linear-algebra

[long blade]

just a bit confused

why zero vector doesnt work

ohh

nvm

i think its referring to a5 about u + (-u) = 0

do u think it meant that?

[dusky epoch]

uh

[gilded nebula]

it asks if the set of polynomials with positive coefficients is a vector space
the zero polynomial, while is a zero vector for the standard polynomial space, is not in this space

[dusky epoch]

yeah ok so

i was wrong with saying our set had 0

if anything, x^2 + 1 wouldn't be in V either because it has a zero coefficient and zero isn't positive

sorry for misleading you there

[long blade]

??

why wouldnt x^2 + 1

not be in V

isnt all the coeff positive?

oh

do u mean

like

x^2 + ox + 1??

that 0??

@dusky epoch?

[dusky epoch]

yes that's exactly what i'm talking about

[long blade]

hmm

but

isnt that a bit strange

i mean

if it was like that then

we can assume

0x^3 + x^2 + x + 1

wouldnt work

either

since theres a 0

[dusky epoch]

nah that's a trailing zero

if we counted those then V would be empty

[long blade]

kk

i was just thinking

maybe its becasue

0 isnt considered a positive

number

and it fails no 4 because it says object 0 in V which it fails

but anyway ty

i gained a better

understand

ing

[sour jetty]

Hello, I'm having trouble understanding how they get rid of second column, first row and change it by multiplying the determinant with -(-1). Could anyone tell me what this technique is called in English? I can't find any helpful resources about it online, not sure what to search for

[keen kiln]

Ok saying

It's easy let me explain

[storm python]

cofactor or laplace expansion

[sour jetty]

Appreciated!

[storm python]

np

[keen kiln]

See of yo can understand

Wait. Wait ..

Here you go

See if you can understand

[storm python]

i mean. you showed a bunch of work but without actually stating the method

[keen kiln]

Ooouu

Wait a bit...

Basically solve it using the basic 4*4 expandion

Then you get 3 matrix with zeros as a column which equals 0

This u remain only the one with no columns zero

Got in?

[minor rose]

YO

ANYONE ON HERE

thAT CAN PLZ HELP

IM LITERALLY DYING

[storm python]

no

[minor rose]

brah ive been on one question

for the past 3 hours

plzzz

[storm python]

read #❓how-to-get-help and #rules

[wary terrace]

need some help

[viscid kernel]

@wary terrace do you need to use Parseval’s formula, I think I found another way for solving, however Im not 100 percent sure cuz Im not an expert at linear algebra

[wary terrace]

nah i dont need to

but i think its jsut a hint that it could be easiesr if i do use it ya know @viscid kernel

[viscid kernel]

Hmm ok wait let me write my “ way of solving it “ Ill send a picture of it

[wary terrace]

thanks @viscid kernel

[icy osprey]

Can anyone help me with this one?

[dusky epoch]

what's giving you trouble here

[wary terrace]

also need help with this one haha

[viscid kernel]

@wary terrace

And so on.....

[icy osprey]

I honestly have no idea where to start @dusky epoch

[dusky epoch]

@wary terrace @viscid kernel you two please move to another channel

[viscid kernel]

Now you have to do substition.... and the norm of w is the square root of 270

[dusky epoch]

@icy osprey do you know the list of axioms for a vector space

[viscid kernel]

@dusky epoch ok

[dusky epoch]

also baklava

you just said some bullshit right there

this isn't R^4

[viscid kernel]

Ah I see

[icy osprey]

i googled these ann

[dusky epoch]

did you not get taught these in class

[viscid kernel]

@wary terrace I made a mistake. I thought it was on R^4

[icy osprey]

not really, I am doing a midterm review rn

[dusky epoch]

[icy osprey]

or i just forgot everythinh

online is hard

[dusky epoch]

i mean

okay so

this isn't shown in that list

but your S is not closed under addition

(1,0,0) and (0,1,0) are both in S but their sum (1,1,0) is not

[wary terrace]

oh

[icy osprey]

so that is one property ?

of the vector space that S does not meet right?

[dusky epoch]

yes

there are others but this is one of them

[icy osprey]

so that is specific to my problem right

the one i originally sent

how did you get (1,0,0) and (0,1,0)

[gray dust]

take 5secs to verify that (1,0,0) and (0,1,0) are vectors in S

take another 5secs to verify that (1,0,0)+(0,1,0) is NOT in S

[icy osprey]

im confused what you are saying

i understand the sum is not s

but how did you get (1,0,0) and (0,1,0) when S is not a vector space

[gray dust]

my point is are you able to check that (1,0,0) and (0,1,0) are in S?

[icy osprey]

no you are not

[gray dust]

you very much can

[icy osprey]

whaaaat

[half ice]

S is a set that contains elements of the form (a,b,c) where a,b,c follow a certain rule.
But S is not a vector space

[icy osprey]

ok understood

[gray dust]

S is defined as the set of 3 tuples where the sum of the squares of each component is less than or equal to 1

[icy osprey]

yes

ohhhh

the set being (1,0,0) and (0,1,0)

each one less than or equal to 1

[gray dust]

no that is NOT the set S itself. (1,0,0) and (0,1,0) are members of S

[half ice]

I like to think of it as "all points within the sphere of radius 1"

[icy osprey]

ok

[half ice]

Clearly (1,0,0) and (0,1,0) are in the sphere, as well as (0.5, 0.5, 0.5)

[icy osprey]

so since the sum of the members is not equal to s itself it fails addition

[gray dust]

that's poor wording. you'd say the sum of those two vectors is not in S

[half ice]

The sum of two elements of S is not in S, so S is not closed under addition.

[icy osprey]

ok, now I understand

so i can literally just write that as the answer?

i could also show the addition

[half ice]

Yeah pree much lol. Every vector space needs to be closed under addition, and S isn't, so it's not a vector space

[icy osprey]

ok cool

I have more midterm review questions I will ask later if you all are around

I appreciate the help

[half ice]

Np! Feel free to ask

[icy osprey]

cool!

honestly have another question after this, lmaooo....online got me dumb

[gray dust]

@wary terrace just bash through the computations for vector projections

for vectors $a,b\in V$, $\proj ab=\frac{\ip{a}{b}}{\ip{b}{b}}b$

[stoic python]

RokettoJanpu:

[icy osprey]

for this problem, can I do rref and find where the pivots are

and those vectors will be the basis

[gray dust]

that'll get you a basis but the q also wants the basis vectors to be orthogonal

[icy osprey]

i would have to use gram-Schmidt right?

[gray dust]

sure

[icy osprey]

wouldn't that give me an orthonormal basis?

orthonormal is orthogonal with length 1

[gray dust]

orthonormal is also orthogonal

[icy osprey]

ok, gotcha

[gray dust]

also you don’t need to normalize in the GS process

[icy osprey]

ok

[quartz compass]

I abnormalize

[icy osprey]

can i normalize if i want to?

or will that be incorrect

it is more work

[gray dust]

i did say orthonormal is also orthogonal

[icy osprey]

so we good to normalize!

[gray dust]

i wont stop you

[icy osprey]

FIGHT ME

😆

[pallid rampart]

Why do work when you can avoid work

[icy osprey]

what is up whoever!

you helped me a few days ago

[pallid rampart]

Ok

[icy osprey]

ya, that is about it

😦

[pallid rampart]

I don’t remember

[icy osprey]

story of my life

my dad doesn't remember me either

[pallid rampart]

Damn

[grizzled jasper]

F

[icy osprey]

Assistance very much necessary

[gray dust]

write a 3x3 example

[icy osprey]

ok let me come up with one

lets do... [1 5 2;6 9 4]

sorry that it is in matlab form

then I will do A^tA

oops i need another row

[gray dust]

make the cols mutually orthogonal

[icy osprey]

so something like this would work?

ok now i did A^tA

[icy osprey]

nvm i think i figured it out

[junior agate]

Describe geometrically (linear, plane or 3-space)
(1,0,0) and (0,2,3)
As per me it should be 3-space
2,3 in y-z plane and 1 will give the 3-space
But the book answer says it's a plane. How? 👀

[half ice]

Are they asking you about the two vectors, or are they asking you about the space that the two vectors span?

Each vector is in three space, yes. Together they span a 2D subspace @junior agate

[steady fiber]

ya, it's a 2D subspace embedded in 3-space

[wintry steppe]

whats the difference between a module and a vector space?

is it just that one is over a ring and another is over a field?

[sonic osprey]

Yes

But it's better to think of vector spaces as being modules

Since fields are special types of rings

[wintry steppe]

ok

[quartz elk]

Am I being slow or should the rotation not be in the opposite direction

[uneven wadi]

damn im in my first semester of linear rn and reading this chat makes me feel even more out of touch with the subject rip

[bitter glade]

Don't worry about it, linear algebra is like the first abstract class

a lot of the ideas sound very foreign at first, but if you haven't yet i highly reccommend 3blue1browns essence of linear algebra videos @uneven wadi

[uneven wadi]

yeah ive watched that a lot, he makes great videos. he really helped me out back in the day when learning calc

idk when discussing the concepts i feel like i understand but when it comes to doing a proof or problem on my own its rough haha

ik i just need more practice, linear just aint my thing so its not coming as quickly as other math

[shadow drift]

They asked me suppose A is a 3x5 matrix. How many solutions are there for Ax = 0 (which is homogenous)

And I said "zero or infinite", depends on the matrix

But apparently the answer was "infinite"

[slow scroll]

well its not zero. there is always at least one solution

[wintry steppe]

yea

[shadow drift]

oh yea thats what i mean lol

arent homogenous either 1 or infinite

[wintry steppe]

no

but if the number of rows is less then the number of colums

there is always a solution

[slow scroll]

*non trivial solutions

[shadow drift]

so if number of rows is less than columns, its infinite?

[wintry steppe]

yeah

[shadow drift]

So 3x5, Ax=0 has infinite solutions?

I would guess either only or infinite?

[wintry steppe]

yes because the rank of A is 3

which means the null space has dimension 2

well actaullly

*less than or equal to 3

[slow scroll]

well you cant saying anything in particular about rank and nullity

yea

[wintry steppe]

so the null space is greater than or equal to 2

which implies there are infinite solutions

[shadow drift]

Oh rip, so there are 2 extra trivial solutions

And infinitely more

[wintry steppe]

non trivial

[shadow drift]

Rip, so if my thinking is correct.
There is 1 trivial

And

Since we have essentially 2 free variables

We now have infinite

Because of the two empty rows

[wintry steppe]

at least 2 free variables yes

[shadow drift]

big rip, cost me 1 point on exam.

well thanks pals

now that clears it up

[viscid vale]

(both pages are the same problem, 2nd pic is just a continuation)

so im trying to diagonalize this symmetric matrix and somewhere along the way i think either my normalizing or my forming an orthogonal eigenbasis step went wrong (my eigenvectors are for sure correct)

but i have no clue what went wrong, because each step makes sense to me here

[wintry steppe]

quick question, a linear map between $\wedge^m V \rightarrow \wedge^m V$ is the same as a multilinear alternating map between $V\times ... \times V \rightarrow \wedge^m V$

[stoic python]

David Kubin:

[long blade]

hello

i just started learning about vector space

and im still struggling understanding the aspects about it

if someone could guide through me this question

it would be helpful

[dusky epoch]

ok well

[long blade]

heey

Ann

i talked to u yesterday

[dusky epoch]

you're gonna want the list of axioms handy again

[long blade]

thx for ur help before

kk

1 sec

[dusky epoch]

bc this is, once again, a matter of going through them all

[long blade]

kk

https://cdn.discordapp.com/attachments/540211747613704221/698391274016866355/fsafsdf.jpg

im a bit confused as to why theres 2

equations

[dusky epoch]

what do you mean

[long blade]

the x + y = xy

and k * x = x^k

are they

[dusky epoch]

those are the definitions of $\oplus$ and $\odot$

[stoic python]

Ann:

[long blade]

okay

is it a rule?

[dusky epoch]

what do you mean

[long blade]

u said definitnio

like

is

[dusky epoch]

no it's not a general rule.

[long blade]

x + y = xy

a rule

??

[dusky epoch]

these are specific to this one problem

[long blade]

ohh kk

[dusky epoch]

and do not exist outside it

[long blade]

cool

kk

https://cdn.discordapp.com/attachments/540211747613704221/698391274016866355/fsafsdf.jpg

[dusky epoch]

so axiom 1, closure under funky addition

[long blade]

im just posting the question so we dont need to scroll up

[dusky epoch]

axiom 1 boils down to "if x > 0 and y > 0, then xy > 0"

is that clear

[long blade]

yeah

i understand

that

but um

[dusky epoch]

and it's just as clear, then, that this axiom is satisfied

[long blade]

why do we assume

that y > 0-

y>0

[dusky epoch]

what if i said "if $x \in V$ and $y \in V$, then $x \oplus y \in V$"

[stoic python]

Ann:

[long blade]

yeah

[dusky epoch]

V here is defined as the set of positive real numbers

[long blade]

yeah

[dusky epoch]

y ∈ V

[long blade]

oh

lol

[dusky epoch]

means y is a positive real number

[long blade]

yeah

i didnt

see that

kk cool

i understand

stupid question

[dusky epoch]

ok

[long blade]

https://cdn.discordapp.com/attachments/540211747613704221/698391274016866355/fsafsdf.jpg

[dusky epoch]

axioms 2 and 3 boil down to the commutative and associative laws respectively for real number multiplication

is that clear

[long blade]

yeah

[dusky epoch]

ok now the zero vector

this is the interesting part

if our "funky addition" is actually multiplication, what's the identity element

i.e. what can you funky-add to something to leave it unchanged

[long blade]

0

?

[dusky epoch]

no

the real number 0 isn't even an element of V

[long blade]

ohhh

1

[dusky epoch]

it's not a positive real number

yes 1

[long blade]

wait

why did we think it was multiplcation

a bit confused

https://cdn.discordapp.com/attachments/540211747613704221/698391274016866355/fsafsdf.jpg

[dusky epoch]

$x \oplus y := xy$...

[stoic python]

Ann:

[long blade]

yeaah

where did the identity

come in

[dusky epoch]

$1 \oplus x = x$

[long blade]

like why did we suddenly

[stoic python]

Ann:

[dusky epoch]

so the real number 1 is actually the zero vector here.

[long blade]

yeah

wait

what does the zero vector

represent

exactly

like im confused

how

[dusky epoch]

the zero vector is the vector that when added to anything else in the vector space leaves that thing unchanged.

where "added" is to be understood in terms of the operation that the vector space calls addition.

[long blade]

okay

https://cdn.discordapp.com/attachments/540211747613704221/698391274016866355/fsafsdf.jpg

[dusky epoch]

is it clear that funky-adding 1 to something leaves it unchanged

[long blade]

yeah

it does

[dusky epoch]

great

so now axiom 5

additive inverses

what can you multiply x by to get our zero vector, a.k.a. 1

[long blade]

um

x?

i mean

x ^-1

[dusky epoch]

x^-1 there we go

you've also answered the two additional questions in the problem along the way

[long blade]

ohh

[dusky epoch]

okay so can you check the other five axioms on your own? they will each boil down to a law of real number algebra so it should not be terribly hard

[long blade]

in the

axioms

it said

that

u + (-u) = 0

but its not actually - u

were meant

find the vector that multiply it becomes teh zero vector

[dusky epoch]

the - is not to be interpreted as the real number additive inverse

and the + sign should, in your case, be replaced by the funky plus sign that the problem used

[long blade]

okay

just one thing

u konw the second equation

um

k * x = x^k

what were we meant to use this for

we didnt seem to use this equation did we

[dusky epoch]

well we didn't get to any of the scaling axioms yet.

so of course we didn't talk about scaling at all.

[long blade]

ohh okay

ty

for ur help

much appreciated

@dusky epoch i have one question

u know in the axioms

they bring in um other vectors

for example

w

how to check those

[dusky epoch]

wdym

[long blade]

axioms

[dusky epoch]

all the boldface letters in the axioms denote vectors

[dusky epoch]

arbitrary vectors from your vector space

[long blade]

in axiom 3

it says a W

[dusky epoch]

there is nothing special about the name w

[long blade]

yeah i just mean

[dusky epoch]

it's just another arbitrary vector from the space

[long blade]

ohhh okay

kk

cool

ty

[fading lily]

Let's say we have a linear map $$T: V \rightarrow V$$

[stoic python]

Circulation:

[dusky epoch]

uh huh

[fading lily]

give me a moment i need time to formulate this coherently lol

[dusky epoch]

ok

[fading lily]

We also have the linear map $$P:V \rightarrow V$$.

[stoic python]

Circulation:

[fading lily]

Denote the null space of T and P as $$N_T, N_P$$ respectively

[stoic python]

Circulation:

[fading lily]

Let the set $$T(N_P) = {T(v) : v \in N_P}$$

[stoic python]

Circulation:

[fading lily]

I want to prove that $$dim(T(N_P)) \leq dim(N_P)$$

[stoic python]

Circulation:

[fading lily]

what i actually want to show is that $$dim(N_T) = dim(N_P)$$ but this is one part of the puzzle

[stoic python]

Circulation:

[dusky epoch]

$\dim T(W) \leq \dim(W)$ is true no matter what kind of subspace $W$ is

[stoic python]

Ann:

[dusky epoch]

just pick a basis of T(W) and pull back along T to get a LI set in W

in your case W = N_P aka ker(P) i guess

[fading lily]

what does pull back along T mean?

let's define a basis of T(W) to be ${v_1, v_2, \cdots, v_n}$

[dusky epoch]

ok that was a bit of a fancy turn of phrase where i pretended to be an algebra professor

[fading lily]

how would i go from there

[dusky epoch]

"the" basis

[fading lily]

lmao

[dusky epoch]

a space has many bases

[fading lily]

a basis

[dusky epoch]

anyway

[stoic python]

Circulation:

[dusky epoch]

for each i between 1 and n, let w_i be such that Tw_i = v_i

{w_1, ..., w_n} will then be LI in W

[fading lily]

yes

but how do i know that that is a basis?

i dont know the dim of W

or N_T in this case

[dusky epoch]

you don't need to know that it's a basis just to prove dim(T(W)) ≤ dim(W)

[fading lily]

u right

basis is like the lower bound

the smallest li set

[dusky epoch]

also, if all you have are two arbitrary linear transformations T and P with seemingly no relation between them, then proving dim(ker(T)) = dim(ker(P)) will be impossible

no, a basis is a maximal LI set.

[fading lily]

what does maximal mean

{w_1, ..., w_n} will then be LI in W
@dusky epoch also why would this be LI exactly?

[dusky epoch]

why don't you try to prove it yourself lol

[fading lily]

because if it wasnt LI then ${v_1, \cdots, v_n}$ wouldn't be LI?

[stoic python]

Circulation:

[dusky epoch]

\{ ... \}

but yes that's precisely why

[fading lily]

ok it took me a hot minute lol

and we aren't sure if ${w_1, \cdots, w_n}$ generates $N_T$

[stoic python]

Circulation:

[fading lily]

but $n \leq dim(N_T) $

[stoic python]

Circulation:

[fading lily]

oh i get why u said maximal

[wintry steppe]

yoooooo I need help rn my h.w is due in 13 minutes

and this stupid discord made me wait 10 minutes to talk

is there quick easy way to do this

[pallid rampart]

Just, use the definition

[wintry steppe]

not even sure what C[a,b] is

could you explain

[pallid rampart]

Set of continuous functions on [a,b]

[dusky epoch]

C[a,b] is the vector space of all continuous functions on [a,b]

[wintry steppe]

oh

so the area of that is zero

is it sin or something

[dusky epoch]

you're overthinking it

you need to prove that if int[a,b] f(x) dx = 0 and int[a,b] g(x) dx = 0 then int[a,b] (c1f(x) + c2g(x)) dx = 0

for any scalars a and b

[wintry steppe]

so uh

[dusky epoch]

don't overthink it

[wintry steppe]

how about you show me what to do, i turn it in, then u explain it

clock is ticking

[dusky epoch]

do you know anything at all about integrals

seriously

like

integration is linear

[wintry steppe]

yes

[dusky epoch]

okay bad choice of constants

[wintry steppe]

so g(x) is int[a,b] of f(x) for what u just said

and u can just keep doing it

[dusky epoch]

no

no no non onon on on on ono no no

[wintry steppe]

g(x) is just another continuous function

[dusky epoch]

$\int_a^b (c_1f(x) + c_2g(x)) \dd{x} = c_1 \int_a^b f(x) \dd{x} + c_2 \int_a^b g(x) \dd{x}$

[stoic python]

Ann:

[dusky epoch]

g(x) is a function WHOSE INTEGRAL OVER [a,b] IS ZERO

bruh

[wintry steppe]

oh

[dusky epoch]

were you not able to arrive here on your own

[wintry steppe]

the problem never mentioned g(x)

[dusky epoch]

did you even read what i wrote.

you need to prove that if int[a,b] f(x) dx = 0 and int[a,b] g(x) dx = 0 then int[a,b] (c1f(x) + c2g(x)) dx = 0

[wintry steppe]

it was hard to decipher

[dusky epoch]

how

[wintry steppe]

i like the bot format better

this makes sense

[dusky epoch]

you need to prove that \textbf{IF} $\int_a^b f(x) \dd{x} = 0$ and $\int_a^b g(x) \dd{x} = 0$ \textbf{THEN} $\int_a^b (c_1f(x) + c_2g(x)) \dd{x} = 0$ for any scalars $c_1, c_2$

[stoic python]

Ann:

[dusky epoch]

there. happy?

[wintry steppe]

yes, thanks

[fading lily]

is this a logically sound proof to the question i was asking earlier?

Let the basis for $T(N_P)$ be defined as
[\alpha = {w_1, w_2, \cdots, w_n}]
Let $v_i \in N_P$ be such that $T(v_i) = w_i$. Define
[\beta = {v_1, v_2, \cdots, v_n}]
Notice that $\beta$ is a linearly independent set in $N_P$.
This is because if $\beta$ was not linearly independent, then $\alpha$
would not be not linearly independent. Furthermore, $|\beta|= n \leq dim (N_{P})$
because the basis for $N_{P}$ is a maximal linearly independent set.
Therefore, $dim(T(N_{P}))\leq dim (N_{P})$.

[stoic python]

Circulation:

[fading lily]

@dusky epoch

[dusky epoch]

h

yeah ig that works

[fading lily]

lol you dont sound very convinced

[dusky epoch]

i don't feel like wading through jargon rn. i'm too tired for that

[fading lily]

okay thanks for the help 🙂

[agile wolf]

What matrix transformation would offer this result?

[quartz compass]

set a=1 and the rest to 0

then if you imagine the matrix multiplication, that will just give you the first column

walk through it a bit, write it out and see for yourself

[agile wolf]

Alright thanks

One question, will the matrix have components that use the variables a,b,c,d,e?

oh nevermind

Could I get a tip for solving this problem? I'm not sure how a matrix can manipulate the X variable, rather than just the coefficients

[quartz compass]

expand it out

so that you have coefficients on 1, X, and X^2

[agile wolf]

ah I see

thanks

[quartz compass]

just try stuff first

[spiral sonnet]

How can I tell if a matrix is one to one and on to?

[keen mirage]

using the definitions of one-to-one and onto

and you say the linear transformation represented by the matrix is 1to1/onto

a matrix is just an array of numbers

[spiral sonnet]

okay so I think I know how to tell if it's one to one, but I'm not sure exactly what on to is

[keen mirage]

onto means everything in the codomain has a preimage

say T: R^n -> R^m is our function, then T is onto if for every v in R^m, there exists a u in R^n such that T(u) = v

[viscid kernel]

So as I was rewatching the series of " the escence of linear algebra the second time ". I kind of got confused at some point.

https://www.youtube.com/watch?v=BaM7OCEm3G0

Why do those coordinates P1 P2 and P3 correspond with the unitvectors ?

yall can watch from 8:05

[dusky epoch]

are you asking why these equations are true

[viscid kernel]

yup

I knew the answer but I forgot since its been a long time I watched that video

[dusky epoch]

$p_1x + p_2y + p_3z = (v_2w_3 - w_3v_2)x + (v_3w_1 - v_1w_3)y + (v_1w_2 - v_2w_1)z$ is an equality meant to hold for \textbf{ALL} possible values of $x, y$ and $z$

[stoic python]

Ann:

[dusky epoch]

so in particular you can set x=1, y=0, z=0 and get the first equation

x=0, y=1, z=0 to get the second

and i'll leave it to you to see what values of x, y and z will give you the third equation directly in the same way

[viscid kernel]

ur my hero

@dusky epoch do you know any simulation in which I can like create vectorspaces ( like 3blue1brown ) in a 3D dimensional space ?

[dusky epoch]

uhh

you mean his illustrations?

i think he put out a library or something called manim

but i've never used it so idk

[viscid kernel]

Hmm

[slender yarrow]

yeah it's on github https://github.com/3b1b/manim

[viscid kernel]

@slender yarrow thanks

Hmm I clicked on that link, everything looks different. Idk what to do? @slender yarrow may you help me out ?

[slender yarrow]

i never used it myself so i don't really know either

but yeah you gotta code the scenes yourself

[limber sierra]

it's a github page

it includes installation instructions, but assumes youre familiar with programming stuff

it relies on python, conda, and latex iirc

[viscid kernel]

I can only code in python

I dont know if its enough

[slender yarrow]

well the lib is for python

so you're lucky

[viscid kernel]

lib ?

[slender yarrow]

the library, manim

[viscid kernel]

aight

[smoky lagoon]

is the dimension of the corresponding set of eigenvectors for a certain eigenvalue just the number of free variables if i were to set up a matrix using each eigenvector?

so if i had vector (1,0) and vector (0,1) the dimension would be 2?

[smoky lagoon]

figured it out realized i already knew the answer

[half ice]

I get that e_i is a list where each component is a basis vector of V.

In the lemma they say for I = J and I have zero idea what equality for a multi-index notation means

a^I is the wedge of every member of the dual basis. So a^I seems to be like it can only be one thing. But they go on to say there's many of these?

[covert skiff]

woah, that looks like some pretty difficult linear algebra. Here I am, can't figure out how to find out a determinant of a 4*4 matrix 🤦‍♂️

[half ice]

The course is really differential geometry, but this part is all vectors

[covert skiff]

lol, i hate vectors tbh

[half ice]

Looked on Wikipedia, and it is able to state what the basis is in a way that was clear and made sense to me. I still don't understand what this part of the book is trying to say. How horrible.

[grizzled jasper]

can someone help quick times

[storm python]

sounds like an exam

[grizzled jasper]

shhh

[grizzled jasper]

its exam review

i got it anyways

[dry spear]

im trying to take the determinant to find the eigenvectors

but i can't find anything that simplifies it

other than a 0

then i just get a huge expression

[quaint marlin]

Lol

[covert skiff]

Can someone tell me if this is in REF?

Ah, should've rotated it

My b

Row echelon form

i want to find out the determinant, by doing REF

[loud flame]

No, you can't have non-zero on the bottom row except for the number farthest to the right

[covert skiff]

hmm

how do i make it that? I can't think of a way

and i've been told that 4 in the first row needs to be 1

I can do R1-4R4 for that

but can't think of a way about row 4

[loud flame]

Some say that the first number of each row must be 1, but I'm not sure if this is an official requirement

You have to use multiple operations

R1-4R4, then newR4-5R2, then newnewR4-R2/2

for example

[covert skiff]

I said it rong, if we do R1-4R4, the first element of the matrix becomes 0

why R4-5R2?

[loud flame]

Do all that for the last row

to get 0 on the first three places on that row

To get 1 one the first row, you can just divide it by 4

[covert skiff]

all three for R4?

[loud flame]

yep

[covert skiff]

Alright. It's just, I can't wrap my head around this. I can do like max 2 calculation in my head

Ok, I'll do that rn, and let you know how it goes. Thanks for the help

[loud flame]

Just write them down, takes some time but worth it in the end

np

[covert skiff]

I don't get R4-R2/2

I've got 3 6 -1 18 in my last row

[loud flame]

what did you get after R1-4R4?

for that row

[covert skiff]

1 2 0 1

for my last row

no, my bad

3 11 9 -7

and then i did R4-5R2, after which i got 3 6 -1 18

[loud flame]

You need to remember to multiply the last row with 4 before you subtract it from the first row

[covert skiff]

0 5 9 -10 is what I get after doing that for the fourth row

[loud flame]

that seems right

Now you have to find a way to get the second number of that row to be 0

[covert skiff]

R4-5R2

[loud flame]

that's right

[covert skiff]

i get 0 0 -1 -60

now, need to get rid of -1

[loud flame]

yep

[covert skiff]

R4-R3/2?

r3 i have 0 0 -2 10

[loud flame]

yes, but look out for the sign on your last number during the operations

It seems like you mixed up the 3rd and 2nd row for the last number during the R4-5R2 operation

[covert skiff]

hmm, i get +15 now

Nicely spotted 👏🏻

[loud flame]

In this exercise you will get the same result either way as the last row will be reduced to one number in the end. But this is not usually the case, it is very easy to do small mistakes like that

[covert skiff]

yeah, tbh I hate REF, and RREF

thanks for the help. I made another mistake but caught it 🙂

[loud flame]

nice

[covert skiff]

So, I've got another question that I can't seem to solve 😦
I've found that the determinant of I2 = 1; -3I2= -3
Determinant of I3 = 1; -3I3 = -3
Where I is the Identity matrix
can you conclude about det (−3𝐼n) an, in general, det (𝑘I𝑛) for any 𝑘ER ?

kER = k is an element of real number

[dry spear]

anyone here good with eigenvectors/values?

[half ice]

@covert skiff
There's definitely a pattern in those four examples

[dry spear]

oh sorry ill wait

my bad

[covert skiff]

There's a pattern, like whatever you multiply it by the answer is the same what you'll get

[half ice]

Wait, check your determinants again

@dry spear
Go ahead I'm good with it

[dry spear]

ok

If $A\vec{x}=\lambda\vec{x}$, then is $A=\lambda$?

[stoic python]

Spectex:

[half ice]

A is a matrix, λ is a scalar

[dry spear]

wait no

lambda*identity=A

is that true?

[half ice]

No, usually not

x is a vector and has no inverse, so you can't "divide both sides by x"

[dry spear]

ok

so i have a question then

how would I do this

[half ice]

"If λ is an eigenvalue of A"
Can you write that as an equation?

[dry spear]

$A\vec{x}=\lambda\vec{x}$

[stoic python]

Spectex:

[dry spear]

is that what you were asking?

[half ice]

Yes exactly!

"Then 2 - 3λ + 5λ² is an eigenvector of 2I - 3A + 5A²"

Is the statement
(2I - 3A + 5A²)x = (2 - 3λ + 5λ²)x

You have the first statement, you want to prove the second one is true

[dry spear]

yeah

so how would i prove that though

replace lambda with the expression?

[covert skiff]

kaynex @cosmic dune when you're done helping spectex. Sorry for interrupting

[dry spear]

oh i think i got it

@covert skiff ur good to go

[covert skiff]

Hmm, so I got 9 as the determinant when I do -3I2

For -3I3 i don't get it. Since it is both an upper and lower triangle, how do we calculate the determinant then?

Can we do, the product of the diagonal?

For -3I3:
-3 0 0
0 -3 0
0 0 -3

How to calc det? And I can't figure out what the pattern could be

[loud flame]

https://www.khanacademy.org/math/algebra-home/alg-matrices/alg-determinants-and-inverses-of-large-matrices/v/finding-the-determinant-of-a-3x3-matrix-method-1

Shortcut for calculating determinants of 3x3s

[dry spear]

im kinda confused about the last eigenspace

there's no eigenvalue for it

[slow scroll]

im not really sure what this question is asking, but to compute the last eigenvalue, you can just use the fact that the sum of eigenvalues equals the trace of the matrix

[dry spear]

yeah i figured i had to compute it

I just showed that all of the given eigenspaces are eigenspaces for A

for this one how would I find C, if there is a C that holds this statement true

[covert skiff]

@loud flame ok, I got it. So, I get the det = -27

Now, can you tell me the pattern?

For even number it's the square which results in a positivie (for I2), for odd numbers its the exponent to the number it is (for I3).

Uhh, too confusing. Can someone tell me some clear explanation?

[limber sierra]

the determinant of a triangular matrix is the product of its diagonal entries

that's the whole pattern

note that this only works for triangular matrices (otherwise calculating determinants would be much easier)

but of course, scalar multiples of the identity matrix are always triangular

[dry spear]

waxwing could you help me with mine?

[limber sierra]

do you have any theorems on similarity?

there are a lot of theorems that make your life easier

but if you havent seen those yet, then just

set up a system and reason on what the values must be

[dry spear]

the only thing we learned about this was how to diagonalize a matrix and find the other ones

but B isn't a diagonlized matrix

[limber sierra]

then yeah, you can suppose a $C = \begin{pmatrix}c_1&c_2&c_3\c_4&c_5&c_6\c_7&c_8&c_9\end{pmatrix}$ exists

[stoic python]

Namington:

[limber sierra]

and then figure out what the system $A = CBC^{-1}$ would be then

[stoic python]

Namington:

[dry spear]

the C^-1 is throwing me off though

[limber sierra]

(hint: multiply by C on the right to make your life easier)

[dry spear]

oh

so CA=CB?

[limber sierra]

on the right

order matters with matix multiplication

[dry spear]

how do you know if it goes on the right or left

[limber sierra]

$A = CBC^{-1}$ implies $AC = CBC^{-1} C = CB$

[stoic python]

Namington:

[limber sierra]

because your C and C^{-1} don't cancel out

otherwise

[dry spear]

ohh

so if it was $C^{-1}BC = A$, then $CA=BC$?

[stoic python]

Spectex:

[limber sierra]

right

[dry spear]

ah ok

so then multiply out all the entries and see if the equations hold true?

[limber sierra]

yeah, well

more like: see if the system is solvable

[dry spear]

yeah

ok makes sense

ty

[limber sierra]

[and note that, if a solution matrix for C exists, it needs to be invertible]

[dry spear]

yeah

there was no solution

:)

[gloomy arrow]

Because it has to be less than or equal to

[raven tiger]

anyone know anything about the union and intersection of intervals?

[slow scroll]

what does that have to do with linear algebra lol?

[raven tiger]

i dont know which one to ask it to so i just keep clicking algebra lol

[slow scroll]

ill ping you somewhere else.

@raven tiger go to #help-1

[pallid rampart]

Correct

The theorem is: if V is a finite dimensional space and H is a subspace, then H is finite dimensional with dimension less than or equal to the dimension of V

[gloomy arrow]

Is there a quick way to do this one?

There are 8 pivot columns right?

So would the dimmension bew 8

[half ice]

Yes

[pliant fox]

<@&286206848099549185> know how matrix inversion can be done with series of matrix multiplication? is there a way to get matrix transposal via matrix multiplication/addition, or is it not possible?.

[steady fiber]

first you say inversion, then you say transpose

pick an operation first

also don't ping before 15 min

[pliant fox]

transposal. can it be obtained with matrix multiplication?

[warm stirrup]

Think of it as a linear transformation

[pliant fox]

?

[gaunt geyser]

hey I don't get their little statement about assuming A is symmetric here

they say "if not, we can replace A by (A+A')/2"

isnt this only true if A is symmetric

[dusky epoch]

they aren't claiming that A = (A+A')/2

they're claiming that $x'Ax = x' \paren{\frac{A+A'}{2}}x$

[stoic python]

Ann:

[gaunt geyser]

hmm how does that work out if A =/= (A+A')/2

[quartz compass]

look at x'Ax as a matrix

what are its dimensions m by n

[dusky epoch]

$x' \paren{\frac{A+A'}{2}}x = \frac12 (x'Ax + x'A'x) $

[stoic python]

Ann:

[gaunt geyser]

@quartz compass it should be a single equation right?

[quartz compass]

?

x'Ax is a matrix

what are the dimensions of this matrix?

it's an m by n matrix

what's m, what's n?

[gaunt geyser]

I mean the dimensions of A are nxn

I believe x would then be nx1

and x' would be 1xn

[quartz compass]

and so their product is a matrix

[gaunt geyser]

so x'*A would have dimensions nx1

and x'Ax would be 1x1

or a single equation

[quartz compass]

it's an expression

there's no equal sign

it's not an equation

[gaunt geyser]

oh yeah

[quartz compass]

so are 1x1 matrices symmetric?

[gaunt geyser]

yes

[quartz compass]

x'Ax = (x'Ax)'

must be

now A is not A' necessarily

but x'Ax = (x'Ax)' is

now work through (x'Ax)' by rules of transpose to try to simplify that a bit

[gaunt geyser]

there I just get (x'Ax)' = xA'x'

[quartz compass]

no not quite

(BC)' = C'B'

you gotta change the order

[gaunt geyser]

oh shoot right

so x'A'x = (x'Ax)'

[quartz compass]

yeah

and since that thing on the right is x'Ax

we have x'Ax=x'A'x

so now we can add it to itself since it's the same thing, and divide by 2

[gaunt geyser]

ah I see

thats just really weird intuitively

that any square matrix in linear form can be assumed symmetric

[quartz compass]

the end result being a scalar (1x1 matrix) is why we can do this

multiplying by x and x' is basically "erasing" any kind of extra information about the matrix's structure

since it's the same vector multiplying both sides

[gaunt geyser]

right

[quartz compass]

like if you look at the result, it's gonna be some quadratic form like

Ax^2+Bxy+Cy^2

which would come from a 2x2 matrix which has 4 entries

but here there are only 3, which correspond to the fact again

just slightly different perspective on it

err here A,B,C are scalars and x,y are scalars as components of the vector x earlier

maybe I overloaded the notation a bit too much my bad

[gaunt geyser]

yeah I just did the multiplication for a 2x2 matrix [[1,2],[3,4]]

because the 2 and 3 end up 2xy+3xy

[quartz compass]

yeah

[gaunt geyser]

when you take (A+A')/2

its just 2.5xy+2.5xy

[quartz compass]

it's nice because now this gives an actual reason to care about symmetric matrices

they're not just "looks nice when flip flop"

it will turn out symmetric matrices have real eigenvalues with orthogonal eigenvectors

and you can then do stuff like diagonalize the matrix representing your quadratic form so that you can get a change of basis so your quadratic form has no cross terms, it'll just be like Ax^2+By^2

and this sort of thing will generalize to the complex case with hermitian matrices and so on

[gaunt geyser]

ah I see

I took a linear algebra class that was very calculation based

running through this textbook

http://www.janmagnus.nl/misc/mdc-ch18.pdf

as a more fleshed out overview of linear algebra

[quartz compass]

sure looks good

[gaunt geyser]

any intuition on why B*B' or the reverse in quadratic form is always positive semidefinite

been playing with the math for a bit now

with a 2x3 matrix B

[stoic python]

gREEnOnions:

gREEnOnions:

[gaunt geyser]

where

[stoic python]

gREEnOnions:

[gaunt geyser]

Here B can be negative

a and c are positive

so I am trying to figure out why

[stoic python]

gREEnOnions:

[gaunt geyser]

for negative b

I've found this

but don't know why the left hand side is always true

right*

[dusky epoch]

presumably A^Tx is known not to be the zero vector?

[gaunt geyser]

why can't it be less than zero

[dusky epoch]

why can't what be less than zero

(A^Tx)^T (A^Tx)?

[gaunt geyser]

yeah

[dusky epoch]

it's equal to $|A^Tx|^2$

[stoic python]

Ann:

[gaunt geyser]

oh I see

[agile wolf]

does the linear transformation that reflects a vector in a 3d plane have eigenvectors?

[dusky epoch]

sure

any vector in the plane is an eigenvector with eigenvalue 1

and any vector perpendicular to the plane is an eigenvector with eigenvalue -1

[agile wolf]

So would the transformation have eigenvalue of only -1?

[dusky epoch]

reread what i just wrote

[agile wolf]

Oh. Correct me if I'm wrong the eigen value of 1 is just like a 2d reflection? I'm a bit confused by that

[dusky epoch]

the eigenvalue of 1 is an eigenvalue of 1.

[agile wolf]

Oh nevermind I was confusing myself

tyty

Can I say Tv = T x e1+T y e2? Where T is some transformation matrix

Or does this only hold for certain transformations

[dusky epoch]

is T a linear transformation

[agile wolf]

yeah this is the entire problem

[dusky epoch]

well

$T(xe_1 + ye_2) = T(xe_1) + T(ye_2)$ is simply an instance of the DEFINITION of linearity, is it not?

[stoic python]

Ann:

[hollow finch]

How can I prove that for any real nxn matrix A, the matrices I,A,A^2,... are linearly dependent? This is apparently possible without knowledge of eigenvalues/the characteristic equation.

[dusky epoch]

I, A, A^2, ...

how long does this list continue

[viscid kernel]

This is the first dimension theorem. I was able to prove this in 3D, but the amount of basis vectors tell me that this is also valid in 3 + n dimensions. Why can you assume that this is also valid in 3 + n dimensions ?

[hollow finch]

@dusky epoch It is not specified. This is the problem. It is from an Abstract Algebra textbook, but I think this is more of a linear problem than an abstract one.

[dusky epoch]

well

$\bR^{n \times n}$ is a vector space of dimension $n^2$

[stoic python]

Ann:

[dusky epoch]

so ${ I, A, A^2, \cdots, A^{n^2+1} }$, a set containing $n^2+1$ matrices, is guaranteed LD.

[stoic python]

Ann:

[hollow finch]

Well

That's a lot easier than I expected

It did not specify a degree so yeah if we have a set with more elements than the dimension of the space the set cannot be LI.

Thanks

[wintry steppe]

I, A, A^2, ..., A^{n^2+1} is n^2 + 2 matrices

you can just do I, A, A^2, ..., A^{n^2} as well i suppose

[hollow finch]

As well as A^{n^n} too, but yeah up to A^{n^2} would be the minimum to guarantee LD.

[dusky epoch]

oh. yeah

silly me

still guaranteed LD tho.

[barren void]

$\pi = {(x,y,z)\in \mathbb{R}^{3} : x - y - 1 =0 }$

[stoic python]

Tomás Sentagne:

[barren void]

I have this implicit equation of a plane and I need to find an lets sey explicit function z = z(x,y) wich graph is the plane

$z = z(x,y)$

[stoic python]

Tomás Sentagne:

[barren void]

Any idea?

[vital swallow]

but z is a free variable there. e.g. the points (2,1,0) and (2,1,5) both satisfy the implicit equation

so z = z(x,y) would not be a function

[barren void]

All right thats what I was thinking

There must be something wrong in this excercise

[vital swallow]

normally you would just solve for z

ax+by+cz = d --> z = (d-ax-by)/c is your function

[barren void]

Thats right

[wintry steppe]

I need help understanding if I'm doing solving one of my hw questions properly

They want us to determine if a given set is a subspace of Pn for an appropriate value of n. (I don't really understand what they mean by an appropriate value of n)

I think they're talking about polynomials when they say a a given set is a subspace of P sub n

The question gives us an example...

Oh, hello kevypoo

how is this computed

I seen you in typology, sup

hello peepee

NinjaPeeps

y r u in doge server

Was just looking for an MBTI server to chill in

uh, going back to Linear Algebra. They gave us an example that reads "All polynomials of the form p(t) = at^2, where a is in R"

[gloomy arrow]

8 because 8 pivots?

[coral tinsel]

what's the relation between Simplex, Gaussian Elimination, and LU Decomposition?

Are the first two the same while the last is one possible algorithm to solve the first two?

feel free to ping me if you can help

[agile wolf]

What will a matrix for a linear transformation look like if the basis consists entirely of eigenvectors?

[dusky epoch]

entirely of the transformation's eigenvectors?

it'll be diagonal.

[agile wolf]

Like an identity matrix but instead of 1s theres the eigenvalues?

[dusky epoch]

uh

yes

[stoic python]

RestlessQ:

[dusky epoch]

??

[fathom flax]

I have tried - 70

And - 15

[gray dust]

show work for computing f(v)

[flint flicker]

https://youtu.be/kYB8IZa5AuE @covert skiff

I remember this video being really good

[magic light]

https://i.imgur.com/zIK49IC.png

What is the meaning of M(m x n) (R) in this context?

[flint flicker]

I'm pretty sure that means the set of all m by n matrices of real numbers

[ocean flame]

@covert skiff Do you speak any other languages than english? I believe I have pdfs in swedish regarding transformations

[quartz compass]

a couple nitpicks first

it's not equating it, it's making an augmented matrix

and this works for n*n not just 4*4

you can do the exact same thing without the matrices and with variables

it's very cumbersome but

what it basically amounts to is doing elimination on a system of equations

since you have both matrices what you're doing is kind of like taking your steps of how to turn a matrix into the identity matrix and at the same time doing those on an identity matrix

to give you the reverse steps

so that's what's giving you that [A | I] to [I | A^-1] trick

depends on what better means

computationally speaking, doing the row reduction is pretty much the fastest

although algebraically it's nice to have things like the cofactor and determinant stuff floating around

sometimes there are algebraic tricks to exploit the form that allows you to skip actually computing certain things

like a diagonal matrix, to invert that you just have to take the reciprocals of all the entries

just saying there are sometimes tricks, roughly like this or more complicated

yep

[vague fulcrum]

can anyone help me prove this identity

i cant seem to get it

im trying to prove that ker(BC) is a subset of ker(B) + ker(C)

so i let x be in ker(BC)

so (BC)(x) = 0

and i know i need to show that x = b + c such that b is in kerB and c is in kerC

but idk how

[dusky epoch]

is ker(BC) ⊆ ker(B) + ker(C) even true

[vague fulcrum]

that's the whole question

so according to my homework set, yea

i just wanna know why they say that in the hint