linear-algebra

[hexed mural]

well i can show you

what I am doing

[rose coral]

Sure

[hexed mural]

https://www.youtube.com/watch?v=xd8WlYrmPQg

I am gonna release the source code for free

[rose coral]

Woah that's really cool

[hexed mural]

I had to use quaternions to get that rotating properly

that took me a week to figure out the math for that

[rose coral]

Hahaha nice

Yeah that's pretty complicated

[hexed mural]

quaterions are wild

[rose coral]

Did you use some 4x4 matrices or something?

[hexed mural]

I didnt use a rotation matrix

i wrote out

it's actually really clever how i did it

i made a rotation Vector

and made a quaterion

out of it

[rose coral]

Cool

I was just asking how you multiplied the quaternions

[hexed mural]

ah

you do quaterion math

[rose coral]

Cause one way is to represent them as matrix

[hexed mural]

the non comunitive

way

Q1Q2

The engine has it

[rose coral]

Yeah so you coded in how to mutliply quaternions?

[hexed mural]

for toy

no

it was written

[rose coral]

Oh ok

[hexed mural]

I had to figure out how to make the quaterion

properly

that was a pain

[rose coral]

I see

[hexed mural]

basically turn a rotation velocity

in to a quaterion

[rose coral]

Yeah they're pretty wacky

[hexed mural]

so, the goal is to learn how to take any physics movement

make the collision and netcode it

from scratch as much as possible

people have been complaining about it

alot on how to do it

so i figured I'd do some simple plane game

you see in the video I am using a box for collision

I have not wrote it yet tho

[rose coral]

Mmm ok

But that's a sick goal

[hexed mural]

the geometry stuff

is a bit of my weakness

all this vector and geometry stuff is a pain it's been a long time since i've used any of it

You can play test what I have

if you wan

t

I'll add you

[rose coral]

Yeah it can be pretty unintuitive

[hexed mural]

if you're interested

[rose coral]

Sure, but I can't right this moment

[hexed mural]

no worries

[rose coral]

Yeah that sounds good

Thanks

[wintry steppe]

How would I find |u + v|^2 - |u|^2 - |v|^2, given u * v = 4?

write out |u + v|^2, probably

[quartz compass]

is that supposed to be a dot product?

[wintry steppe]

yea

no, they're working on an algebra and multiplying the vectors

/s

[quartz compass]

lol

if that's all you're given, then since they're saying it works for all vectors that satisfy that, just pick some specific vectors and compute it 😎

(jk don't do that, do it the way TTerra said it's much faster lol)

[wintry steppe]

wait

i can't say they're working on an algebra unless 4 is also in their vector space

uh oh

unless they're on some shit like V ⊕ R

[rose coral]

If it's a unital algebra then it should contain a copy of R

[wintry steppe]

@wintry steppe Hmm yea, but im only getting (u_0 + v_0)^2 + (u_1 + v_1)^2...

you don't need to write the vectors' components out

(they may not even have any, but i do not know the context)

it should go something like this: |u + v|^2 = <u+v,u+v>, and simplify using properties of the inner product

(if you don't know what an inner product is, then i'd wager you're working in R^n and so probably <a,b> = a^T b = a dot b)

Yeah I assume so

As far as im aware, inner product is a generalization of dot product, but most of the times my teacher is talkin bout dot prod

Im not aware of the properties tho

I don't see the link between the two tho

|u + v|^2 and a dot b

|u + v|^2 = (u +v) dot (u + v)

As far as im aware, inner product is a generalization of dot product, but most of the times my teacher is talkin bout dot prod
Im not aware of the properties tho
largely the same: linear, symmetric, positive definite (ignore people who say words like "sesquilinear" or "nondegenerate" for now :) )

So then the original question: |u + v|^2 - |u|^2 - |v|^2, given u * v = 4?
(u + v) dot (u + v) - |u|^2 - |v|^2

yeah

I feel like (u+v) dot (u + v) can be decomposed further but im not entirely sure

it can

And how would I evaluate |u|^2 too?

the same way

?

|u|^2 = u dot u?

we wrote |u + v|^2 as (u + v) dot (u + v), so we can write |u|^2 as

yes

(u + v) dot (u + v) - (u dot u) - (v dot v)

Okay this looks better than before

now try messing with that first term a bit

see if you can make something nice pop out

(v dot v) + 2(v dot u) + (v dot v)

Ah i see

Answer was 2(v dot u) whole time.

i think the first term should be u dot u

but

yes

Oh yep sorry typo

Thats interesting

So: |x|^2 = x dot x

yes

this is often taken as a definition

I've always thought of |x|^2, as |x| as sqrt(x0^2 + x1^2 ..), so |x|^2 = x0^2 + x1^2 ...

well they are the same

if you write out x dot x, what do you get?

Oh lol

Thanks heaps TTerra, I think I have better understanding now

no problem

[native rampart]

As far as im aware, inner product is a generalization of dot product, but most of the times my teacher is talkin bout dot prod
@wintry steppe inner product is same as dot product

You just use a different basis

[dusky epoch]

"inner product" is a more general term

[native rampart]

An orthonormal basis

[dusky epoch]

tho yeah all inner products on a findim vector space are equivalent up to a change of basis

[native rampart]

Well,If two things are equivalent, they are "same"

[dusky epoch]

but still if your space is not just R^n it might help not to be locked in the idea of dot product being the only inner product

[native rampart]

True

[muted holly]

but still if your space is not just R^n it might help not to be locked in the idea of dot product being the only inner product
@dusky epoch hi Ann probably being rly silly rn but can u gimme an example please =D

[dusky epoch]

oh idk polynomial vector spaces

like $P_2$ with the inner product $\ang{p,q} = \int_{-1}^1 p(x)q(x) \dd{x}$

[stoic python]

Ann:

[muted holly]

Ohhh right ok thanks

Should've thought of that

[old flame]

Suppose that V is finite dimensional and T ∈ L(V ). Prove that T is a scalar multiple of the identity if and only if ST = TS for every S ∈ L(V ). I could prove the forward direction starting from T, but lost in the other direction, pls help :3

[amber bay]

have you tried using the contrapositive

[spiral star]

think about the properties of the scaled identity. if you apply it to any vector it only gets scaled by a constant factor. so can you show that if TS = ST, then for all v in V: Tv = cv or in other words: {v, Tv} is linearly dependent? when you have that, you just have to show that the same factor c applies to all vectors. but that can be done by inspecting how a basis of V gets mapped.

[native rampart]

Or you can brute force it by taking appropriate S, belonging to L(v)

That almost always works

[old flame]

I'm just a little lost on the meaning of "a scalar multiple of the identity" cause like at the end don't we have to show something like $TI=aI$ for some constant a ?

[stoic python]

Otoro:

[real sable]

Hey I wasn't there from the beginning. What does T represent ?

[spiral star]

you have to show $T = \lambda I$

[stoic python]

Flow:

[spiral star]

for some lambda

[gray dust]

@real sable operator on findim V

[real sable]

Findim ?

[gray dust]

finite dimension

[real sable]

Right, sorry I don't learn Maths in English haha

[native rampart]

Try substituting S such that S(e1)=e1 and S(ei)=0 else

And do the same for other basis vectors,now take S(e1)=e2 and S(ei)=0 and repeat and you are done

[real sable]

What is the exact question ?

[spiral star]

just scroll up

[native rampart]

Suppose that V is finite dimensional and T ∈ L(V ). Prove that T is a scalar multiple of the identity if and only if ST = TS for every S ∈ L(V ). I could prove the forward direction starting from T, but lost in the other direction, pls help :3
@old flame

[real sable]

$Also, please ignore this message: I'm just trying the LaTeX bot$

[stoic python]

PrixMortDiable:

[native rampart]

#bots

[spiral star]

when i proved this, i went with a dirty contradiction argument 😎

[native rampart]

when i proved this, i went with a dirty contradiction argument 😎
@spiral star What did you do?

[spiral star]

what i suggested above

show that for all v in V, {v, Tv} is linearly dependent

and then showed that the factor is equal for all v

by observing how a basis gets mapped

i used contradiction for the linear dependence, because i was lazy and it only took one line

[old flame]

Here is my attempt for the proof of the first part. Let $(v_1,...,v_n)$ be a basis of V. Take $v \in V$, then $v=a_1v_1+ \dots +a_nv_n$ for $a_1,...,a_n \in F$. Apply T, so $Tv=a_1Tv_1+\dots+a_nTv_n$. Since $(Tv_1,...,Tv_n)$ spans range T and $(Tv_1,...,Tv_n) \in V$, since range T is a subspace of V. Since each $Tv_i$ is a vector in V, they can be written as a combination of basis vectors in V. $Tv_1=n_{i,1}v_1+\dots+n_{i,n}v_n$, so $Tv$ is a linear combination of the basis vectors of V. Thus, $Tv=cv$ for some constant c.

[stoic python]

Otoro:

[spiral star]

i dont think you can make that last conclusion

what you said before is true for every linear map, not just scaled identity

you have to show that the same constant c applies for all vectors

also, at some point you will have to use ST = TS

[old flame]

I thought you said I need to show that ${v,Tv}$ is linear independent ?

[stoic python]

Otoro:

[spiral star]

no, linearly dependent

that's a property of the identity map

(and the scaled identity)

[old flame]

so what you're saying is that I can't just use T, in some point I need to make use of the identity ?

then that concludes the linear dependence ?

[spiral star]

your goal is to show that if ST = TS for all S, then T is the scaled identity

so you can step by step show that T maps vectors like the scaled identy

one of the properties of the scaled identity is that it maps v -> cv for a fixed constant c

and {v, cv} are linearly dependent for all v

and if ST = TS for all S, then T will map vectors in that way

so you can split your proof into two steps

[old flame]

wait does range S = range T help in anyways ?

[spiral star]

step 1) if you take any vector v in V, then ST = TS implies that there is some c such that Tv = cv

step 2) in (1) you have shown that for each vector you find such a c. now you show that each c was the same

and then you are done

there are different ways of proving this as well, that's just if you want to follow what i did

[old flame]

I'm just having trouble trying to use ST=TS to be honest, but thanks for clarifying the steps, let me try again then

[native rampart]

Try taking a particular S

[spiral star]

yea, and then apply it to v

[native rampart]

Most linear algebra concepts are better understood if you try taking a case,and try to make sense of the whole thing that way

[deep flicker]

how can I find the function of graphs such as this?

or even this

[old flame]

@native rampart Something like $S : V \rightarrow V$, where $Sv=av$, for $v \in V$ ?

[stoic python]

Otoro:

[native rampart]

Yes, But take a more useful case like S(e1)=e1 ,S(ej)=0 for other basis vectors.(Your case reveals no information about T)

Think of transformations in terms of basis vector transformations

[old flame]

@native rampart What about this, $S: V \rightarrow V$, $Sv=b\sum_{i=1}^{n} a_{i}T^{-1}v_i$ where we supposed $\dim V=n$

[stoic python]

Otoro:

[old flame]

so if we write $v \in V$ as $v=\sum_{i=1}^{n}a_iv_i$ then S transforms it into that above and T brings it back to just bv, thus its a scalar b multiplied with the original vector v

[spiral star]

you cant assume that T is invertible

[stoic python]

Otoro:

[old flame]

Well I showed that range T=V, so T is invertible, does that work ?

[spiral star]

i dont think you showed that

[native rampart]

If it worked,what would it tell you about T?

Look for a use first,and then check existence.

For example,If I wanted to know something about T,knowing T(ei) would give me all the info I need

Therefore look for what S,you can take so that you know something about T(e1) for some basis vector

(You can get info about other basis vectors in a similar way)

[old flame]

Let me show you. I'm not sure but here goes. $ST=TS$, $T(Sv) \in range T$ while $S(Tv) \in range S$ \implies $Sv \in V$ and $Tv \in V$. So $range T=V=range S$ ?

[stoic python]

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

[native rampart]

Ok,Then what do you know of T you didn't know before

[old flame]

T is invertible ?

[native rampart]

Well,Range T is not V

It is a subset of V

Anyway,if that was true,would you have gained any knowledge about T(v)?

For some v

[old flame]

oh so my proof is not valid ?

[native rampart]

It is not valid

For that,you have to show an arbitrary element of V belongs to image of T

[old flame]

oh, so I need the other direction as well

[native rampart]

Anyway,That is not required for this question.

[old flame]

ah ok, don't wanna get sidetracked

[native rampart]

I mean,You do prove that,but this approach doesn't help you,as far as I am aware

Look for how you can know something about T(e1)

Like if $T(e1)=a1e1+a2e2+....$

How do you get info about a1,a2..?

[stoic python]

DrunkenDrake:

[old flame]

linear dependence or independence ?

[native rampart]

e1,e2.. are basis vectors of V

[old flame]

then the coefficients besides a1 would be 0 ?

[native rampart]

This is a linear transform

It takes in a vector,and spits out a random vector

Basically,a function

[old flame]

but Tv is in V, so it could be written as a combination of basis vectors nonetheless

[native rampart]

Yes

Exactly,Why a1e1+a2e2...

[old flame]

So are you saying to define $S: V \rightarrow V$ such that each basis vector gets mapped to a linear combination of the basis vectors. Something like this then, $S(v_i)=a_{1,i}v_1+...+a_{n,i}v_n$

[stoic python]

Otoro:

[native rampart]

That's how T is

We have to show $\a{n,i}v_n$ is the only non zero term in $T(vi)

Hopefully,you get my point

[old flame]

but then aren't we ignoring S ? since we are directly defining T ?

[native rampart]

We want T to be this

And we use S to get what we want

[stoic python]

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

[old flame]

So what you're trying to say is to define T as the transformation of a basis vector into a linear combination of the basis vectors, but with only that vector itself i.e others are zero

And with this in mind, we can see what S could be to do the trick ?

[native rampart]

We have to show this is the only valid definition

Such that the condition is satisfied

That's where we use S

[spiral star]

can you guys ping me when you worked through that approach? i'd like to share my solution in the end 🙂

[old flame]

sure thing, I am more than excited to see your solution too once I hopefully got it lol

[native rampart]

Just share your solution. He will learn from that.

Or should we continue?

[old flame]

let me try for a bit please 🙂

[native rampart]

Ok, try on your own. i won't disturb. Ping him,when you are done

[spiral star]

yea just continue. i think a proof without a contradiction argument in it would be preferable anyway

but it's nice to have some different approaches

gives some more insight

[native rampart]

Shouldn't the elementwise analysis be the most common?

Considering this is linear algebra

[spiral star]

yea i think so

but it was too cumbersome for me

so i kind of took a shortcut

[native rampart]

How is 2 steps too cumbersome?

[spiral star]

mostly writing out the linear combinations i guess lol

or maybe i overcomplicated it

[native rampart]

Probably latter

[spiral star]

true 🙂 yea it was just as short lol

[devout void]

Hello so our book says for 3 points to be in a line there are 3 and only 3 multipliers x1 x2 x3 which aren't =0 at the same time (x1^2 + x2^2 + x3^2 > 0)

although we don't have any logical reasoning or proof that explains and

and idk how to really use this statement on any exercise

anyone can help me

[wintry steppe]

can you post what the book says

[devout void]

@wintry steppe its albanian

and it says kinda what i said

anyone has any clue ?

[rose coral]

Ok I think I understand what it is saying

You're confused as to why it's true?

[devout void]

ye read it and makes sense

although the way it proofed it seems weird

idk

[rose coral]

Oh ok

Well without seeing where this is used, I couldn't tell you why they want the conditions $\lambda_1^2 + \lambda_2^2 + \lambda_3^2 > 0$ and $\lambda_1 + \lambda_2 + \lambda_3 = 0$

[stoic python]

The_Vman:

[devout void]

look

there are

3 points

[rose coral]

Yeah

[devout void]

and we need to prove that they are in a straight line

and the book says

if we take an origin point O

and 3 points ABC which create the line L:

we can create 3 vectors OA=r1 OB=r2 OC=r3

and now the book says

if we prove that $\lambda_1^2 + \lambda_2^2 + \lambda_3^2 > 0$ and $\lambda_1 * r1 + \lambda_2 * r2 + \lambda_3 * r3 = 0$

[stoic python]

Julinka:

[devout void]

then 3 points are in this straight line

[rose coral]

I see

[devout void]

wanne see how the book proved it ?

[rose coral]

Sure

I think I might understand it from the symbols

[devout void]

ill translate

[rose coral]

I have some ideas as to what they might do

Sure

[devout void]

im totally a rookie at linear algebra and idk if im stupid or it's just cause im learning the basics still

so the book splitted lambda into 2 other numbers

like i see what it's going to get at

but i would never come up with that solution

idk

i feel stupid when i see these op proofs

[marble lance]

I think that's not true. Consider the points (0,0), (1,0),(0,1) in R^2. Then your three vectors are 0, i, j. And 1(0) + 0(i) + 0(j) = 0 while 1^2 + 0^2 + 0^2 > 0, and yet the three points do not form a line. Or am I missing something and being dumb like always?

[rose coral]

Well it does take a while to be able to come up with these proofs, but it's all just practice. Practice doing proofs, talking to others, reading through proofs, spending time understanding

[devout void]

i see

well i guess rn i just need to understand it

@marble lance origin doesnt count

[rose coral]

You are missing the condition that 1 + 0 + 0 =/= 0

[devout void]

origin is a forth point

[rose coral]

No I think picking the origin is fine

But you want the scalars to add up to 0

[marble lance]

Ah, I wasn't looking that far up so didn't see the extra condition, thanks

[devout void]

although ye those won't be in a straight line cause condition is not met

aswell

[rose coral]

No problem

[devout void]

yea yea after some reading i get the conditions now

these vectors weird af

[rose coral]

Lol

After a while you get used to them

[devout void]

i hope so

[rose coral]

The best way of understanding them is visually in my opinion

[devout void]

i rly need to nail this exam

yeeee tru af

that's what im trying to do

[rose coral]

So whenever you can drawing some diagrams helps

That's good!

[devout void]

thanks Vman

you are the g

[rose coral]

Lol no problem but I don't feel like I did much. I can still help with the proof if you want

[devout void]

alright

hit me

[rose coral]

Ok but I need a sec to go over the proof myself

[devout void]

ye ofc

[rose coral]

They start by assuming that A, B and C are on the same line right? And they prove that there are real numbers $\lambda$ that satisfy the conditions

[stoic python]

The_Vman:

[devout void]

ye

[rose coral]

Ok so if you have two vectors AC and CB (with at least one not zero) that go in the same direction, the there are real numbers $k$ and $\mu$ such that $\mu * AC = k * CB$ with at least one of $\mu$ and $k$ not zero

[stoic python]

The_Vman:

[rose coral]

Right? Being able to scale vectors so that they are equal means they point in the same direction and vice versa

[devout void]

yes

[rose coral]

Ok so then they split $AC = r_C - r_A$ and $CB = r_B - r_C$

[stoic python]

The_Vman:

[devout void]

ye

[rose coral]

And that gives them the linear combination between the r vectors that equals zero

By rearranging the vectors as they did

[devout void]

ehm

wait wait

[rose coral]

Sure

[devout void]

so we have these 3 points

[rose coral]

Yes

[devout void]

and the book says

if those are in a straight line

ofc we can say that

AC = k*CB

ok till there

srry

lambda

same thing but follow the book

[rose coral]

Yeah

[devout void]

now the book randomly splits lambda into k/mu

and im like why

anyway so

i keep on reading

(k+mu)rc=mura+krb

this is after some simplification

?

[rose coral]

Yes

[devout void]

ahh

can i try real quick

to do what they missed on the book

and send the pic

[rose coral]

Of course

Yeah I didn't see how they got that either and I had to calculate it myself

[devout void]

oh

that was short

so just splitting AC

into OC-OA

[rose coral]

Yes

[devout void]

and OC = r3 and OA = r1

or Rc and RA

Ra

you get rc - ra = rb -rc

oops

rc - ra = k/mu()

(rb-rc)

k i messed it uo

up

ye ye i got it

[rose coral]

No worries

[devout void]

so that's the reason why they split it

to get k + mu?

damn

k i got it

[rose coral]

Yeah I think so

[devout void]

big iq xd

[rose coral]

Lol

[devout void]

okay so

they reversed the stuff there

rather finding the multipliers for the points

they found a generalization for multiplayers lambda

for specific point a b c

if that makes sense

after the part $\lambda_1 = \mu$

where is da bot xd

[rose coral]

Yeah it's taking a really long time

[devout void]

@stoic python

where u at xd

$\lambda_1 = \mu$

anyway

[stoic python]

Julinka:

[rose coral]

Lol there it is

[devout void]

ez

so am i correct?

[rose coral]

Yeah

That solves the first part

[devout void]

well the part where i get anxiety is when i think that although i got it for now and i can rewrite it without looking i cant memorize it for long till the exam along with 999 other proofs like this

[rose coral]

Mm ok

[devout void]

how did you deal at your time when you faced these types

ye and on the other part the book says let's prove the second condition where $\lambda_1^2 + \lambda_2^2 + \lambda_3^2 > 0$

[stoic python]

Julinka:

[devout void]

so he says lets just say $\lambda_1 =not 0

[rose coral]

I did a good amount of proofs, but also I tried to understand how they worked when I could. There are also some concepts that the book/professor emphasizes more and those are usually more important, in that they help with the rest of the concepts

[devout void]

did you excel all your exams

im on EE btw finishing first year

[rose coral]

Sometimes

[devout void]

we did some stuff online or lets say all of it online

and im not sure if our prof emphasized anything

so i need to learn everything

[rose coral]

Hmm ok

[devout void]

yee sad

[rose coral]

Even still, there are some things that come up more often

[devout void]

well tbh

we just did these for intro

what i need to remem is that 2 vectors are colinear if they fulfill the condition a = kb

and vice versa

the dot product, cross product

etc

which i have learned

kinda

[rose coral]

Ok yeah those are good to know

[devout void]

ye anyway Vman do you do online tutoring

like for pay etc

[rose coral]

Um noo I haven't done any

[devout void]

ic

well ye thanks man

[rose coral]

No problem

[devout void]

you are doing god work helping people

like that indian dude who always saves your azz before the exam xD

[rose coral]

Hahaha thanks

[devout void]

imma head to learn

[rose coral]

Sounds good

[gaunt field]

can someone help with this? https://gyazo.com/455ef39c4b812501af75578444f93970

[vague cedar]

@gaunt field

[gaunt field]

I don't see the connection 😦

Okay, so if I use the first inequality, do I have to prove it first?

[vague cedar]

maybe?

dunno what your professor wants

[gaunt field]

me either LOL, it's the first homework so i dont know how he grades

[vague cedar]

also for the second one you use this definition of a dot product

$v\cdot w = |v||w|\cos\theta$ where $\theta$ is the angle between the vectors

[stoic python]

jacob:

[gaunt field]

oh yeah I'm familiar with that one

[vague cedar]

so $\min v\cdot w = |v||w|\min \cos\theta$

[stoic python]

jacob:

[quartz compass]

similarly for the first part |v-w| you should expand |v-w|^2 as a dot product (v-w)*(v-w) and you can get an answer that way

[vague cedar]

oh true

[gaunt field]

similarly for the first part |v-w| you should expand |v-w|^2 as a dot product (v-w)*(v-w) and you can get an answer that way
@quartz compass can you quote which one your referring to? bit confused where the |v-w|^2 comes from

[quartz compass]

I would but I gotta go, I think jacob can help

[gaunt field]

ok np!

[prisma pier]

what's the question?

[gaunt field]

@vague cedar So assuming I proved it I can get | | v - w | | > = 2 ? and that's it?

[vague cedar]

if you proved the inequality sure but i think merosity's way is better tbh

[gaunt field]

oh i have to find largest possible

can you explain what merosity was talking about?

[vague cedar]

$|v-w|^2 = (v-w)\cdot (v-w) = v\cdot(v-w) - w\cdot(v-w)$

[stoic python]

jacob:

[vague cedar]

squaring it and then taking the square root after you're done with your work is just a helpful trick

[gaunt field]

how come I'm squaring it though?

[vague cedar]

to express it in terms of |v| and |w|

[gaunt field]

oh rightt

[vague cedar]

$|v-w|^2 = v\cdot v - v\cdot w - w\cdot v + w\cdot w$

[stoic python]

jacob:

[vague cedar]

does that make sense from the previous expansion

lemme know once ur good to move on

[gaunt field]

yep makes sense

[vague cedar]

the dot product is commutative ie. a dot b = b dot a

$|v-w|^2 = |v|^2 - v \cdot w - v \cdot w + |w|^2 = |v|^2 + |w|^2 - 2v\cdot w$

[stoic python]

jacob:

[vague cedar]

$|v-w| = \sqrt{|v|^2+|w|^2-2v\cdot w} = \sqrt{|v|^2+|w|^2-2|v||w|\cos\theta}$

[stoic python]

jacob:

[vague cedar]

now you have |v-w| as a function of a single variable (theta) and you can optimize it as you'd like

you might recognize this as reminiscent of the law of cosines from geometry 🙂

[gaunt field]

hmm

[vague cedar]

lemme know if you have any other questions

or anything you want me to elaborate on

[gaunt field]

so the min value would be when theta is 0?

i mean for v-w

[vague cedar]

yup

[gaunt field]

its kinda hard to figure out what the angle would be at max

[vague cedar]

you want cos theta to be as small as possible right?

cuz you're subtracting that

[gaunt field]

yeah

[vague cedar]

cos theta = -1 at theta=pi

and min cos theta = -1

[gaunt field]

okay i see i think

[vague cedar]

alternatively you can take some derivatives and plug them back in and see what's a larger number

optimizing $|v-w|$ for some value $\theta^$ is the same as optimizing $|v-w|^2$ for some value $\theta^$ so you could do $\frac{d}{d\theta}(|v|^2+|w|^2 -2|v||w|\cos\theta)=0$ and solve for $\theta$

[stoic python]

jacob:

[vague cedar]

which boils down to $\theta=0$ or $\theta=\pi$ for $0\leq \theta \leq 2\pi$ and you can plug in the two angles to compare some values (including the endpoints $0$ and $2\pi$)

[stoic python]

jacob:

[vague cedar]

first derivative test is what i think it's called from calc 1

[gaunt field]

okay so my answer isn't an actual number but an expression right? so like the smallest for | v - w | would be = sqrt(|v|^2 + |w|^2 -2)

oh nvm

i can plug it in

lol

thanks!

[prisma pier]

I saw that you can write the determinant of a matrix in terms of the trace of a matrix, with:

for the determinant a 2x2 matrix

does anyone know if there's a nice pattern for larger matrices written in terms of traces like this?

I think for 3x3 it works out to:

actually nvm, found something on Wikipedia

[gaunt field]

https://gyazo.com/316941bd883abcf8613a84abe930e2c2 how does one do this 😦 i skipped it earlier

[limber sierra]

so you want $u_1 v_1 + u_2 v_2 < 0$, $u_1 w_1 + u_2 w_2 < 0$, and $v_1 w_1 + v_2 w_2 < 0$, right?

[stoic python]

Namington:

[limber sierra]

suggestion: make one vector all positive, one vector all negative, and one vector one positive/one negative

youll have to adjust the magnitudes to make it so that the negative term in each dot product dwarfs the positive term

but this shouldnt be too hard with some trial-and-error

[gaunt field]

so I just plug in stuff till it works?

[limber sierra]

well, there should be a bit more thought than that

like, if you have, idk

(-1) * (-1) + 1 * (-1)

it would make sense to make one of the factors of the second term larger

since the second term is the negative one

so you want that to be the larger term

[gaunt field]

makes sense yeah

[lucid cedar]

Hello, This is the Solution to the question "Find two distinct square roots of i."

My question is to ask that about this process of adding 2pi to the power of e at the top of the solution. does this mean that I can keep adding 2pi repeatedly and create distinct roots every time? It would not make sense to me that sqrt(i) has infinite solutions.

Oh wait, I see that it wouldnt have infinite roots, but i dont really understand how I am supposed to know that adding 2pi(i) would get me the other root. To me it seems more intuitive that adding pi would do that, but I dont have a good grasp of what is happening here

[merry cave]

e^(i*θ) represents the how the argument changes in an anti-clockwise direction and we can get other roots by rotating 2π/x in the plane where x is the number of roots.

Adding 2πi/2=πi to the angle will rotate the number π radians in the complex plane.

Another way to think of it is that e^(πi/4) is equal to cos(π/4) + i*sin(π/4) = 1/sqrt(2) + i * 1/sqrt(2) and adding πi to the angle results in it rotating π radians in an anti-clockwise direction so e^(πi/4+π)=e^(5πi/4) = cos(5π/4) + i * sin(5π/4) = - 1/sqrt(2) - i * 1/sqrt(2) which makes sense since the number would have rotated 180 degrees in the complex plane.

Like the black dot would be the first root that was calculated and the second root was the π radian rotation of that number in the complex plane

and that is represented by e^(iπ)

so then you have e^(iπ/4) * e^(iπ) = e^(iπ/4 + iπ) = e^(5πi/4)
I hope that makes sense

[lucid cedar]

it does

I appreciate your answer, thank you

[merry cave]

you're welcome

[lucid cedar]

I have another question, more conceptual perhaps this time. The book im using, Linear Algebra Done Right, talks about vector spaces saying they map something to something else. So in this specific example that confuses me Axler says "If S is a set, Then F^S denotes the set of functions from S to F."

So I can easily agree given the rules of combining functions that they classify as a vector space. The part that I dont understand is what is meant when we say "to F". F is just a stand in for field If I read Axler right. Is F^S just a formal declaration that the set S is a field?

[half ice]

What page?

[lucid cedar]

14

chapter 1. section 1.B

[half ice]

F is assumed to be a field, but S is just a set

Then a function S → F is just a mapping that says where each element in S gets mapped to

[limber sierra]

F^S is just the set of functions with domain S and codomain F

you'll see this notation all the time; for example, the set of all real-number sequences is often denoted R^N

[lucid cedar]

so S->F because the functions inside of S must resolve to some number? IE: S_1 = f(x) = x^2

[half ice]

Like, take S to be a set with just one element.
S = {sunshine}

Then F is any field
Why not R

One such function we could put on this is mapping sunshine to 1.

Another is sunshine gets mapped to π

[lucid cedar]

Oh so like a bijection from elements of S to the real numbers or the complex numbers, thus we use F for either?

[half ice]

My example is very specifically not a bijection haha

A function needs not be a bijection, is what I mean

[lucid cedar]

okay that clears it up i think

[half ice]

Now, F^S is a set of all such functions

[lucid cedar]

oooooo

my brain is smooth like a marble. This is actually simple

I miss understood and thought that S was a set of functions. IE S := {x, x^2, x^3,...}

[half ice]

Rofl that's a new expression

S could be a set of functions!

But those functions aren't being used

[lucid cedar]

right right whew. thanks so much

[half ice]

Np. Feel free to ask if you need anything else!

[viscid kernel]

How does linear mapping between complex planes work ?

Any intuition yall can give me ?

[dusky epoch]

wym complex functions

[viscid kernel]

I mean complex planes

Sorry Ima change it rn

[dusky epoch]

that doesn't make much more sense

i mean definition-wise it really is the exact same except all the real scalars are now complex scalars instead

yeah you lose some of that geometric visualization but that's just the cost of going more general

[viscid kernel]

Hmm if its the same, then its ok. But I thought it would be like something else

Exactly

I was actually wondering the geometric visuals

[dusky epoch]

look, most of linear algebra that doesn't rely on inner products can be reformulated to work over any field

not just R or C

[viscid kernel]

Linear mapping keeps the gridlines paralel and origin fixed, so these properties are both true between reel/complex planes ?

[dusky epoch]

i

honestly think you're over-relying on the geometry rn

[viscid kernel]

How Am I supossed to understand it if I cant visualise it tho. 😦

[dusky epoch]

the definition of a linear mapping is that a mapping L is linear iff for every vector v, w and scalar c it is true that L(v+w) = Lv + Lw and L(cv) = c * Lv

like there are examples of linear operators that are not just matrix multiplication on R^n or C^n

for example the derivative

that's a linear operator right there

and other differential operators, like the second derivative, or the laplacian if you're doing multivariable, or integration...

idk. you gotta be comfortable with some amount of symbol-pushing

[viscid kernel]

Hmm, ok. I think I know which way you are going now. Is there anything you can recommend me ( book,video,... ) to understand like the essence of these things ?

[dusky epoch]

i might've recommended essence of linear algebra before but it sounds like you've watched that already

[viscid kernel]

Yeah

Tbh I watched that series 3-4 times

I dont imagine myself proceding this much in college if those videos didnt exist 😄

Anyways Ima continue my work, thanks @dusky epoch

[night citrus]

is there a unit vector notation that can be used to indicate "unit vector of (a × b)" without having to say something like "let unit vector of (a × b) be equal to û"

[native rampart]

(axb)^?

[night citrus]

i thought about making it (a × b) with a hat but it looks too wonky to be real notation

it is the actual notation for it?

[native rampart]

I don't know

[rare swallow]

@night citrus Just normalize it? That is, say û = (a × b)/||a × b||

[night citrus]

that's what i was trying to avoid, but if that's the correct way then its fine

[vague cedar]

i would write $u_{a\times b}$ for what you're trying to achieve

[stoic python]

jacob:

[vague cedar]

@night citrus

[night citrus]

ah neat! thank you

[hollow finch]

say we have a 4x6 matrix A which has a rref of R and we find a 4x4 matrix T for which TA=R. Is T unique?

[gray dust]

A=4x6 0 matrix. rref(A)=A. ANY T satisfies TA=A

[hollow finch]

I mean a specific A matrix. But I see your point. I suppose there are some conditions which would determine if T is unique (if that is possible). The A i am considering is rank 3

[half storm]

Well you could prove it, right? So suppose that there exists a 4x4 matrix U such that UA = R.

A U has the associated left-multiplication transformation L_U. We know that linear transforamtions are determined uniquely about how they act on a basis.

So maybe the next thnig that you would need to show is that $L_T = L_U \iff Tx = Ux$ for all $ x \in \mathbb{R}^4 $. Just a thought.

[stoic python]

JohntheDon:

[half storm]

T is necessarily the product of elementary matrices yea?

[hollow finch]

yes T is invertible

[half storm]

So I guess the question is the sequence of elementary row operations is unique.

I'm pretty sure there are plenty of different row operations that you can do in order to end up with the same result, so I'm gonna go ahead and say that the answer is no.

T isn't unique.

[hollow finch]

i think im inclined to agree

but i wonder if that changes if A is rank 4

then wouldnt we be able to let T be the inverse of the matrix composed of the pivot columns of A?

[half storm]

I'd venture to say no to that aswell.

As long as you end up in the rref at some point, then it doesn't matter how many row operations that you do, if you get what I'm saying.

There is definitely a set of matrices that correponds to a set of row operations where you end up with the least amount of elemntary matrices that gives you the rref of A i.e. R.

that is a unique number of matrices i.e. the last number of row operations/ elementary matrices needed to give you the RREF of A, but you can get to the RREF by doing a bunch of different row operations, so long as you end up with the RREF of A.

[hollow finch]

if A is 4x6 and rank 4, then wouldnt T to get to rref have to be the inverse of the pivot columns though?

since the columns are separate and the pivot columns together become the identity matrix

[half storm]

inverse of the pivot columns, I'm not sure I know what you mean.

[hollow finch]

the columns which have leadings 1s in R

[half storm]

oh yea I know what you mean by pivot columns

what do you mean by their "inverses"

[hollow finch]

if we label the pivot columns c1,...,c4 even if thats not their order in A

then then if we let B=[c1...c4]

T would necessarily have to be the inverse of B

since in R Tc1=e1...Tc4=e4

where ei is the ith column of the identity matrix

sorry for not doing latex and subscripts im in physics class rn lol

[half storm]

O.k. I see what you're saying. You just mean that T of the columns gives you standard basis vectors.

And what you said is definitely true, but I don't think that contradicts the point that I've made.

[wintry steppe]

focus on lecture

[half storm]

You can use any row operations that you want so long as you, as you end up in the RREF of the matrix, there a bunch of different operations that you can use to get there.

some of them would actually be redundant - it wouldn't necessarily be the quickest way to get there.

[gray dust]

@wintry steppe hypocrite

[hollow finch]

hey its being recorded so i can always go back

[half storm]

but you could still get there is what I'm saying, it may be hard to put in word what I'm saying without giving an example.

[hollow finch]

T=E1...Ek=F1...Fm where Ei and Fi are elementary matrices

they dont have to be the same i gotcha

but they would still end up being the same matrix because the inverse of a matrix is unique

[half storm]

yup yup

So yea if the rank changes to 4, then they have to be the same matrix.

[hollow finch]

interesting. so then i would conject that if an mxn matrix has maximum possible rank(m and/or n depending on which is larger), then the transformation to rref is unique.

ill see if i can prove that. but i think it would really be exactly what i said above using that an inverse matrix is unique.

[wintry steppe]

[half storm]

Yea I think so too.

I guess it kind of depends on what you define as transformation. Like, the number of row operations required to reduce to the RREF wouldn't be, but if you think of the transformation the matrix such that TA = I_n, then yea, it definitely is.

[hollow finch]

right. any invertible transformation can be decomposed in an infinite number of ways as the composition of of smaller invertible transformations, but the final result can always be the same.

[wintry steppe]

man doing linear algebra selfstudy is too hard, can someone explain me in easy terms how do 2 matrices relate to eachother relative to different basis?

or something to help me starting how to tackle this problem

[half storm]

What exactly do you mean "how do 2 matrices relate to each other relative to different bases"

[wintry steppe]

it means can we predict how the matrix would look like from someone that chose a different basis.

i donmt understand the question all the well either, hoped somone could give me some insight

look into change of basis matrices

[half storm]

That's a slightly better wording of the question.

but yea look into what change of basis.

If I have a matrix then how the resultant vectors look from someone elses perspective depends on what basis that they're looking at the resultant vectors from.

[wintry steppe]

the idea is that with respect to a choice of basis on the domain and codomain, we can represent a linear map by a matrix. since we can always pick multiple bases, we want to figure out how to relate the two matrices we get when we make different choices of bases. this is (for a linear operator at least) given by conjugation by a "change of basis matrix"

the math behind it isn't terribly difficult, but the idea (as in the "why?") can take a bit to wrap your head around

picking a different basis can make computations a lot simpler, so knowing how you change from one basis to another is important

e.g. if your problem is to find the expression for the linear operator on R^2 projecting to a line, then you can use the line to get a new basis of R^2 in which projection has a very simple form, then use your change of basis matrix to get the expression of that linear operator in the standard basis (what you originally wanted). thus, making computations easier

[wintry steppe]

I see, appreciate the clarification

[copper pulsar]

How can I solve this matrix equation $ (X-B)^T(X+A)=X^T\cdot A+X^T\cdot X+B^T\cdot X$

[stoic python]

sqrt(x):

[wintry steppe]

expand the left side, i guess

[copper pulsar]

How?

I simplified the right side $X^T(X+A)+B^T\cdot X$

[stoic python]

sqrt(x):

[copper pulsar]

I could multiply both sides by $(X+A)^{-1}$

[stoic python]

sqrt(x):

[copper pulsar]

yes, they are all regular matrices

They are all invertible

I know that $B^TX=(X^TB)^T$. But this does not really help me

[stoic python]

sqrt(x):

[half storm]

Transposition of matrices is a linear property.

[copper pulsar]

You mean $(A+B)^T=A^T+B^T$?

[stoic python]

sqrt(x):

[half storm]

yea

[copper pulsar]

How does that help me?

[half storm]

Oh my bad I checked the end result.

What you are trying to show.

[copper pulsar]

I have to solve $(X-B)^T(X+A)=X^T A+X^TX+B^TX$ for $X$

[stoic python]

sqrt(x):

[half storm]

You lost a negative sign in there.

I think.

[copper pulsar]

Nope, that's the original question

[half storm]

Oh wow

nvm.

I'm just using some inductive/plausible reasoning

[copper pulsar]

It's a little awkward

[half storm]

just seems weird that's what the end result has to be.

[copper pulsar]

I don't know how

Is (X-B)^T(X+A)=(X^T-B^T)(X+A)?

I see

yeah, then its fine

[limber sierra]

remember that transposing is just swapping the rows and columns

it doesnt matter whether you swap rows or columns before or after doing the subtraction

[copper pulsar]

So X^TX+X^TA-B^TX-B^TA

well thanks, i was a little uncertain whether $(X-B)^T(X+A)=(X^T-B^T)(X+A)$

[stoic python]

sqrt(x):

[half storm]

Yea transposition is a linear map.

[copper pulsar]

btw. I guess it's more elegant to $(X^T-B^T)(X+A)=X^T(X+A)-B^T(X+A)$ looking at the right side

[stoic python]

sqrt(x):

[copper pulsar]

It cancels

Is $X=-0.5A$ correct?

[stoic python]

sqrt(x):

[pulsar turret]

@wintry steppe I’d like to come back to our question few hours ago. Could you explain why?

[wintry steppe]

the "why" is what i wrote, if that's what you mean

i dont mind elaborating on anything else

[pulsar turret]

Well I’m having trouble visualizing the proof

I’ll show you how it’s wittten although it’s not in english so I hope you can somewhat make sense of it

So I guess you’ve got 1 object seen from 2 different basis. And you work it out so that you have an basis e’<k> written as a summation of another basis e<k> multiplied by a matrix of some sort

[wintry steppe]

hm

i don't understand this language so i'm not sure i can help too much

[pulsar turret]

Hmm I see

[rose coral]

What is the question? Why the change of basis is represented as a matrix?

[pulsar turret]

I’m not so sure lol

[rose coral]

Ok

Well, I can talk a bit about changing basis if you want, and if it doesn't make sense just stop me

Or if you know what kind of thing you're confused about let me know

[pulsar turret]

It’s a proof of how to know you’re looking at the same object (vector or matrix?) when the basis is different. So you and someone else have different basis. How to know you’re looking at the same object

[rose coral]

Ok

So you have a vector $v$ in your vector space. If you choose a basis $e_1, \dots, e_n$ then you can write $v$ as a linear combination $v = \sum_{i = 1}^n a_i e_i$

[stoic python]

The_Vman:

[rose coral]

And then you get the "coordinates" $a_i$ which let you represent the vector as $\begin{bmatrix}a_1 \ \vdots \ a_n\end{bmatrix}$

[stoic python]

The_Vman:

[rose coral]

But these coordinates depend on the basis you chose

[pulsar turret]

Yes makes sense

[rose coral]

And so if you had chosen a different basis $e_1', \dots, e_n'$ you would have $v$ as a different linear combination $v = \sum_{i = 1}^n b_i e_i'$

And that gives different coordinates

Ok so now the question is, how do you know if you have the same vector? Or maybe, if you know the coordinates in one basis, what are the coordinates in the other?

[stoic python]

The_Vman:

[rose coral]

So if you know nothing about the bases, then you can't do anything

You need some information that relates the basis

In particular, it is enough to know what the vectors $e_i$ look like in the basis $e_i'$

[stoic python]

The_Vman:

[pulsar turret]

Okey, that’s what was proven at the end right?

[rose coral]

I'm not completely sure, but that's how you find the change of basis matrix

Then, if for example $e_1 = 2e_1' - e_2'$ and $e_2 = 3e_2'$ then if you know $v = 3e_1 - e_2$ then you can find what linear combination of $e_i'$ works for $v$

[stoic python]

The_Vman:

[pulsar turret]

Makes sense

[rose coral]

Namely $v = 3e_1 - e_2 = 3(2e_1' - e_2') - (3e_2') = 6e_1' - 4e_2'$

[pulsar turret]

Any idea why automorfisme was introduced to try and prove this?

[stoic python]

The_Vman:

[rose coral]

Hmm let me think

[pulsar turret]

Automorfisme can be represented by invertable matrix Q. That’s where the Q comes from

[rose coral]

Yeah

I mean the way I think about it, is that you aren't doing anything to the vector

So it's the identity automorphism

But you are representing it in a weird way

Because to represent the identity automorphism as a matrix, you need to chose some bases.

And you get to choose a basis for the input and the output

So normally, you only deal with one basis and so you use that basis for the input and output, and the identity automorphism looks like the standard diagonal identity matrix with 1s on the diagonal

But if you are working with different basis, then the identity map can look different.

For example $\begin{bmatrix}1 & 1 \ 0 & 2\end{bmatrix}$ could represent the identity automorphism if you are going from a basis $e_1, e_2$ to a basis $e_1', e_2'$ where $e_1 = e_1'$ and $e_2 = e_1' + 2e_2'$

[stoic python]

The_Vman:

[pulsar turret]

I see

[rose coral]

As when you input the basis vector $e_2$ for example, which is written in the basis $e_1, e_2$ as $\begin{bmatrix}0 \ 1\end{bmatrix}$, you get $e_2$ back (as you applied the identity automorphism which does nothing) but now you are writing the $e_2$ vector in the $e_1', e_2'$ basis where it looks like $\begin{bmatrix}1 \2\end{bmatrix}$

[stoic python]

The_Vman:

[pulsar turret]

Okey

It says here that Q is the “transition matrix” just like you said applying identity automorphisme

Matrix Q contains de coords in koloms of the basis B’, expressed in the basis B.

[rose coral]

ok

[pulsar turret]

That’s what you’re doing right here right

[rose coral]

Yes, I think so. If the columns are basis vectors of B' written in the basis B, then it should take a vector written in the basis B' to itself written in the basis B

The columns are the outputs (each one is the output of putting in basis vectors of B' written in B'), and so if you write the columns using the B basis, that's what the basis you end up in is

That's how I try to remember it, how am I writing the columns (using which basis)? Because that's what all outputs will be written in

[pulsar turret]

Makes sense tbh

[rose coral]

Glad to hear it

[pulsar turret]

So to make a summary of this. You’ve chosen a basis b ( e1, ... , en) and someone else choose b’ (e’1 , ... , e’n) because their both isomorphism, it’s actually automorphism. This automorphism can be represented by an invertable matrix Q.

This matrix Q, is based on the coords of b’ and b. So the coords of the [V]b = Q * [V]b’. This means that matrix times the vector V using basis b’ gives us coords of vector V using basis b

[rose coral]

Yes exactly

[pulsar turret]

Theres a last part, we didn’t talk about it yet, but it’s just substituting some variabeles and numbers

[rose coral]

The important thing to remember is that whatever numbers you use to represent a linear transformation, whatever bases you use, the things intrinsic to the linear transformation stay the same. The rank, the dimension of the nullspace/kernel, if the map has an inverse, etc. won't change from basis to basis

Sure go ahead

[pulsar turret]

Thanks for reminding me

At the bottem stelling 6.8

Q is the transition matrix from basis B to basis B’

The formula that inside the box, how to interpret it?

Wait. So if basis B is (e1, e2, ..., en). Is this 1 basis. Or n basis?

[rose coral]

No here there are only 2 bases, B and B' each one with n vectors

[pulsar turret]

Ah right got confused for a sec

[rose coral]

The formula gives how you write some e' vector in terms of the basis vectors e from B

[pulsar turret]

I think I understand

[vague sundial]

Hello, is that okay to ask a question about vectorial spaces in this room?

[limber sierra]

yes.

(though the usual English term is vector space)

[vague sundial]

Thank you!

I'll take the chance to ask - how do you call it in english a collection of things?

[rose coral]

I think I understand
@pulsar turret Awesome I have to go for a bit but just ask if you have more questions, in this room if free or one of the questions rooms

[limber sierra]

a set?

[pulsar turret]

@rose coral thank you, I appreciate the help and step by step thought process really helped me.

[vague sundial]

Yeah that must be a set. So I have this already solved problem, but I can't figure out how they did it.

It reads: Prove that the set P (...) is not a vector subspace of R^3

I don't understand why did they assume x(2) was x^2(1)

[latent ledge]

Suppose $T\in\mathcal{V}$ is diagonalizable. Prove that $V = null T \oplus range T$. the problem doesn't say the dimension of V, if V is finite, use V is a direct sum of eigen spaces. What if V is not finite, how do I argue about that

[wintry steppe]

that's what the elements of P look like. the set P is defined by those points (x_1, x_2, x_3) satisfying x_2 = x_1^2 @vague sundial

[stoic python]

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

[wintry steppe]

this is a weird "proof without words" that seems like it's supposed to be accompanied by someone speaking (to francish)

[vague sundial]

I cut that part out

[limber sierra]

$x_1^2 - x_2 = 0$ implies $x_1^2 = x_2$

texit?

[wintry steppe]

bot dead maybe

[limber sierra]

$test$

😦

[wintry steppe]

:(

[stoic python]

Namington:

[wintry steppe]

oh

[vague sundial]

Oh now I understand

Btw, is u and v always the same? I know I'm making really dumb questions. I stopped having math for a really long while

[limber sierra]

no

the point of this proof is to show that this space is closed under addition

(since that's one of the things we need to have a subspace)

what that means is, for any u, v in the space P

u + v should also be in P

so u and v might not be the same

[vague sundial]

I mean when we're making this kind of proof, how do I pick u and v values?

I get that we got u from looking at the condition, but what about v?

[limber sierra]

i'm not sure what you mean

u and v are totally arbitrary

they're symbols that represent any vectors from P

not a specific one

[vague sundial]

What about v_3 and u_3, what happened to them in the end of the calculus? They were not used

[limber sierra]

all you need to show is that u + v is a member of P

that is to say, it needs to satisfy this

or in others words:

$(u_1 + v_1, u_2 + v_2, u_3 + v_3)$ satisfies $(u_1 + v_1)^2 - (u_2 + v_2) = 0$

that is what you need to show

we dont need to show anything with the 3rd coordinate

since that doesnt affect whether our vector u+v is a member of P or not

[stoic python]

Namington:

[limber sierra]

typo corrected^

[vague sundial]

Why doesn't it affect?

I appreciate in advance how patient you've been

[limber sierra]

this is what we need to check.

note that we dont care about x_3

(as long as its a real number)

we just need to check that x_1^2 - x_2 = 0

x_3 doesnt come into play at all

that is to say, no matter what value x_3 has

it won't affect whether this vector is a member of P

[vague sundial]

Okay I think I understood now

[limber sierra]

$\begin{pmatrix}2\4\x_3\end{pmatrix}$ is a member of $P$ no matter what value $x_3$ takes

[stoic python]

Namington:

[limber sierra]

since it satisfies $2^2 - 4 = 0$

[stoic python]

Namington:

[limber sierra]

x_3 could be 50, or -12305.394, or pi, or 0, or whatever

and it wouldnt affect membership in P

[vague sundial]

Yeah I understood, thank you very much for explaining, I was struggling here

[latent ledge]

by definition of eigen space, it is null(T-lambda I), what is range(T-lambda I)

the basis of null(T-lambda I) is an eigen vector I guess

[wintry steppe]

if null(T - lambda I) is nonzero then range(T - lambda I) isn't all of V

that's right off the top of my head

idk, it's isomorphic to V / ker(T - lambda I) (think about what this space looks like)

(first isomorphism theorem)

feels like kind of an open ended q

unless im missing something

[rose coral]

I think the range should be something related to the other eigenspaces

If you have null(T - lambda I) = null(T - lambda I)^2 then the image should be precisely the other eigenspaces (for finite dimensions)

If my linear algebra isn't that rusty

For example if you have a linear map T that is diagonalizable with eigenvalues 2, 3, 7 then range(T - 2 I) is exactly null(T - 3 I) + null(T - 7 I)

If the map isn't diagonalizable you need to look at things like the characteristic polynomial and higher powers of the (T - lambda I)

[latent ledge]

thank you, i think I get it know

For example if you have a linear map T that is diagonalizable with eigenvalues 2, 3, 7 then range(T - 2 I) is exactly null(T - 3 I) + null(T - 7 I)
@rose coral this clearifies my another question about why null(T-lambda_n I) subset sum_{i=1}^{n-1} range(T-lambda_i I)

[rose coral]

No problem, glad I could clarify that for you

[turbid raptor]

i need some help

can someone point me in the direction of a video explaining how to solve an equation like

8d-2=6d-10

[wintry steppe]

#prealg-and-algebra

[dreamy iron]

I'm playing around with elements of the set of all periodic functions from R to R.  Let's call it $\mathbb{S}$.

$f(x) = \sin (x) \in \mathbb{S} \\$
$g(x) = \sin (\pi x) \in \mathbb{S} \\$ 
$h(x) = \sin (x \sqrt{2}) \in \mathbb{S}$.
$\\$ 

Any two of the linear combos of the above functions is a counter example to the notion that $\mathbb{S}$ is a linear subspace of $\mathbb{R}^\mathbb{R}$.```

[stoic python]

ninnymonger:

[dreamy iron]

How do I show that f+g, or g+h, or f+h is NOT a period function?

This is my playing around on Desmos with these functions: https://www.desmos.com/calculator/lkeoihmryh

By inspection, they're not "periodic"

How about this, since they're "periodic" they should pass through y=0 at "regular" intervals?

[half storm]

wouldn't they be periodic though

$sin(x + 2\pi) + sin(\pi x + 2\pi) = sin(x) + sin(\pi x)$ ?

[pallid rampart]

Um no

[half storm]

oh no my bad

I see what's wrong

x + 2pi

[pallid rampart]

The x+2pi is inside parenthesis

[half storm]

yea

I guess you can try a proof by contradiction

but still seems hard

[dreamy iron]

This is what the internet says.

I am being asked to show that S is not a subspace of R^R.

and showing that f+g \notin S seems sensible.

[pallid rampart]

How does that imply sin(sqrt(2)p)=0 and cosp=1 tho

[dreamy iron]

I don't understand the internet's solution......but root(2) \in Q seems like a good contradiction.

[pallid rampart]

It is

But like 1=cosp+... does not imply the sin(sqrt(2)p)=0 and ...

[gray dust]

you may have an easier time using indicators instead of sins, eg indicator on integers & indicator on multiples of e

[dreamy iron]

indicators....as in indicator functions?

this is from Axler's LADR......

[gray dust]

yes

[dreamy iron]

i don't know how to use indicators.

[gray dust]

the defn on its wiki page is all you need to know

indicator on integers & indicator on multiples of e
their sum is 2 at 0 but nowhere else

[half storm]

What's wrong with this proof?

[pallid rampart]

At the highlighted point is an instance where cos(p)+sin(sqrt(2)p)=1 but cos(p)≠1 and sin(sqrt(2)p)≠0

[half storm]

I see now

[dreamy iron]

so the internet proof is wrong?

[half storm]

I have no idea how they conclude that cos(p) = 1

[dreamy iron]

1 = a + b = a - b \iff a = 1 & b = 0

[pallid rampart]

Ah

Oh yeah that makes sense

Ok that's actually kind of big brain

If they explained it a bit more

[half storm]

oh snap

that's slick af

The rest of it follows I think.

[pallid rampart]

Yeah

[storm ravine]

my teacher hasnt taught anything and im lost on this

pls send help

[half ice]

You can write:
(7 - λ)x - y = 0
-6x + (8 - λ)y = 0

Then it comes down to reducing a matrix @storm ravine

[storm ravine]

ive never done that before :/

[old flame]

@Flow#8160 Ok so I took a day off ystd, here's what I got so far. Let basis of V be $(v_1,...,v_n)$.

Define $S: V -> V$, where $S(\sum_{I=1}^{n} (a_iv_i) = a_iv_i$, where $i \in {1,...,n}$. So $TSv= T(a_iv_i) = a_i Tv_i$ and now I'm stuck on STv, since S transforms basis vectors

[stoic python]

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

[spiral star]

@old flame you need to be a bit more precise with that notation, since it's kinda misleading here. take any v_1 in V \ {0}. extend to a basis (v_1, v_2, ..., v_n) of V. then define S by S(v_1) = v_1 and S(v_i) = 0 for the other basis vectors v_2, ..., v_n. now, what happens when you consider STv_1 and TSv_1?

[old flame]

Both are the same ?

Ah ok I will edit that in my final proof

[dusky epoch]

there is some bad latex in there

[spiral star]

yea that too lol

[dusky epoch]

instead of $S: V -> V$ write $S: V \to V$

[stoic python]

Ann:

[dusky epoch]

looks nicer doesn't it

[old flame]

Oh LOL sorry I didn't edit that, cause I was just typing on my note pad

So TSv_1 would become Tv_1, while I'm not sure for STv_1, since Tv_1 is not a basis vector, however we could express it in terms of basis vectors and continue with it ?

[spiral star]

Since ST = TS we get S(Tv_1) = T(Sv_1) = Tv_1 and therefore Tv_1 is in range(S).

now, what is range(S)?

[old flame]

range S is also range T

range S is a subspace of V

[spiral star]

range S is not range T

but its a subspace, but you can define it explicitly

we constructed S, so you can read off its range

[old flame]

Wait could you explain why ? Doesn't TS=ST means elements from both ranges are in the same space ?

[spiral star]

range(ST) = range(TS) but not range(S) = range(T)

that assumption would completely contradict what you want to prove

your goal is to show that T is a scaled identity

if range(S) and range(T) were the same, then range(T) would be a 1D subspace of V

you cant derive it from what you know anyway

so again, what is range(S)

we defined S by S(v_1) = v1 and S(v_i) = 0 for the other basis vectors

what is range(S)

[old flame]

Ok let me see

[spiral star]

range(S) = ....

[old flame]

$range S = v_1$ where $v_1 \in V$ right ?

[stoic python]

Otoro:

[spiral star]

almost, but range is a vector space

it cant be a vector

but range(S) = span(v_1)

by construction, range(S) = span(v_1) = { λv_1 | λ in F }

hence, Tv_1 = λv_1 for some λ in F.

[signal lagoon]

hello I see you right here can you help me with assignment

I'm too smol brain for this

[spiral star]

@old flame with this construction we showed that T maps any vector v in V to some scaled v. the next step is to show that every v gets scaled by the same factor

[hollow finch]

if a vector w is in the span of v1 and v2 where v1 and v2 are elements of R^n with n>=3, what is the "correct" way to get the coordinate vector c such that [v1 v2]c=w (alternatively c1v1+c2v2=w)
is it just elimination or is there some other method?

[spiral star]

if you only have to find this for a single w, then elimination is the standard way

if you want to do this for all w in some space spanned by v1 and v2, then you can calculate the one sided inverse of [v1, v2]

and then calculate the coordinates by multiplying by the change of basis matrix

[hollow finch]

by one sided matrix you mean (lets say U i guess) U[v1 v2]=I_2?

[spiral star]

yea

(of course that requires v1 and v2 to be linearly independent)

[hollow finch]

im almost certain U would not be unique, but then c would be c=Uw right?

yeah definitely thats a given. otherwise the question is trivial

[spiral star]

yea, then (v1, v2) is the basis of a vector space, and you are basically doing a change of basis here