the usual addition

[radiant yarrow]

Yeah it's making sense

Let c,d be two scalars belonging to R

[reef sierra]

is every diagonalizable matrix invertible?

[teal grotto]

what are you're thoughts

[radiant yarrow]

$(c+d)(a_1,a_2) = (ca_1 , a_2) + (da_1, a_2) = (ca_1 + da_1,a_2 a_2)$

[stoic python]

Researcher in Pre-algebra

[radiant yarrow]

Is this right?

[reef sierra]

Looks right to me

[radiant yarrow]

What if c+d = e where e belongs to R, and $e(a_1,a_2) = (ea_1,a2) \neq (ca_1 + da_1,a_2 a_2)$?

[stoic python]

Researcher in Pre-algebra

[radiant yarrow]

@lavish jewel

[reef sierra]

@teal grotto I dont think they are

[teal grotto]

why

[drowsy flower]

isnt zero matrix a counterexample for this?

[teal grotto]

yes. just wanted him to get to that lol

any diagonal matrix with at least one 0 on the diagonal works as a counter example, since the determinant of a diagonal matrix is the product of its diagonal entries

[drowsy flower]

nice

okay I am going to ask a question now

[stoic python]

meguuuuu

[radiant yarrow]

<@&286206848099549185>

[drowsy flower]

I am given a vector $x \in C^n$ and $b \in C^m$ and $x$ minimizes $\norm{Ax-b}$ and $A$ is $m \cross n$ complex matrix

[stoic python]

meguuuuu

[quasi oxide]

@drowsy flower do an orthogonal projection onto the image of A

[lavish jewel]

yeah ortho projections will do the trick

if you know the SVD, you can use that too

[drowsy flower]

okay let me see, also idk svd so

[lavish jewel]

you did it backwards, but yes

[radiant yarrow]

Backwards?

[lavish jewel]

if you have (c + d), treat that as a single element

so ( (c+d)a1, a2)

[radiant yarrow]

Okay

Got it

then use distribution and check with the original

So this can't be a vector space?

[lavish jewel]

you cannot do the step in the middle cuz of the coefficients

i think not, cuz you'd get a2^2 yeah?

[radiant yarrow]

Yeah

[lavish jewel]

sounds about right

[radiant yarrow]

step in the middle?

[lavish jewel]

this here

the chunk in the middle is false

it would be true iwth usual addition, but not here

[radiant yarrow]

Okayy

[radiant yarrow]

Then how is it done?

Won't that mean scalar distributivity holds if that chunk is false

[lavish jewel]

backwards

as we just did above

you have (c+d) (a1,a2) = ( (c+d) a1, a2)

and you wanna check if this is equal to c(a1,a2) + d(a1,a2)

but the latter is equal to ( (c+d) a1, a2^2)

[radiant yarrow]

okayy

[lavish jewel]

so it isn't distributive

[radiant yarrow]

I don't understand this question

[wintry steppe]

Has someone tried to imagine what is the determinant of a non-square matrix?

I know that if the matrix is not square, the determinant doesn't exist

But idk if there exists any extension of what would be the determinant for non square matrices

[lavish jewel]

it's asking you if you have addition and multiplication by real scalars, do we still have that the a_i are in C

[radiant yarrow]

Oh yeah

Do I just say "yes because adding real number to a complex number, it still has complex number as codomain"?

[lavish jewel]

that's not a requirement though

[radiant yarrow]

Okay

just yes then

[lavish jewel]

in fact, there is no addition of vector and scalar

[radiant yarrow]

Okay, they said coordinate addition and multiplication

[lavish jewel]

remember you'Re looking at addition V x V -> V and multiplication F x V -> V

[radiant yarrow]

Yes got it

So scalar multiplication holds

So yeah

[lavish jewel]

right. we have usual addition, usual multiplication, and these are all associative, commutative, etc, and the product of a real and a complex number is complex

[radiant yarrow]

Let V = {(a1, a2, . . . , an): a_i ∈ R for i = 1, 2, . . . n}; so V is a vector
space over R by Example 1. Is V a vector space over the field of
complex numbers with the operations of coordinatewise addition and
multiplication?

V can't have R as a codomain so it's not the same vector space right

[lavish jewel]

right

not for scalar multiplication

addition is R^n x R^n -> R^n, and scalar mult is C x R^n -> C^n

[radiant yarrow]

Got it

Changing the vector space itself

got it

Should I move to next topic?

There are 7 more problems

[lavish jewel]

i'd recommend you try a few more, especially if there are harder ones at the end

[radiant yarrow]

Okay

[nocturne jewel]

$V=\mathbb{R}\cup{\infty} \ u+v=min(u,v) \ c\times u=c+u$ Verify if this is a $\mathbb{R}$-vector space or not

@radiant yarrow if you want one to do

[stoic python]

Mosh

[sinful valve]

can anyone here help explain the derivation of the perspective projection matrix in 3D graphics.

[radiant yarrow]

R union with element infinity?

[nocturne jewel]

yes, so the vectors are all the real numbers and infinity (which behaves as you expect it to)

ie it's the "biggest real number"

[marble lance]

It behaves as follows: min(u, infty) = u and u+infty = infty for any real u

[radiant yarrow]

But Aren't vector spaces defined in finite numbers?

[marble lance]

That's all you need

[nocturne jewel]

what do you mean?

[marble lance]

Vector spaces can be arbitrary sets

You've had matrices, sequences, polynomials, etc. as examples

Not just numbers.

[radiant yarrow]

Okayy

my bad

[nocturne jewel]

Im also not saying if it's a vector space or not.. the question was to verify if it is or not

Also giving this one cause this was the one that fucked me over when I learned about vector spaces

[radiant yarrow]

c x u = c+ u
u+v = min(u,v)

[nocturne jewel]

yes, for scalar c and vectors u and v

[radiant yarrow]

So real numbers are vectors

also infinity

[nocturne jewel]

Yes, F is a F-vector space

[radiant yarrow]

c is a real number

won't that make c+ u = min(c,u)?

[nocturne jewel]

no

[radiant yarrow]

Okay

Since c is a scalar

[nocturne jewel]

c isnt a vector, it's a scalar, so you cant add a scalar and a vector

[radiant yarrow]

not the same element

Wait so c+ u isn't possible

[nocturne jewel]

it is

[marble lance]

In u+v, the + refers to vector addition which is being defined to mean min(u,v). In c+u, the + refers to normal addition of real numbers. Same symbol, different meanings.

[radiant yarrow]

Okay

[nocturne jewel]

$1\times u = 1+u$

[stoic python]

Mosh

[radiant yarrow]

give me a min

[nocturne jewel]

for example

Now my prof did use the O symbols for the operations but I couldnt be asked

[marble lance]

c+u isn't possible where + is vector addition. But the + here is normal addition which is possible because it's the scalar multiplication here.

[nocturne jewel]

$u\oplus v = min(u,v) \ c\otimes u = c+u$

[stoic python]

Mosh

[radiant yarrow]

Not a vector space

since 1u=/= u

That's what I think

[marble lance]

That works

Several axioms fail

[radiant yarrow]

There's only one additive inverse for all the vectors

in this set

[marble lance]

What do you mean by that?

[radiant yarrow]

Oh wait

my bad

only for positive real numbers

u+0 = min(u,0)

[marble lance]

Are you saying they have or don't have an additive inverse?

[radiant yarrow]

where as no additive inverse for negative numbers

[radiant yarrow]

positive numbers have one inverse

negative ones don't

[marble lance]

Before you can find inverses

You must identify the identity

What’s the additive identity if there is one?

[radiant yarrow]

u itself

[marble lance]

No

[nocturne jewel]

which vector specifically is the identity

[marble lance]

An identity is one element in the whole space

[radiant yarrow]

u+u = min(u.,u)?

Okay

[marble lance]

It cannot be dependent on u

Each vector doesn't have its own identity

[radiant yarrow]

Got it

I don't understand how sequences are defined in this book

[nocturne jewel]

same way they always are..?

it's just a list of numbers with some general rule

[lavish jewel]

pog infinity as 0 vector moment

[nocturne jewel]

[nocturne jewel]

Yeah.. what's the problem?

[radiant yarrow]

what does a_n mean?

is it a single element?

[nocturne jewel]

it's the general term

${a_n}={a_1,a_2,a_3,...}$

[radiant yarrow]

okay

[stoic python]

Mosh

[pale coyote]

Did you see example 5?

[forest quiver]

Hey quick concern

I have the urge to do x_1 * v_1 + x_2 * v_2 = (240, 2824)

This would be the overall production of the company to get this much resources

But when it says each, wouldn't that be 2 separate equations?

x_1 * v_1 = (240, 2824)
and
x_2 * v_2 = (240, 2824)

This would be the number of hours, x_1 and x_2, it takes mine 1 and 2 respectively to get this much resources

[fleet orbit]

Need some help on how to approach this problem

[teal grotto]

the middle two columns are linearly dependent

start by looking at A(0,1,0,0) and A(0,0,1,0)

also the first and last columns are the same. that should be a big hint

[fleet orbit]

Sorry I'm trying to figure out what you are leading to, are you referring to the identity matrix for A(0,1,0,0) and A(0,0,1,0)?

[teal grotto]

@fleet orbit (0,1,0,0) is a column vector. A(0,1,0,0) is just regular matrix multiplication on the left by A

[fleet orbit]

this is what I could make out for w and w'

[teal grotto]

awesome. now look at the first and last columns

[fleet orbit]

got it!

thanks so much!

[teal grotto]

yw : )

[fleet orbit]

yeah its so hard to find these methods though, wish my professor showed us more examples

I keep ending up trying to do something like A*(w1,w2,w3,w4)=A*(w1',w2',w3',w4') because im so used to just going to the coefficient matrix and setting up a linear system

[last holly]

try to think what free means and it'll give you a clue

it might help clarity if you get the pivots to be 1, though i don't think all profs require that

[clever merlin]

Trying to see if there’s a way to find K from just the information present

[last holly]

how are you trying this

[honest zealot]

(Speaking on a friend’s account) Initially i tried to find the inverse matrix and premultiply but you cant find an inverse of 2x1 matrix

[last holly]

ok you can see it's a linear transformation though right?

can you think idealistically?

like B' is a little far, if it were close that's one you already know

(and maybe figuring that out will trigger the next bit)

[honest zealot]

Chief imma be real wit you, I’m a year 11 student and I got no clue what linear transformation is

[last holly]

oh no problem, hahahha

ok but visually you do

you know flipping stuff?

rotating stuff

[honest zealot]

Yes

[last holly]

cool, matrix multiplication represents those types of trasnformations

so you have 1 for a flip,

one for just repeating the image (identity)

etc

what i'm saying is, find a simpler matrix first (for example, for a perfectly symmetrical flip)

and then worry about the next step

[honest zealot]

Trial and error?

[last holly]

not quite

informed exploration hhahahah

i always think of trial and error as just random

[teal grotto]

can you see how A and C’ are related in the picture

[honest zealot]

Yes

[teal grotto]

which other ones just visually look related by that same reasoning

[honest zealot]

C and A’

[teal grotto]

and then B and B’ have to be related, just by process of elimination

so K is a two by two matrix. can you see how the second column has to be (2,1)?

oh let me correct myself. i was looking at the image to idealistically.

A and A’, B and B’, and C and C’ are all related

[last holly]

but your still right about the column

[teal grotto]

the one thing that’s great is that we know the second column of K has to be (2,1)

[clever merlin]

Roight

[teal grotto]

since K is a small enough matrix, you could just try to solve for the first column of K directly

[last holly]

even when you get it though i'd play around transforming the thing

the more intuition you build the less you have to stress about abstract problem solving

[clever merlin]

The answer is in the textbook but the teacher posed the question but he didn’t even know the answer

[last holly]

that's a bit sad

but linear algebra is amazing, you can leverage it for a lot of things

[clever merlin]

The context of the question is to find A’, B’ and C’ with the known transformation matrix but he posed it as how can you find the transformation matrix itself just through the image and the end result

[last holly]

right

but that's also not very helpful hahaha

the flavor of the transformation is the key

[clever merlin]

Yeah I know hahahahaha

[last holly]

the name for this

is a shear @clever merlin

it's a very badass type of trasnformation

shear: a strain in the structure of a substance produced by pressure, when its layers are laterally shifted in relation to each other.

[clever merlin]

That’s very funky

Determinant is always 1

[teal grotto]

i don’t think that’s true

[last holly]

not a shear?

[teal grotto]

i stand corrected nvm

yea nvm. that makes sense. volume doesn’t get scaled by a shear

[last holly]

i honestly forgot haha

i'm not shearing enough in my life...

clearly i'm doing something wrong

[clever merlin]

🥲

[teal grotto]

lol

[zealous junco]

lol how do u show

$|u+v|^2 = |u|^2+|v|^2$ of complex numbers

[stoic python]

Anticipation

[zealous junco]

implies orthogonal, i can only say real part <u,v>=0

nvm i found counterexample this is not true

u = (1,1,i) v = (1,-1,1) i think

lol or just u = 1 v = i

[dusky epoch]

the norm of a vector $(z_1, \dots, z_n) \in \bC^n$ is $\sqrt{\sum |z_k|^2}$ not $\sqrt{\sum z_k^2}$

[stoic python]

Ann

[dusky epoch]

also |1+i|^2 = |1|^2 + |i|^2

and also the inner product in $\bC^n$ is $\ang{z,w} = \sum z_k \overline{w_k}$

[stoic python]

Ann

[zealous junco]

yea so what i said above is not true for C right

our problems set i think is asking us to prove for R so the fwd direction is. not true for C

cuz this is for L^2 space

[sonic osprey]

I don't think your counterexamples work, 1 and i aren't orthogonal for example

[ocean sequoia]

also

[forest quiver]

$p(x)=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5$

[stoic python]

Tim O'Brien

[forest quiver]

Let velocity values be x, and force values their corresponding outputs

[ocean sequoia]

oh

you are solving for a

not x

ok

[forest quiver]

Yeah you got it

[ocean sequoia]

yea nvm

i see

[forest quiver]

It would be hell to type out all of that

But yeah if we do that

[ocean sequoia]

I thought you were solving for x in rref

[forest quiver]

then we get teh exact coefficients

[ocean sequoia]

which

wouldnt be possible

ok

[forest quiver]

yeah we would have too many unknowns

but yeah computational/applied stuff like this is really chill

Do you know where I can find more stuff like this?

like using computers to fit polynomioals/approximate/etc..

[ocean sequoia]

yea polynomial regression would be doing something else

hm

no

actually

[forest quiver]

IM not quite sure how poly regression works

[ocean sequoia]

look up splines

[forest quiver]

alright

This looks right up my allex

but its not a function

this website says its a function

xD

[ocean sequoia]

what isnt

[forest quiver]

These circles

[ocean sequoia]

/what website

[forest quiver]

But this seems like stuff I like

[ocean sequoia]

id check out introduction to statistical learning

its free

and read the section on this

and it comes with code\

[forest quiver]

ok cool

[ocean sequoia]

or it should i think

[forest quiver]

Statistical learning?>

a book?

[ocean sequoia]

https://www.statlearning.com/

[forest quiver]

holy cr*p

this is nice

seems like a lot of reading

and not a lot of exercises though

[ocean sequoia]

its like the baby bible for this stuff

[forest quiver]

baby bible?

[ocean sequoia]

yea the big boy bible is elements of stat learning I struggled with this one

but I couldnt self study that without a professor or someone to rouintely talk to

you actually might be able to now given #advanced-probability

here

[forest quiver]

This seems like really important stuff to learn

[ocean sequoia]

if you want to do ML

yea

[forest quiver]

but like

looking at it

lots of words

complicated

no exercises

doesn't seem that fun to self study

[ocean sequoia]

intro

should have ex

both should

have a ton

[forest quiver]

oh shit yeha

theres R

oh yeah nevermind

there are exercises

but still I want to get through linear algebra first im havin ga lot of fun with this

[ocean sequoia]

I would

[forest quiver]

do I need calc 3 for stats?

[ocean sequoia]

nah not really

[forest quiver]

I might wnat to do that first as well

idk theres so much to learn

[ocean sequoia]

some vector calculus is helpful for like neural netsa

but even then

its mostly just notational heavy imo

[forest quiver]

what math do you do

[ocean sequoia]

none

atleast by this channel

im a DS major

[forest quiver]

DS?

[ocean sequoia]

with a focus in neuroscience

datascience masters

student

[forest quiver]

Oh that's why you know this so well

You going to be makign big bucks

in the future

I still don't know what I want to study

[ocean sequoia]

hahahahha i actually just want to do research

[forest quiver]

me too but its hard

[ocean sequoia]

yea :/

im stressed about applying for phd programs

[forest quiver]

do you have internships

and projects etc.. ?

[ocean sequoia]

im doing some research now

on alzheimer's

[forest quiver]

oh neuroscience right

[ocean sequoia]

btw

i do want to point out that we werent solving a polynomial

in rref

[forest quiver]

we were getting coefficients

[ocean sequoia]

yea

[forest quiver]

of the polynomial

yes

[ocean sequoia]

i thought you wanted the x terms lmao

[forest quiver]

no I understnad this p well

[ocean sequoia]

ok cool

i think python can do rref

might be worth taking a look at

[forest quiver]

yeah I learned how to do that as well

[ocean sequoia]

oh nice!!!

[forest quiver]

it doesn't give fractional answers tho

it gives some BS decimal

[ocean sequoia]

oh

hahahah

[forest quiver]

that idk when they cut it off

[ocean sequoia]

it shouldnt matter bthg

tbh

[forest quiver]

yeah maybe not

[ocean sequoia]

the rounding error should be super minimal

[forest quiver]

I used my TI84 tho and it gave me fractions

IIRC

man RREf is so useful

[ocean sequoia]

hhahah it is

[forest quiver]

i feel like I can control data so muhc better

now

[ocean sequoia]

🙂

wish I wouldve been more clear that was rref

couldve helped more

[forest quiver]

nah no worries

it's good to learn alone

[ocean sequoia]

[forest quiver]

i come here way more often than I should

[ocean sequoia]

i do too

lmao

[zealous junco]

oh yea i was trying to give an example norm(u)^2 + norm(v)^2 = norm(u+v)^2 but the functions are not orthogonal

[leaden tide]

Yeah, that wasn't gonna work

[shut marlin]

Is anyone able to critique my (potentially dodgy) linear algebra proof? Thanks

I'm not that good with Latex so it might be hard to read

[native rampart]

In the fundamental theorem step,it should be "dim null ST_1+dim range T_2"

[shut marlin]

@native rampart I see, thanks for pointing that out for me

@native rampart Sorry for all the questions, but does that make the proof invalid? Or could it still be ok if were to fix that up?

[native rampart]

Yea looks good

Both parts are correct

[native rampart]

Nah, just change it. Typos aren't that big of a deal

[shut marlin]

Yay! I was feeling adventurous and decided to deviate from the typical techniques the textbook used to solve its problems. It's good to know that the proof worked out 😄

[native rampart]

Welp,there is one problem you haven't directly addressed. What if v is in null(T_1) but not in null(T_2)

If you address that,you can just straightaway define S and you are done

[shut marlin]

Good point

That's a clever way to go about it

[native rampart]

After you address that you get rank(ST_1)=rank(T_2) which doesn't mean T_2=ST_1

It means T_2=A ST_1 for some invertible operator A

[shut marlin]

Thanks for the heads up, I'm going to try and improve the proof now and see if I can make it a bit nicer

Thanks for the help!

[native rampart]

If you want a nicer proof for only if part:
Let {$e_{1},e_{2}...e_{k}$} be a basis of null($T_{1}$). Now extend this to a basis of V. i.e.,basis of V is {$e_1,e_2...e_k,e_{k+1}...e_n$}

[shut marlin]

That's a much nicer proof technique than the one I was using.

[native rampart]

You see that ${T_1(e_{k+1}),T_1(e_{k+2})...T_1(e_{n})}$ will form a basis for range($T_1$)

[stoic python]

Buncho Dragons

Buncho Dragons

[native rampart]

So you can just choose a S such that $ST_1(e_i)=T_2(e_i) \forall i \in {k+1,k+2...n}$

[stoic python]

Buncho Dragons

[shut marlin]

I like how elegant it is too

[native rampart]

Now it turns out that T_2(e_i)=0 for the remaining elements in {1,2,3...k}

So such a S exists

[shut marlin]

Thanks for showing me this method

I'm going to try to replicate it in some of the other exercise problems

[wintry steppe]

Linear algebra brrrrrr

[turbid trellis]

When solving Ax=0 and Ax=b

Why are the terms of the free variable's the same ? Because reduced echelon form is always the same ?

The only thing which makes the result differ are the coefficients I solved for, however the terms associated with the parametric vector form in the linear combination stay the same

The book I'm reading says something about it however it slightly confuses me

[lavish jewel]

the thing is that 0 + b = b

so any x that solves Ax = 0 can be added to any x that solves Ax = b, and you still get b

just by linearity, yeah?

let's call them x_0 and x_b so we don't confuse them

Ax_0 = 0, Ax_b = b

[turbid trellis]

Ah I see, so all solutions in a consistent system with free variables are parallel lines ?

[lavish jewel]

just add both sides of the equation

A(x_0 + x_b) = 0 + b = b

[lavish jewel]

i'm not sure this is true

pretty sure it isn't

[turbid trellis]

In a R^2 system *

[lavish jewel]

ah

uh...

still no

[turbid trellis]

Okay then I'm confused

[lavish jewel]

in R^2, the domain is a plane

[leaden tide]

If there is 1 free variable, the solutions form a line

If there are 2, they'd form a plane

the plane

[turbid trellis]

Okay so if there is 1 free variable

The solutions are parallel lines

and with 2 parallel planes ?

[lavish jewel]

hmmm aight i see what you're trying to do

1 free var gives you just 1 line

you have x_b + t*x_b, which is of the form of a line

just one line

[turbid trellis]

Ah I see

[leaden tide]

you don't get two parallel lines unless you go non-linear i'm pretty sure

[turbid trellis]

But at the end it is all parallel because each time you just change the vector you add to the free variable terms

[lavish jewel]

uh

it's parallel, sure, but it's the same line

[turbid trellis]

Yes I understand

[lavish jewel]

ah you mean for different bs

?

[turbid trellis]

yes different base

If that's the correct term

[lavish jewel]

i meant b vectors, idk what you meant by base there

[turbid trellis]

a + x_1(b)

For example, and then the a changes

Depending on the right-side variables of the system of linear equations

[lavish jewel]

i'm not sure you can have that in 2D

if you change only a and nothing else, it's the same line

[turbid trellis]

I understand but it's parallel to the other a's right

[lavish jewel]

lets say we have Ax = b, and A is rank 1. the system has a solution if b is in the span of A

[turbid trellis]

Yes

[lavish jewel]

so b is of the form c*a, for some vector a in the columns of A

[turbid trellis]

Yes

[lavish jewel]

so the solutions for generic b would be c*a + t * x_0

for scalars c and t

[turbid trellis]

c sub zero ?

[lavish jewel]

if you change c (which is the same as changing the vector b), the resulting line need not be parallel to the previous one

[turbid trellis]

Ah okay I understand

[lavish jewel]

i'm pretty sure we can make a simple counterexample

[turbid trellis]

and t is the same in all solutions of the same matrix, when changing the rhs

[lavish jewel]

the t is arbitrary, since x_0 is in the nullspace, yeah

doesn't depend on the rhs

[turbid trellis]

Okay cool thanks, I understand now

[lavish jewel]

a simple example would be a matrix
1 0
0 0

let's pick b = [1;0] first

we have solutions of the form [1;0] + t[0;1], yeah?

then change b to [2;0]

the solutions are now of the form [2;0] + t[0;1]

those two lines are not parallel

hmm

or are they

[turbid trellis]

The vector a moves them right

[lavish jewel]

i guess they are lol

[turbid trellis]

doesn't change the direction

[lavish jewel]

i'll blame it on being hungry

[turbid trellis]

Yeah that's what I was confused about haha, if they're parallel and why

[lavish jewel]

then yeah, parallel lines depending on the value of b

[turbid trellis]

Hmm I always get tired after eating haha

And energetic without eating

Thanks for the help

[lavish jewel]

aight

[turbid trellis]

I don't understand 19 and 20

What do they want me to get ? B is not a line

[dusky epoch]

a line can be parallel to a vector

[turbid trellis]

Ah I see, so a + x_1[5,3]

[dusky epoch]

-5

but yes

[turbid trellis]

Ah cool, thanks that's stupid I did not think of that myself haha

A line can be parallel to vector, how could I forget lol

[harsh wolf]

if a tuple is linearly dependant

will other tuples of the same "length" be also linearly dependant ?

[teal grotto]

tuple meaning (v1,v2,...,vk) where k is the length of the tuple?

[hard drum]

(1,0),(2,0) in R^2 are linearly dependent, (0,1),(1,0) are not, providing a counterexample (if I understand you correctly when you say tuple)

[harsh wolf]

oh right

thats what i meant

[hard drum]

oh sweet

[coarse sandal]

How do i start this problem

cause I know how to transform a basis into a orthogonal basis using Gram Schmidt

but im not sure how to start this problem

I figured it out

[dire thunder]

you are assumed not to use gram schmidt tho

[coarse sandal]

The projection can be found using A(A^TA)^-1*A^T * b

[midnight kayak]

hello

if a matrix is found to not have nonzero rows by the time it's reached row echelon form

then we can say for sure that it won't have any by the time it's reached reduced row echelon form right

please ping me for an answer

[teal grotto]

this is correct

[midnight kayak]

pog

also

the reason we often compute the rank via gaussian elimination

is something to do with how that process makes clear how many rows at most are linearly independent right

[teal grotto]

@midnight kayak yes. row reduction can be thought of as multiplication on the left my elementary matrices, each of which is invertible.

so say A is an n x m matrix and E is an invertible n x n matrix. the kernel (or null space) of A is going to be the same as the kernel of EA. so by the rank nullity theorem, rank(EA) = rank(A).

this applies directly to row reduction, since if you row reduce and n x m matrix A to its row reduced form A’ by multiplication on the left by elementary matrices, say
(Ek… E1)A = EA = A’
with Ek…E1 = E, then E is invertible so
rank(EA) = rank(A) = rank(A’)

this means that the number of linearly independent rows of A’ is the same as the number of linearly independent rows of A

[zealous junco]

can someone help with SVM

I don't understand what happens if C is large, I know theres large penalty put into misclassification in this case but how does that affect maximizing the margin?

why will people choose to use small C?

[wintry steppe]

You might be better off asking in a computer science discord

What book is this?

[dusky epoch]

large C means your model will try its hardest not to misclassify anything

which will make it sensitive to outliers or noise in the data

[zealous junco]

so is overfitting the main problem? or

[dusky epoch]

well that's the big one i can think of rn

[zealous junco]

and this is pattern recog & ML

pretty good book imo

[dusky epoch]

youre in luck because i happen to have studied SVMs to some extent

but yeah overfitting

the bigger C is, the closer results you get to hard separation (where mistakes are forbidden)

[zealous junco]

I don't see any other major problem with big C, as overfitting is not too big of an issue in classification, but unless it is computationally harder to do?

like as the constrained minimization problem

and i feel like on the other side if C -> 0 then wouldn't you just be almost always drawing the parameters to 0? so seems pretty boring there

[dusky epoch]

if C is too small you get margins that are too wide and misclassify too much and are essentially meaningless

i.e. w unreasonably close to zero

[zealous junco]

ic, thanks

actually yea another question was what it meant by "controlling model complexity"

so like if they mean trade-off on the highlighted line there then they implying large C = greater model complexity?

and is that "complexity" in the sense of how hard and/or well numeric solvers will be able to do the minimization?

[wintry steppe]

I think it’s a reference to the fact that larger C will allow your weights to get bigger

[dusky epoch]

large C = more overfitting

[lavish jewel]

complexity there refers to the overfitting. it will prefer curves with large total variation as opposed to smooth ones, since it picks up any random variations as being part of the model

[zealous junco]

👍thanks

[wintry steppe]

Hey everyone, I have a basic linear algebra question

I am quite confused by this, why is the third side (the hypotenuse) represented by $$\textbf{v} - \textbf{w}$$?

[lavish jewel]

make a drawing

\, not /

[stoic python]

SMILEYYY

[lavish jewel]

make a drawing

let v and w be position vectors pointing from the origin to some point

[wintry steppe]

Ok, give me a sec

V or W could be either two legs of the triangle, yet the length still doesn't satsify the hypotenuse

[lavish jewel]

why do you say that?

v - w here is the vector (-3,2)

what's the length of that vector?

[wintry steppe]

sqrt(13) = sqrt(2^2 + 3^2)

ok, but what is v and w in this case

[lavish jewel]

mhm

v is (0,2) and w is (3,0)

it would also work the other way

[wintry steppe]

shit

thanks

ohhhh I was thinking like of the length of the vectors (one of the components, because conveniently one of the components is the length of the whole vector), like the length of v - the lengh of w, but in reality you were suppose to think of the vector (including all components)

[ocean sequoia]

The real trick is to never use SVM

[ocean sequoia]

Also be careful with this I’m GBM overfitting happens a lot even in classification

[ocean sequoia]

@zealous junco tho SVM with a guassian kernel can outperform but if its just a linear SVM they generally arent ~great~

https://jmlr.csail.mit.edu/papers/volume15/delgado14a/delgado14a.pdf

https://stats.stackexchange.com/questions/111654/adaptive-boosting-vs-svm

[wraith patio]

how do you prove a transformation is onto?

would you do it by contradiction? show there cant be a vector outside its range or smthn

[wintry steppe]

do you have a specific question?

there probably isn't a completely general method to do so (of course, what you said may always work, but it also may not be the best method)

some ways work better in certain situations, etc

How do you justify that $det(A)= det(A^t)$ if A is an elementary matrix of type 3, that is, adding to any row in the identity matrix of order $n$ another row multiplied by a scalar $k$?

Oh no, it wasn't that

Now it's what I wanted to ask

I understand that for the other two types of elemental matrices this holds as they are symmetric so $A^t = A$, but with the third type $A^t \neq A$ so...

[stoic python]

kuro

[wintry steppe]

can you argue that A and A^t both have to have determinant 1

[stoic python]

kuro

[wintry steppe]

I had to correct a word

[wintry steppe]

They would both have 1111...*1

Okay

A chain of 1s multiplied until we get to the row or column with an additional k

Oh

And you never multiply by that k

er maybe not always 1

but you should be able to just argue directly by calculating each's determinant

Because in that row/column you just simplify it when applying $\alpha_{ij} = (-1)^{i + j}\cdot det(A_{ij})$, where $A_{ij}$ is the A matrix without the row i and column j

[stoic python]

kuro

[wintry steppe]

And as it's the main diagonal, that $(-1)^{i+j} = 1$ always

[stoic python]

kuro

[wintry steppe]

Well...

When you get to the column with k, you would have $k\cdot\alpha_{ij}$

[stoic python]

kuro

[wintry steppe]

In the sum

So it doesn't have to be always 1

btw

When you calculate it for the transposed matrix, it's the same

The determinant computation will be the same as it will be in both cases:

$1\cdot 1\cdot \dots\cdot (1\cdot \alpha_{ij} + k\cdot \alpha_{pj})\cdot\dots\cdot 1$

[stoic python]

kuro

[wintry steppe]

Where $p$ is another row

[stoic python]

kuro

[wintry steppe]

And in the column case (which is basically the transposed matrix... I understand what I'm writing, don't worry about this comment)

$1\cdot 1\cdot \dots\cdot (1\cdot \alpha{ji} + k\cdot \alpha{jp})\cdot\dots\cdot 1$

[stoic python]

kuro

[wintry steppe]

So that solves it

And $det(E_{ij}(k)) = det(E_{ij}(k)^t)$

[stoic python]

kuro

[wintry steppe]

Well, it makes sense and I proved it

So that makes me understand everything

Thanks @wintry steppe

i didn't really do much but i'm glad you seem to have got it

I got that question because in my LA book there were proofs about some determinant properties and in order to prove that $det(A) = det(A^t)$ they used that the property holds for elementary matrices, but they didn't justify it and I understood that for types I and II it holds for elementary matrices as they are the symmetric, but as 3rd type wasn't symmetric, it wasn't obvious

[stoic python]

kuro

[ocean sequoia]

see the real trick is to go over axler

so you dont have to do determinats

[sinful valve]

is there any good books on linear algebra which arent boring as hell

[teal grotto]

nah. all of em are super boring

[forest quiver]

Two textbooks of mine have conflicting ideas

[stoic python]

Tim O'Brien

[forest quiver]

But also

[stoic python]

Tim O'Brien

[forest quiver]

Does orientation not mean anything? This is quite confusing

[teal grotto]

the dots represent different ‘multiplications’.

the first dot is the dot product. the second dot is matrix multiplication

[forest quiver]

Hmm

Are you sure?

[teal grotto]

that’s what it looks like

[forest quiver]

so dot product = matrix multiplication theN ?

[teal grotto]

sort of. v•w = v^T w

if v and w are in R^n, for example

[forest quiver]

Im p sure

they are interchangable

[teal grotto]

the standard inner product in R^n is only defined for vectors. not sure how you’re suggesting to interchange them. the entries of matrices after multiplying can be expressed in terms of dot products, which is what the above picture is saying

[forest quiver]

yeah

so is dot product the same as matrix multiplication

it's the same process

[teal grotto]

yes

if you multiply 1xn and nx1 matrices, on the left and right respectively

[forest quiver]

ok that's quite confusing

[teal grotto]

in what way

[forest quiver]

one sec I will show you why it's confusing to me

1 min

actaully nvm

it just works out

1xn nx1 are interchangable b/w dot product and normal multiplication

It doesn't matter which sequence I use, I will get the same result

[last holly]

@forest quiver "normal" multiplication is kind of unhelpful in linear algebra

[forest quiver]

alright

[last holly]

dot product and cross product help distinguish what's happening

[forest quiver]

Im getting to linear transformations next section

so I will learn about dot procduct more then

[last holly]

i know it's confusing naming conventions

[forest quiver]

yeah :(

[last holly]

i got confused too at first

[forest quiver]

do you think that the first tiem around I go through

I shoudl look for 100% understanding?

[last holly]

no

[forest quiver]

or do you think I should just learn as best I can

ok

[last holly]

it might not even be possible

since linear is so deep

[forest quiver]

really?

[last holly]

so unless you're a genius haha

yea it comes back

[forest quiver]

alright

I will take it with a professor in 2 years as well

so looking forward to that

[last holly]

if you go into other maths, or computer science it has a lot of applications

[forest quiver]

Its the mos tuseful math I have every done

ever done

the thing I like most about it probably

is what I can do with data

now

[last holly]

it is indeed very useful

[forest quiver]

like sure you can see vectors as arrows or whatever

but i see them as data points

(data1, data2, data3, etc...O

)

[zenith mauve]

I'm trying to determine a Basis for a subspace whose vectors are all orthogonal to (-1, 1, 1), and my textbook gives the solution B = {(1, 0, 1), (0, 1, -1)}

which is algebraically written as:
0 = (-1, 1, 1) * (a, b, c)
0 = -a + b + c

• c = -a + b
  therefore vector forms (a, b, a-b) define a valid basis for W

algebraically, I got:

0 = (-1, 1, 1) * (a, b, c)
0 = -a + b + c
a = b + c
therefore vector forms (b+c, b, c)

is this not also a valid Basis?

[sonic osprey]

@zenith mauve It is also valid yes

[ocean sequoia]

let S the set of all numbers such that x,y are an element of the real numbers and x^2 + y = 0 does this form a subspace of R

no right because (1,-1) and (2, 4) are in the set but (1,-1) + (2,-4) = (3,-3) which isnt in the set

am I wrong?

[wintry steppe]

(2, 4) isn't in the set

oh u meant (2, -4)

uh but then you should get (3, -5) from the sum

but the argument still works

:catThink:

[ocean sequoia]

Yea meant (2,-4) lmao so it’s not a sub space right cause you get (3,-5) which isn’t in the set @wintry steppe so it’s not closed under addition

[wintry steppe]

yep

also i guess you meant to write subspace of R^2

[ocean sequoia]

Yea sorry

Was doing a brief refresher and wanted to make sure

also I stumbled upon this question am I bit confused

https://math.stackexchange.com/questions/2216776/linear-algebra-polynomial-subspace

why is that not closed under addition here

im a bit confused as to what it means by not in the form

[wintry steppe]

to be in the desired form means that it's a quadratic in t with leading coefficient 1 and linear coefficient 0

the thing they got has leading coefficient 2

[ocean sequoia]

ahhhh that makes sense

thank you

[wraith patio]

im not sure how to start this

my thought was to isomorph a basis for Ker(T) to get a basis for null(A) and show the nullity is the same, then use the dimension theorem to show the rank must be the same too but idk if thats the best way

[stoic python]

coycoy

[teal grotto]

@wraith patio

*equal to L_A

[wraith patio]

arent we given that with $A=[T]_\beta^\gamma$?

[stoic python]

taxminion

[teal grotto]

if you know that $[T]_{\beta}^{\gamma}$ is already equal to the composition above, then what i have told you is kind of pointless

[stoic python]

coycoy

[teal grotto]

otherwise it’s kind of helpful

[wraith patio]

i think im a bit confused on the proof in general bc it seems like the matrix being equivalent to the transformation is such a basic fact idek how to start proving it

i mean the proof is about the rank/nullity specifically but it just seems too obvious to me idk

[teal grotto]

it is pretty obvious, but if you want to prove it you need to use isomorphisms

[wraith patio]

ig. but it probably doesnt help im not super sure how to prove a transformation has a certain rank or nullity in general

my instinct is to start with the dimension of a basis for the null space but idk

[lavish jewel]

you have to show that the rank of the composition of the isomorphisms (or their inverses, rather) and T has the same rank as T

they did tell you that A = [T]_\beta^\gamma, but nothing about how exactly A is made from T, which is rather trivial, but necessary

[teal grotto]

your instinct was sort of right. but you are given a basis already, so i don’t know how choosing one is really going to weave into this proof.

once you have the fact that i have given above, then you can show that $\iota_{\mathbf{v}}$ when restricted to the null space of A is an isomorphism between null(A) and the kernel of T

[stoic python]

coycoy

[wraith patio]

yeah doing an isomorphism between null(A) and ker(T) is where i went originally. i got bogged down trying to prove that the dimension was preserved but maybe i didnt need to since its an isomorphism

[teal grotto]

if U and V are vector spaces over some field F, then U is isomorphic to V if and only if dim U = dim V

[wraith patio]

right but can i say for sure that the isomorphism of a basis for ker(T) will be a complete basis for null(A)?

[teal grotto]

yes

but you don’t need to here

[wraith patio]

bc its a isomorphism?

i dont even need to say it you mean? like its understood?

[teal grotto]

no, you just don’t need that fact, unless you are actually trying to show this as a preliminary lemma:
if U and V are vector spaces over some field F, then U is isomorphic to V if and only if dim U = dim V

[wraith patio]

maybe im just not understanding what you mean. i thought we were trying to prove that the dimension of null(A) and ker(T) are the same, so how could i use that? or do you mean its using that null(A) and ker(T) are isomorphic to prove they have the same dimension?

[teal grotto]

i’m trying to guide you to an isomorphism between nul(A) and ker(T). by showing that, then dim null(A) will be equal to dim ker(T)

[wraith patio]

right thats what i meant

so how's this

suppose $\operatorname{nullity}(T)=k$ and a basis for $\ker(T)$ is $\eta={v_1,\ldots,v_k}$.

let $\phi_\beta:V\to F^n$ be the isomorphism such that $\phi_\beta(\beta_i)=e_i$ for $1\leq i\leq n$.

since $A=[T]\beta^\gamma$, $\phi\beta(\ker(T))=\operatorname{null}(A)$, making them isomorphic. therefore, $\dim(\ker(T))=\dim(\operatorname{null}(A))$ and $\operatorname{nullity}(T)=\operatorname{nullity}(L_A)=k$.

since the dimension of the codomain of $L_A$ and $T$ are equal, by the dimension theorem, their ranks must be equal as well. qed

[stoic python]

taxminion

[zealous junco]

actually about my SVM question yesterday, why in soft SVM that all non-support vectors are correctly classified points?

or "most"

[lavish jewel]

what are non-support vectors

[zealous junco]

Those with Lagrange coeff that’s zero

Like the coeff in the complementary slack condition

[wintry steppe]

BTW guys I am learning Linear Algebra on my own and I have finished reading and solving exercises of the books Linear Algebra by Sheldon Axler and Steven Roman. What book for Linear Algebra do you recommend for reading next? I am not looking for something that is much more advanced, but I am looking for a book that links linear algebra and other fields of math, like Graph Theory + Linear Algebra or something. If you know good books that combine two or more fields of math, please tell me. Thanks.

[lavish jewel]

i'd have to look at the cost function to answer you more clearly, but off the top of my head, it has to do with the interpretation of complementary slackness

having a "lagrange coeff" (not really lagrange, these are from KKT) of 0 means the point is already feasible

you need mu_i * g_i(x) = 0, yeah? for conditions g_i(x) <= 0

if you are on the boundary of the feasible set, i.e. when g_i(x) = 0, then you can penalize this by having a nonzero cost, a mu_i different from 0

if you're inside the feasible set, g_i(x) is strictly smaller than 0, and then there is no penalty for this term

so you set mu_i = 0

so if the point has a weight mu_i of 0, it means the inequality was already satisfied

which in your case presumably means the point was on the correct side of the hyperplane

waving my hands wildly

[wintry steppe]

See that I'm reading a linear algebra book and I'm not going to read another, and I know Axler. It's difficult, so I won't read him until I finish a lot of books.

[zealous junco]

yea thanks i got it, sort of

my understaning of hard SVM (linearly separable data) is fine i think, its that they normalize so w^t phi(x) ≥ 1 and equality is achieved whenever x is a point on the margin, and so if the KKT coefficients mu ≠ 0, i.e. x is a support vector, then it must be on the margin and contrarily if w^t phi(x) > 1, i.e. a vector not on the margin then mu = 0, i.e. x is not a support vector

but I think im not perfectly understanding soft SVM yet since you have the slack variables so could it ever happen that mu = 0 but the x is classified incorrectly here..

[lavish jewel]

not, that's the point of complementary slackness

well

if you are using interior point methods, at least

for soft svm, this would be like being past the hyperplane, but still being within some distance of it

if you're on the correct side of the hyperplane, mu = 0

if you're on it or on the wrong side, mu gets larger

the inequality constraint is applied to a function of the vectors that is not just dependent on the hyperplane and the point, but rather some transformed distance

soft svm tries to minimize the number of incorrectly classified points and how far away they are from the hyperplane

the only thing you have to look at is the inequality constraint, really

[zealous junco]

yea thanks i sort of get it now

thanks for explaining, i guess the only main difference here is the support vector no longer required to live on the margin, i.e. when mu = gamma (gamma being the parameter)

but it remains true that those points s.t. mu = 0 is not contributing to making prediction since they are better classified than the support vectors

since for those points we have tn*yn ≥ 1- xi which means it is at least better than equality

[lavish jewel]

that depends on the cost function

we were only discussing the inequality constraints

the original optimization target may include all the points

the inequality constraints only involve misclassified points due to slackness when written in KKT form

[zealous junco]

oh

ill think about it later, thanks

[lavish jewel]

yeah, remember you're optimizing some f(...) + sum (mu_i * g_i(x))

[wintry steppe]

There's an exercise in a book that says:

1. $A$ is symmetric $\longleftrightarrow$ $A^{-1}$ is symmetric.

2. $A$ is antisymmetric $\longleftrightarrow$ $A^{-1}$ is antisymmetric.

They show that:

$A$ is symmetric $\longrightarrow$ $A^{-1}$ is symmetric because $A = A^t$ and $(A^{-1})^t = (A^t)^{-1} = A^{-1}$ so $A^{-1}$ is symmetric.

$A$ is antisymmetric $\longrightarrow (A^{-1})^t = (A^t)^{-1} = (-A)^{-1} = -A^{-1}$ and $A^{-1}$ is antisymmetric.

And finally the book says that the reciprocal is obtained bearing in mind that $A = (A^{-1})^{-1}$, but I don't get it

[stoic python]

kuro

[wintry steppe]

I get that if $A^{-1}$ is symmetric then $A^{-1} = (A^{-1})^t = (A^t)^{-1}$

[stoic python]

kuro

[wintry steppe]

Oh

Then $A$ is symmetric

[stoic python]

kuro

[wintry steppe]

But I didn't apply that $A = (A^{-1})^{-1}$ as the book says

[stoic python]

kuro

[wintry steppe]

Well, and in the antisymmetric case

If $A^{-1}$ is antisymmetric, then $(A^{-1})^t = (A^t)^{-1} = -A^{-1}$ and $((A^{-1})^t)^{-1} = ((A^{-1})^{-1})^t = A^t$

[stoic python]

kuro

[wintry steppe]

And $-(A^{-1})^{-1} = -A$

[stoic python]

kuro

[wintry steppe]

So then it's proved

Okay

And what about $A$?

[stoic python]

kuro

[wintry steppe]

Well, we should have that if $A^{-1}$ is symmetric, then $A^{-1} = (A^{-1})^t = (A^t)^{-1}$ and $((A^{-1})^t)^{-1} = ((A^{-1})^{-1})^t = A^t$ and as $A^{-1} = (A^{-1})^t$ then we should have that $A = A^t$

[stoic python]

kuro

[wintry steppe]

So then $A$ is symmetric

[stoic python]

kuro

[wintry steppe]

Okay, I don't have more questions

i see you've taken the daminark approach to problem solving

[wintry steppe]

Well, before just asking for a solution, I wanted to be sure I wasn't understand what was following... but then I noticed I could deduct it and I finally got it the way the book wanted

And now the book gave an exercise that was

Show that the product of 2 lower triangular matrices is a lower triangular matrix and that the product of 2 upper triangular matrices is an upper triangular matrix and then deduct a formula for the powers of a diagonal matrix.

The book doesn't give you how to get a formula for the powers of a diagonal matrix, but I observed you can get by induction

That if $A\in\mathfrak{M}_n(\mathbb{K})$, then $A = \begin{pmatrix}
d_1 & 0 & \dots & 0 \
0 & d_2 & \dots & 0 \
\vdots & \vdots & \ddots & \vdots \
0 & 0 & \dots & d_n
\end{pmatrix}$ and

$A^p = \begin{pmatrix}
d_1^p & 0 & \dots & 0 \
0 & d_2^p & \dots & 0 \
\vdots & \vdots & \ddots & \vdots \
0 & 0 & \dots & d_n^p
\end{pmatrix}$ where $p\in\mathbb{N}$

[stoic python]

kuro

[wintry steppe]

And you get that the product of two diagonal matrices is a diagonal matrix because as a diagonal matrix is both an upper and lower triangular matrix, and knowing that the product of two upper triangular matrices is another upper triangular matrix and the product of two lower triangular matrices is another lower triangular matrix, then the product of two diagonal matrices must be a diagonal matrix

[wintry steppe]

And then you see it's easy to prove by induction this

This is very funny

[wintry steppe]

Who is Daminark?

moderator here who frequently makes long posts to channels to solve their problems out loud / have gomez come in and solve for them

Oh

[zealous junco]

so @lavish jewel your saying that here, whenever a_n=0 i.e. the constraint is not activated, it means the t*y(x)-1+xi is already in the correct side of the margin/good enough s.t. its misclassification error can be ignored?

[lavish jewel]

yea

[ocean sequoia]

Honestly I don’t do that but I have found that sometimes asking questions in channels helps cause it forces me to frame the problem

[wintry steppe]

Yes

[zealous junco]

also is it true that whenever xi_n is not 0 (the point x_n is beyond the margin) then the kkt constraint is active? otherwise i feel like its not achieving the minimum?

i.e. is it possible for points x_n where the inequality constraint is not active to have positive xi_n

[lavish jewel]

i would hope not, otherwise kkt wouldn't work

this is all just the definition of complementary slackness

[nocturne jewel]

Previous test question that I messed up previously but think I have the answer, give me a minute to type it out

$\langle (I-T^2)[w],w\rangle=\langle w,w\rangle - \langle T^2[w],w\rangle \ =\langle w, w\rangle - \langle T[w],T[w]\rangle$ So want to show that $\langle T[w],T[w]\rangle \geq \langle w,w\rangle$

[stoic python]

Mosh

[nocturne jewel]

actually just gonna write it out

[wintry steppe]

Has anyone worked with the Vandermonde determinant?

[dire thunder]

@wintry steppe not much but i used it

iirc it is something with rows as
x^n x^(n-1) ... 1

?

[wintry steppe]

The proof of the formula for $V_n$... it's being like a pain in the ass for me

[stoic python]

kuro

[dire thunder]

,w Vandermonde determinant

[stoic python]

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Requested by Commander Vimes#4407

[dire thunder]

so yes

[wintry steppe]

Yes, that is the determinant

[dire thunder]

use properties of determinants

i mean

[wintry steppe]

I mean, I tried that

When I get home I'll give the work

[dire thunder]

iirc it requires factoring x_n's out etc

and using jordan expansions

[wintry steppe]

I still haven't worked with Jordan expansions

When I studied linear algebra I didn't understand most of the course and this summer I'm trying to learn everything

I almost finished the matrices part and I was doing the book exercises but the Vandermonde determinant is very difficult to work, btw, I need to keep working with it because I must be able to prove it by myself or I won't learn much linear algebra

[dire thunder]

wait how you calculate determinants then

oof

sorry

[wintry steppe]

With the Laplace expansiom

[dire thunder]

i meant laplace

[wintry steppe]

Laplace and properties

[dire thunder]

ok so

no need even in laplace

[wintry steppe]

I know you start by doing what you do in the Gauss method to make the first column $\begin{pmatrix}
1 \
0 \
0 \
\vdots \
0 \
\end{pmatrix}$

Which doesn't change the determinant

[dire thunder]

well

ye

but not row

[stoic python]

kuro

[wintry steppe]

I have to give more context: In my book they give the transpose matrix of that

[dire thunder]

ah

well t hen yes

use that addition of multiple of column to column does not change determinant

[wintry steppe]

Well, it's the same determinant as $det(A) = det(A^t)$

[stoic python]

kuro

[dire thunder]

basically

[dire thunder]

if you do this

then determinant would be expressible in terms of determinant for smaller matrix

[wintry steppe]

Yeah, I was trying to do that

Okay

I got it

By induction

But you should do like something weird by columns

[dire thunder]

not really weird

[wintry steppe]

In order to get the $V_{n-1}$

[stoic python]

kuro

[wintry steppe]

I'm not used to work with columns, just rows

[wintry steppe]

Then you say that the determinant is that product by hypothesis

And you get it

Well

[dire thunder]

well wait

havent you learnd

that det(A^T)=det(A)

[wintry steppe]

I'm not at home right now so I can't write it down

[wintry steppe]

Yes. I proved that yesterday

[dire thunder]

then use it if you are not comf with column ops

[wintry steppe]

Yes, I think I'll use that property, so it's easier for me

Thanks @dire thunder , you are nice

[dire thunder]

yw

[placid plaza]

Does anyone know what this is asking?

[wintry steppe]

without knowing what "your parabola" is, no

this probably belongs in #prealg-and-algebra or #precalculus, not in #linear-algebra

[turbid trellis]

Where does linear dependence come from ? Like it's a non trivial solution

How is that related with dependence?

Or how should I actually make some sense of that twrm

[nocturne jewel]

a set of vectors is said to be dependent if they span to the 0 vector non-trivially

[hard drum]

The point of 'dependence' is that if, say, vectors $v1,\dots,v_n$ are linearly dependent then there exist $\alpha_1,\dots,\alpha_n$ not all zero such that $ \alpha_1 v_1 + \dots + \alpha_n v_n = 0$ and so if (wlog) $\alpha_1 \ne 0$ then we may write $v_1 = -\frac{1}{\alpha_1}(\alpha_2 v_2 + \dots + \alpha_n v_n)$

[stoic python]

nuclearpotat

[hard drum]

so we can see that the vector v1 in some sense is depnedent on the others as whenever we write v1 we could instead write some combination of all the others

[last holly]

if you think of it visually it helps a ton

all the formal stuff comes easily after (at least it a little)

the formal is necessary for higher dimensions so you can't skip it

[undone shuttle]

Hi. I have a question and I'll describe it a lil bit long. So I got a question like this:

and here's a text solution I have solved myself

Now I have to imply this into Matlab. Here is what I've done so far:

in which: f is the matrix where I write down the transformation f in matrix form, ff is the first column of f and temp is the first row of E, transposed. The terminal looks like below:

What I want to do is to do the exact same thing in the text solution: I'll take the first row of E, then substitute those values into the matrix f, by column, respectively x1, x2 and x3 (to both of its columns), then calculate the new pair, just like this:

And repeat for the following two rows to get 2 more pairs.
As in the terminal result, I got an error. May I ask what cause the problem and how do I solve it? Or do you have another "smarter" solution for this particular question?

Thank you in advance.

[umbral birch]

i messed up somewhere, not sure where could someone help?

[last holly]

i mean it's simple to test u2+u4 to see if you get the first row value for v correct (which it doesn't add up here)

[small glen]

Hello! I've created my vector and parametric equations, but I don't know what to do from there. How do I solve this sort of question? No answer needed!

[umbral birch]

@last holly i used the formula v x u1 /u1 x u1. is this not correct?

[last holly]

i mean formulas still have to add up in the end

i just picked u2 and u4 because they have 0's so it's easier to check

[umbral birch]

ive been redoing it for a while now and its still not adding up

[last holly]

(in a test you could verify quickly this way)

[umbral birch]

it didnt add up

but im not sure where i messed up because i recalculated and double checked?

[last holly]

are you applying it correctly?

[umbral birch]

i can show u my work @last holly

[silent sandal]

Hi