linear-algebra

[stoic python]

John Doe Smith:

[torn hornet]

yeah basically

[paper egret]

ye i did that

and i have to show that 0,0 is unique

🤔

[torn hornet]

well translate the left eqn to a matrix eqn

and use your knowledge of null spaces

[paper egret]

wtf is a null space

👀

[jagged saffron]

null space is solutions to ax = 0

[torn hornet]

hmm so null space of a matrix is the set of all vectors that go to 0 when acted by it

Ax=0, then x in null(A)

[paper egret]

LOL I DONT THINK I LEARNED THAT YET HAHAHAHAHAHA

[torn hornet]

rip

but we can still easily show it, so if the matrix is invertible (0,0) is the only solution right

[paper egret]

um wat

invertibe?

im a lost potato lmao

[torn hornet]

has an inverse

$A^{-1} A=A A^{-1} = I$

[stoic python]

John Doe Smith:

[jagged saffron]

you could also just show it has non zero determinant

[paper egret]

ok yea we did NOT touch that yet, i know what an identity is and stuff, but i dont think we can use that

[jagged saffron]

[paper egret]

no detemrinants

[jagged saffron]

ok well

row reduce?

[paper egret]

LAWL

[jagged saffron]

simplest way

[paper egret]

NEVER LEARNED THAT EITHER

[torn hornet]

yeah i was gonna use determinants lol

[jagged saffron]

huh?

gaussian elimination?

[paper egret]

NOP

[torn hornet]

ok just solve the equations then

[jagged saffron]

[paper egret]

oh ok

[torn hornet]

by elemination lol

[paper egret]

a + 4b = 0
2a + 9b = 0 i guess i just solve normal

and then i just state something like

dis shit only got one solution

so its linearly independent

[torn hornet]

yeah basically

[paper egret]

kk

sry for breaking everyone's braincells here

[torn hornet]

although in a general problem you would want to do determinants lol.i think they might be trying to lead into that

[paper egret]

professor said no determinants

or else we die

[torn hornet]

ok then the problem just stupid lol

[paper egret]

im just gonna assume he's okay with a very informal proof of unique solution

[jagged saffron]

can someone help with my question

im p sure the homework question is wrong

[torn hornet]

i mean its like asking someone to formally prove x+1=2 implies x=1

also yeah, repost?

[jagged saffron]

#2

pick v1 = v2 and v 3.. vn whatever

then its dependent

OH WAIT

I AM

[torn hornet]

whats F^N here

[jagged saffron]

BLIND

it says linearly dependent

no independent

ok whatever i got it

[torn hornet]

rip

[jagged saffron]

also

usually F^n is field in n dimensions

arbitrary field

[torn hornet]

oh just some random field

[jagged saffron]

[torn hornet]

so you still need help?

[jagged saffron]

nope

[torn hornet]

ok gl

[paper egret]

i understand what it means for a set of vectors to be linearly independent, but what does it really mean conceptually?

cuz atm the only thing i'm picturing is a bunch of lines in different dimensions

and see if there is an intersection LO

L

[gray dust]

@paper egret does it help if i say "if at least one vector in the set can be "reached" by multiplying the other vectors by certain scalars and adding them up, the set is linearly dependent"

[paper egret]

i kinda had my own idea, but im not sure how well it goes

in short, if each vector isn't really a multiple of another vector, then they are inependent from each other

cuz if u had a vector of 2,4 and 1,2, they are basically identical, and you cant really do anything with them

cuz one is a multiple of another

[gray dust]

yeah it's not that good

[paper egret]

but if they arent multiples of each other, independent

hmmmm i figured

[gray dust]

let's say you have a plane in 3d space, spanned by two vectors

[paper egret]

yups

[gray dust]

you add a third vector to the set, and it lies in that plane, and it doesn't point in the same direction as either of the two original vectors

you can't express that third vector as a multiple of just one of the original vectors

[paper egret]

now that you speak of direction, it's starting to make a bit more sense

[gray dust]

but you CAN express the third vector as a linear combination of the two original vectors

[paper egret]

yup

im sorta animating it in my head LOL

[gray dust]

aka you can multiply the first vector by some scalar and the second vector by some scalar and add em up in order to reach the third vector

[paper egret]

gotchu fam

[gray dust]

there you go, linear dependence

[wintry steppe]

Basically, what you're trying to say, independence is what makes us build upon our basis vectors

[gray dust]

blitz, we want to explain linear independence in really simple terms

[wintry steppe]

I'd love it explained in general

because I was thinking, how fundamentally it comes to play

[paper egret]

its almost like having a standard lol

you can build better stuff as long as you have some standard

[wintry steppe]

Suppose that F(v_1), ..., F(v_n) are independent. What can we say about vectors v_1, ..., v_n?

F - a linear map

[paper egret]

wat

can u write inl atex plz

[wintry steppe]

we can say that every linear combination of v_1, ..., v_n lies in the kernel of F

[paper egret]

ok thats too far, never touched kernel

if that is a topic

[wintry steppe]

I was thinking out loud

sry

[paper egret]

o np

[thorn robin]

@goon like you said, if you want to see if two vectors are independent, just ask if one is a multiple of the other
now assume you have two independent vectors v_1, v_2 so they span a plane a_1 v_1 + a_2 v_2, and say you want to ask whether a third vector v_3 is independent from the first two
it's easy, just check whether v_3 lies in that plane you formed from v_1 and v_2 (unless v_3 is 0 of course)

[wintry steppe]

If v_1, ..., v_n are independent, then we have some system of coordinates, because two linear combinations are uniquely determined by the coefficients. I. e. by introducing linear independence we introduce coordinates

so this is why linear independence is important, it enables us this uniqueness

[noble swallow]

^^

However F(v_1), ..., F(v_n) lin indip. implies v_1, ..., v_n lin indip.

And if I am not wrong F injective iff for each k>=1, v_1,..,v_k lin indip implies F(v_1),...,F(v_k) lin indip

[paper egret]

👀

[wintry steppe]

Yes. I omitted easy proof that F(v_1), ..., F(v_n) being lin indep. implies that v_1, ..., v_n is lin indep.

my bad. As for the other one, it's true as well, one can check

If we have a linear combination of F(v_1), ..., F(v_k) mapping to 0, then from injectivity, we have a linear combination of v_1, ..., v_k mapping to 0, i. e. every coefficient must be zero and so F(v_1), ..., F(v_k) are lin indep.

for the other direction, F(x) = 0 implies x is linearly dependent and so must be 0

[jagged saffron]

I’m trying to prove that the set of commutators of n x n matrices form a subspace

Where commutator is defined to be [a,b] = ab-ba

I cant seem to show its closed under addition, any hints?

I know that [a,b] has trace 0

But i dont know how to show the set is equivalent to all matrices with trace 0

[pallid swallow]

hmm...

But to show that you might need to come up with a basis

[jagged saffron]

If i narrow the matrices down to 2x2 is it easier to prove?

[pallid swallow]

well, let's not narrow too quickly

[jagged saffron]

cause the question is on 2x2

[pallid swallow]

[a, b]+[a, c]=[a, b+c]

hmm...

welp if question is 2x2 maybe we should narrow

who knows if it works for 3x3

[jagged saffron]

It does

[pallid swallow]

If we have a basis, we need only 3 elements

[jagged saffron]

The bonus is for n x n

[pallid swallow]

so just find some examples, and hopefully you would stumble on a basis

[jagged saffron]

In general

[pallid swallow]

hmm I'm not finding something yet

trying to express [a, b]+[c, d] as a commutator is strange

[wintry steppe]

every commutator has zero trace right

so i suppose what you need to do is prove that every matrix of zero trace is a commutator

hmmmst

[noble swallow]

In general though that is hard to do, if I recall correctly

[torn hornet]

could we use [SAS^(-1),S B S^(-1) ] =S[A,B]S^(-1)?

[noble swallow]

@jagged saffron are you sure you are not simply asked to prove that the span of commutators is the set of trace zero matrices?

[torn hornet]

actually hmm, if we can show [A,B] can have any eigenvalue a trace zero can have, using the above the proof would be complete right?

[jagged saffron]

this is the question

[wintry steppe]

lol

[jagged saffron]

K is arbitrary field

[wintry steppe]

is this some lie theory class

[jagged saffron]

nope

some grad linear algebra class

not smart enough for lie theory

[torn hornet]

ok lol grad LA, i think thats my que to exit

[jagged saffron]

i dont think this is that hard

its the first week lol

maybe n x n is hard in general

[torn hornet]

2x2 shouldnt be too hard i think

just like expand two matrices out and bash out a solution

[jagged saffron]

o god

[torn hornet]

or better yet, try forcing a zero trace matrix to be a commutator

[jagged saffron]

I've tried looking up how to prove zero trace <-> commutator but its apparently pretty hard in general

[torn hornet]

well in 2x2 it shouldnt be too hard

[jagged saffron]

yea

i think i know how to do it for 2x2 now

still curious about n x n though

[wintry steppe]

https://people.eecs.berkeley.edu/~wkahan/MathH110/trace0.pdf

[jagged saffron]

oh

[wintry steppe]

lo

[jagged saffron]

idk if i can understand this

[noble swallow]

Lol a true challenge

[jagged saffron]

ill figure it out tonight

[noble swallow]

@jagged saffron I retrieved a solution for 1), I need some time to latex the matrices now lol

[jagged saffron]

oh :o

i tried writing the matrices out but i didnt really get anywhere

for 2x

2

[stoic python]

Mat:

[noble swallow]

@jagged saffron

There is still to consider the case where c=0 I suppose, but it shouldn't be too difficult maybe

[jagged saffron]

OH

[noble swallow]

However above it is used the idea that John was suggesting before

[jagged saffron]

[noble swallow]

[jagged saffron]

i dont understand for generic 2x2, why is the first equality true?

my LA is kinda rusty

[noble swallow]

I tried to find a similar matrix which had 0s on the diagonal

I omitted that part of the calculation

[jagged saffron]

oh similar matrices

i havent done that before

they are similar?

wait

i meant det

[noble swallow]

Basically similar matrices are matrices associated to the same transfrormation with respect to different bases

And S operates the change of basis

[jagged saffron]

so basically you just solved for S to get closed form solution for any matrix?

2x2 with trace 0

I think i kinda get it but not too sure about the similar matrices part

[noble swallow]

I searched for a basis v,w such that L_A(v)=Av=λw and L_A(w)=Aw=μv

So when we write the matrix associated to L_A wrt this basis we have only 0 on the diagonal

This matrix is similar to A and S is just the transition matrix between that basis and the standard basis

[jagged saffron]

uhhhh

gimme a min

ill ping you if i have questions

[foggy lodge]

Is there a good book for elementary set theory?

[sonic osprey]

I enjoyed Velleman's How to Prove It

[foggy lodge]

My linear algebra professor is going over sets and maps

ok, thanks!

ill check it out now

[feral mountain]

Is it a good idea to drop the geometric interpretation of linear algebra cuz we can't think in higher dimensions?

[sonic osprey]

Not really no

Much of what happens for lower dimensions happens for higher dimensions

[feral mountain]

Yes but I don't like how there isn't much rigour to extending our geometric intuition to higher dimensions idk

[sonic osprey]

Think geometrically, work algebraically

[wintry steppe]

Honestly, no. Few times in probability geometric intuition was essential to work with linear algebra

[feral mountain]

😮 probability, can you give me any examples?

[wintry steppe]

you don't understand probability anyway

[feral mountain]

😦 is 2nd year probability not enough?

[sonic osprey]

I don't understand what kind of examples you're looking for

[half ice]

Know what happens geometrically, be able to work algebraically. Then when the geometry fails, you can go the other way - that is, you can use algebra to solve higher dimensional geometry problems

[feral mountain]

Yeah I just don't get why learn the geometry when it will eventually fail when you move to higher dimensional things

[wintry steppe]

I don't remember exactly, but it was about a random vector with normal distribution and how you can use rotations so that its every coordinate is independent

I think

[half ice]

Even worse, if you get into more abstract vector spaces, you can't attach any geometric idea onto some

The vector space of polynomials for example, is infinite dimensional. It's more useful to think in terms of the real line in that case.

[wintry steppe]

It's a bad idea to drop interpretations in general

unless maybe to have a second view

[half ice]

^
You have a tool available to you. Yes it isn't always useful. Why drop it, when it sometimes is?

[sonic osprey]

Yeah and there are plenty of examples where a geometric view is essentially to understanding how to solve a problem

[feral mountain]

Oh right I did basically picture 3 dimensions to basically reason everything in vector calc, I guess it's useful in the real world when you usually have dimensions less than 4

[sonic osprey]

Yeah that's a huge example

[native lodge]

some of the most informative examples you will probably ever encounter happen in the lower dimensions

elimination and various views of matrix multiplication are easily seen in 2 dimensions

once you move higher up, you take what you know and then apply it there

even though we can't imagine the intersection of hyperplanes or other things in 5 space, we can still do the elimination to find a solution point, just like when it was lines meeting at a point in 2 space

[jagged saffron]

@noble swallow oh ok thanks i understand it now

[undone garnet]

any idea for this problem?

xA^2=A

yB^2=B

well

[dusky epoch]

is O supposed to be the zero matrix

[undone garnet]

it is

[pallid swallow]

hmm

really feels that A, B are diagonal matrices

A has rank(A) entries 1/x, B has rank(B) entries 1/y and rank(A)+rank(B)=n

Anyone can find a counterexample to these claims?

Well, xBA+yB^2=B, so that means BA=0

hey, A and B commute

also note that det((A+B)^n)=det(A^n+B^n)

(in this case)

and because xA^2=A, thus multiplying out by A^k, gives xA^(k+1)=A^k, so x^mA^(m+1)=A

det((A+B)^m)=det(x^(1-m)A+y^(1-m)B)

I'm guessing that similar matrices might also be a solution, so for invertible P
xP^-1 AP + yP^-1 BP=I as well
P^-1 ABP=0 as well, so the matrix doesn't need to be diagonal

similar matrices would have the same rank, and determinant is multiplicative so that's all fine

so A, B DO NOT have to be diagonal matrices

but maybe they are similar to such matrices

[gaunt mulch]

anyone got any idea about this

[sonic osprey]

What have you tried?

Or what are you confused about?

[dusky epoch]

this isn't linear algebra

[gaunt mulch]

wait a lil' bit

[sonic osprey]

That's not really what (a) asks for

[gaunt mulch]

in later stage we have have to find eigen value so i wrote like this

[dusky epoch]

uh

no

this isn't the matrix the problem asks for at all

like, for one, the specification only allows the entries to be 0 or 1. there's literally no other value possible for any entry of your matrix.

[gaunt mulch]

the whole question

[dusky epoch]

this

isn't

the

matrix

the

problem

asked

for

at

fucking

all

[gaunt mulch]

but my fcking teacher has asked for it

[dusky epoch]

then tell us what the teacher is asking for, since they're clearly asking for something OTHER than what the problem asks for!

[sonic osprey]

Read part (a) slowly and carefully please

[gaunt mulch]

teacher has told us to complete all the question

[sonic osprey]

The matrix you wrote down is not what they want in part (a) at all

Yeah, and we're telling you that you're not doing what the question is asking for

[dusky epoch]

why is the (1,1) entry in what you wrote equal to 3, when the problem says all entries are either 0 or 1?

[gaunt mulch]

yeah i know

[dusky epoch]

clearly, 3 is not 1 and 3 is not 0.

[gaunt mulch]

i was tring to make a 4*4 matrix

so that i can calculate eigen value from matrix

[dusky epoch]

of course your matrix is gonna be 4 by 4

[gaunt mulch]

but how?????

[dusky epoch]

read the definition

[gaunt mulch]

ok let me try

This

[dusky epoch]

why did you not read the problem the first time around

[gaunt mulch]

i also dont know

[stoic python]

Mat:

[noble swallow]

@undone garnet

Maybe like this, I don't know if there is a quicker way to conclude though

[strong bison]

Hello friends

Could someone help me with this problem

I got until here (C) but don't know how to continue from here on

Sorry for the bad handwriting

[half ice]

Next column! You want a 1 on the second row there

So divide that row by 5

Then add it into the other rows as necessary

[wintry steppe]

how is this a vector space

like where is the unique 0 element?

it satisfies the addition operation and scaling operation but isnt that not good enough?

or does it mean that the 0 element is in F and since it is a sequence on F, it is a vector field?

[quaint heart]

Lmao it's the zero sequence

@wintry steppe

Consider the sequence of all zeros

That's the zero element in the vector space

[wintry steppe]

interesting

{0} this thing?

[quaint heart]

(0, 0, ...)

[wintry steppe]

how do you arrive at the 0 sequence from the definition?

[quaint heart]

Think about how it interacts with the other sequence

And which sequence would be the additive identity

[wintry steppe]

ohhh is it because V consists of ALL sequences in F?

[quaint heart]

Yeah

[wintry steppe]

interesting I understand

thank you sir

[gaunt mulch]

what to do??

[half ice]

Notice you can write the system as such:
[x'] = [0 2][x]
[y'] [3 1][y]

[gaunt mulch]

ok

[half ice]

The eigens of that matrix are everything the solution needs. Can you find them?

[gaunt mulch]

yup

but what about x at 0 is equal to 3

[half ice]

Once you get the general solution, you can use that to find the coefficients

[gaunt mulch]

ooooooo

thx

[half ice]

Np, think you've got it from here? Feel free to ask if you need

[wintry steppe]

suppose I had the set of all differentiable real valued functions defined on the real line

and I wanted to prove this set is a vector space

to show it has the zero element, is it sufficient to say "there exists a unique x \in R s.t. f(x) = 0"

or how else can I fix this/make it better

[sonic osprey]

What no

I think you're very confused about what you're trying to do here

[wintry steppe]

interesting tell me more

[sonic osprey]

The elements of your vector space are not numbers or anything, the elements of your vector space are functions

The addition of "vectors" in this space is the addition of functions

[wintry steppe]

so I need to show the set has the zero function?

[sonic osprey]

Well, is that the identity under addition?

[wintry steppe]

the function that for all x, f(x) = 0

that function must be in this set?

[sonic osprey]

If you're sure that's the identity then yes

[wintry steppe]

kaynex what do you think

[half ice]

Zoph has got this lol. I'm not stealing that thunder

[wintry steppe]

interesting so if the set is V, I should say "there exists f \in V s.t. f(x) = 0 for all \x in R?"

I think thats what my teacher wrote for the continuous functions proof

[sonic osprey]

That's right

[wintry steppe]

thank you zoph

[half ice]

If you're new to linear algebra, you may not know the term "identity" yet. Specifically, your "zero function" should not change any other function when you add them together

Which you may not need to prove in depth, it's probably enough to note that you have a zero function

[wintry steppe]

I see I understand

thank you guys v much

[heavy holly]

Hey

[wintry steppe]

If V is a vector space, then for v_1, v_2 in V, v_1 + v_2 is in V

this is correct right?

[sonic osprey]

Yes

[wintry steppe]

thank you

[dusky epoch]

why would it be otherwise

[strong bison]

Thank ye @half ice

[wintry steppe]

I used two inductive proofs for this

first I showed a1w1, a2w2, ..., anwn is in W

then I showed their sum is in W

is there a faster better way?

[quaint heart]

I don't think so

But this is the sort of thing you do once

And never again

So it's fine

[wintry steppe]

I see thank you

[blissful vault]

a system with 7 variables and 5 equations always has infinite solutions if there is at least one solution
why is that

?

[sonic osprey]

This isn't true

[blissful vault]

???

why

i was thinking that if there are more variables than equations then there are 000000 | 0

rows at the bottom

because in theory we were taught that zero rows = infinite solutions

[dusky epoch]

each row represents an equation

zero rows are a symptom of there being too many equations

[blissful vault]

what if there are not enough equations?

[sonic osprey]

Ah just kidding, it is true, my bad

[blissful vault]

but why?

i was taught in class that zero rows = infinite sol'n

but here we are lacking equations

[sonic osprey]

Well you want your system to be square, to have the same number of equations as variables

So you add in two zero rows

At least that's one way of thinking about it

[blissful vault]

okok

thx

[rose parrot]

Can someone explain example d) here? I don't really understand what it's saying, cannot really visualize it

Here's btw where $\bR^{[0,1]}$ and similar notation comes from (idk if it's a universal notation tbh, so just to be sure):

[stoic python]

KKZiomek:

[pallid swallow]

$\mathbb{R}^{(0, 3)}$ is the set of functions from $(0, 3)$ to $\mathbb{R}$

[stoic python]

Element118:

[pallid swallow]

If we just add two such functions, we have a function where $f'(2)=2b$. For our function to remain in the set, $2b=b$, so $b=0$.

[stoic python]

Element118:

[pallid swallow]

@rose parrot

[rose parrot]

Ohhh okay

Thanks for clearing that up

Not only that, but also we need to have an additive identity - the zero function

Which clearly cannot have a slope other than 0 at x=2

Since it's a constant

Unless I'm missing something

[sonic osprey]

That's the point

That's why it can only be a subspace if b = 0

[rose parrot]

Ok

Thx

[uneven fog]

Does anyone has any simple way to calculate rank of word as in dictionary

I mean i can do it it one by one but its very time consuming.
Eg find rank of ZENITH as in dict
Ans-616

[paper egret]

rank?

[uneven fog]

After how many words

The word will come

If they are arrenged as in dictionary

[paper egret]

that depends on how many words the dictionary have tho

[uneven fog]

No...

[paper egret]

im not understanding ur question

[uneven fog]

I will send u ques

[pallid swallow]

Do you have the entire dictionary?

Do you need an efficient way, like O(length)

[uneven fog]

@pallid swallowno

[paper egret]

binary search LUL

[pallid swallow]

ah, no wonder

[uneven fog]

We just arrange the word alphabetically then
Start to find how many words are possible

[pallid swallow]

the dictionary is all permutations

this is combinatorics

[uneven fog]

Unless i find the req wors

[paper egret]

oh

ooOh

[pallid swallow]

that explains

[uneven fog]

But it takes too much time

I was wondering if there is a trick

[paper egret]

i dont get it, but it does look like perms coms

[uneven fog]

Or smth like that

[pallid swallow]

Well, there's a way

but I think it's more efficient to just simply do that method

you can't go wrong that way

if the word is really long, I have a O(length * log (length)) method

but argh

might be less efficient due to the numbers being big

[uneven fog]

Prolly then i will keep doing it that way
That ques usually comes for 4 marks....

[pallid swallow]

yeah, that method is the cleanest way to do it

if you want to make it more efficient, well, imo not worth it

[uneven fog]

Lmao
Well....its life😂

[pallid swallow]

unless you are dealing with a string of length 50 or more

wait, make that 200 or more, by hand

[slow scroll]

what a strange dictionary to only contain the permutations of z e n i t h

[pallid swallow]

and not telling us about it until later

[rose parrot]

and it defines them all

What a linguistic revolution

[uneven fog]

@slow scroll its asking how the words will be arranged in sequnce simalr to in dict. The words are not in dict.

[indigo cradle]

can anyone help enlighten me on the proof of this theorem

i'hv tried following through it with a random vector but to no avail

[sonic osprey]

Which part are you confused about

[indigo cradle]

but it really doesnt make sense to me how you can come up with eigenvalue to "some nonnegative vector x"

when it may not be an eigenvector in the first place

[stoic python]

Zopherus:

Zopherus:

[indigo cradle]

@sonic osprey

sorry i still dont get it

[livid falcon]

Can someone explain to me what we did here? I know f(e1) = alpha1 * u1 + alpha2 * u2 , then we solve the system with f(e2) and f(e3) but I don't know how they did that here

[sonic osprey]

@indigo cradle Sorry I had to go to the grocery

@livid falcon Ask in one of the question channels

@indigo cradle No, you're not supposed to fix a single x

[indigo cradle]

@sonic osprey from my understanding of what u wrote previously, i fix a vector to get a tmax which should give me a eigenvector

but i was tryna show that tmax isnt an eigenvalue so that wouldnt be possible

[sonic osprey]

Because you're not supposed to fix a vector

Each t has a different vector

[indigo cradle]

but if i dont fix a vector

how would i find a tmax

[sonic osprey]

I mean it might not be easily findable

It doesn't have to be findable

[noble swallow]

The text is sort of condensed imo so I don't know if I'm getting it right, but I think what they are claiming is that if\ $T={t \in \bR: \exists x \neq 0, x \geq 0 \text{s.t.} ~ Ax \geq tx} $ \
then $\exists \max(T)=t_{\text{max}}$

[stoic python]

Mat:

[sonic osprey]

It might help you to think of it as $T = {t \in \mathbb{R} : \exists x_t \geq 0 : Ax_t \geq t x_t}$

[stoic python]

Zopherus:

[sonic osprey]

In other words, the choice of x depends on t

[indigo cradle]

thanks guys

[noble swallow]

But I don't understand how they would easily prove that T is bounded and sup(T) is in T

[indigo cradle]

i wud however need to take some time to decipher the set notation

im not familiar

[sonic osprey]

I think it's pretty clear that T is bounded?

Like I think you can show that ||Ax||/ ||x|| is bounded which would imply that T is bounded

[noble swallow]

@sonic osprey mm I don't know much about this, I found this on wiki

[sonic osprey]

Yeah that makes sense

[noble swallow]

But were we given that A is a continuous linear map?

[half ice]

Bounded operators don't have to be linear

I'm not sure about continuous.

[dusky epoch]

[noble swallow]

In this case A is linear however

[paper egret]

mmmm anyone know wtf linear transformation means?

according to my notes, it something to do with changing from nth dimension to mth dimension

wtf is that supposed to mean

[dusky epoch]

uh

[sonic osprey]

linear transformation just means a function that is linear

[dusky epoch]

linear transformations preserve addition & scaling

ie given two vector spaces V and W (over the same field, obv), a map L: V -> W is called linear if for every v1 and v2 in V and scalar c, you have L(v1 + v2) = L(v1) + L(v2) and L(c*v1) = c*L(v1)

[paper egret]

lmao i dont get thtat, but that's in my notes

👀

yea my notes say the same thing, but i'm not understnading wht the map is

T: V -> W

what does that really mena?

[sonic osprey]

T is a function from V to W

[paper egret]

V is a vector space of, for example, F^n, while W is a vector space of, for eample, F^m

[dusky epoch]

it doesn't matter what V and W are

beyond them being vector spaces

this is like the most basic defn in linear algebra

i mean @paper egret look

[paper egret]

👀

[dusky epoch]

you've got this new type of mathematical object

a vector space

[paper egret]

yes

i get that

part

[dusky epoch]

it's a set with some structure on it

and where there are sets

there are maps between them

but not any ol' map will do, because if your maps don't interact in any way with the structure you've placed on your objects

then your objects ain't much more than just sets

[half ice]

A function is linear if it splits over addition, and scalar multiplication.

[dusky epoch]

a linear transformation is just a transformation that plays nicely with the structure that makes a vector space a vector space

that's why you set these apart from all the other maps

and study those in detail

[paper egret]

lol you got any examples to show this

[dusky epoch]

examples of what

linear transformations?

[paper egret]

linear transformations

its not getting in my head

[half ice]

Any matrix

[dusky epoch]

i'm pretty sure that, unless your book is really shit-tier, it follows the definition with at least 2 or 3 examples.

[broken hawk]

watch the 3blue1brown video series

[half ice]

That's an important feature of linear algebra, a matrix can be seen as a linear transformation on a vector it is multiplying

[dusky epoch]

...or that, actually.

[paper egret]

is this the 3b1b lin alg thing?

series

[dusky epoch]

yes

[broken hawk]

yes

[dusky epoch]

EoLA

[paper egret]

aight will take a look at that

[broken hawk]

it’s the perfect supplement to a linear algebra course because of all the visual motivations

[paper egret]

the only things i have is professor's notes, and it just looks like outright hieroglyphics

[broken hawk]

but it sticks to a very low level of complexity (it pretty much only does stuff in ℝ² and ℝ³, which is fine for the most part but there’s of course more to linalg than that)

so it really can’t replace a course, but it’s a good supplement

[paper egret]

damn if this is low level complexity, i'm fucked

[broken hawk]

🙄

[paper egret]

AHAHAHAHHAHA

[broken hawk]

what I mean is linalg works with a decently broad variety of things, and while 3b1b’s videos give great motivation and visuals for low-dimensional linalg over ℝ, they won’t really help e.g. with infinite-dimensional stuff, or with linalg over finite fields…

but you can worry about those things later

[paper egret]

now im very worried

[broken hawk]

I mean, infinite-dimensional linalg is usually a follow-up course or coupled with functional analysis, so you probably don’t have to worry about that

and linalg over finite fields is not even interesting but sometimes gives counterexamples n shit

[paper egret]

👀👀👀

[broken hawk]

how rigorous is your course?

are you doing mostly computational stuff, or did you start right away with axioms of vector spaces over arbitrary fields n stuff?

[paper egret]

2nd day of lecture and we started doing proofs for linear transformations, thats all i know

[broken hawk]

seems reasonable

[paper egret]

we left off on homogenous linear equations

[half ice]

What's your major?

[paper egret]

idk how rigorous that means

lmao im thinkin about dual majoring in CS and mathematics

so i find this course to be PRETTY IMPORTANT

[broken hawk]

well then you’ll get the pleasure of seing all sides of linalg!

[paper egret]

WHICH MEANS IF I DONT GET IT, IM GONNA TRY MY BEST TO GET IT

[broken hawk]

cause math majors don’t care much about the computational side, and cs majors don’t care much about the proofy side of it ^^

[paper egret]

getting the best of both worlds LOL

[broken hawk]

but linalg’s fun

don’t be scared of it

[paper egret]

no im already scared

[half ice]

I'm a fan of doing a quick computational course before doing a more rigorous one, as lin alg can be covered in a chapter for a quick course

[paper egret]

mom take me hom im scared

[half ice]

3b1b is also great for computations

[paper egret]

welp im spending the whole dya today figuring this out, awesome

[broken hawk]

I say watch the videos first, at least the first few

[paper egret]

ye doing that rn

[broken hawk]

(though you might wanna revise them at some point)

[paper egret]

his voice is nice

[half ice]

They're not strong enough for a proof based course, but it's nice for getting intuition if you're missing it

[paper egret]

gotchu

[broken hawk]

a lot of linalg boils down to “there’s this intuitive thing we can do in low dimensional spaces… can we extend this to arbitrary spaces?”

(e.g. rotations, measuring angles, …)

the answer often turns out to be “yes, but it gets a tiny bit less intuitive”

“but the computations stay the same so who cares”

[paper egret]

ive kinda noticed that in some proofs

i guess thats where i might be getting confused on

cuz our professor just starts with n'th dimensions

and ur just like wtf

[broken hawk]

meh, an n-dimensional space is just a space where positions have n coordinates

and everything else stays the same¹

¹as long as you stick to linear algebra

[paper egret]

thats hard to think about it 👀

[broken hawk]

only if you rely too much on visuals

[paper egret]

waaaaaait.... wait wait waitttt

if i'm understanding this correctly

u change ur standard basis to some shite, and that acts as ur "function"

so lets say ur in 2 dimensions

and you have two vectors

[broken hawk]

you’re going to have to be a bit more precise about what you’re talking about

[paper egret]

oh lol

say ur in 2 dimensions, and u got 2 vectors, and then u have some transformation, solve that system and u get ur ok i dont know what im talking about now

OH, when it says linear transoformation from T:V->W, the dimensions are the same

damn idk how to exaplain it

[broken hawk]

calm down, take it slow, work through some examples

[paper egret]

you punch in vector

and it spits out anotha vector

[broken hawk]

write “you”

[paper egret]

sry

[broken hawk]

(idk why I care about that, I’m usually all about abbreviated writing but u for you just kinda annoys me. can’t explain why)

[paper egret]

god this is frustrating, i think i get it, but at the same time, i don't know how to put it to words

$T_A(x)=b$

[stoic python]

MemesPlease:

[paper egret]

ok this, really means that you have some matrix A, and you multiply by by some column matrix of x, to get another matrix b right?

and (x) is your transformation

or how it transforms

[half ice]

You can make a function T: V → W
That takes in vectors and spits out vectors.

This function is linear if the function:

• Splits over addition T(x + y) = T(x) + T(y)
• Splits over scalar multiplication T(ax) = aT(x)

[broken hawk]

the notation $T_A(x)$ to me would indicate this is a function which is equivalent to multiplying the vector $x$ from the left with the matrix $A$

[stoic python]

Sascha Baer:

[half ice]

Note the output need not have the same dimension as the input

[broken hawk]

that is $T_A(x) = Ax$

[stoic python]

Sascha Baer:

[paper egret]

wait how come the output doesn't need to have the same dimension as the input?

[broken hawk]

matrices aren’t always square

[paper egret]

HUH

yes

[half ice]

Who ever said it should have the same dimension?

[broken hawk]

what I wrote above is not quite rigorous yet

but I really don’t want to confuse you rn

so I’ll leave it at what I said

[paper egret]

i thought it had to have the same dimension

how can you transform a 2 dimensional vector to a 3 dimensional one

[half ice]

With a 3×2 matrix

[broken hawk]

you won’t be able to make a pretty animation of it

[half ice]

Or 2×3 idunno

[broken hawk]

2×3

[paper egret]

ok

ye

[broken hawk]

but here’s an example

[paper egret]

so 2x3 matrix means, there are 3 2x1 matrices

or u can split it up that way

[broken hawk]

wait what

[paper egret]

wait am i wrong?

[broken hawk]

okay, let me just make an example of a linear transformation without reference to matrices

take a 2-dimensional vector

and your function $T$ takes that vector, and just adds a 0 at the end. so $$T\left(\begin{pmatrix} 1 \ 1 \end{pmatrix}\right) = \begin{pmatrix} 1 \ 1 \ 0\end{pmatrix}$$

[stoic python]

Sascha Baer:

[broken hawk]

this is a function that takes a two-dimensional vector to a three-dimensional one

(this particular one is called an “embedding”)

[paper egret]

whoa that's a thing?

thats cool

[broken hawk]

check on a piece of paper that this function is linear

using the axioms of a linear function

[paper egret]

wait how do i check that?

[half ice]

That is actually a really cool example

[broken hawk]

This function is linear if the function:

• Splits over addition T(x + y) = T(x) + T(y)
• Splits over scalar multiplication T(ax) = aT(x)

verify that these two things hold

[paper egret]

uhhh

ok

WAIT

WAT

[half ice]

Where the input is any 2D vector
(So work with some vector [a b])

[paper egret]

THAT'S POSSIBLE?

YO MY BRAIN

ITS SHRINKING

[broken hawk]

the proof might actually be hard to write out for you simply because there’s almost nothing to prove

[paper egret]

wtf how

ok the constant one might be easy to show

but the addition one

[broken hawk]

lemme do the constant one for you

and then you can try the harder one yourself

[paper egret]

ok

[broken hawk]

so, let’s say $a$ is some real number, and $\begin{pmatrix} x \ y \end{pmatrix}$ is any 2D vector

[stoic python]

Sascha Baer:

[broken hawk]

now, $$T(a \begin{pmatrix} x \ y \end{pmatrix}) = T(\begin{pmatrix} ax \ ay \end{pmatrix}) = \begin{pmatrix} ax \ ay \ 0\end{pmatrix}$$, right?

[stoic python]

Sascha Baer:

[broken hawk]

with me?

[paper egret]

yes

[broken hawk]

on the other hand, $$aT(\begin{pmatrix} x \ y \end{pmatrix}) = a\begin{pmatrix} x \ y \ 0\end{pmatrix} = \begin{pmatrix} ax \ ay \ 0 \end{pmatrix}$$

[stoic python]

Sascha Baer:

[broken hawk]

also happy with that?

[paper egret]

yes

[broken hawk]

now you can immediately see that those are the same thing

so we get $$a T(\begin{pmatrix} x \ y \end{pmatrix}) = T(a\begin{pmatrix} x \ y \end{pmatrix})$$

[stoic python]

Sascha Baer:

[paper egret]

oh, now that i've seen it, the addition one doesn't seem to obad either

[half ice]

You type that very quickly

[paper egret]

right?

[broken hawk]

copy-paste

^^

[paper egret]

now, for the addition one, you just do the multiplication all on one line, and show that they are equal

[broken hawk]

and not bothering with \left and \right

[paper egret]

ok that makes so mcu hsense

linear transformations satisfies the two rules of addition and constant multiplication

[broken hawk]

while I’m not quite sure what you just wrote, let’s just roll with believing you got it

[paper egret]

so in short, T really

[broken hawk]

cause I have a point to make here still

[paper egret]

is a function

👀

[half ice]

These always boil down to "Simplify left side, Simplify right side, show they're equal"

[broken hawk]

now consider the matrix $$\begin{bmatrix} 1 & 0 \ 0 & 1 \ 0 & 0 \end{bmatrix}$$

[stoic python]

Sascha Baer:

[broken hawk]

what do you get if you multiply a vector from the left with that matrix?

a 2D vector that is

[paper egret]

like

(a,b)

for now?

[broken hawk]

ya

[paper egret]

you'd just get

a,b,0

[broken hawk]

seem familiar?

[paper egret]

matrix multiplication

is al i see

[broken hawk]

in short, multiplying from the left with that matrix is the same as applying the function we just looked at above!

so if we call this matrix $A$, the function from before would’ve been $T_A$ using your notation from before

[stoic python]

Sascha Baer:

[paper egret]

OH, So this is our magic thingy that makes things work

our transformation

[half ice]

If a transformation is linear, there's a matrix for it

(For finite dimensional stuff anyway)

[broken hawk]

(in finite-dimensional spaces)

yea

[paper egret]

dis some black magic

[broken hawk]

I should tutor linalg at some point

[paper egret]

+1

[broken hawk]

I think I’d be alright at it

[paper egret]

please do

[half ice]

Yeah Lin alg has a lot of cool results

[broken hawk]

I keep getting roped into tutoring other subjects

[paper egret]

OH WAIT, THATS WHY THE NOTATION IS $T_A(x) = b$

it makes sense

[stoic python]

MemesPlease:

[paper egret]

its almost like the function, with base A

A is the one that causes the transformation

[broken hawk]

and I’m not one to say no to a secured job to gamble on a potential other job that I might enjoy a tiny bit more but might instead just not get

[paper egret]

AAAAAAAHHHHHHHHH

[broken hawk]

so I’m tutoring algorithms and complexity next semester and not linalg I

[paper egret]

u must be better at algorithm and complexity than lin alg

👀

[half ice]

That sounds pretty okay too!

[broken hawk]

oh it absolutely is

pays less tho since it’s less hours

since linalg is 2h/week tutorium, algo is only 1h/week

and less homework to correct, which I’m not really gonna complain about

also I don’t think I would’ve wanted to tutor linalg under this particular prof anyway

he taught us methods of mathematical physics and his homework was ridiculous

I wouldn’t wanna correct that shit, and he gave bonuses for people who do a lot of homework so people would actually do it

and then I’d have to work

[paper egret]

ok one question, now wtf is this supposed to mean, every linear transformation is
$T = T_A$ for a unique matrix A

[stoic python]

MemesPlease:

[broken hawk]

that’s the thing kaynex just said before

[paper egret]

o

[broken hawk]

basically, if you have a linear transformation

which as we saw before can be described in creative ways

you can always find a matrix

such that the tranformation is really just multiplying with that matrix

and different matrices correspond to different functions (that’s the “unique” bit)

[paper egret]

oh that makes sense

T just means transformation then, and we can always find a matrix A that'll give our result

[broken hawk]

you should be pretty sceptical about it

it’s a powerful result

[paper egret]

ok that makes sense

the proof

is really funk

[broken hawk]

and not really obvious imo

like, in 3D space, I can describe to you the following function:
take the line going through (0,0,0) and the point (1,1,1). for your given input in the space, find the point on that line which is closest to it. Take that point and double all coordinates, and then swap the first and third coordinate for good measure. that’s the output

this is a linear function

is it obvious that you can find a matrix representing that?

[paper egret]

u basically gave the instructions

so i would think it's possible

[broken hawk]

if you know how to map each of those instructions to matrices then yea, you can do it

it’s definitely not an impossible thing to do

but it’s not, a priori, obvious to me that the step “find the point on that line which is closest to it” has a matrix associated with it, for example

(it does, it’s called a projection matrix)

[paper egret]

ic ic...

[broken hawk]

basically what I wanna get at is that while linear functions are kinda simple there’s still a lot of them and you should not just take that matrix result for granted

here’s another one

you know what derivatives are?

lemme run you through a funky example

[paper egret]

ye

[broken hawk]

I’mma assume yes. let’s take the space of all polynomials with degree ≤2

so constant, linear and quadratic polynomials

this is a vector space, because if you add any of those you get another one, and if you mutliply by scalars too, and stuff works nicely yadda yadda

feel free to verify if you don’t believe me but youc an just take my word for it

now, take the function D which maps a polynomial p to its derivative p‘

so D(p(x)) = p‘(x)

[paper egret]

uh huh

[broken hawk]

this is a linear function:
D(p + q) = (p + q)‘ = p‘ + q‘
D(ap) = (ap)‘ = ap‘

agreed?

(I left off the (x) because it’s kinda pointless, just imagine it)

[paper egret]

yes

[broken hawk]

your theorem states there’s a matrix associated with the derivative

and there in fact is

[paper egret]

yup

[broken hawk]

if we write the polynomial ax² + bx + c as a vector (c, b, a)

[paper egret]

😮

[broken hawk]

wait sorry

I want them in that order

[paper egret]

uh okay

[broken hawk]

it doesn’t really matter I just started writing the matrix that way

then the matrix $$\begin{bmatrix} 0 & 1 & 0 \ 0 & 0 & 2 \ 0 & 0 & 0 \end{bmatrix}$$ takes $\begin{pmatrix} c \ b \ a \end{pmatrix}$ to $\begin{pmatrix} b \ 2a \ 0\end{pmatrix}$, which corresponds to the polynomial 2ax + b

[stoic python]

Sascha Baer:

[paper egret]

black magic again

[broken hawk]

indeed

[paper egret]

that actually makes a ton of sense though

wait quick question

[broken hawk]

,rotate

[paper egret]

the n>m is fucking with me hard

[paper egret]

but i think i have an idea on what matrix A is

is this the right idea?

it should be like close to an identity matrix, with some messed up dimension change thingies

[broken hawk]

the matrix won’t be square

[paper egret]

yes, but im having trouble figuring out the dimensions

and where the 1's g

go

the stupid n>,

n>m

[broken hawk]

essentially what it has to do is “forget” about the last few coordinates

[paper egret]

it has to forget n-m coordinates

so n can transform to m

[broken hawk]

let’s say instead what if it kept it n-dimensional but the last few coordinates turned into 0s instead

could you work that one out?

I believe if you can work that out, then you can take it from there

think about the shape of the matrix too

[paper egret]

wait but wouldnt that make it so that its (1, ..., x_m, 0, 0, 0) instead of like (1,..., x_m)?

[broken hawk]

how many rows and columns does it have to have

yes, it would. it would be a different function altogether

[paper egret]

cuz im trying to avoid the 0's that come after x_m

[broken hawk]

it’s just a smaller side-exercise

[paper egret]

wait, hold on

how do you ignore those 0's???

[broken hawk]

but I believe you can solve the actual exercise if you do that

I can’t give you any tips without just spoiling the solution

but try to figure out the shape of A

maybe those things together will show you the way

[paper egret]

from what i got, it should be m * n

[broken hawk]

yes

[paper egret]

but from what i've drawn, it's more like n * n, cuz you got those 0's

[broken hawk]

if you make it n×n then you get the solution to the exercise I gave you

[paper egret]

this is WEIRD

[broken hawk]

now you just have to “lob off” those 0s

[paper egret]

n*n is a solution, but that would leave you with the matrix (x_1, ..., x_m, 0 0 0 0 0 0)

now to get rid of those 0's

[broken hawk]

actually how does your matrix that gives that look?

just to make sure

[paper egret]

ok now im getting an identity rectangular matrix, this is weird

wtf

wait wait hol on

gimme a sec

ok yea something aint right here

what i have for now

is a nxn matrix

but like the diagonals should all be 1, except the last couple of rows, but that wouldnt make any sense

cuz n * n matrix is square, and when the diagonal is all 1

it should reach the two corners

but it's not

[broken hawk]

yea sorry I can’t follow you

[paper egret]

i dont think im following myself either

UHHHHH

ok lets start back from the beginning

we know that n > m

[broken hawk]

I’ve heard worse ideas

[paper egret]

👀

n > m, means we are essentially shrinking it

x_1(1, 0, 0, 0,... 0) + x_2(0, 1, 0, 0, .... 0)

etc etc

[broken hawk]

yea. you’re taking a large vector and cutting off the last few coordinates

[paper egret]

let me work with a smaller example, only cuz im dumb

let n = 5, and like m = 3

[broken hawk]

alright, and actually, take a specific example too

let’s say the vector is (1,2,3,4,5)

[paper egret]

ok

[broken hawk]

you take it from here

[paper egret]

oh as in the resulting vector is (1,2,3,4,5)?

oh waitn o

starting vector

im dumb

can i use a different example

i did it on paper,

x_1(1,0,0,0,0) + x_2(0,1,0,0,0) + ... + x_5(0,0,0,0,1) = (x_1, x_2, x_3)
n = 5, m = 3 is what i wrote

[broken hawk]

uh, what you wrote here is false

[paper egret]

wait rly?

[broken hawk]

you can't sum up 5-dimensional vectors to get a 3-dimensional one

[paper egret]

how do you cut that off properly?

[broken hawk]

well that's your job to find out innit

[paper egret]

cuz its n = 5, and we have to make that into m = 3

ah shit

well for starters, we can make matrix A, column size m

wait

WAIT

hol up

[broken hawk]

okay lemme give you a hint... or maybe not I wanna hear what you hsve to say first

[paper egret]

we shud just make

it an m* m+n matrix

where the m*m part is identity

and the remaining m*n is just 0

GOT IT

it is a rectangular matrix

m * (m+n)

m rows, m+n columns

m rows * m columns is identity, and m rows * n columns is just all 0's

[broken hawk]

can you write out the actual result tho

[paper egret]

ye

dis what i got

[broken hawk]

,rotate

[paper egret]

wait actually hold on

it shouldnt be m*n

it should be more like

wait no

its not m+n

it should be

just n

m*n

[broken hawk]

sorry cat on my lap gonna be harder to answer

[paper egret]

where # of columns is n

OH WAIT I GOT IT

Lemme revise

gimme a sec

[broken hawk]

yea alright I can barely type now

[paper egret]

,rotate

[paper egret]

this is more like it

goteeem

[broken hawk]

well the top line is still wrong but the matrix is right

[paper egret]

wait

rly

which top line

OH

I SEE

Lemme fix it

aight fixed it

[broken hawk]

how it should actually look is $$x_1 \begin{pmatrix} 1 \ \vdots \ 0\end{pmatrix} + \dots + x_m \begin{pmatrix} 0 \ \vdots \ 1\end{pmatrix} + x_{m+1} \begin{pmatrix} 0 \ \vdots \ 0\end{pmatrix} + \dots + x_n \begin{pmatrix} 0 \ \vdots \ 0\end{pmatrix}$$

[stoic python]

Sascha Baer:

[paper egret]

yes

[broken hawk]

where the vectors have m coordinates

[paper egret]

ok ngl running thru a simple example actually helped me a shit TON

proof is hard

😢

[broken hawk]

you’ll get used to it

[paper egret]

ah shit now i gotta do it for m>n, welp shudnt be that bad

[broken hawk]

this shit’ll be trivial to you in a few weeks

but you have to work through it first ^^

[paper egret]

hopefully

[broken hawk]

I’mma go now tho

[paper egret]

aight

@broken hawk @half ice @dusky epoch thanks a shit ton for the help man, really appreciate it

❤

[broken hawk]

ah now the cat goes away

[paper egret]

[wintry steppe]

How do I find the slope y’all

[paper egret]

a little bit more context? @wintry steppe

[wintry steppe]

Ok

I’m in high school and can not remember how to find the slope. For example (4,7), (-3,6)

[paper egret]

wrong channel, but here's the formula
$\frac{\delta y}{\delta x}$

[stoic python]

MemesPlease:

[gaunt mulch]

I have question that can we interchange the rows before finding eigen values

[gray dust]

a matrix's eigenvalues are not guaranteed to be the same after you apply row operations

[dusky epoch]

\Delta @paper egret

@gaunt mulch no

[broken hawk]

in fact, even the simplest possible example you could think of, a permutation matrix, doesn’t have the same eigenvalues as the identity

[rose parrot]

I have a soft question regarding the methodology of doing exercises from Axler's book (or really any book tbh). Should I do half of the exercises for each section, and then do the other half some long time later, to have problems ready to refresh my mind on the topic if I need to?

Or should I do all of them at once

Because there's not that many per each section

One section has 10 while another 25

[broken hawk]

you could do that; you could also just revisit problems you did once half a year later, by then you’ll probably have forgotten parts anyway and can judge which are even worth redoing

(or, if all fails, just find some more exercises in another book)

I think it’s worth doing more exercises as long as you’re not finding them too easy

as long as they’re challenging, you’re learning stuff

[rose parrot]

Ok that sounds good