#linear-algebra

2 messages ยท Page 94 of 1

wintry steppe
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they'd just laugh

narrow mortar
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lol

half ice
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This is hs. There's no control over bad questions

narrow mortar
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hey so its not b-c?

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its b+c?

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(8,1) = b+c

wintry steppe
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b - c gives you the point on the far right

half ice
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I said |b| + |c|

wintry steppe
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I already did all of the vector math to compute the diagram

narrow mortar
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oh ok

half ice
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Very different thing than b + c

narrow mortar
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oh

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i meant

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|b|+|c|

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which is

wintry steppe
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tbh

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I'd just write a quack answer on it

narrow mortar
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(8,1)
= sqrt 65

wintry steppe
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and then just turn it in to the head of the math department if your teacher demands that you solve an impossible problem

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it's like squaring the circle

narrow mortar
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|b|+|c|= sqrt 65

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lol

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what

half ice
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Uh no?

narrow mortar
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b=(3,-8)

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c=(5,9)

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b+c= (8,1)
|b+c|= sqrt(65)

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@half ice

half ice
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@narrow mortar

narrow mortar
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yes?

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hi ๐Ÿ˜›

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๐Ÿ™‚

half ice
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So |b + c| is the process of getting the vector b + c, then taking the magnitude.

|b| + |c| is getting the length of each vector separately, then adding the lengths

narrow mortar
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oh

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b+c

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is

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sqrt 73 + sqrt 106 tho

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and a was sqrt 5

half ice
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Actually I guess either |b| + |c| or |b + c| should be usable

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But since they are not colinear, each will give a different answer

narrow mortar
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hm

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|b| + |c| = 18.8
|b + c|=8.06

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they are quite different tho

half ice
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Oh derp, c goes the other way doesn't it?

narrow mortar
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yes

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so sqrt 106- sqrt 73?

half ice
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Then I mean
|b| + |c| โ‰ˆ |b - c|

narrow mortar
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oh

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lol

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thats what i did

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b-c = (-2,-17)

half ice
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I'm a dorp, don't mind me

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Yes you got it

narrow mortar
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which is sqrt 293

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oh then the height was

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|b-c|

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whats a dorp lol

wintry steppe
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bruh can I roast your teacher

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please

narrow mortar
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how?

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sure u can in ur mind

wintry steppe
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I applaud you for actually doing a bad problem

narrow mortar
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thanks

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took me..

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hmm

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HOURS

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ok ima scan and submit

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coz im done with this

wintry steppe
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feelsbad

narrow mortar
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lol

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@wintry steppe

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can u look over my assignment before i submit?

wintry steppe
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looks right

narrow mortar
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yay

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ok ima submit this trash

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jk its math its beautiful

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i โค๏ธ math

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@wintry steppe I THINK I FAILED

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THE LAST QUESTION THO

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im scared to submit

wintry steppe
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who cares

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it's not much of a grade anyway

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and the last question was

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in my honest opinion

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garbage

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and you can tell him that pretty much all of the math undergrads and grad students you asked said it was wrong

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and yeah I'm pretty familiar with linear algebra, so I'm pretty confident about basic things like this

narrow mortar
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dude whos more right

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me or my friend

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she did

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SHE CHANGED LAST MIN if she gets it right

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ima call her a traitor

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LOL

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she had the same thing as me first

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i did that

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she did

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@half ice

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which seems more right

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lol

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@wintry steppe

wintry steppe
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lemme say this for the last time

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nobody's "more right"

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you can't be "more right" about a shitty problem

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this is not something that math can answer

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you could literally write anything there

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and mathematically, nobody would be more right than anyone else

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so just turn it in

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and accept that you have a bad teacher and will lose points

narrow mortar
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lol

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i want a 90 in calccccccc

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ahhhh

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awwwww

wintry steppe
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and if you don't want to accept that, then send the problem to the department chair or your principal, but that's overkill

narrow mortar
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nah im not sending it to anyone

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tho i think shes right hmm

wintry steppe
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then you have to accept that whatever you turn in will most likely be wrong

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she's not

narrow mortar
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lol

wintry steppe
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she's not

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she's not

narrow mortar
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LOL

wintry steppe
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please

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stop

narrow mortar
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lol okkkkk

wintry steppe
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fretting over a single problem

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as I said before

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nobody's right

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because the problem is wrong

narrow mortar
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oh well its annoying when u spent 10000000000000 hours on it

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and get it wrong anyways

wintry steppe
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yeah well next time you shouldn't waste time like that

narrow mortar
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it hurts :c deep down

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lol

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but i like math

wintry steppe
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should've taken the advice that the problem was wrong and left it at that

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send it with whatever you tried

narrow mortar
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i did

wintry steppe
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at least it shows that you tried

narrow mortar
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lol

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ok

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i mean kaynex helped so hes prob right lol

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and u all helped

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so like

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I MEAN NOT RIGHT BUT RIGHT DEEP DOWN

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right in my eyes I guess

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lol

wintry steppe
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it's admirable to like math enough to try to work on hard problems, but working on impossible garbage is wasting time

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the truth became pretty evident that it was flawed

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and at that time, you should just give up any notion of "correctness" and write down an attempt

narrow mortar
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ye

torn silo
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yeah math should be about feelings not hard logic, I mean who wouldn't want planes made with love instead of mathematical correctness

narrow mortar
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oh ok

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lol

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Sorry

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I understand it was wrong

wintry steppe
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and if the teacher really doesn't see his mistake after being shown a diagram and even acknowledging that b and c are not collinear

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then he's incompetent too

narrow mortar
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but when I said was it right i meant right as in "right" - the right my teacher thinks

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whatever lets just move on

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from this

wintry steppe
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and we're not psychics

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we don't read minds, and what you're pretty much asking is to read his mind

narrow mortar
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ye but i told u what he explained

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so i asked u what he wants me to do

wintry steppe
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1+1 = 3, what number am I thinking in my head?

narrow mortar
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OK LETS MOVE ON

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2

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coz ur smart

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LOL

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and sensible

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THANK GOD IM DONE THO

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@wintry steppe so how are u?

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are u annoyed with me? ;-;

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just asking

wintry steppe
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nah more annoyed at your teacher

narrow mortar
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oh ok

pale coyote
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it's safe to come back?

torn silo
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You mean when you calculate the eigenspace?

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Well you can't really do something like

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$A - \lambda$

stoic pythonBOT
torn silo
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if lambda comes from the reals

narrow mortar
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@pale coyote LOL yes its safe

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to come back

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are u scares of me now ;-;

pale coyote
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yes

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$A$ is a matrix, $\lambda$ is a number

stoic pythonBOT
narrow mortar
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noo plz dont be scared of me ๐Ÿ˜ฆ

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@pale coyote

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:c

torn silo
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Yes you multiply lambda with the identity matrix because that's what you want to do with the lambda

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its how lambda becomes something you can use in matrix operations

pale coyote
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$\lambda v$ is the same as $(\lambda I)v$ so $Av = \lambda v$ is the same as $Av - \lambda v = 0$ which is the same as $Av - \lambda Iv = 0$ or $(A - \lambda I)v = 0.$

stoic pythonBOT
narrow mortar
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@half ice are u also scared of me

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;-;

pale coyote
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;-;

narrow mortar
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noooo u cant be scared of meeeeeeeeeeeeeeee ;c

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@pale coyote

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that makes me sad ;c

quartz compass
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I don't think it's very interesting to consider the 0 vector as an eigenvector

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probably why

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0 eigenvalues on the other hand can be interesting though

pale coyote
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@narrow mortar :{

narrow mortar
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yes?

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;-;

pale coyote
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L-L

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@DISC allowing 0 eigenvector gives no useful information about a transformation/matrix

pallid rampart
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If you consider 0 to be an eigenvector, then a lot of theorems will break

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Like, a set of eigenvectors are linearly independent

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Each nxn matrix has at most n eigenvalues

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Sure

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Wait

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it's not exactly eigenspace

torn silo
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then you did something wrong

pallid rampart
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I'm not sure what eigenspace is, but it seems like to be all the eigenvectors corresponding to a eigenvalue

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There's no way that set can be linearly independent, since it's closed under multiplication

torn silo
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you calculate base vectors and then the space generated by them is the eigenspacce

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so it's the base that needs to linearly independent

gray mason
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Any set with the zero vector can not be linearly independent

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Yup

wintry steppe
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Hey

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I need help finding the equation of the plane that has 3 points P(0, 1, 1) Q(1, 0, 1) and R(1, 1, 0)

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I started by creating 2 vectors: PQ(1, -1, 0) and PR(1, 0, -1)

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Then I did crossP(PQ, PR) = i + j + k

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And finally the equation x + y + z = -2

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But it's not right. Anyone know why ?

slow scroll
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the RHS is not correct. plug in points

wintry steppe
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RHS ?

slow scroll
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right hand side

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plug in P for example 0 + 1 + 1 = 2 which is not equal to -2

wintry steppe
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Oh I see

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But when I graph x + y + z = 2

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I only have 2 points fulfilled

gray mason
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Is your cross product correct?

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Your norm isn't correct then

wintry steppe
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crossP([1, -1, 0], [1, 0, -1])

gray mason
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you got 1,1,1?

wintry steppe
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Yeah

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oh bruh

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I graphed (1, 1, 1) instead of (1, 1, 0)

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as point C

gray mason
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lol

wintry steppe
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thx guys

slow scroll
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npnp

wintry steppe
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Last thing tho

narrow mortar
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DUDE

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GUYS

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@pale coyote

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u there?

wintry steppe
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(x - 0) + (y + 1) + (z + 1) equals x + y + z = -2

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So I get -2 instead of 2 :/

slow scroll
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it should be (x-0) + (y - 1) + (z - 1) = 0
subtract coords

wintry steppe
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Ok great, thx

slow scroll
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np

narrow mortar
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@wintry steppe

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DUDEEEEEE

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U THERE?

slow scroll
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d0000d

narrow mortar
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lol

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funny

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@half ice

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KAYNEX AND @wintry steppe

slow scroll
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you mean not a subspace?

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true, it does not have the zero vector, but it also fails the other axioms

narrow mortar
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@half ice GUESS WHATTTTTTT

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KAYNEXXXXXXXXXXXX

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MY MIND HAS EXPLODED

slow scroll
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Let f be in W, and define g = 2f
then g(0) = 2f(0) = 2 so g(0) != 1 and g is not in W

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you could just say: let $f \in W$. Since $2\in \bR$ and $2f(0) = 2$, it follows that $2f \notin W$.

stoic pythonBOT
slow scroll
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Either is fine

gray dust
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this is not a big deal

slow scroll
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it doesn't even have to be a generic f \in W. we know in particular that the function f(x) = 1 is in W. you can take that function, multiply it by 2, and since 2f(x) = 2 we have that 2f(0) = 2, and conclusion follows.

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nope, and that's one reason why i defined g = 2f. To show that the new function you get out of scalar multiplication / addition needs to also be in the set W.

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np

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try not to think of it as anything different from applying the definitions. If we are showing that the set $W = { f: \bR \to \bR : f(0) = 1}$ is NOT closed under scalar multiplication, we are showing that there exists some element $f \in W$ and some $c \in \bR$ such that $(cf) \notin W$. \ \ What does it mean for $cf$ to not be in $W$? It means that either: $cf$ is not a function from $\bR$ to $\bR$ (of course $cf$ is a function from R to R), OR $(cf)(0) \neq 1$.

stoic pythonBOT
pale coyote
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What @narrow mortar

narrow mortar
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do u really wanna know

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ok

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answer was..... 9.5..or 8.5...

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@pale coyote

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thats sad well rip

slow scroll
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as I looked at a dozen so far and they were not correct
lol becuz the problem aint well defined

torn silo
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your teacher seems like a douche

pale coyote
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your teacher is a moron

narrow mortar
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whos?

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@torn silo

pale coyote
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yours

torn silo
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yours

narrow mortar
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I GOT 77 I THINK

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for my area

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lol

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wow

pale coyote
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almost surely has no idea what theyre doing

narrow mortar
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he said hes gonna post

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solutions for it

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so illl send u when he does lol

cold topaz
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we use the same method to geth the orthonormal vector as we use to get the unit vector right?

torn silo
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orthogonal vectors are gram schmidt

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unit vectors are just vectors with length 1

cold topaz
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i know that. my question is about the method we use to get them

torn silo
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orthonormal vectors are orthogonal and have length 1

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gram schmidt for orthogonal, simple math the norm a vector

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(1, 2) is 1/sqrt(3) (1,2) normed

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no gram schmidt needed

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thats assuming Euclidean norm mind you

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different norms means you need to work differently

cold topaz
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orthonomrmal basis is formed from the norm vectors. roight?

slow scroll
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right, an orthonormal basis is just an orthogonal basis where the basis vectors have unit length.

cold topaz
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so if we have v1, v2, v3, their NORMS form the orthonormal baisis, and not themeselves. Right? so {norm of v1, norm of v2, norm of v3}.

torn silo
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the normed version of those vectors

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v1/||v1| |

gray dust
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use the normalized versions of v1,v2,v3 if they aren't already unit vectors

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you only need to show the set fails at least one of em, so double good job

mystic sentinel
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So I'm reading through a supplementary linear algebra book

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That has to be a typo, right? You can't have a vector space over F[x], since F[x] is a ring, not a field.

slow scroll
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Im pretty sure F[x] is referring to the field of rational functions

mystic sentinel
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No, F(x) is the notation they gave for rational functions.

slow scroll
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so you're taking F[x] to be polynomials? That doesn't sound right

mystic sentinel
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Yeah, exactly.

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So would this be a module then, instead of a vector space?

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Since the set of coefficients forms a ring?

urban spear
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doesn't a module qualify as being a vector space?

mystic sentinel
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Other way around.

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A vector space is a module over a field.

urban spear
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ah

narrow mortar
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@pale coyote u there?

pale coyote
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o god

half ice
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o god

unborn zenith
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o god

dreamy fiber
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o god

narrow mortar
#

guys

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@pale coyote @half ice

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posted answers

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@wintry steppe

feral mountain
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o god

half ice
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Oh snap I was wrong, he did take the "real" answer

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Wait no, wtf

quartz compass
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@mystic sentinel which vector space axiom(s) does it violate?

half ice
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Wat the fuq

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I think he still doesn't know it's not a triangle, Star

narrow mortar
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๐Ÿ˜ฆ

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yes i also messed this up ๐Ÿ˜ญ

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and now im crying coz i messed it up

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he posted all the answers

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ugh

half ice
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He just found some random side to call a "base"

narrow mortar
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i found 2 extra points that i wasn't supposed to

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coz i labelled the paaraallelloogram wronggggg

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but i did it according to coordinates :c

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oh no nvm..

half ice
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I imagine there's a lot of people in your class that are VERY confused.

narrow mortar
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i mixed them..

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by accident...

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rn im more sad about #4 ๐Ÿ˜ฆ

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what a stupid mistake i made ๐Ÿ˜ฆ

mystic sentinel
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@quartz compass Not all elements of F[x] are inverible.

narrow mortar
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ugh

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@half ice is what he did reasonable

quartz compass
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seems like that should be part of the vector space axioms but it's not listed here

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the only inverse required is for there to be a negative

mystic sentinel
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It says in the axioms that r,s โˆˆ F

quartz compass
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but it does seem weird to me that it's not a field

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yeah it's just plain wrong, no getting around it

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the entire example they give is wrong

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it's not really a typo, they just don't know what they're talking about haha

mystic sentinel
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Well great ๐Ÿ˜›

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I guess the Humble Bundle textbook sale wasn't that great an idea lol

quartz compass
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I guess I doubted myself for a second, just seemed too weird

mystic sentinel
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I've noticed so many typos in terms of typesetting

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So like ... I wonder if I keep going and learn what I can from it or just find a different book ๐Ÿ˜›

quartz compass
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I guess so long as it's not a well known LA book

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I don't want to come across it in the future just in case

mystic sentinel
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Mercury Learning is where it's from, as well as a bunch of other books I got in this bundle

narrow mortar
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@half ice kaynexx

half ice
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Wat

mystic sentinel
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@quartz compass Any other recommendations for more advanced linear algebra books?

narrow mortar
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LOOK

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so i did this

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and he got these answwers

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DID I FAIL #3

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AND #4

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i failed 6 for sure but

quartz compass
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no recs from me

mystic sentinel
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Protip: allcapsing is probably not a great way to get people to want to respond to you, @narrow mortar

narrow mortar
#

?

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Oh no sorry i wass stressed

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so i just tend to type in caps

mystic sentinel
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Just keep in mind.

narrow mortar
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sorry yes

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i will

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@mystic sentinel

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๐Ÿ˜ฆ

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sorry

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im sad i think i failed

mystic sentinel
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Yeah I get you. :< Was that for a final?

narrow mortar
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no it was an assignement

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for unit 6

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@mystic sentinel

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like a test assignement

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tho can u compare my work with his

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and tell me if u think i failed ;-;

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@half ice can u wow why am I stressinggggg

eternal finch
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Oh, I misread.

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Got rid of that garbo.

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Let's see.

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(x) + (y) = (-a) + (-b) = (-a - (-b)) = (-a + b).

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Wait, yeah, I don't think I misread.

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It doesn't matter what + normally means.

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You just forget what "+" ever meant and define "+" as they did.

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The vector space would make it (-a - (-b)) = (-a + b) instead, which would prove that vector addition doesn't hold.

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Right.

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There's nothing wrong with the definition of addition, but you still have to show that the addition satisfies the properties we want.

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For example, the "addition" should be associative.

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In which case it might fail.

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Oh, actually, it doesn't just fail commutativity. It does in fact fail associativity, too.

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However, showing that it fails commutativity is enough to make the whole thing fail.

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Commutativity is the easiest thing to show goes wrong, too.

cursive narwhal
#

Perhaps it might be better for you to write the defined operations as $+_V$ and $\cdot_V$, where V is this specific vector space. This might prevent you from confusing yourself.

stoic pythonBOT
eternal finch
#

Hm. I mean, it all depends on how recently you learned this.

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I would say it's easy, but that's because the way I learned linear algebra most recently begins with notions of vector spaces.

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Whereas you started with solving systems?

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Ah, ok. Yeah, but as far as vector spaces go, this is a basic question. There might be tricky questions where the addition or the multiplication is weird. I mean, you see here the addition is what's normally a subtraction.

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You just need to drill for these types of questions.

limber sierra
#

yeah this is just "do you understand how vector spaces work"

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if you do, the question is "easy"; if you dont, its impossible

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it's just a check that you understand how to reason from definitions

eternal finch
#

I also like Abhijeet's suggestion.

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If you got confused by the fact that the addition was different, you can notate it differently to remind yourself it's some vector addition rather than the "normal" vector addition.

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Yeah.

cursive narwhal
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In this specific problem, it should be noted that standard addition and multiplication in R are involved. They're used in defining these new operations

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So you cannot use the same symbols for them

eternal finch
#

That's a fair point.

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I think I used a plus in a circle and a dot in a circle before I found out about direct sums, which used a plus in a circle.

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In my numerical analysis class, we used a plus in a circle to denote addition that a computer does, subject to rounding off to some number of digits, so it seemed natural at the time.

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Ah, lol.

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You know, it might be, too.

cursive narwhal
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Xor looks different, i believe. That's a cross in a circle

eternal finch
#

The important thing is no two symbols collide.

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A xor can in fact be written as a plus in a circle.

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Til.

cursive narwhal
#

Oooh

humble oak
#

hello, given some homogenous system that is 3x5 and has rank 2 what conclusions can i make about it?

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pretty sure one is that the last row is a bunch of zeroes

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is it possible to determine the number of basic solutions it might have or tell whether it's consistent or inconsistent?

slow scroll
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"last row is a bunch of zeros." In the rref maybe, but not necessarily in the original system

humble oak
#

true

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so i can basically make no assumptions about it?

slow scroll
#

also, wdym by homogenous system. How does a system have rank

humble oak
#

mmm in this question over here it says homogenous system

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might've been a typo

limber sierra
#

uh

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whats wrong with a homogeneous system

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????

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it just means all the constants are zero

slow scroll
#

Ax = 0 does not have rank. A has rank

humble oak
#

interesting

limber sierra
#

oh i see what you mean

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ok sure

slow scroll
#

you could also be talking about the augmented matrix [A|0] i have no idea

humble oak
#

uh

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would you prefer

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i send the screenshot

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here

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of the question

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i know it has non-trivial

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bc it has a row of zeroes in rref

eternal finch
#

What is a "basic" solution, btw?

humble oak
#

not a trivial one

eternal finch
#

I see.

humble oak
#

in my case

eternal finch
#

It means no solution to the system.

slow scroll
#

i think that basic means linearly independent solutions

humble oak
#

inconsistent = no soln

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but homogenous

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always has trivial one

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oh here's a definition for basic soln

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hm

eternal finch
#

Hold on... if Ax = 0, then isn't Akx = 0 for any scalar k? Could you elaborate on what they mean by basic solutions?

humble oak
#

that's their definition up above

slow scroll
#

he means linearly independent

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dim of null space

eternal finch
#

I see, so by basic solutions, they mean a basis of the kernel?

humble oak
#

woah this terminology seems out of scope for me haha

eternal finch
#

That just seems like a weird way to phrase it. 2 basic solutions...

humble oak
#

what is kernel and null space?

slow scroll
#

its just their terminology

eternal finch
#

Yeah.

humble oak
#

huh interesting

eternal finch
#

Ok, so, putting aside terminology.

#

so you could rewrite the rest of the pivot column variables in terms of the 2 free variables; so 2 basic solutions?
This would be what I'd think.

humble oak
#

oh wait a minute i'm pretty sure you're right

#

i didn't rly notice this until now but on the previous question telling to find the basic solutions given an actual matrix

#

the combinations use the numbers of the free variables

#

so there must be 2 basic solutions OOO:

slow scroll
#

rank 2 does not mean 2 basic solutions tho

humble oak
#

i meant like two free variables

eternal finch
#

Oh, you right.

#

The rank is the dimension of the image...

#

@.@

slow scroll
#

nope, not 2 free variables either

humble oak
#

oh

#

gg

quasi vale
#

rank is 2(no of pivot variables)

#

we have 5 variables here, so we have 3 free variables

eternal finch
#

^

quasi vale
#

or 3 basic solutions

humble oak
#

TRUE

eternal finch
#

lmao @.@

humble oak
#

but again

eternal finch
#

Yeah, same.

humble oak
#

this would assume that the system is in rref

eternal finch
#

Rank is preserved by elementary row operations.

humble oak
#

oh

eternal finch
#

So, right away from the rank, you can tell how many pivot variables you'll have.

humble oak
#

ohhh so it shouldn't matter whether or not it's in rref or not?

#

bc rank is PRESERVED OOO:

quasi vale
#

you can tell rank from echolon form

#

or RREF both, but it'll stay the same

#

won't change

humble oak
#

alrighty, i believe i got it

#

thanks yall, i will probably be back shortly ๐Ÿ˜ฆ

elfin ingot
#

quick basic question

#

how would u describe the linear operator on some vector space V over field F

#

T(x_1,x_2) = (x_1,0)

#

geometrically?

#

how do i say it makes the point go to x axis XD

eternal finch
#

Yeah.

elfin ingot
#

thats it?

cursive narwhal
#

Looks like a projection

eternal finch
#

People say you project onto the x-axis.

elfin ingot
#

project

#

okay tysm

eternal finch
#

If we're talking two-dee space, in general, you can project a vector onto any line.

#

In other words, make the point go to that line.

elfin ingot
#

cool tysm

cursive narwhal
#

Well, it's not exactly a projection map exactly

#

But it definitely looks like it

#

And i don't think there's any issue with calling it one

elfin ingot
#

let T be the linear operator on C^3 for which T_ep1 = (1,0,i) , T_ep2 = (0,1,1) ,and T_ep3 = (i,1,0)

eternal finch
#

I guess my description is vague. What I meant was something to the effect of an orthogonal projection of a vector onto a line.

dusky epoch
#

"two-dee"

#

pls

elfin ingot
#

is T invertible

#

cant i just define

#

a function that just swaps these 2 coordinates?

dusky epoch
#

what are ep1, ep2 and ep3

elfin ingot
#

standard basis

dusky epoch
#

well

#

are (1,0,i), (0,1,1) and (i,1,0) linearly indep

elfin ingot
#

idk let me check ig

#

yea

#

no

dusky epoch
#

well there you have it

elfin ingot
#

wait

#

the operator must perserve lienar indep

#

right?

#

well there you have it
@dusky epoch can u clarify this abit

dusky epoch
#

you can translate the linear dependence of {(1,0,i), (0,1,1), (i,1,0)} into a vector v โ‰  0 such that T(v) = 0

elfin ingot
#

yea okay

dusky epoch
#

(-i, -1, 1) seems to be in ker(T)

elfin ingot
#

noo

#

oh

#

okay

#

wow ur so fuckign smart

#

yea

#

got it

eternal finch
#

Besides the fact that column rank equals row rank, are there any interesting facts involving dual vector spaces and systems of equations? Or facts about systems of equations that you can show using dual vector spaces?

iron belfry
#

hello guys i'm self-teaching myself this topic
any suggestion for a good book of exercises/problem sets with a comprehensive solution & explanation that comes with it
many books are good but i cant check if i'm right or wrong because they don't give answer to every exercise

cursive narwhal
#

Not sure about how you're approaching the subject but Linear Algebra Problems Book by Ikramov is great

#

It has solutions but I don't really refer to those for the proof problems. I usually post my proofs on stackexchange and hope that other people will give their critique.

torn silo
#

ya solutions don't always get you that far, because you most likely proved it differently, but that doesn't mean the proof is invalid

#

So asking others for help is the smartest thing you can do

autumn kraken
#

When you have a K-dimensional vector and say "lambda element K", does that mean "lambda element { 1, 2, 3 }" if K=3?

dusky epoch
#

what?

#

do you have some more context?

elfin ingot
#

T is a linear operator on vector space V

#

suppose rank(T^2)=rank(T)

#

prove that the range and nullspace of T intersect trivially in the zero vector

dusky epoch
#

first prove im(T^2) = im(T)

elfin ingot
#

how can i do that

dusky epoch
#

prove im(T^2) โŠ† im(T)

elfin ingot
#

let y = T^2(a) ---> y = T(T(a)) = T(b) for some b = T(a)

#

thats so trash but not sure

dusky epoch
#

the wording is trash, you're right

elfin ingot
#

what wording D

#

XXD

#

i said nothing

dusky epoch
#

"for some"

elfin ingot
#

whats bad about that

autumn kraken
#

They say "V is a K-vector space", and I have to prove that if v element U and lambda element K, then lambda*v element U

elfin ingot
#

yea

#

okaay

#

for some is useless

#

okay

autumn kraken
#

and I dont know what to make of "lambda element K"

dusky epoch
#

ฮป is a scalar

#

K is your field

elfin ingot
#

now let y = T(c) ---> y=T(T(a)) for T(a) = c ---> y= T^2(c)

#

---> y in img T^2

#

good?

dusky epoch
#

hrngh

#

ok im out

elfin ingot
#

whats wrong?

autumn kraken
#

ohh I thought K just means the amount of dimensions

#

thanks

elfin ingot
#

lmao wtf is wrong

dusky epoch
#

O4GTKLP'SDFGKLNXGFLMFHGADSFGHa

#

everything is wrong

elfin ingot
#

can u explain ?

#

or are u going to just crash type

dusky epoch
#

i'm trying to make myself not go through the 100th mental breakdown this week

#

so NO i can't explain

#

and if you keep asking me i'm gonna yell at you to fuck off

elfin ingot
#

u can just chill i mean there is nothing after u

#

lmaao its not like ur getting payed

dusky epoch
#

FUCK OFF.

elfin ingot
#

fuck you

dusky epoch
#

THAT'S WHY I SAID I WAS GONNA TAB OUT.

elfin ingot
#

wtf is wrong with you

dusky epoch
#

but you pulled me back

elfin ingot
#

fucking mental

dusky epoch
#

YES I'M MENTALLY ILL WHAT OF IT

elfin ingot
#

tab out

autumn kraken
#

so if I have a vector space in Rยณ then K is R?

#

in order to define vector with scalar multiplication?

wintry steppe
#

They say "V is a K-vector space", and I have to prove that if v element U and lambda element K, then lambda*v element U

#

what does this even mean

#

what is U

#

a subspace?

#

@autumn kraken

elfin ingot
#

yo mart any help?

#

i rly dk whats wrong with what i said

wintry steppe
#

also learn to use texit or something because what you wrote is pretty much unreadable

elfin ingot
#

the poster understands why img(T) = img(T^2)

#

i dont can u help with that?

autumn kraken
#

yeah exactly, U is a subspace of V

#

and I am supposed to prove that

wintry steppe
#

lol wrong ping

#

@elfin ingot

#

um

#

you don't need to prove anything

#

that's just by definition

elfin ingot
#

?

#

how

autumn kraken
#

argh it's hard to articulate myself without texit lol

wintry steppe
elfin ingot
#

whats RT?
i dont understand htis notation

wintry steppe
#

range

elfin ingot
#

thank you

#

i get it

#

but i still dont know why what i said

#

was that bad

#

but i get this proof

wintry steppe
#

just learn to use the bot pandaOhNo

autumn kraken
#

sorry for my basic questions

wintry steppe
#

so do you remember what the definition of a subspace was?

autumn kraken
#

Yeah I have it on my screen right now.

#

I am supposed to determine if this is a subspace or not

#

I think I understood now that lambda is just in R

#

for scalar multiplication

wintry steppe
#

yes

#

so do you see why that'd be a subspace?

#

specifically as you said it's closed under scalar multiplication

autumn kraken
#

I am trying to show that it's closed under scalar multiplication

#

I know why it is but I can't write it down formally lol

wintry steppe
#

ok so you're checking that for every $v\in U_2$ and $\lambda\in\bR$, $\lambda v\in U_2$

stoic pythonBOT
autumn kraken
#

exactly

#

since when x + y + z = 0 then for any lambda, x * lambda + y * lambda + z * lambda = 0 right?

elfin ingot
#

yea

wintry steppe
#

^

elfin ingot
#

yo mart whats wrong with what i said?

#

for the proof img(T^2)=img(T

wintry steppe
#

i didn't read it

elfin ingot
#

1 sec

autumn kraken
#

but is this enough? my proofreaders are strict af

elfin ingot
#

let y = T^2(a) ---> y = T(T(a)) = T(b) for some b = T(a)

#

for showing img(T^2) = img(T)

autumn kraken
#

like how do I go from the first to the second, or is this not necessary

wintry steppe
#

Fix $(x,y,z)\in \bR^3$ with $x+y+z=0$. Take any $\lambda\in\bR$. Then $\lambda(x,y,z)=(\lambda x,\lambda y,\lambda z)$ has $\lambda x+\lambda y+\lambda z = 0$.

dusky epoch
#

you missed a z

stoic pythonBOT
wintry steppe
#

ty lol

#

@autumn kraken

autumn kraken
#

I see thanks

#

thought there'd be more steps necessary

dusky epoch
#

i mean

#

you can go full on formal

#

write out the proof that anything multiplied by zero is zero

#

write out the proof that the distributive law is true for three numbers

#

write out explicitly every single use of the transitive property of equality

#

this rabbit hole is quite deep

autumn kraken
#

yeah I see

#

lol

#

a thing I dont understand is that to prove it's a subspace you need to show that 0 is element of U, or that U is not empty. But why are they the same?

#

if U contains vectors how can it contain 0?

#

do they mean the neutral element of addition like (0, 0, 0) in Rยณ ?

wintry steppe
#

what

#

to prove it's a subspace you need to show that 0 is element of U, or that U is not empty.

#

this is not true at all

autumn kraken
#

that's what my script says lol

#

at the bottom the prof says "1. could be replaced by U is not empty"

torn silo
#

that means theyre are the same thing

#

that's not the same as or

autumn kraken
#

ah yeah sorry

#

but that's what I dont understand, why they're the same thing

wintry steppe
#

you need the other two axioms too

torn silo
#

Do you understand what a vector space is?

#

I mean as in the definition?

autumn kraken
#

kind of, but not really because I still have a hard time reading formal stuff

torn silo
#

You should read those kinds of defintion vector spaces are kinda important in la

autumn kraken
#

what I know is that elements of a vector space are vectors

#

which is why I dont get how 0 can be an element of it

wintry steppe
#

to understand that you just have to notice since U is closed under scalar multiplication and nonempty we can guarantee there is a v in U so that 0(v)=0

autumn kraken
#

ohhh

#

I am seeing it in the definition

#

it's (V, +, 0) and the scalar multiplication function

#

so when you're saying "something element U", it doesnt have to be a vector, it could refer to something else

torn silo
#

no U only contains vectors

#

the zero vector is (0, 0, 0) in Rยณ

wintry steppe
#

that might be a bit confusing yes

autumn kraken
#

so 0 and (0, 0, 0) are just synonymous in the definition?

#

depending on the context?

torn silo
#

0 is the zero element

wintry steppe
#

when there is ambiguity like then we write $0\cdot v=0_V$

torn silo
#

if you do addition a + e = a

stoic pythonBOT
torn silo
#

then e is 0

#

the zero element

autumn kraken
#

okay I see

torn silo
#

assuming you're in a group

autumn kraken
#

so in (V, +, 0) they mean the zero-element or neutral element (?) and not just the scalar 0

wintry steppe
#

they mean the additive identity yes

autumn kraken
#

thanks a lot

split heart
#

Hi, I've a question.
A=[
3 1
1 3]
A -4 =[
-1 1
1 -1]

#

Can somebody explain me what's happening?

dusky epoch
#

uhh

#

do you have a picture of what this is supposed to be

split heart
#

I'm just confused with A - 4

dusky epoch
#

it's $A - 4I$, not $A - 4$.

stoic pythonBOT
split heart
#

I'm blind, sorry

#

And where does this I matrix come from?

dusky epoch
#

it's the identity matrix

#

in some sense it's the matrix equivalent of the number 1

pale coyote
#

A is a matrix and 4 is a number, so A-4 doesn't make sense

autumn kraken
#

Is a simple scalar also a polynomial?

pale coyote
#

Sure

#

Degree 0

autumn kraken
#

thanks

#

I would write down two functions as polynomials and add them but I don't know where to go from there

#

or rather how to make sure (f1 + f2) is also in U3

wintry steppe
#

My assignment is to simplify this

#

but i dont see any way

dusky epoch
#

just a hunch but maybe it's the cube of a sum
also, wrong channel

wintry steppe
#

what channel then

autumn kraken
#

not sure

pale coyote
#

Madmike

#

You don't know they're polynomials

storm python
#

I mean vector space are additively closed and they're checking exactly that

pale coyote
#

All you know is one property, namely f(0) = 7

#

So take two things that satisfy this: f and g

#

Does their sum satisfy this as well?

autumn kraken
#

Is it possible to say $\\f_1(x) := a_0 + a_1 * x + ... + a_n * x^n = a_0\\$ because in $U_3$ it's defined that $f(0) = 7$?

stoic pythonBOT
pale coyote
#

No

autumn kraken
#

ok ๐Ÿ˜„

pale coyote
#

Oh sIt

#

Wait

#

They are polynomials

#

Sorry I misread

#

But either way you don't have to do that

autumn kraken
#

np

pale coyote
#

It's even easier

storm python
#

Just let f(x) = 7 and g(x) = 7 and add them

pale coyote
#

What would (f+g)(0) be?

autumn kraken
#

oh

#

lol

#

is that valid notation btw?

#

can I write it down like you did moonside

#

(f + g)(x) ?

storm python
#

Yeah it's commonly used

autumn kraken
#

thanks

#

got to be careful with that stuff because if I use things that were not mentioned in our script, then my proofreader will delete all my points lol

#

it's so stupid

pale coyote
#

The sum of two functions f and g is a new function (f+g)

autumn kraken
#

just to be sure, in this case (f+g)(x) = 14 right?

pale coyote
#

But when you evaluate this new function at a point x, its value is equal to the sum of the values of the original functions: (f+g)(x) = f(x) + g(x)

autumn kraken
#

oh I see

pale coyote
#

Yes

autumn kraken
#

I see so it's not closed under addition because (f + g)(0) = 14 and not 7

#

If I find out something is not closed under addition, do I nonetheless have to include the check if it has a neutral element or not? Like is it a prerequisite to checking additive close?

pale coyote
#

no

#

once it fails any of the conditions youre done

autumn kraken
#

thanks a lot

#

are you fine with me asking for help here btw?

#

since it's probably easy stuff for you

#

or is this not a good channel for that

pale coyote
#

its linear algebra related, so should be good

autumn kraken
#

by the way

#

are you sure it's correct this way? Because it's only saying that $f(0) = 7$ and not $f(x) = 7$

stoic pythonBOT
autumn kraken
#

or is it because of what I posted above about $f_1(x) = a_0$?

stoic pythonBOT
torn silo
#

hmm closed under addition means you leave the space by using addition

pale coyote
#

i dont understand your question @autumn kraken

narrow mortar
#

hi moonside

autumn kraken
#

as far as I understand it, there's an infinite amount of polynomials in $R[x]$ and $U_3$ contains the ones where $f(0) = 7$, but does that really also mean $f(x) = 7$ ? Hope that makes sense lol

stoic pythonBOT
autumn kraken
#

why can I say f(x) = 7 and g(x) = 7 then?

#

ahh okay thanks

placid oracle
#

How do I calculate LX of {(1,0),(0,0)} if LX : V โˆ’โ†’ V defined by the formula LX(A) = XA.

placid oracle
#

<@&286206848099549185>

wintry steppe
#

Given 2 planes: 5x-2y-2z = 1 and 4x+y+z = 6

#

How can I find a point on the line where these two intersect?

#

My idea is to set x and y to equal 2 and then solve for z. But I'm not sure how to do that if there is a = 1 and = 6 in the equations.

#

I tried subtracting 1 and 6 respectively so they both equal 0, but I got the point (2, 2, 1/3) which is not on the intersection line.

#

I also tried setting Z to 0 and then that would mean that y = 4x-6 and then solving 5x-2(4x-6) = 1 for x, but it gave me (11/3, 26/2, 0) which is also not on the line.

#

Oh got it, it's y = 6 - 4x not y = 4x - 6

molten pumice
#

Pick a variable and solve each equation for it separately. Then, since those will both be equal to the same variable, set them equal to each other. That will eliminate one of the variables and give you the line where the planes intersect

still agate
#

as an addendum to the previous question -- what is a good way to conceptualize row space vs column space?

#

im getting a little bit lost in the abstraction

elfin ingot
#

the row space is the space spanned by the row vectors of a matrix

#

column space is the space spanned by column vectors of a matrix

odd kite
#

@still agate the column space is the set of possible outputs when you apply the matrix to a column vector. This is called the image. The row space is the orthogonal complement to the kernel. This means that the row space determines what directions the matrix is "sensitive to" and what directions it just gives 0 for.

cold topaz
#

I tried to got a QR decomposition for my matrix A, but the last row came out all 0s. That indicates that it doesnt exist. correct?

still agate
#

@odd kite what is a kernel?

odd kite
#

@still agate It's the set of vectors that get sent to 0 by the matrix. It's also called nullspace

still agate
#

Oh gotcha thanks!

wintry steppe
#

in linear algebra if you have Ax = 2x, how does that become (A-2I)x = 0

#

from where did the I come from

wintry steppe
#

its not equal to 2x

#

cuz 2 is a scalar

#

but if it were BIx, it would equal to Bx, because B is a matrix

sharp merlin
#

How do i do this

slow scroll
#

Ax = 2x
IAx = 2Ix
Ax = 2Ix

wintry steppe
#

hmm i see

eternal finch
#

@sharp merlin Look for rank-nullity theorem.

wintry steppe
#

@sharp merlin r u in my university

#

lol

sharp merlin
#

lol

eternal finch
#

What book do you guys use? Probably commonly used.

sharp merlin
#

where do u go

wintry steppe
#

but that was before midterm

sharp merlin
#

ah

wintry steppe
#

r u in CA?

sharp merlin
#

yeah

#

UCR

wintry steppe
#

well im in UCI

#

similar

sharp merlin
#

oh cool

wintry steppe
#

ive seen that problem

#

dim Col A = rank A, dim Col A + dim Nul A = # of cols of A

sharp merlin
#

Yeah

wintry steppe
#

dim Row A? ive never heard of that

sharp merlin
#

yeah thats what im stuck on

wintry steppe
#

dimNulA is easy i told u the solution

sharp merlin
#

yup

#

i got it

slow scroll
#

dim row is dim col A^T

sharp merlin
#

if dim col A = 2

#

whats dim col a^T

#

also 2?

slow scroll
#

well, there is a theorem that says row rank is always equal to column rank

#

so yes

sharp merlin
#

why is dim nul A = 2

wintry steppe
#

mylab

sharp merlin
#

@slow scroll

wintry steppe
#

right?

sharp merlin
#

yea

wintry steppe
#

rank A = dim of the basis of col A = how many pivot columns of A

#

dim nul A = # of cols - rank A

#

or u can just solve Ax=0

#

and find out

slow scroll
#

when you put A in REF, look at the pivots. The pivot columns of the REF form a basis in the original matrix

wintry steppe
#

yh

#

my teacher in ratemyprofessor has the lowest rating ever

slow scroll
#

oof

wintry steppe
#

the midterm was ez though

#

but had to do row echeleon reduction 5 times

#

lol

slow scroll
#

ya i got pretty good at row echelon last semester

wintry steppe
#

@sharp merlin r u guys behind or something?

sharp merlin
#

wdym

wintry steppe
#

right now we're past this stuff ur doing

sharp merlin
#

we still have 3 weeks left

wintry steppe
#

u done determinants?

sharp merlin
#

starting rn

#

like nxn determinants?

wintry steppe
#

determinants for 3 by 3

#

4 by 4

sharp merlin
#

3 by 3 we already did

#

we doing nxn rn

wintry steppe
#

u did the midterm yet?

#

have u covered inverse of a 3 by 3 matrix?

#

other than row reducing to identity matrix

sharp merlin
#

yeah we did inverse

#

and yes finished midterm

#

Why is my answer wrong for last one

#

@slow scroll

wintry steppe
#

did u solve Ax = 0

sharp merlin
#

wait

#

for nul do u have to convert to rref

#

or just ref

wintry steppe
#

rref

#

Ax = 0

sharp merlin
#

oh thats why

#

got it thanks

hearty cliff
#

Our teacher for machine learning wants us to brush up on linear algebra, but i didnt have to take it for Comp Sci.

elfin ingot
#

i understand shit with representing linear transformations with matrices

#

how can i find the matrix relative to a certain basis?

#

for a transformation

#

i get the idea i think but still

#

can anyone guide me through?

eternal finch
#

Very comprehensive.

hearty cliff
#

oooo ok nice thanks

wintry steppe
#

i hate compsci

hearty cliff
#

Could someone send me a question w/ work similar to this one?

#

Its from an example problem set, not even the actual problem set

eternal finch
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@hearty cliff Here's my attempt at an intuitive explanation. Keep in mind I've not had a course in matrix calculus.

You know that the derivative of f is defined by the ratio (f(x + h) - f(x)) / h. So, if you had no idea what the derivative of vector-valued functions looked like, you might just try just using that ratio.

f(x + h)
= (x + h)^T A (x + h)
||= (x + h)^T (Ax + Ah)
= (x + h)^T Ax + (x + h)^T Ah
= (x^T + h^T) Ax + (x^T + h^T) Ah
= x^T Ax + h^T Ax + x^T Ah + h^T Ah||

Then,

f(x + h) - f(x)
||= x^T Ax + h^T Ax + x^T Ah + h^T Ah - x^T Ax
= h^T Ax + x^T Ah + h^T Ah||

At this point, you might try to divide by h, but uh, how do you divide by a vector? Doesn't make much sense. Instead, consider writing the ratio as

(f(x + h) - f(x)) = derivative * h

So,

||h^T Ax + x^T Ah + h^T Ah = D * h||, where D is the gradient vector in our case.

Now, here's where I go from not very rigorous to not rigorous at all. If we pick h to be very small in magnitude, then the ||h^T Ah doesn't really matter||.

Then,

||h^T Ax + x^T Ah = D * h||

||h^T Ax = x^T A^T h, so we get x^T A^T h + x^T A h = D * h. Or, x^T(A^T + A)h = D * h.||

So, ||D = x^T(A^T + A)||? Well, D * h is a product of a vector with a vector. It should be a dot product. So, what we have is really ||x^T(A^T + A)h = D^T h||. Then, ||x^T(A^T + A) = D^T||.

#

I blotted stuff out so you can go through it step by step.

elder robin
#

Is this the right way to calculate the null space of A?

#

Span of one vector seems weird. Is it just a line?

tall moon
#

@elder robin, how come your null space is four dimensional when your matrix is 3 by 3?

gray dust
#

why are there 4 variables?

limber sierra
#

nothing wrong with having a span of one vector, but uh

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yeahhhh you cant multiply that matrix by that vectr

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so its certainly not correct

tall moon
#

at least the rref is correct.

elder robin
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Oh there shouldn't be an x_4

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I was using the columns for some reason

limber sierra
#

i'm confused why you're augmenting this matrix at all?

elder robin
#

But if x_1, x_2, x_3 =0, then what is the span? span([0,0,0])

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I just watched a khan academy vid and the way he did it is augment the matrix to solve Ax=0

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so {A|0}

limber sierra
#

alright, sure

#

well, if they're all zero, then the span is just {0}

elder robin
#

Then we need all x's such that Ax=0

limber sierra
#

and a basis for this space is just the empty set

elder robin
#

what is span{0} geometrically?

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is it just a point

dusky epoch
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it's just {0}

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yes it's just a single point

#

the origin

elder robin
#

oh yeah I guess that isn't much of a span

cold topaz
#

orthonormal set is different from orthonormal basis. right?

limber sierra
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but the empty set spans it

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well, an orthonormal basis is a basis sm

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whereas an orthonormal set isnt necessarily a basis

cold topaz
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orthonormal basis is made of norms. but whats set?

dusky epoch
#

"made of norms" wat

limber sierra
#

...

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"orthonormal" is a property a set of vectors can have

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if that set forms a basis

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then we call it an "orthonormal basis"

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which is really an abbreviation for "orthonormal set that is a basis"

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or "basis made of a set of vectors that are all orthonormal to each other" or however you want to phrase it

cold topaz
limber sierra
#

maybe.

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depends on what A is

cold topaz
#

i have A

dusky epoch
#

an orthonormal set per se need not be a basis for the vector space it lives in.

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it just so happens that if the number of vectors in your set matches the dimension of your space, then your set will actually span the space and hence be a basis, and hence be an orthonormal basis.

cold topaz
#

i calculated u1u2 = u2u3 = u1u3 = 0 and ||u1|| = ||u2|| = ||u3|| = 1
then what?

limber sierra
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i'd assume that first line is the dot product?

cold topaz
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i want ti prove that A is orthogonal

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it is dot product

limber sierra
#

then you've shown the columns are orthonormal

cold topaz
#

so im done with the columns? no more work needed?

dusky epoch
#

what is your definition of an orthogonal matrix?

cold topaz
#

AA^-1 = I
The screenshot above.
The same screenshot but for rows.

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or only AA^-1 = I is enough?