linear-algebra

[vast torrent]

depends on domain and codomain

[solar osprey]

How can we even asses the dimension of a transformation

[crystal pagoda]

isomorphic

implies?

[solar osprey]

Same dim

[vast torrent]

L(V,W) has dimension dim(V)dim(W) for finite dimensional vector spaces

[solar osprey]

My question has nothing to do with this btw lol

[vast torrent]

but yeah what Rixia said

[crystal pagoda]

yeah but it's easier to consider M mxn

[vast torrent]

since that L is a bijection

that answers your question for this case

[solar osprey]

To prove its onto

I just used the proof of part 3

[crystal pagoda]

yeah

that's fine

[vast torrent]

but in general $\dim \mathcal L(V,W) = (\dim V)(\dim W)$ for finite dimensional spaces

[stoic python]

gfauxpas:

[crystal pagoda]

finite dimensional is always nice...

[solar osprey]

Saying that any vector X can be written as a linear combination of the columns of some vector A

[crystal pagoda]

then you get to functional analysis and nothing is nice

[solar osprey]

Speaking of finite

The rank-nullity theorem

Stating that dim(V) = rank(L) + nullity(L)

[crystal pagoda]

what about it

[solar osprey]

Ive been thinking about how to prove it doesnt hold for infinite dimension

[crystal pagoda]

it does hold though?

it's just not very useful

[solar osprey]

It holds for infinite?

Ahh wait

[vast torrent]

you have to set up rules for addition of infinite numbers

[solar osprey]

If kernel(L) = {0}

And L: V —-> V

Then L is an isomorphism

[crystal pagoda]

yes

[solar osprey]

it doesnt hold for infinite dimension

[crystal pagoda]

ImL=V

[vast torrent]

if you say that $(+\infty) + (+\infty) = +\infty$ then okay things should be fine but it's not very yseful as Rixia said

[stoic python]

gfauxpas:

[crystal pagoda]

you would also have n+inf=inf

[vast torrent]

sure why not

[crystal pagoda]

again, not very useful

so i guess you would resort to cardinals

[vast torrent]

just don't try to subtract infinite from itself, that breaks things

in this naive system we just discussed

[crystal pagoda]

now of course you need axiom of choice to say that it holds

but i would say that's almost always assumed

[vast torrent]

if you don't assume a.o.c. then dimension isn't well defined for all spaces

[crystal pagoda]

finite dimensional vector spaces are still fine

infinite ones break to an extent

but back on the topic

basic idea of the proof is the same as the finite dimensional case

you consider a basis of ker T, a basis of imT

[solar osprey]

Lol

[crystal pagoda]

then the union of basis for ker T and basis for preimage of imT forms a basis for V

[solar osprey]

Part 5-6

[crystal pagoda]

infinite dimensional vector space breaks the implication that kerT={0} implies surjective

[vast torrent]

they tell you which spaces you should use for your coutnerexamples you just have to figure out what operators are linear that break it

do you know the notation

[crystal pagoda]

consider T{x_i}={x_i-1}

then consider T{x_1,x_2,..}

T is surjective but kerT is not {0}

[vast torrent]

wait who asked the question I lost track

[crystal pagoda]

hm?

[vast torrent]

that's the converse rixia

[crystal pagoda]

ah right

[vast torrent]

im saying who is helping the question asker and who is asking the question I lost track

[solar osprey]

yea but still

KerL = {0} is equivalent to

[crystal pagoda]

im trying to help

[solar osprey]

1-1

its a double implication

if its finite dimensional

wait wtf am I even saying

1:00 am

x (

[crystal pagoda]

ok let me clear my thoughts

[vast torrent]

do you know the notation R[x}

R[x]

[crystal pagoda]

part 3 and 4 tells you in the finite dimensional case, injective implies surjective

and the other way

but that's not true in infinite dimensional vector space

that should be enough to say that it breaks

use the left shift example i stated earlier

or right shift, if you so wish

ohhhhhhhhh hold up

i think i know what your professor is getting at

@solar osprey consider derivative and integration as linear operators

in R[x]

[solar osprey]

yes

yes

[tulip oasis]

does anyone know what the U is?

found it in the book:

Actually I will zoom in:

do I just treat U as B?

[crystal pagoda]

U is a basis

it should be stated

[solar osprey]

Im back to strike again

Let V be a vector space of dimension n. Show that V is isomorphic to R^n

[tulip oasis]

okay cool

[steady fiber]

if dim(V)=n, then it has a basis of n vectors, {x1, x2, ... , xn}

[tulip oasis]

can I show my work?

[steady fiber]

map those n basis vectors

to the standard basis of R^n

[solar osprey]

x1e1 + ....+ xnen

[steady fiber]

mapping one basis to another of the same size is an isomorphic map

you should probably word it better

[solar osprey]

Yes I should

[steady fiber]

but that's the essence of the solution

[fallow jolt]

How do you prove that Null(a) is orthogonal to Rowspace(a)?

[barren panther]

thats an interesting question

is the row space the same as the transpose of the column space

[steady fiber]

no

[tulip oasis]

For the question I posted, I picked #2 for the answer

sorry i'll let you finish

[barren panther]

sorry I meant column space of the transpose of the matrix*

[fallow jolt]

Row(a) = Col(A^T)?

[barren panther]

yrs

yes*

[steady fiber]

assume x in null(A), then Ax=0

let a_1, a_2, ..., a_m be the rows of A, then Ax=0 implies a_1 * x = 0, a_2 * x = 0, etc. So each vector in the basis of the row space is orthogonal to any vector in the nullspace

so Null(A) is orthogonal to Row(A)

[barren panther]

ok

and can I modify this slightly to say that the null space of the transpose is orthogonal to the column space?

[fallow jolt]

intriguing proposition...

[steady fiber]

ya

[fallow jolt]

I believe so! @steady fiber, do you concur?

[steady fiber]

null space of the transpose is orthogonal to the row space of the transpose, and row(A^T)=col(A), so ya

[barren panther]

ooooh

nice

ok that makes a lot of cents thank youu

[fallow jolt]

@steady fiber, what if instead of Row(a), we had col(conjugatetranspose(A))?

Would this still hold?

[steady fiber]

for a real matrix yes :^)

[barren panther]

what if there were complex numbers?

[fallow jolt]

in A?

[barren panther]

yea

[steady fiber]

ya idk, I'd have to think about that

[fallow jolt]

ooo, that tickles my brain...

[steady fiber]

idk my conj transpose properties

[fallow jolt]

this is conj transpose

[steady fiber]

ya I know what it is

just haven't used it in forever

definitely haven't used it in a pure math context in a hot minute

physicslife

nah I dont think it holds

[barren panther]

why dont you think so?

[fallow jolt]

It was on my homework a while ago, I do think it holds

They said "prove" not "disprove"

[crystal pagoda]

Consider it in an inner product space?

Then conjugate transpose is the adjoint operator

But that only applies to linear operators

Hmm

[solar osprey]

Good night

[barren panther]

goodnight

[crystal pagoda]

@fallow jolt i think i got it.

consider <x,A*y> for x in null A

<x,A*y>=<Ax,y>=0 since Ax=0

then <x,A*y>=0 for all x in Null A and y in V

[fallow jolt]

What is this < > notation?

[crystal pagoda]

inner product

then Null A is orthogonal to Im A*

but Col(A*) is Im A*

thus we are done

[tulip oasis]

I think I made a mistake in calculating an invertible matrix

Here's what I got:

actually nvm

here we go

I get this strange thing

k = l and k = -l

or k = +- l

so then I get an S matrix of just 1's

but I checked this on wolfram and I saw that it wasn't suppose to be this

Did I make a mistake in my computation?

[shrewd slate]

If T is a linear map from V to W, is it always possible to define a basis B = {v_1, ..., v_n} for V s.t
T(a_1 v_1 + ... + a_n v_n) = a_1 w_1 + ... + a_n w_n

∀a_j and for some w_j ∈ W

I’m asking for a proof I’m trying to cook up

I suspect the answer is yes, but I worry that I suspect that out of convenience for myself

[crystal pagoda]

yeah

you can do that

check your dimensions though

might not be a basis after the transformation

[shrewd slate]

You mean it might not be a basis for W?

I don’t mind that

If that’s what you mean

[crystal pagoda]

yeah

[shrewd slate]

Epic, thanks

[crystal pagoda]

you can always split a vector into a linear combination of basis vectors

and T is linear, so everything is convenient

[shrewd slate]

So it works out for all a_j, that’s p neat

[crystal pagoda]

a_j are scalar so they just get pulled out anyway

[shrewd slate]

Ahhh

[slow scroll]

you want to show that if Tx = Ty then x = y.
what is the natural thing to do, starting with Tx = Ty

yep

npnp

[bold blade]

i need to prove a vector identity but im stuck

this is the one i need to prove

[slow scroll]

@pastel saffron R(T) = N(T) is definitely not true in general: i.e. when dim(V) is odd since you necessarily have R(T) != N(T)

np

[half ice]

That is, |u| = √[u•u]
@bold blade

[slow scroll]

.. and thats because of rank nullity, dim R(T) + dim N(T) = dim(V)

[half ice]

What are R(T) and N(T)?

[slow scroll]

R(T) is Im(T) and N(T) is ker(T) i believe

[half ice]

Ker(T) is a subset of the domain
Im(T) is a subset of the codomain

Unless the domain and codomain are the same, it doesn't make sense to ask about relations between ker(T) and im(T)

[slow scroll]

well in inyhaks case, T^2 = 0 which would imply im(T) \subset ker(T)

do you know that <v, w> = 0 for all w in V implies that v = 0?

yeah, as long as T is an operator.

im not sure i see what the "for all" quantifier is for here. It seems like as long as there is a non zero vector a for which <a, Ta> = 0, then Ta = 0 and a is in the null space.

This would hold on arbitrary fields, yes

and umm, by operator, i mean T maps from a vector space V back onto itself.

wdym both instances?

well, i shouldn't say arbitrary fields, since we are talking about inner product spaces, but we didn't assume one or the other here, so it should work for both

@pastel saffron no im wrong

<a, Ta> = 0 for all a in V says nothing about the null space. The situation we have is not the same as the "<v, w> = 0 for all w in V" situation i described earlier

probably, yea.

Ta is orthogonal to a, but not to V. (think about it, a cant be in V and be orthogonal to V, unless ofc a = 0)

i thought it might help to check the case when a is an eigenvector (these always exist when T operates on a complex space), but i don't see it going anywhere hm

if no one else helps in 15 minutes, you can ping @ Helpers 🤷‍♂️

[slim laurel]

Hey guys, could someone explain to me why this is true?

a bit unsure of why the last bit is true

[slow scroll]

AA^T qi = 0
qi^T AA^T qi = qi^T 0 = 0

[slim laurel]

oh that makes a lot of sense...

thank you

[slow scroll]

np

[wintry steppe]

I fail to see how an arbitrary inner product can say anything about invertibility

[quartz compass]

sorry you feel that way

[bold hearth]

If i have this matrix

and i have to find the eigenvalues

can i get it to upper or lower triangular form

or does it have to start out that way

can't quite figure it out

leaning towards it having to start as this

[bold hearth]

Because the eigenvalues of a triangular matrix are its diagonal elements, for general matrices there is no finite method like gaussian elimination to convert a matrix to triangular form while preserving eigenvalues

[silent dune]

Going based off my notes from last semester just wondering if you do make it a upper triangular matrix per say and then get the determinant to use to solve for the eigenvalues?

@bold hearth

[bold hearth]

hmm

i dont get quite what you mean

[silent dune]

one sec

So for example this matrix we wanna figure out the real eigenvalues so we do this

Any of this look familiar?

[bold hearth]

Yeah

but when you change the rows

in the first step

[silent dune]

ye one sec

[bold hearth]

does that not alter the values of the determinant

[silent dune]

Read that

[bold hearth]

Ahh thank you

But this can take a very large number of steps

[silent dune]

Ye thats why I was thinking you were maybe shown a different way

[bold hearth]

if its a really large matrix

ok now i understand

the (greek letter - 1) part was clever

i hadn't thought of that

[silent dune]

ye subtracting the greek letter multiplied by the identity matrix

[bold hearth]

no not that part

in the row operation

[silent dune]

oh

LOL

i wouldn't have thought of that anyway

[bold hearth]

that one is obv ofc but the row operation one was the one i was wondering about

Ok so its always possible to get it to this form

but i may take infinite number of steps

i think

if its super large

or not finite

[silent dune]

but yea it seems long so just maybe see if there is another way 😂

[bold hearth]

Because the eigenvalues of a triangular matrix are its diagonal elements, for general matrices there is no finite method like gaussian elimination to convert a matrix to triangular form while preserving eigenvalues

yeah since i had a question i was trying to answer

if its always this easy (as a upper triangular matrix) to find the eigenvalues

and i thought yeah its always this "easy" you just get it to this form, which may be very hard, but once that is done its easy to find the eigenvalues

[crystal pagoda]

You can resort to finding the minimal polynomial of the matrix

[pseudo stone]

how do we determine the uniqueness of a function?

[crystal pagoda]

Depends on context

But usually you assume they're not unique and say f1 and f2 both qualify

And show f1=f2

[real sable]

In the real vector space V of the continuous functions of R on R, we study the linear map A taking any function f in V to the function A(f) defined by (see picture).
What are the eigenvalues and eigenvectors of A?

How can I approach such a problem?

[pallid rampart]

Well, a function $f$ is an eigenvector of $A$ if $Af=\lambda f$ for some $\lambda$. Therefore we can write that $\int_0^xf(t)\dd{t}=\lambda f(x)$. Now notice a few things: substituting $x=0$ yields $f(0)=0$; the left hand side is differentiable by fundamental theorem of calculus therefore any eingenvector $f$ must be differentiable; differentiate both sides and obtain a differential equation with initial conditions

[stoic python]

Whoever:

[limber sierra]

this is not linear algebra

see #prealg-and-algebra or a generic questions channel

linear algebra is the study of vector spaces

so vectors, matrices, linear systems, norms, eigenstuff, determinants

[pallid rampart]

see #prealg-and-algebra or a generic questions channel

[torn hornet]

lol that eqn isnt even linear

[pallid rampart]

Lmao

[wintry steppe]

my bad lol

[slow scroll]

two in a day within 15 minutes of each other whoop

[keen frigate]

How do I do question 1?

[slow scroll]

this is also... not linear algebra xd

[keen frigate]

Bruh

I don’t get maths wtf

Do you know how to do it anyway ? 😂

[slow scroll]

its in y=mx+b form, just compare the m's.
a+2 = 3a - 1

i.e. since they're parallel their slopes are the same

[keen frigate]

Ohh

[pallid rampart]

At least this is linear

😂

[slow scroll]

i mean, not really

its affine

[pallid rampart]

Ok fine

[slow scroll]

[torn hornet]

lmfao

[ivory tulip]

Hey, guys, if v is a unit vector, would ∥(−1)v∥ just be 1?

[pallid rampart]

👀

[keen frigate]

Wtf is a affine ? 😂 you guys are too smart for me Jesus

An affine*

[slow scroll]

y=mx is linear. y=mx+b is affine

[torn hornet]

oh finally an actual LA question

yeah lapisguy

[pallid rampart]

I mean yeah, ||-v||=sqrt(<-v,-v>)

and you take out the -1

@slow scroll where b≠0

[slow scroll]

yep, thnx

[keen frigate]

No ones on the pre algebra section so can I ask my question here ?

[fair light]

Reading up on vector transform. How would I go about getting the transformed point given an angle

Say.... transforming a point 90 deg

[shy atlas]

u can think of that point(lets say in 2D) as a vector

u can multiply it by the rotation matrix

theta is the angle u wanna rotate ur vector by

u multiply by R and u get a new vector which is gonna be rotated by theta

@fair light

[fair light]

if the question asks... (1, 2) rotated ccw 90deg. how should I do that?

cos(90) -sin(90)?

[shy atlas]

yeah

plug in 90 degree for theta

[fair light]

R x that matrix?

[shy atlas]

that matrix is R

[fair light]

You have R =

Where would (2,1) get input ?

[shy atlas]

the dotted stuf fis ur R

if u multiply these together, its gonna yield another vector which is gonna be rotated 90 degrees

[fair light]

I have a particle and I wanna rotate it. Unfortunately I have a single line input

[shy atlas]

wdym a single line input ?

[fair light]

The "textbox" where I can add the formula

[shy atlas]

as i said, the end result of this matrix multiplication is gonna produce a vector in 2D

like (3, 5) lets say

u shouldnt have any problems inputing that

[cursive narwhal]

"I have a particle and i wanna rotate it." What is the general context for this? If you're working with a particle, you don't care about the rotation of the particle no?

[shy atlas]

[fair light]

lol.. sorry.. it's for visual effects

So I would have to write it as : 2x0 + 1x-1 for X

I think I got it

Thanks @shy atlas

[shy atlas]

[fair light]

How would that matrix change if it's 90 cw?

[slow scroll]

rotate -90

[fair light]

ahh.. coolz thanks!

[slow scroll]

npnp

[fair light]

what about a 3d point?

what does that matrix look like?

[slow scroll]

its more complicated. It depends on the axis of rotation

[fair light]

oh ok. What is this called? How would I look it up?

I don't need a 3d point, I'm just wondering

😅

[slow scroll]

https://en.wikipedia.org/wiki/Rotation_matrix

[stoic python]

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

[slow scroll]

in general, you need a few more rotations beforehand that will "align" the axis of rotation with one of the x, y, or z axes so that you can apply one of the rotation matrices I listed above (R_x, R_y, or R_z)

and then you have to apply the inverse of those initial rotations to get the axis of rotation "back where it belongs"

[fair light]

Ahh.. I see

I see what's going on there

Thanks again!

[slow scroll]

np np.

@fair light ah yea i found the part of the wikipedia article that talks about this

but as you can see, the matrix is nasty in general lol

[fair light]

Sweet.. I was just reading through it 😃

[limber sierra]

how are you defining dimension?

actually wait

thats probably not the best question to ask here

first off, you can assume that every vector space has a basis, right?

and if it's f.d. then it has finite basis

well ive got to go soon but i'll give a vague, but hopefully helpful, hint

consider row operations on a given basis (particularly helpful for intuition may be to consider a standard basis)

intuitively, you want to, for every vector

find a basis such that removing any vector from that basis

now excludes your original vector

[this works because basis vectors are linearly independent]

[slate mauve]

basic question: if I have two vectors A and B in R2, how do I represent the line formed by A and B?

[quartz compass]

"the" line?

two vectors generically will span a plane so long as they're not parallel

or oh maybe I see a way, like you mean pick the two points defined by the vectors A and B and use that to define a line?

draw a picture I think that will help. Take one point on the line, such as A, and then take all multiples of the vector parallel to your line to it, like this:

(B-A)t+A

[south wadi]

how did they say 4A - 9 * that matrix = 2A - 5 * that matrix

see how they switched the 4A -9 with the 2A - 5

on the 2nd line

[slender yarrow]

they just simplify (2A^t - 5....)^t

that's it

[south wadi]

oh ok

i see that

how bout here now

so they just set

2A - 5 * that matrix = 4A - 9 * that matrix

then solved for A right

[slender yarrow]

yeah

[south wadi]

yeah ok

also

I don't really understand this I stuff

so they took the inverse

i understand

then wehere did the I go?

and also

[cursive narwhal]

It's just matrix addition

[slender yarrow]

$I$ is just $\begin{bmatrix} 1&0\0&1\end{bmatrix}$

yk

[stoic python]

emeric75:

[slender yarrow]

in this case at least

it's just calculation yeah

[south wadi]

why is 1 divided by -1/2

1/2

shouldnt it be -1/2

on the top left

u seee it?

[slender yarrow]

yeah sounds like they fucked up

[south wadi]

ok

wait

no

it is right

$I$ is just $\begin{bmatrix} 1&0\0&1\end{bmatrix}$

[stoic python]

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

[south wadi]

WTF

u said I is just 1 0 01

but how did u know that

[slender yarrow]

there's double backslashes in the middle

and they disappear when you copy paste

$I$ is just $\begin{bmatrix} 1&0\\0&1\end{bmatrix}$

[south wadi]

$I$ is just $\begin{bmatrix}\ 1&0\0&1\end{bmatrix}$

[stoic python]

emeric75:

[south wadi]

so when ever I see an I

thats what I is

ok

[slender yarrow]

well it could be of another dimension

[south wadi]

so if we are in R3

I is just

rref

1 0 0
0 1 0
0 0 1

[light fable]

can someone help me

they didnt teach this yet since we are in quarantine

imma solve the first one but idk if im right

so what i think the answer for number 1 is a = 5 square root of 2 and b = square root of 46

[pallid rampart]

So what is the dot product?

[light fable]

do i need the dot product?

for a and b

[pallid rampart]

Lol yeah

[light fable]

so the comma is meant for the dot product

[pallid rampart]

Don’t you see? $\norm{a}=\sqrt{a\cdot a}$

[stoic python]

Whoever:

[pallid rampart]

The dot there is the dot product between a

Oh sorry I didn’t mean to be aggressive

[light fable]

sorry im more of a hands on

they teach this usually like normal but classes are suspended

so my problem is, what approach should i take

to solve it

[gray dust]

do you know how to compute dot products

[light fable]

am i in the right track? or what am i suppose to do with the exercise above

not yet

all i got is this

[gray dust]

that's the one

if $a=(a_1,a_2)$ and $b=(b_1,b_2)$ then their dot product is given by
$$a\cdot b=a_1b_1+a_2b_2$$

[stoic python]

RokettoJanpu:

[light fable]

so the first element multiplied and added to the second elements of both?

until to n

is that right?

[gray dust]

sorry your wording is just butchering what i typed above

[light fable]

lol sorry

[gray dust]

here's an example

[light fable]

can you do an 3dimensional one as example

[gray dust]

let $a=(1,2,3)$ and $b=(4,5,6)$ then their dot product is computed as
$$a\cdot b=1(4)+2(5)+3(6)=32$$

[stoic python]

RokettoJanpu:

[light fable]

ok that make sense but how do i get the | |a| | and | |b| |

[gray dust]

||a|| is called the norm of a and is defined as

$\norm{a}=\sqrt{a\cdot a}$

[stoic python]

RokettoJanpu:

[light fable]

so in your example will that a*a be a 32 * 32

so that it gets a nonnegative

sorry if im bad at this

[gray dust]

no what i did previously was take the dot product of a & b

in the definition of norm of a, inside the square root you dot a with itself

[light fable]

ah so thats why its square root

it multiplies with itself

[gray dust]

for now, for you there's no reason why there's a square root, you were just given a formula and your hw is to follow it

to find the norm of a, you take a dot a, then square root the resulting product

[light fable]

so a dot a

3(3)+5(5)+(-4)(-4)

then square root that

[gray dust]

you should tell me when you're doing the hw, you just threw a pile of numbers at me

and especially since i already used a in my example

[light fable]

oh ok

ok let me use your example then

from a={1,2,3} and b ={4,5,6,} becomes 1(4)+2(5)+3(6)

then square root?

[gray dust]

first, what's your goal?

finding ||a||?

[light fable]

idk, the exerices only states find | | a | | and | | b | |

i wish it has more exact intructions

[gray dust]

i'd say you were given everything needed to do the hw

i'll repeat the definition of norm

$\norm{a}=\sqrt{a\cdot a}$

[stoic python]

RokettoJanpu:

[gray dust]

to find ||a|| which is called the norm of a, you first dot a with itself ie take a dot a, then square root the resulting number

[quartz compass]

@light fable you're not fishing for answers are you, hoping that he'll eventually get exhausted and give you the solution

[light fable]

yep sorry im just so confused

[pallid rampart]

Well

[quartz compass]

I think you're thinking too hard and reading too little

[pallid rampart]

Here is a super quick example of how to calculate norm product

[quartz compass]

it says it on the stuff you sent

just plug and chug to formulas

[light fable]

welp i think i got it

but just unsure about myself

[pallid rampart]

$\norm{(1,2,3)}=\sqrt{(1,2,3)\cdot(1,2,3)}=\sqrt{1\cdot1+2\cdot2+3\cdot3}=\sqrt{1+4+9}=\sqrt{14}$

[stoic python]

Whoever:

[quartz compass]

show your best guess and working if you're unsure, it's nicer to the people helping you if you put forth the effort of at least making some attempt

[light fable]

my only question is, will a and b separated since it says find | | a | | and | | b | |

[gray dust]

you find those separately

[light fable]

oh ok

then i thinks thats about it

i think the first one is a = 5 square root of 2 and b is square root of 46

welp thats what i think

9+25+16 = 50

then squareroot it it becomes 5 squareroot of 2

idk

is that the end

[gray dust]

we'd say "||a||=5sqrt(2)" not "a=5sqrt(2)"

[light fable]

ah ok

sorry if i took so long to get it

im just so unsure if im doing whats asked

[gray dust]

don't worry, but what i'd do is read over everything i just said a little more thoroughly, and if you can help it, don't overthink most things

[light fable]

im just so confused with the second part

it says to show a new vector to be a unit vector

[gray dust]

a vector is called a unit vector if its norm is 1

if a is a nonzero vector, then a/||a|| (which is a divided by its own norm) is a unit vector, and your notes gives a hint on how to show that

[light fable]

oh ok i think im getting it a little

for now imma sleep

thank you for your time

[gray dust]

you're very welcome, have a good night

[grave anvil]

hey is this the right place to ask about latex?

to do with linear algebra

[gray dust]

go ahead

[grave anvil]

im trying to get a matrix <x_1,x_2,x_3> and im starting off with \begin{pmatrix} x_1 \ x_2 \ x_3 \end{pmatrix}

[stoic python]

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

[grave anvil]

oh yikes

it's giving me a product when i compile it

https://gyazo.com/3d87470aff98686f690ec4e80dc01a85

first time using latex

[gray dust]

try surrounding the line with dollar signs?

$$\begin{pmatrix} x_1 \ x_2 \ x_3 \end{pmatrix}$$

[stoic python]

RokettoJanpu:

[grave anvil]

https://gyazo.com/8095df855e64b8a4ab21837766a7cd8a

[slender yarrow]

sounds like you didn't close the math mode or something

[grave anvil]

i get this

what does $$ do?

[gray dust]

enter math mode iirc

[slender yarrow]

same as \[ pmuch

[grave anvil]

tbh i dont know what [ does either

[slender yarrow]

well you didn't have a \] closing math mode tho before

[grave anvil]

im taking it from a really confusing guide

would you have any recommendations on how to learn latex

seems my knowledge gap may be too big

[slender yarrow]

the latex wikibooks is quite good

• there's a small guide in #resources

[grave anvil]

alright thank you i'll try use this

[slender yarrow]

but yeah try $$\Pi_1 : \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}$$

[stoic python]

emeric75:

[slender yarrow]

it works on my overleaf

i don't see why it shouldn't on yours

[grave anvil]

https://gyazo.com/7144cb838fe72eca44c328959af869e0

possibly my uni has some preloaded settings?

im not familiar with overleaf

[slender yarrow]

hmm

what's your preamble?

maybe you don't have amsmath imported or shit like that

[grave anvil]

https://gyazo.com/484ef898971d9473f9f0a087928671a2

is this the preamble?

[slender yarrow]

yeah the beginning

[grave anvil]

i think the top two came already in

i dont know what usepackage is

[slender yarrow]

yeah add \usepackage{amsmath} in the beginning

it's just to import a library in your document

and those matrix thingies are in amsmath

[grave anvil]

do you know what utf8 would be?

[slender yarrow]

it's just the encoding you use in your text

if you want to support french accents,etc..

[grave anvil]

ok the switch worked

thank you very much

for future, do you know where i should ask for latex help?

im assuming linear alg channel is bordering on misuse

[slender yarrow]

usually ppl just go in #computing-software (or #math-discussion sometimes)

or just plain #❓how-to-get-help channels

[south wadi]

how the fuck is this wrong

this makes no sens

e

i am trying to get determinant of this

i reduce

1 1 1 1
0 2 0 0
1 1 -1 1
1 1 1 -1

R1 - R2

but aparently

that 2 has to be ngative

because u only get the right answer

[hoary agate]

yeah it should

[south wadi]

from doing R2 - R1

WHY

[hoary agate]

you're doing R2 -> R2- R1

[south wadi]

No i did R1 - R2

[hoary agate]

-1-(+1)=-1-1=-2

[south wadi]

why cant u do that?

Why cant u do R1 -R2

why does doing R1 - R2 get u the wrong answer

[hoary agate]

because you're multiplying R2 by -1 first

[south wadi]

shit makes no sense so frsutrating

[hoary agate]

that means the determinant changes sign

instead if you do R2-> R2-R1, there's no sign changes

and everything's k

[south wadi]

I never multiplied by -1

I'm just getting it into REF

and aparently u cant do R1 - R2

because that gets u the wrong answer

like wtf is this shit class

[hoary agate]

you're doing R2-> R1-R2 which is actually two manipulations R2-> -1×R2 and then R2-> R1+R2

you're implicitly multiplying a single row by -1

resulting in the determinant changing sign

so if u keep ur manipulations with only positive coefficients of each row, then you won't face such errors

[south wadi]

huh

[hoary agate]

or you should do R1-> R1-R2

in which case you'd get 0 +2 0 0 in first row

and you can simplify

[south wadi]

why do u have to multiply

by -1

[hoary agate]

you dont have to.. you just did it lol

like you're doing R2-> R1-R2

[south wadi]

????

1 - 1 = 0

[hoary agate]

see you're multiplying R2 by a -1 in that row transformation

[south wadi]

im subtracting it

[hoary agate]

yes exactly

you're adding R1 + (-R2)

[south wadi]

yeah ok

[hoary agate]

there's a ×-1 which crept in there

[south wadi]

wht's the issue

u cant mutiply rows ?

[hoary agate]

if you change sign of one row, you flip sign of the determinant

only multiply with positive numbers, to preserve determinant

that's why R2-> R2-R1 works

because R2 sign isn't being changed

[south wadi]

but it's the same shit

[hoary agate]

it's not..

[south wadi]

ur multiply R1 by -1

[hoary agate]

but I'm not transforming R1

I'm transforming R2 - > R2 - R1 but R2 sign isn't changed

[south wadi]

holy shit this class sucks

[hoary agate]

you're just not understanding it.

read some texts on row reductions

[south wadi]

nah it's just some stupid rules

[hoary agate]

and how it affects the determinant

it's not just random rules, it's derived results from the properties of determinants

you could figure them out yourself

by testing on simple matrices like 3x3

[south wadi]

couldn't be bothered honestly

3 more weeks of this shit and im out lmao

[hoary agate]

yeah then don't blame the class for your apathy

[south wadi]

im free from this class

this class

is 100% useless

the only thing u use it for

[hoary agate]

put effort get results man that is all it is

[south wadi]

is solving equations

the class is useless

[hoary agate]

rip

[south wadi]

i dont see any

real life applciations

how can people major this stuff

[hoary agate]

linalg is useful in QM

and all the application oriented extensions

quantum computing

ML in general too

[south wadi]

half the stuff that comes out of my profs mouth is bs

never will i use this crap

after i graduate

[hoary agate]

what are you majoring in?

[south wadi]

Engineering

[hoary agate]

oh rip

[south wadi]

yeah this is just plain useless to us

[hoary agate]

oh

ye then

they must be

except if any entry is 0

if one row is all 0s

and the other is all 2s

you can't write the row of 2s as a scalar multiple of the row of 0s

idk

[wintry steppe]

uh

linear algebra useless in engineering?

yikes

not just ML and quantum computing; there's electrical engineering, where it's used extensively to analyze circuits, statistics, control theory, mechanical engineering

[tulip reef]

what do you mean? engineers would never linearly approximate anything

\sin x = x

[hoary agate]

lmao

sorry, wasn't aware of the EEE applications

[cursive narwhal]

Yo mamma wasn't aware of the EEE applications

[hoary agate]

well, yeah that too

[pallid rampart]

Prank

Relax

Memes

Chill

[smoky lagoon]

I have hard time with finding the standard matrix of transformations

[cursive narwhal]

$T(a_2x^2+a_1x+a_0) = (x^2-1) \cdot 2a_2 + x(2a_2 x +a_1) +a_2x^2+a_1x+a_0$

$T(a_2x^2+a_1x+a_0) = 5a_2x^2 + 2a_1 x + (a_0-2a_2)$

[stoic python]

Abhijeet Vats:

[cursive narwhal]

So, that's your new polynomial after applying the transformation

[smoky lagoon]

OH I see that

[cursive narwhal]

In other words:

$T(a_0,a_1,a_2) = (a_0-2a_2,2a_1,5a_2)$

[smoky lagoon]

but I don't see how I can get a matrix out of that

[stoic python]

Abhijeet Vats:

[smoky lagoon]

I know you can but i'm slow

[cursive narwhal]

So, find the images of the standard basis vectors and they'll form the columns of your matrix

@smoky lagoon Do you understand that?

[smoky lagoon]

sort of

images of this?

[cursive narwhal]

Yes

Those are your canonical basis vectors in $\mathbb{F}^3$.

[smoky lagoon]

oriented differently obviously

[cursive narwhal]

Wait what the fuck

[smoky lagoon]

so would I just put that basis in r3 into the polynomial?

[stoic python]

Abhijeet Vats:

[cursive narwhal]

No, you already have a formula describing the transformation

[smoky lagoon]

okay so what's my next move

[cursive narwhal]

Like, you know exactly how the coefficients change, by virtue of the derivation I did earlier

[smoky lagoon]

right right that makes sense

[cursive narwhal]

So, find out how the basis vectors change and you're done. You don't have to worry about the polynomials they make up

As far as we're concerned, those polynomials are entirely described by the coefficient vectors.

[smoky lagoon]

so what's my first step here?

sorry that I'm dense

[cursive narwhal]

T(1,0,0) = (1-0,0,0) = (1,0,0)

[smoky lagoon]

hold on

[cursive narwhal]

Sure.

[smoky lagoon]

so would T(0,1,0)= (0,2,0)

?

[cursive narwhal]

I mean, you're using the formula above, right? So just make sure you're computing things correctly and it should be correct

[smoky lagoon]

I sure hope I am

so T(1,0,0)= (1,0,0)
T(0,1,0)=(0,2,0)

T(0,0,1)=(-2,0,5)

[cursive narwhal]

Yeap. Now, these will form the columns of your matrix

And you're done

[smoky lagoon]

cool! could you explain the logic on finding the polynomial in the first place?

[cursive narwhal]

Now, just check if the initial transformation was correct. I did it rather quickly and it's 3am here, with me falling asleep so try to check it through thoroughly.

Finding the polynomial?

[hoary agate]

sleep mate, sleep

[smoky lagoon]

the inital transformation

[cursive narwhal]

Well, a transformation is nothing but a function

That's all it really is

So, your function was defined in a particular way in the problem above

[smoky lagoon]

ah you use the transformation as your y and then do the derivatives as such

and t hen put it in for y* and y**

[cursive narwhal]

I took the most general polynomial that I was allowed to use and found the image of that general polynomial.

Yes, precisely.

[smoky lagoon]

interesting

[cursive narwhal]

That allowed me to get the polynomial at the very end. Now, we initially agreed to talk about polynomials in terms of a vector that would have the coefficients of the polynomial as its components

So, that's why I got the transformation into the vector format above

Once that happens, finding the columns is easy

[smoky lagoon]

the first COLUMN is (1,0,0) and the second would be (0,20) and the third would be (-2,0,5)

or would those be the rows?

[cursive narwhal]

The former

There's a beautiful treatment of precisely this topic in Klaus Janich's Linear Algebra textbook, where he derives the fact that the images of the basis vectors form the columns of the associated matrix. That discussion was a delight and you should definitely check it out.

[smoky lagoon]

I might have to check it out then

what's my game plan here?

wait

I can use the original transformation

I know that a0=1, a1=1, and a2=1

[cursive narwhal]

[tulip reef]

[gray dust]

it's a rare hankel

let the unknown quadratic be y=a_2x^2+a_1x+a_0. rewrite y as the column vector (a_0,a_1,a_2) and x^2+x+1 as (1,1,1) (imagine these are column vectors). let A be the standard matrix of the map T. part c is solving the system T(a_0,a_1,a_2)=A(1,1,1)

[shrewd slate]

If the quotient space V/U of a vector space V with respect to subspace U is finite dimensional, can we assume that V is also finite dimensional?

Probably, right?

Might just say that it’s clear and call it a day tbh

[slow scroll]

It feels pretty false to me, but idk

[pseudo stone]

Guys i dont understand why Det (u - xIn) = 0 is equivalent to X belongs to sp(u) ?

[shrewd slate]

Uhh idk I feel like V/U couldn’t be fin dim if V wasn’t

I think V/U would have to fill up all the infinite dimensions if that were the case?

[pseudo stone]

HELP

[slow scroll]

what does u - xln mean

and sp(u)

[brittle juniper]

If U is a vector hyperplane of V, dim(V/U) will be 1 @shrewd slate

regardless of dim(V)

[shrewd slate]

Shidd

That’s not good

Thanks

[brittle juniper]

You're welcome

[shrewd slate]

Proof machine broke

[brittle juniper]

idk what you're calling Jordan, but the minimal polynomial of T divides X²-1, so it has only simple roots @pastel saffron

[steady fiber]

ya, that works as a pretty simple proof

you do have to specify that the minimal polynomial of T has distinct roots so it's diagonalizable

[slow scroll]

yeah, i mean, invertibility doesn't really say much about diagonalizability.

[shrewd slate]

Back to Bunchuong’s example where dim(V/U) = 1, would it be appropriate to say that if B_1 is a basis for U and B_2 is a basis for V, then

B_2 = B_1 ∪ {v}

For some v ∈ V?

[brittle juniper]

if you take two arbitrary bases like this, it's very unlikely that B_2 has all the vectors of B_1

from U being a hyperplane of V, you can have the existence of v in V such that B_1∪{v} is a basis of V

[shrewd slate]

I was thinking of B_2 as an extension of B_1, I should’ve mentioned they’re not both arbitrary

But yea, what you said is what I meant, thanks!

Looks like my proof can be saved after all

[novel ether]

help pls do I just dot product them, then what do i do with the dt

[dusky epoch]

do you know what a derivative is

[novel ether]

yeap

but isnt that when its multipled by dt not divided by

[dusky epoch]

...

are you not able to take the derivative of a scalar quantity

$A \cdot B = 5t^2 \sin(t) - t \cos(t)$

[stoic python]

Ann:

[dusky epoch]

$\dv{t} (5t^2 \sin(t) - t \cos(t)) = ; ?$

[stoic python]

Ann:

[novel ether]

ah okay its just that thank you i appreciate it

im a muppet

[sonic osprey]

what exactly are you asking?

If that is true for all vectors a, then T must be self-adjoint

what are you even asking

[dusky epoch]

on the field of R, is <a, Ta> = <Ta, a>
inner product is symmetric

so yes, <a, Ta> = <Ta, a> for all vectors a and operators T

[wintry steppe]

this is fun for making machine learning algos

[long lion]

Need help

[slow scroll]

#prealg-and-algebra or #precalculus

[long lion]

This algebra 2

[slow scroll]

yea. its not linear algebra tho

[long lion]

Then where do I go

There’s not any

[slow scroll]

#prealg-and-algebra or #precalculus
@slow scroll

[long lion]

It’s not calculates

or pre algebra

[slow scroll]

those channels are where high school algebra goes, yes

[long lion]

Oh

Ok

[maiden silo]

yo can somebody help me with linear algebra

[slow scroll]

[maiden silo]

who??

[slow scroll]

type your question in the channel, and someone might help

[maiden silo]

Does every nonzero linear transformation from Rn to itself has at least one nonzero eigenvalue?

[slow scroll]

what are the eigenvalues of a rotation matrix?

[maiden silo]

are you asking me?

it depends where you're rotating

[slow scroll]

think about it: what points are fixed under rotation about the origin in R2

[maiden silo]

the origin

[slow scroll]

so.. 0 is an eigenvalue. are there any others?

[maiden silo]

uhhh

[slow scroll]

thats it. So a rotation matrix from R2 to R2 has only one eigenvalue: 0

[maiden silo]

this is the rotation matrix

to get the eigenvalues, you do the rotation matrix subtracted by yI

then set the determinant equal to 0

correct

[slow scroll]

yes. Note: it DOES have eigenvalues in C, but not in R\{0}

[maiden silo]

that means

y^2 - 2ycos +1 =0

0 can't be an eigenvalue

there are no eigenvalues in the rotation matrix

@slow scroll

[slow scroll]

yeah youre right. I was assuming 0 could be an eigenvector, which is false.
but yeah, you can show that the characteristic polynomial never has real roots.

[maiden silo]

Let A=PDP−1, where D is a diagonal n×n matrix. If B is the basis for Rn formed from the columns of P, then D is the B-matrix for the transformation given by left multiplication by A.

What does this question even mean

[slow scroll]

I don't see a question, but its talking about diagonalization. Have you heard of that?

[maiden silo]

It's asking if it's true or false

Yea I have

[slow scroll]

yeah, on second look, im not really sure what its talking about either 🤷‍♂️

it could be saying that D is just A in the basis of B, which would be true

[maiden silo]

Can Two matrix representations A and B of a linear transformation T can have different eigenvalues?

@everyone

[limber sierra]

could you give an example of a linear transformation T with two distinct matrix representations?

[maiden silo]

Is this shit true or false

@limber sierra

[slow scroll]

let $\begin{pmatrix}a&b\c&d \end{pmatrix}$ be some real matrix without complex eigenvalues. Then let $z = a + bi$ and $w = c + di$.

[maiden silo]

@slow scroll what

[stoic python]

kxrider:

[maiden silo]

@dry maple what in the word

@slow scroll

[slow scroll]

wat?

[maiden silo]

what's the answer lol

is it true or false

i need help explanation

[slow scroll]

Look closely at what I wrote: in other words, the problem is reduced to: is there a matrix with real entries (nonzero in the second column) that has no complex eigenvalues?

[maiden silo]

huh

@slow scroll i need more of an explanation i still confuzzled

[slow scroll]

So... lets take a matrix like $\begin{pmatrix}1&1\1&1 \end{pmatrix}$ and compute its characteristic polynomial.

[stoic python]

kxrider:

[slow scroll]

what would its eigenvalues be/

[maiden silo]

1

@slow scroll

[slow scroll]

hmm that aint right

[maiden silo]

what..

[slow scroll]

what did you get for its characteristic polynomial?

[maiden silo]

oh wait

one sec lol

0 and 2

brainfart sry

[pallid rampart]

[slow scroll]

npnp. so there are no complex eigenvalues.
All that is left to do is let z = 1 + i and w = 1 + i and there is your counterexample

[pallid rampart]

brainfart

That's an interesting expression

[maiden silo]

wait @slow scroll how would that work

it would be a 2x2 matrix filled with 1+i

[slow scroll]

So recall this. Re(z) = 1, Im(z) = 1 and Re(w) = 1 and Im(w) = 1

Re(z) and Im(w) are always real numbers

[maiden silo]

the matrix is

z z

w w

wait

what is re(z) and lm(w) mean

[slow scroll]

Re(z) = real part of a complex number z
Im(z) = imaginary part of a complex number z
so if z = a+bi
then Re(z) = a and Im(z) = b

[maiden silo]

oh

wait

wouldnt Im(z)=bi

[slow scroll]

nah, Im and Re are always real numbers

[maiden silo]

wait but

Im is the imaginary part

[slow scroll]

yea, but any complex number has the form a + bi where a and b are real.

[maiden silo]

oh

so the answer is false

because if you have a matrix filled with 1s

then you get eigenvalues 0 and 2

[slow scroll]

yep i.e. it fails when z = w = 1 + i

[maiden silo]

bruh this so confusing

[slow scroll]

you just need to find a basis of eigenvectors of A. Then Q is the matrix which takes vectors from the standard basis to that basis of eigenvectors

[maiden silo]

i hate coronavirus no teachers to help me

[slow scroll]

ripperoni

i.e. Q is the P^-1 in PDP^-1

[maiden silo]

i got 1,0

and 1,1

for the eigenvectors

[slow scroll]

this is because you have $\vec y_{k+1} = D\vec y_k = DQ \vec x_k$ where $D$ is diagonal. It should at least remind you of $PDP^{-1}$ diagonalization stuff.

[stoic python]

kxrider:

[maiden silo]

wait

i got

1 1

0 1

i def did somehting wrong

that's not Q

[slow scroll]

how did you get that?

[maiden silo]

the eigen vectors are 1, 0

and 1, 1

standard basis vectors are 1, 0

0, 1

so i just formed the transformation matrix with that

[slow scroll]

close: that is the matrix that sends vectors in eigen basis to the standard basis: i.e. that would be Qy = x. You need Qx = y.

all you have to do is take the matrix you came up with and invert it. Then that matrix would send vectors in the standard basis to vectors in the eigenbasis (1,1), (1,0)

[maiden silo]

1, -1

0 1

[slow scroll]

i agree

[maiden silo]

wait hold up

why did i do it flipped

like why was my way flipped

[slow scroll]

flipped?

oh

[stoic python]

kxrider:

[maiden silo]

ohhh i see

can i pm u

[slow scroll]

i suggest watching this video as well https://www.youtube.com/watch?v=P2LTAUO1TdA&t=

[maiden silo]

I think the fourth one is wrong

But can anybody help confirm that the answers to these are

T, T, T, F, T

I think I've logic-ed through it but I may have made some mistakes in my thinking

<@&286206848099549185> I didn't know you couldn't ping helpers until after 15 minutes until I just looked over the faq again, so I removed my original tag. Sorry guys

[novel ether]

hey team i have set up the intergral but I dont know what to make the intergral limits, was looking for some help

[dusky epoch]

what interval does t run through

isn't it [0,1]

[hidden arch]

hi, I am new here and my math is very basic(I presume) where do I send it ?

[novel ether]

@dusky epoch i have no idea

[dusky epoch]

(0,0,0) corresponds to t=0, (1,1,1) to t=1.

@hidden arch if you're not sure, send it in one of the ten questions channels that aren't occupied

[hidden arch]

Thank you and sorry for doing that here

[novel ether]

@dusky epoch if the second point was say (1 , 2, 1) would it sitll be t= 1

i dont understand the second part

[dusky epoch]

it wouldn't even be on the curve {x=t,y=t^2,z=t^3}

[novel ether]

o right i kind of see what you mean now