linear-algebra

[winter reef]

🥖 👌

[slender yarrow]

lewd

[sand hedge]

[wintry steppe]

i have 7 conditions that a function s(a,b,c,d) needs to satisfy.Find a formula for that function:

s ∝ a/(a+b)
s ∝ c/(c+d)
s ∝ 1/[b/(a+b)]
s ∝ 1/[d/(c+d)]
0 <= s <= 1
s = 0,if and only if a=c=0
s = 1,if and only if c=d=0

What's ∝?

proportional to

[broken hawk]

that’s not very linear algebra, is it

[wintry steppe]

well in which one should i post it in?

i still don't understand
does s ∝ a/(a+b) mean that s(a,b,c,d) = k(c,d) * a/(a + b) for some function k?

no it just means that s = k*a/(a+b) i think

then obviously your conditions are inconsistent
s(a,b,c,d) = k1 * a/(a + b) = k2 * c/(c + d) cannot be true everywhere unless s is zero

i think you multiply the two to make a new constant k1*k2

then I don't understand how your ∝ works
try to explain it more formally

and yes, this is probably not a linear algebra question

try in one of #❓how-to-get-help channels

i posted it there too

[burnt gate]

it doesn't seem inconsistent to me

[wintry steppe]

just like the derivation of the ideal gas law

you just multiply

the proportionality relations together

and construct a equation with a constant R term

https://socratic.org/questions/how-is-ideal-gas-law-derived

anyone got any ideas?

How do i solve $(1 - x)^ (b - 1) (a - x (a + b)) = 0$

[stoic python]

blasphemy:

[nova yew]

for what

[wintry steppe]

for x

[nova yew]

oh for xx

take each part and seperately set equal to 0

so (1-x)^(b-1)=0

and a-(x(a+b)=0

then solve

[wintry steppe]

thank you

[nova yew]

no problem

[stable flower]

Okay I'm having some weird issues, could someone send a lil help? It's not directly for a class, but it'll help me out lmao

I have a three-dimensional vector space V, with pairs (x,y,z), over the field F_4. I'm trying to construct the set of all one dimensional subspaces of V, so I imagine that there are (63/3)=21 of them, but I'm having a really hard time constructing it properly.

I can do it in the F_3 case just fine but the 2s in F4 are throwing me off

[wintry steppe]

there are no 2s in F4

F4 is not 0,1,2,3

[stable flower]

Oh shit. It's 0, 1, a, 1+a isn't it

[wintry steppe]

yes

[stable flower]

Fuck my ass goddammit

Okay hang on

So can I build a basis for a subspace with just one vector (001)?

I think yes since that's a dim one subspace

[wintry steppe]

yes

it will have 4 points
000, 001, 00a, 00(1+a)

[stable flower]

Some of them can definitely have three though, right? Like (01a)

That gives 000, 01a, 01(1+a)

[wintry steppe]

no
every line has 4 points, one of them 000
line with basis p is 0 * p = 000, 1 * p = p, a * p, (1+a) * p

[stable flower]

Ohh

Okay hang on

Gimme a min

[wintry steppe]

what's your minimal polynomial for a?

[stable flower]

x^2+x+1

[wintry steppe]

one dimensional subspace generated by (01a) is { (0,0,0), (0,1,a), (0,a,1+a), (0,1+a,1) }

[stable flower]

Yep, just calculated that! Thank you!

[wintry steppe]

you are welcome

[stable flower]

Okay so all of the ones that start with 0 are generated by 001, 010, 011, 01a, 01(a+1)

[wintry steppe]

yes

[stable flower]

And repeating this for the rest should obtain 21?

I suppose 0 is included

[wintry steppe]

0 is not one-dimensional

[stable flower]

Oh right

Hhhh

[wintry steppe]

it is zero-dimesnional :)

[stable flower]

I'm trash

[wintry steppe]

you can assume that the first coordinate is 1 for other 16
because if the basis is one vector (x,y,z)
then (x,y,z)/x is also a basis and it is (1,y/x, z/x)

[stable flower]

OHHHH RIGBT

okay yeah that makes total sense

[wintry steppe]

then you just need to prove that for two vectors (1,y1,z1) and (1,y2,z2) the one-dimensional subspaces generated are different if the vectors are different

there are 16 of them in total

[stable flower]

Oh, the problem isn't really concerning the spaces themselves, I'm using it for a fairly different problem

But I need to have all the spaces in hand

And. Okay. For the two dim subspaces, do I just take two different generators and smack em together or

[wintry steppe]

basically yes
they must be linearly independent

but maybe you want to read something about projective spaces and grassmannians

[stable flower]

Hmm will do

This is for a design theory problem lmao. Don't typically consider fields but in this case we can bc it's a finite projective plane and hhhh

[wintry steppe]

Hassett "Introduction to Algbraic Geometry" chapter 11 looks better than everything else I have
But may be still too advanced

[stable flower]

Hmm. I might have to look at the book in general. It's probably not too advanced, I just forgot the process for forming fields 😅😅

[stable flower]

okay. So. Just to confirm. The two-dim subspace made by 001 and 010 is
{000,001,00a,00b,010,0a0,0b0,011,0a1,0b1,01a,0aa,0ba,01b,0ab,0bb}

I at the very least know that the size is correct

(oh, shorthand is b=a+1)

[wintry steppe]

can't you just say {0xy} ? 🤔

[stable flower]

yknow

shit

yes

for this one lmao

[wintry steppe]

also, I don't know what notation you've been taught, but we write vectors as {x,y,z} etc... just if that's so, gotta be careful

[stable flower]

Lol, same. but when writing a million vectors and the vectors are sort of a side note, writing it as xyz saves everyone

[wintry steppe]

In which case we write: $
\begin{pmatrix}x \ y \ z \end{pmatrix}$

[stoic python]

taninounet:

[stable flower]

no

literally the text uses xyz in this case

because there's just so many

[wintry steppe]

lol

[stable flower]

I know how to write vectors lmfao

[wintry steppe]

then you could <(0,1,1)>

[stable flower]

s t o p

[wintry steppe]

:p

[gritty mountain]

Hey i need to prove that the vector (1,3,5) was not in the image or solution set of matrix C

So what i did was i put the vector as a column and performed row operations

I got a 0 0 = x situation so i concluded that therefore there were no solutions

Is this correct?

[half ice]

a + b = 1
a + b = 3
5a + 4b = 5

Those first two lines are already puzzling

Indeed, you can get 0 = 2

So no value of a and b will make that work

[gritty mountain]

Okay thanks :)

[wintry steppe]

is there a thing called bilinear algebra?

please dont call me a retard

[slow scroll]

there is multilinear algebra

[wintry steppe]

oh okay

[wintry steppe]

hi do anyone understand how they got this?

[slow scroll]

wrong channel?

[wintry steppe]

nvm

[dull kettle]

Is S = {(x,y,z) | x = y = 0} a subspace?

[slender yarrow]

subspace of what space

[shy atlas]

my space

[slender yarrow]

ur space

[shy atlas]

https://myspace.com/

smh

[dull kettle]

one condition is that the set has to contain zero vector but in this case, the condition is given for x and y only

question didn't mention subspace of what space, but I assume it's r3?

[slender yarrow]

yeah ik i'm just being picky

"the condition is given for x and y only" and?

what's the usual 0 vector in R^3 ?

[dull kettle]

(0, 0 ,0) ?

[slender yarrow]

yeah

are its 2 first components equal to 0?

[dull kettle]

yes

[slender yarrow]

then bam

it's in S

[dull kettle]

but we can assume z to be zero?

i don't know it's just that nothing was given in the condition for z

[slender yarrow]

this means it can be anything

(in R)

since there's no condition/restriction put on it

[dull kettle]

thank you!

[slender yarrow]

[stoic python]

وادو هيك:

[undone depot]

@stoic python help

[stoic python]

A brief description and guide on how to use me was sent to your DMs! Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[unborn gulch]

Are there any recommended readings on learning how to find the inverse of a Linear Transformation?

[ember pilot]

$(-3)^ny_n=[x_{n+1}-2x_n]$\$y_{n+1}-y_n=(-4n-1)(-3)^{-n-1}$\$y_n=y_0-\sum_{k=0}^{n-1}(4n+1)(-3)^{-n-1}$\$(-3)^ny_n=(-3)^ny_0-\frac14(4n-3)(-3)^{2n-1}+\frac14(4n-3)(-3)^{3n-1}$\$x_n=v_n2^n$\$v_{n+1}-v_n=x_12^{-1-n}-x_02^{-n}-(4n-3)2^{-3-n}(-3)^{2n-1}+(4n-3)2^{-3-n}(-3)^{3n-1}$\$v_n=v_0+\sum_{k=0}^{n-1}x_12^{-1-k}-x_02^{-k}-(4k-3)2^{-3-k}(-3)^{2k-1}+(4k-3)2^{-3-k}(-3)^{3k-1}$\$v_n2^n=v_02^n+x_12^n-x_1-x_02^{n+1}-2x_0+\dots$

[stoic python]

Simple_Art:

[ember pilot]

||the remaining is trivially disgusting and left as an exercise for the reader||

[wintry steppe]

https://cdn.discordapp.com/attachments/490971739061354506/549435453346938900/unknown.png

how to

[novel wasp]

I had a few quick questions, simple

I know that a homogenous system of equations that have more variables than equations always has infinite solutions

But does that hold true for non homogenous systems as well?

Like if a question asks "A system of 3 equations and 4 unknowns always has infinite solutions?" Will that be true always?

Or only for homogenous equations?

[maiden dagger]

if there exists a solution there will exist a family of solutions whose dimensionality depends on the rank of the system

but i can give you a nonhomogenous system with 2 equations, in as many variables as you want, that has zero solutions

[stoic python]

tubular:

[maiden dagger]

the lore is that if you have solution $x$ to your nonhomogenous system $Ax = b$ and also a solution $z$ to the associated homogenous system (i.e. $Az=0$) then $A(x+z) = Ax + Az = b + 0 = b$, that is to say, $x+z$ is a new solution to the original nonhomogenous system

[stoic python]

tubular:

[maiden dagger]

but for this to be applicable there needs to be at least one solution to the nonhomogenous system which you can prevent if you get to choose the system

@novel wasp

[novel wasp]

So in general if a question asks "A system has x equations but x+1 unknowns," should I expect there to always be infinite solutions?

[maiden dagger]

either zero or infinitely many

for exactly the reasons i told you above

{a+b+c=1, a+b+c=2} has two solutions and two+1=three unknowns and zero solutions

[novel wasp]

Ah gotcha

Thanks!

[tough socket]

hey here, i just posted some stuff in #groups-rings-fields , if you can check it out

[modest wraith]

I need help with problem #2 on my test correction.

I don't know Math.

[slow scroll]

this is #prealg-and-algebra ffr.
Put the equation in y = mx + b form
6x + y = 24
subtract 6x from both sides
6x - 6x + y = 24 - 6x
y = 24 - 6x
y = -6x + 24

So then what would the m be here?

@modest wraith

[modest wraith]

I see, thank you so much for the help! ☝️

[slow scroll]

np

[dense holly]

lol^

I have a couple questions for an upcoming test if anyone can assist

1. Do you need a free variable to be LD?

tryna think of a counterexample

[slow scroll]

The Columns of a matrix are linearly independent iff there is a pivot in every column when in echelon form. When this isn’t the case, that means you would be able to write one or more of the columns as a linear combination of the pivot columns. Therefore that system of vectors would be LD

[dense holly]

How do we define span in linear algebra

[maiden lintel]

The set of all finite linear combinations

[dense holly]

For a matrix?

Or does range refer to us having a transformation too

[maiden lintel]

Matrices are just representations of linear transformations

[dense holly]

How do you relate range to domain and codomain

That's what my class is doing rn

[maiden lintel]

The range is a subset of the codomain

[slow scroll]

i think range = image
in other words, the domain is the input space, the codomain is the space that is mapped to, and the range corresponds to the elements of the codomain that get mapped to

[dense holly]

how do we know what parts of the codomain are the range

[solid forge]

many of the ideas you first encounter in linear algebra are figuring out if the range = codomain or not

if I give you basis for the domain and a basis for the codomain and then define a linear map between these two bases what does that mean?

do it for R^2 to R^2 or something on your own and play with it, the best way to gain intuition

[slow scroll]

i haven't specifically encountered that yet, but it seems like the codomain would not equal the range if the columns of the matrix corresponding to the transformation didn't span the codomain

[obtuse nova]

I just read that given a mxn matrix A, and a nxm matrix B, AB = I iff m<n
first: is this true
second: how can you prove this

[solid forge]

@obtuse nova do you know how to prove an iff statement?

if it is false how do you prove it?

[obtuse nova]

do you mean contradiction?

[solid forge]

to be clear I asked 2 separate questions

prove this is false: an even number plus an even number is a false number

lol an odd number not false

im tired

[obtuse nova]

are you saying that what I said isn't true

[solid forge]

I am saying if its true, you need to prove the iff statement

if its false you might need to do something else

[obtuse nova]

so do you know the answer by any chance

like is it common knowledge in linear algebra

[solid forge]

yes its common

let m = n

[obtuse nova]

that works I know

[solid forge]

take the identity matrix

I*I = I

but m is not less than n

so we have produced a counter example

[obtuse nova]

oh right

how about if m < n

can that ever produce the identity matrix

[solid forge]

the identity matrix is square

[obtuse nova]

yea what if you had say 3 x 2 times 2 x 3

[solid forge]

and a (mxn) * (nxm) matrix product is (mxm)

[obtuse nova]

yea could you ever get the mxm identity matrix

that's the core of my question

[solid forge]

this is why we require AB = BA = I for the definition of inverses, our question speaks to left and right inverses but not general inverses

your question*

[obtuse nova]

what about A = [1,0] and B = [1;0]

[solid forge]

are those both 1x2?

[slow scroll]

row vector A and column B

i think

[solid forge]

then their product is [1] which is the identity matrix of n = 1

[obtuse nova]

yea that's what I meant

[slow scroll]

I know you can construct a linear transformation A : R^m -> R^n and B: R^n -> R^m m<n such that B(A(x)) = x. I'm not sure how you would prove that such a configuration exists for all m and n tho

[obtuse nova]

do you think you could come up with a 3x2 and a 2x3 matrix that equal the identity matrix of n = 3

[slow scroll]

im dumb. no i don't think you can, but i cant prove it

[gentle knoll]

Do the inclusion map of the smaller space into the bigger space (or any injection), and then define the other map to be the inverse map on the image portion, and anything else on the rest of the space

[placid oracle]

Kind of lost on these two questions...

[slow scroll]

hint: test for b. T(ax1 + bx2) = aT(x1) + bT(x2)

[slender yarrow]

x1*x2

linear

Aaaa

[slow scroll]

on the first one did you get contraction for the answer? I thought the middle one would be a shear but I can't quite remember @placid oracle

[placid oracle]

i think the middle one is a shear too

[slender yarrow]

there's no contraction indeed

[placid oracle]

contraction seems the only one

ok

wb the second one?

[slow scroll]

Test that choice b is a linear transformation and you'll see that it fails

[placid oracle]

o yea

mb, wasnt that difficult

[slender yarrow]

[slow scroll]

[placid oracle]

@slow scroll So I know this is true if x is a scalar, but wb with a vector?

[slow scroll]

Do you mean is T(xy) = T(x)T(y) where x and y are both vectors or something like that?

[placid oracle]

Um no, since c is a scalar. normally its T(xu) = cT(u) where both c and u are scalars => it is a linear transformation, but here u = x which a vector

so idk if the statemtent is true

[slow scroll]

u would have to be a vector. The linear transformation T is a function that takes a vector as an input

[placid oracle]

confused by what ur saying

see, they r using scalars

[slow scroll]

the bold letters are vectors tho

[placid oracle]

ughj

this is what happens when i start studyuing at 2am

sorry mate

[slow scroll]

its all good haha

[placid oracle]

one more question.... what does this mean: the function T(x)=sinx is a linear transaformation of itself from R to R

[slow scroll]

sin(x)? I'm not sure that makes any sense. A linear transformation from R to R is T(x) = ax where a is T(1).

[placid oracle]

yeah, dont really see what its saying

[slow scroll]

[broken hawk]

it's claiming that sin(x) is linear
it is not

also you added an "of itself" on your own, that was not there in the original

don't misquote like that

[slow scroll]

are these like true/false questions or something or is that actually some kind of typo?

[broken hawk]

adds confusion

[placid oracle]

it is true/false

[broken hawk]

what it's claiming is that
sin(x+y) = sin(x) + sin(y)
and
sin(ax) = a sin(x)

[placid oracle]

and mb sascha

ok so clearly false, how do u get that though

[broken hawk]

just find some values where it fails

e.g. 0 = sin(10*pi/2) ≠ 10sin(pi/2) = 10

there, proven nonlinear

[placid oracle]

Not that part, i mean how do u get that assumption sin(x+y) = sin(x)+sin(y) from what is given

[broken hawk]

also for R->R, all linear functions have the form T(x) = ax

that's part of the definition of linear

a linear function T: V -> W (where V and W are vector spaces) fulfills

1. T(u+v) = T(u) + T(v)
2. T(λu) = λT(u)

where u,v are vectors and λ is a scalar

[placid oracle]

yes, i get that

[broken hawk]

R is a vector space

now just replace T with sin

[placid oracle]

Oh

im going to reread this section in the morning, clearly did not do it well

thank u for ur help guys

[slow scroll]

np and gl c:

[forest jay]

I get their logic but how did they arrive at the equation (c+d)b ?

[slender yarrow]

Reuse the fact that Au = Av = b

So cAu + dAv = cb+db

@forest jay

[forest jay]

OH LOL completely missed that thanks

also for something like Ax=b where A is a 3x3 matrix

is it correct to say b spans like R^3?

if there's a solution ofc

Like for part B, I know something like a 2x3 matrix would work, but a 3x2 matrix wouldn't

But I don't know how to formally write out why

[forest yew]

How can I find the intersection of two spans?

[broken hawk]

it’s not really trivial, probably the most straightforward way would be to first find a basis for each (ie just eliminate redundancy in the sets), then manually check for every element in basis 2 whether you can build it by elements in basis 1. All the elements you can will then form a basis for the intersection

[forest yew]

So I am given the two spans and the result of the intersection, I then have to show that the result is true. Would it then somehow be 'easier' to show this than finding the intersection?

[half ice]

Show the result is in both spans

Then, show that anything in both spans is in the result?

Hmm. Doesn't seem easier

[broken hawk]

you can possibly eliminate one direction by instead comparing the dimensions

if you hvae that the dimensions are the same, then either inclusion implies equality

dunno if that’s any easier though

[forest yew]

alright ill try ty!:)

[forest jay]

Could someone help me with the one I sent earlier?

[delicate kayak]

[steel cobalt]

ex: 5A^T - C... Do I transpose A first or multiply by 5 THEN transpose?

[vestal steeple]

If anyone has a few seconds I would really appreciated some help on a few linear algebra problems.

[mighty cargo]

@steel cobalt it shouldn't matter because all the entries in the matrix are multiplied by 5 regardless

[steel cobalt]

thank you

[mighty cargo]

np!

[proper moat]

@forest jay For part B you can think about # of unknowns and equations

if m<n what can you say about the existence of free variables?

and therefore what about the solution set?

of Ax=b

[wintry steppe]

Book question: after many years of hiatus (physics degree), I've decided to get serious about rebuilding my math skills. I'm starting off with an immediate reconstruction of linear algebra, so I am trying to decide whether to go for a computational approach with Strang or do something proof based, like Axler. On one hand, linear algebra is super-important, so it might be wise to make progress as quickly as possible: and looking at Strang, I have no doubt I can work through it or an online course that uses it (18.06 for MIT OCW) pretty quickly. But if I'm to really jack up my mathematical abilities in the long haul, I have to get better with proofs, and I'd probably get deeper insights into the subject from Axler.

What do y'all recommend?

I'd like to do both, but I seriously doubt I'm going to have the time-unfortunately, with a full-time job and a girlfriend, I don't have the time I had when I was an idiot 18 year old.

[slow scroll]

im not too familiar with strang, but you could maybe look into "Linear Algebra" by Kunze & Hoffman. it sounds more theoretical than strang but more matrices and stuff than in "Linear Algebra Done Right" and is probably more relevant to physics. idk sometimes I feel like the best strategy is to just choose something and go for it lol

[wintry steppe]

I think I'm going to go with Axler and see where it goes.

[proper crescent]

Axler's alright for the beginning

[wintry steppe]

If I find that the problems are beyond me to solve, I'll backtrack.

I don't want to rush this: I want to truly get a good grasp on linear algebra this time.

[proper crescent]

But doing characteristic polynomials without determinants is a bit stupid to me

[wintry steppe]

I think being forced to work through it slowly, page-by-page, would help.

I think the temptation to rush through Strang would be too much.

[proper crescent]

But to be fair I think Axler isn't as tough as Hoffman-Kunze

I mean eh

[slow scroll]

I've been using a book called "Linear Algebra Done Wrong" which i do like, but it doesn't have solutions which is really bothering me. Just make sure to choose something with solutions haha

[wintry steppe]

Well, I'm not looking to become a mathematician, so...

[proper crescent]

Hard to say

HK is old-school

[wintry steppe]

Anyway, I'm new here. I'll probably need a lot of help to do this.

[proper crescent]

@slow scroll do you need solutions at this point? I feel like it's fine to just try the proofs yourself and ask on MSE for proof verification/hints if you're stuck

[lean aspen]

do u even sleep

[wintry steppe]

I wish I didn't have to.

[proper crescent]

Pre-internet I feel it was probably more important but in the face of people verifying your proofs and giving you hints based on how far you've gotten (as opposed to having to cover the last few lines or something) is more helpful

I think he's talking about me lmao, my sleep times are fucked

[wintry steppe]

Lol, I really wish I didn't have to. I'm about to embark on a journey in my mid-20s of teaching myself all the math/physics/CS I can.

Youth and the time you have as one is truly wasted on the young.

[lean aspen]

tfw 2old to keep up with kids

[wintry steppe]

It's not a matter of being unable to keep up with kids (well, if you are smarter than I am), it is a matter of having the time when you've got bills to pay and a boss to avoid pissing off.

But I'm ready. 😃

[lean aspen]

I'm not smart tho

[slow scroll]

yea and I do do that sort of thing. Its one of those things where people really need to read the chapter before to get a context for what the problems are expecting. I don't know the format of certain things either i.e. can i reference corollary 2.54 from the chapter before or what. Maybe its just because I'm not too comfortable with proofy stuff yet.

I haven't given up on LADW, but i just feel like it would be a lot faster if I could glance at a solution every now and then to get an idea of what i should be doing rather than reading a chapter and going in cold to the exercises.

[lean aspen]

@proper crescent is there a neat way to define char without matrices even

[proper crescent]

You talk about generalized eigenspaces

[lean aspen]

i feel as though there should be a way to with those yeah

[wintry steppe]

@lean aspen You are probably smarter than me.

[lean aspen]

im just old

[wintry steppe]

Trust me, if I end up succeeding at anything, chances are anybody here can too.

[lean aspen]

okay nvm I think i know how to go about it

[wintry steppe]

I was tested as having a borderline sub-normal quantitative IQ as a teenager, bombed college, and have never done anything consistently constructive with my life until a few months ago.

As I said: if I can do anything, you can too. Don't worry about it. 😉

Time to go learn math!

[proper crescent]

Yeah basically what you'd expect, the rank of generalized eigenstuff corresponds to the algebraic multiplicity

Or well, the dimension

@wintry steppe have fun

[wintry steppe]

I'll be back here when I struggle, I'll be bound

Signing off for now, though. Got to turn off the Internet to focus.

[lean aspen]

☑

[solid forge]

My problem with Axler's book is I could eventually prove most of the problems but I had no real idea what they meant, or what deeper connection I was supposed to draw from them.

[sour garden]

I think axler is pretty transparent in regards to meaning

I didn't read the final third of the book tho

@solid forge could you give an example of a thing axler wasn't clear on

[wicked trellis]

Can anyone help me with (3)

Is it valid to do for example e1 = v1 + 0v2 +0v3

[slow scroll]

it sounds like it wants you to find the linear combination of v1, v2, v3 that equal (1, 0, 0), (0, 1, 0) and (0, 0, 1)

[wicked trellis]

Oh

[proper moat]

just do your standard row reduction

[wicked trellis]

Oki

So just something like this

Or am I tripping balls

[proper moat]

well you can test it rigth

haha

does (3/2)v_1 + (1/2)v_2 + (4/7)v_3 give you [1 0 0]?

sorry [1; 0; 0;] i guess,

[wicked trellis]

Supposedly

[proper moat]

try it

[slow scroll]

@wicked trellis check x_1 maybe?

[wicked trellis]

Think I messed up somewhere

[proper moat]

ye

your v2 coefficient is right

and the last one

for v3

[slow scroll]

i think v_1 is wrong

[proper moat]

^^

[wicked trellis]

Okay I got it

So Ax = 1,0,0 where I solve for x

[slow scroll]

yea thats the idea

[wicked trellis]

And just do that for the others

[slow scroll]

yea

[dense holly]

How do you write solutions in parametric vector form

[slow scroll]

i think thats just when you convert some linear combination of vectors to a system of linear equations

[dense holly]

Is your gave linear dependence or independence, does that imply onto or one to one for either?

[slow scroll]

If you have a linearly dependent transformation T(x), then that means there are non-unique solutions to T(x) = b, which means its not one to one.

[dense holly]

Does independent matrix mean 121 after the transformation?

[slow scroll]

a linearly independent matrix refers to whether the columns of the matrix are linearly independent. Not sure what you mean by "one to one after the transformation"

[dense holly]

If you go from R^2 to R^4, can your transformation be onto

[half ice]

No linear transformation from R² to R⁴ can be onto. A linear transformation of a basis gives a basis for your new space.

Or, actually that doesn't make sense. But the dimension can't be higher

[maiden dagger]

the image of a spanning set of a vector space gives a spanning set for the image of the vector space

[half ice]

Much better way to put that

[maiden dagger]

so if you have a spanning set of size two for R2 to then then it maps to a spanning set of whatever subpace of R4 your vector space lands in

[half ice]

Tubular didn't say it completely wrong like I did

[charred stirrup]

I am having a hard time understanding what the question is fundamentally asking

[jovial crescent]

Is the nullspace of C the union of the nullspaces of A and B? or if not what is the relationship? if anyone could help it would be much appreciated. thank you

[charred stirrup]

I googled ranks and the definition says the set of solutions to a linear equation

[dense holly]

Nonsingular means invertible or not invertible

[proper moat]

A singular matrix does not have an inverse

A nonsingular matrix is a square matrix that is not singular

[dense holly]

Can someone give me an example of a 2x2 matrix that is not invertible

Besides zero matrix

[random wasp]

Well, recall when the determinant is 0, the matrix is not invertible
So given a matrix A = [[a,b], [c,d]],
what solutions are there to det(A) = ad - bc = 0 ??

@dense holly
obviously
a = 0
b = 0
c = 0
d = 0
means ad - bc = 0
but can you think of any others?

||try making all entries 0, except one of them||

[wicked trellis]

For 4 I’m not really sure how to do this, does anyone know

[slow scroll]

yeah so notice that you are given how T changes v1, v2, and v3, and not the standard basis

the matrix [v1 v2 v3] tells us how vectors expressed in terms of the standard basis can be expressed as linear combinations of v1, v2 and v3. Once we know that, we can use T to transform vectors that are now expressed in terms of v1, v2, and v3.

any idea what you would do?

[wintry steppe]

,rotate -90

[wintry steppe]

How do we do Q16

I don't understand what is the question asking

[slow scroll]

numbers like 11 22 33 44 55 66 77 88 99 12 13 14 15 16 17 18 19 23 24 25 26 27 ...

im not sure what that has to do with linear algebra tho

[wintry steppe]

Lol
I dont know where should it go

Algebra?

[slow scroll]

what class is this for? lol

[wintry steppe]

University sample test

[slow scroll]

ehh not here but might as well help.
so if you choose 1 as the first digit, how many options do you have for the second digit?
if you choose 2 for the first digit, how many options do you have for the second digit?
and so on... and add up all of the options

[wintry steppe]

8 for the second digit

[slow scroll]

if you choose 1, you have 9 choices for the second digit

[wintry steppe]

Yea

So 9+8+8+8+...

[slow scroll]

noo, the sequence is 9 + 8 + 7 + 6 + .... + 1

[wintry steppe]

Oh, yea

I forgot about 1

Alright

Thank you!

[slow scroll]

np

[dense holly]

Can I get a good explanation why this is false? I had it on my midterm today and I didnt know the best way to state it

[slow scroll]

yea test if it satisfies $L(\alpha\mathbf a + \beta \mathbf b) = \alpha L(\mathbf a) + \beta L(\mathbf b)$

[stoic python]

bigboy77584:

[slow scroll]

or you could even say that a matrix that maps (x1, x2)^T to (x1x2, 2)^T doesn't exist

[dense holly]

Why not

Does ^T refer to transpose

[slow scroll]

yea just because i dont feel like typing column vectors

[dense holly]

Why doesn't the transpose exist

[slow scroll]

The matrix that maps $\begin{bmatrix} x_1 \ x_2 \end{bmatrix}$ to $\begin{bmatrix} x_1 x_2 \ 2 \end{bmatrix}$ doesn't exist is what i meant

[stoic python]

bigboy77584:

[dense holly]

Yeah and I'm asking why

[slow scroll]

suppose there was a 2x2 matrix 'A' that described this transformation. Then T(x1, x2)^T = A(x1, x2)T = x1 T(e1) + x2 T(e2)
x1 and x2 are not multiplied together => not possible.

Basically, matrices are constructed in a way where they have to satisfy the def of a linear map.

[wicked trellis]

@slow scroll standard basis of e1 to e3?

[slow scroll]

we know T(v1) , T(v2), T(v3). To get the matrix that describes this matrix for some input in terms on the standard basis in R^3, we need to know T(e1) T(e2) T(e3)

@wicked trellis

[wicked trellis]

So t (1,0,0) etc

[slow scroll]

err that explanation isn't quite right, but suppose you had a matrix 'A' s.t. T(A(e1)) = T(v1), T(A(e2)) = T(v2) ...

(hint hint you do)

(hint hint matrix multiplication involved)

does this make sense at all?

[wicked trellis]

Kind of ish

It would be like

Matrix A x v1 = t(v1)

[slow scroll]

i think i may have used some letters from the problem i shouldn't have, but basically the final transformation A needs to be able to take in a vector from the standard basis in R^3, so we need the composition of T with another matrix. The other matrix is the matrix that takes vectors from the standard basis for R^3, and maps them to linear combinations of v1, v2, v3 which our transformation T can understand.

what matrix would take in a vector in the standard basis of R^3 and map to linear combinations of v1, v2, and v3?

the matrix [v1 v2 v3], right?

[wicked trellis]

Ok

Wouldn’t it be the matrix [e1,e2,e3]

[slow scroll]

no thats the identity lol

[T(v1) T(v2) T(v3)][v1 v2 v3](\vec x) is the transformation you are looking for.

[wicked trellis]

But v1 v2 v3 isn’t in r4

[slow scroll]

yea?

[wicked trellis]

Oic

[slow scroll]

i think it will help if you see an example real quick

plug in the vector (1, 0, 0) into the final transformation. this vector gets mapped to v1 which then gets mapped to T(v1) just as it should. If we didn't do this composition, then (1,0,0) would get mapped straight to T(v1) which is wrong

namely, its wrong to just do [T(v1) T(v2) T(v3)](1, 0, 0)^T

[wicked trellis]

I think I get it

[slow scroll]

it takes a while for all of this craziness to make sense haha

[wicked trellis]

Transformations has been bugging me all day

[wintry steppe]

,rotate -90

[wintry steppe]

How do we do Q10 fastly without calculator?

[wise river]

the median of the AP is 36 / 2 = 18

so answer is 18 * 45 ig

[wintry steppe]

Oh

That is great!

Thank you!

[wise river]

np

[wintry steppe]

1min

Why did you times 18 with 45?

[wise river]

cause there are 45 terms..

the formula for ap sum is n * mean

[wintry steppe]

Isn't it

n/2(2a+(n-1)d)?

[wise river]

you can write it as n * (a1 + an) / 2

(a1 + an) / 2 is the mean or median.. both are same

[wintry steppe]

Oh, I see

Okay thank you!

@wintry steppe hi lol

Hey!😄

[wicked trellis]

Ok I’m still confused

For 4

Essentially I’m finding matrix A

But I’m not sure how to get there

[sand hedge]

hmm

$A\begin{pmatrix}1&-1&3\ :2&2&-1\ :3&-1&2\end{pmatrix}=\begin{pmatrix}1&1&0\ :2&-1&1\ :0&2&1\ :-1&0&-2\end{pmatrix} \ A=\begin{pmatrix}\frac{2}{7}&\frac{4}{7}&-\frac{1}{7}\ -\frac{5}{14}&-\frac{3}{14}&\frac{13}{14}\ \frac{11}{7}&\frac{8}{7}&-\frac{9}{7}\ -\frac{13}{14}&-\frac{5}{14}&\frac{3}{14}\end{pmatrix} \ \begin{pmatrix}\frac{2}{7}&\frac{4}{7}&-\frac{1}{7}\ ::-\frac{5}{14}&-\frac{3}{14}&\frac{13}{14}\ ::\frac{11}{7}&\frac{8}{7}&-\frac{9}{7}\ ::-\frac{13}{14}&-\frac{5}{14}&\frac{3}{14}\end{pmatrix}\begin{pmatrix}1\ :2\ :3\end{pmatrix} = \begin{pmatrix}1\ 2\ 0\ -1\end{pmatrix} $

[stoic python]

Colen:

[sand hedge]

factor out 1/14 and it'll look nicer

@wicked trellis

[wicked trellis]

Whoa

Ok let me take a second to digest this

Are u able to walk me through a little bit

[sand hedge]

$\frac{1}{14}\begin{pmatrix}4&8&-2\ -5&-3&13\ 22&16&-18\ -13&-5&3\end{pmatrix}\begin{pmatrix}1\ 2\ 3\end{pmatrix}=\begin{pmatrix}1\ 2\ 0\ -1\end{pmatrix}$

[stoic python]

Colen:

[sand hedge]

there thats nicer

sure what don't you get?

[wicked trellis]

Where did the fractions in the matrix come from

[sand hedge]

Those fractions just came about because thats the solution to the original system?

For A

$\begin{pmatrix}1&1&0\ :2&-1&1\ :0&2&1\ :-1&0&-2\end{pmatrix}\begin{pmatrix}1&-1&3\ ::2&2&-1\ ::3&-1&2\end{pmatrix}^{-1}=A$

[stoic python]

Colen:

[wicked trellis]

Oh I guess I never learned the inverse thing

[sand hedge]

3x3 inverse is not nice, ngl I just asked the computer

[wicked trellis]

😂

Hoooly shit it’s all coming together now

I see now

[sand hedge]

open ur mind 2 the matrices

[wicked trellis]

I see the hard tedious way of doing it

Basically cause e1 to e3 has to be written as a linear combination of v1 to v3 and then apply the transformation to it to get each vector

Rite

Kind of hard to explain

Anyways thanks

[wicked trellis]

Does 2 and 3 look right?

Cause what I got was that there is no vector x in r2 that comes out to (1,2,3)

[frosty pewter]

anyone know how to do part (b) or (c)?

for (b), originally I was thinking it's just a basis for all the matrices with 0's along their diagonal, but now I'm realizing that I haven't thought about negative numbers along the diagonal

[slender yarrow]

well $tr(A) = \sum_{k=1}^n a_{ii}$ and you're studying the vectors in the null space, ie those for which $$\sum_{k=1}^n a_{ii} = 0$$

[stoic python]

emeric75:

[slender yarrow]

you can take out one of the terms in the sum (for commodity, i'll choose the last one), to get $$a_{nn} = -a_{11}-\cdots-a_{n-1,n-1}$$

[stoic python]

emeric75:

[slender yarrow]

so there's only one coefficient in the matrix which depends on others (a_nn in this case)

@frosty pewter

✝ dave

[frosty pewter]

yeah but making a basis for that seems really tricky

[slender yarrow]

it's actually not that hard tbh

[frosty pewter]

because there are infinitely many ways that u can choose your a_ii elements so that they add to 0, right?

could u show me how to do the basis part?

[slender yarrow]

true, but i'm just forcing one term of the diag to be the opposite of the sum of all the other terms

(you're sure trace = 0 like that, and that encompasses your 'infinitely many ways' : i didn't say what the other terms of the diagonal were)

so yeah the equation above gives us a relation between one term of the diag and all the other ones in the diag

(ie tells us nothing about all the other terms not in the diagonal of the matrix)

(seems pretty evident since the trace of a matrix only depends on its diagonal elements)

so first you have one element in the basis for each coeff not in the diag (imma write it on paper, latex's gonna be hell)

[frosty pewter]

yeah i think i get that part, so that's a total of (n^2 - n) elements in your basis - but I'm still a little blurry on how you get the other part of the basis for the diagonals

[slender yarrow]

well yeah ignoring the things not in the diagonal

[neon dune]

For anyone looking to learn about linear algebra there's this cool online book I came across on Hackernews http://immersivemath.com/ila/index.html

[half ice]

I remember seeing that a while ago! They've brought it further

Looks cool

[slender yarrow]

$$A = \begin{bmatrix}a_{11}& \cdots & \cdots & \cdots \ \cdots & a_{22} & \cdots & \cdots \ \cdots & \cdots & \ddots & \cdots \ \cdots & \cdots & \cdots & -a_{11}- ... - a_{n-1,n-1}\end{bmatrix}$$

(oof matrices in tex)

[prime rock]

@neon dune thanks for the link. that's awesome

[slender yarrow]

dem looks kewl indeed

[prime rock]

that texit bot is neat too

[slender yarrow]

you could write this as $$A = a_{11}\begin{bmatrix}1& \cdots & \cdots & \cdots \ \cdots & 0 & \cdots & \cdots \ \cdots & \cdots & \ddots & \cdots \ \cdots & \cdots & \cdots & -1\end{bmatrix} + a_{22}\begin{bmatrix}0& \cdots & \cdots & \cdots \ \cdots & 1 & \cdots & \cdots \ \cdots & \cdots & \ddots & \cdots \ \cdots & \cdots & \cdots & -1\end{bmatrix} + ...
$$

if only discord markdown was less screwed

so basically a sum of matrices of this form up to the a_(n-1,n-1) term

ie n-1 other matrices in the basis

[stoic python]

emeric75:

emeric75:

[frosty pewter]

yeah, but how do you generalize that? Because if the values along by diagonal are like
(1, 0, ..., 0, -1)
then I could also find
(2, -1, 0, ..., 0, -1) and keep going with
(3, -2, 0, ..., 0, -1)
(4, -3, 0, ..., 0, -1) and so on

Do I just say that my basis is infinite where a_nn = -a_11 - a_22 - ... - a_(n-1)(n-1)?

oh wait, actually, i might see where you're coming from

ok that makes sense now, TYSM 😁

[slender yarrow]

c is pretty ez after this

😊

[sour garden]

If A is similar to B and C is similar to D then is AB similar to CD?

[slow scroll]

Hmmmmm. The converse of this is pretty trivial, but I'm not really even sure how to start on this one.

[placid oracle]

Its not one to one, right? since it has a free variable

and it is onto?

[slow scroll]

yea its onto if every row has a pivot and one to one if every column has a pivot

[placid oracle]

ok so its only on to

makes sense

[placid oracle]

This is true, right? wb for one to one?

[slow scroll]

it can't be onto

it can be one to one tho

[placid oracle]

why

[slow scroll]

for problems like that, i think: okay there are 1000 basis vectors in R^1000 and there are 1001 basis vectors in R^1001. If i create a linear transformation T that maps all e_k, a basis vector in R^1000 to v_k, a basis vector in R^10001, i can show that:

T(e_1) --> v_1
T(e_2) --> v_2
...
T(e_1000) --> v_1000

uh oh, i ran out of vectors in R^1000 to map to R^1001. v_1001 did not get mapped to. therefore T:R^1000 --> R^1001 is not surjective

[wintry steppe]

hm

[slow scroll]

similarly, because I can create a linear transformation that maps only one basis vector in R^1000 to one basis vector in R^1001, such that an element of R^1000 does not get mapped to >once, the map is injective

[placid oracle]

injective?

[slow scroll]

one to one = injective
onto = surjective

[placid oracle]

ah

Ok I think I know what you mean

[slow scroll]

(bumping my question again)
https://cdn.discordapp.com/attachments/540211747613704221/550887627956682754/unknown.png

[placid oracle]

wish i could help^

[slow scroll]

i don't think im much farther along than u are, but this question is pretty hard 😦

[placid oracle]

u seem to be understanding it much better than I tho 😛

I think this is false.. but i dont see why?

[inland anvil]

What is the definition of a transformation, I don't know it

Then there is a 75% chance I'll be able to answer the question

[placid oracle]

was looking for a formal definition:

[slow scroll]

im pretty sure the answer is yes
does T(a0 + b0) = aT(0) + bT(0)?

[inland anvil]

I may be wrong, but I think it is false

Correct me if I'm wrong bigboy

but I think that you could define that it map the zero vector onto the zero vector

and then screw around with the rest of it to make it non-linear

[slow scroll]

hmm not sure what you mean. so our goal is to prove that
T(a0 + b0) = aT(0) + bT(0)
lets start with the LHS
T(a0 + b0) = T(0(a+b)) = T(0) = 0
now RHS
aT(0) + bT(0) = a0 + b0 = 0 + 0 = 0
therefore T is a linear transformation im pretty sure??

[inland anvil]

hmm

what is the definition of a linear tranformation

I thought it was more than T(a0 + b0) = aT(0) + bT(0)

also isn't a0 = 0

[slow scroll]

yes

[placid oracle]

bigboy i think ur correct

[slow scroll]

@placid oracle actually i may be wrong

[placid oracle]

hmm why?

[slow scroll]

the domain is made up of vectors in R^n, not just the zero vector. didn't take that into account

so by that logic, its definitely not a linear transformation as we don't know what T does to some arbitrary vector in R^n

typically, a linear transformation is defined by how it transforms the basis vectors of the domain, and then we can represent arbitrary vectors (not basis vectors) as linear combinations of basis vectors:
say we know what T does to the basis vectors (denoted with e_k's)
v = x1e1 + x2e2 + ... + xnen
T(v) = T(x1e1 + x2e2 + ... + xnen) = x1 T(e1) + x2 T(e2) + ... + xn T(en)

But here all we know is that T(0) = 0
so for some arbitrary v, we can say
T(v) = ???

@placid oracle

[placid oracle]

ok i get it, so its just because we dont have enough information about T

[slow scroll]

well we have enough information to know it isn't a linear transformation

[placid oracle]

ok well yeah

[bitter shoal]

Why is it that we can single out the diagonals of a matrix to show it is positive semi-definite. Speaking of which, what's the general process for proving positive semi-definiteness? I know the x and x-transpose parts end up x^2 basically-which is always non-negative.

$x\in R$ ofc.

[stoic python]

bluefrost:

[slim maple]

@keen garden so you want to show that the union of two vector subspaces isn't a subspace, right?

[keen garden]

yes please like prove it mathematically ( with explaination cuz i dont know proofs )

please

[slim maple]

um

so do you have the definition of vector space with you

?

[keen garden]

not the formal definition but like

a bunch of vectors where multiplication by a scalar and addition are defined

[slim maple]

okay, so

what about definition of vector subspace?

[keen garden]

idk

a smaller space

where the same operations are defined?

[slim maple]

haha i guess we'll keep it a bit informal

[keen garden]

ye

[slim maple]

so let's have vector space... A

er

and we'll have two vector subspaces of A called V and W

i'm assuming that neither V nor W is a subset of the other

[keen garden]

ok

[slim maple]

if one is a subset of the other, the union is just going to be the bigger set, which is a vector subspace

is that fair?

[keen garden]

yes

[slim maple]

k, so

if neither is a subset of the other, that means there exists an element v in V that's NOT in W

and there exists an element w in W that's NOT in V

[keen garden]

sure

[slim maple]

okay

so let's assume that the union of V and W, called V U W, is a vector subspace of A

if we can reach a contradiction, that means that this assumption can't be possible

is that fair?

[keen garden]

yes

[slim maple]

k, so we have a vector subspace V U W, which is the union of two vector subspaces V and W

and we have an element v in V that's not in W

and an element w in W that's not in V

but v and w are both in V U W, right?

[keen garden]

yeds

yes*

[slim maple]

if we assume that V U W is a vector subspace, then it's going to be closed under addition. Because v and w are both in there, then v + w is in V U W as well

This means that we have two possible cases:

Case 1: v + w is in V.

Case 2: v + w is in W.

is that fair?

[keen garden]

yes

[slim maple]

k, so let's go with case 1: v + w is in V

V is a vector subspace of A, so each element in V is going to have an additive inverse that's ALSO in V

i.e., if v is in V, then -v is in V

[keen garden]

yes

[slim maple]

k so

We have v as an element in V. So -v is in V. Also, we have v + w as another element in V.

So then -v + v + w = w is in V (because V is a vector space, it's closed under addition)

But this is a contradiction because w is an element in W but NOT in V

[keen garden]

that contradicts that they are not a subset of each other

so the union can be a subspace if and onnly if they are subsets?

cuz if they are a subset then it wouldnt be a contradiction cuz w can be an element in v

right?

[slim maple]

well, we still have case 2

but it's similar logic as to why it isn't possible

we have w in W, so -w is in W, so v + w - w = v is in W

so that's a contradiction

we only proved that

if neither is a subset of the other, then the union cannot be a vector subspace

but your statement is basically right

[keen garden]

yes got it

ty rly ty

was cool

is this is how u prove something?

[slim maple]

there are different proof techniques

but also

if you have two vector subspaces V and W of a vector space A

and we have, let's say

V is a subset of W

then the union is just going to be W

right?

which is already a vector space by our assumption

[keen garden]

ye just the bigger set

[slim maple]

yupyup

[keen garden]

https://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2010/assignments/MIT18_06S10_pset1_s10_soln.pdf am i supposed to know how to solve these?

[sage mantle]

1. Prove that u1 is an eigenvector for S. What is its corresponding eigenvalue?
   i have both u1 and S's value (S being a 3*3 matrix) , how can i solve this? i started by doing Su1=lambdaU1 but it seems way too long

[wintry steppe]

multiplying 3x3 matrix by a vector is taking too long?

you probably need more practice then

[sage mantle]

i didn't fail linear algebra 1 for no reason

but i mean it's S = (1/3)(a 3*3 matrix)

so i multiplied it and it started becoming messy

and then did the = that i wrote above

so i wanted to know whether it's the right method

[wintry steppe]

yes, it's the right method

[sage mantle]

okay , thank you

[restive palm]

Is there a trick for memorising what form your Jordan 3x3 matrix will take if it has 1,2,3 eigenvalues?

[wintry steppe]

There are not many Jordan forms for 3x3

[broken hawk]

(what I just said was wrong)

[restive palm]

Yeah I just get them confused

[broken hawk]

I mean you just have to find the eigenspaces and then check their dimensions

[restive palm]

I can't find a Jordan form without looking it up

[broken hawk]

the only difficult bit about the jordan form are the change of basis matrices

which require finding cycles of generalized eigenvectors, which is hard

[restive palm]

We haven't covered that :L

[wintry steppe]

if you have 3 different eigenvalues, it must be 3 blocks, so only $\begin{bmatrix} \lambda_1 & & \ & \lambda_2 & \ & & \lambda_3 \end{bmatrix}$

[broken hawk]

but the jordan matrix itself you just have to find all eigenvectors and then count them

[stoic python]

Xaositect:

[restive palm]

Yep

Is there no better way than to just learn them?

I just get 1 &2 mixed up

[broken hawk]

it’s a general statement, there’s no need to learn cases

[wintry steppe]

if you have two different eigenvalues, then there are two variants: $\begin{bmatrix} \lambda_1 & & \ & \lambda_1 & \ & & \lambda_2 \end{bmatrix}$ and $\begin{bmatrix} \lambda_1 & 1 & \ & \lambda_1 & \ & & \lambda_2 \end{bmatrix}$

[stoic python]

Xaositect:

[restive palm]

Yup

[broken hawk]

for each eigenvalue, count the number of (linearly indep) eigenvectors you can find

each gets an individual block

[wintry steppe]

in the first case you will have two lin.indep. eigenvectors for lambda1

[broken hawk]

the sizes of the blocks can’t be found that easily, but in 3x3 there’s always only one option

[wintry steppe]

in the second case you will have one eigenvector for both, so here you need something else to choose

[broken hawk]

so you don’t have to worry about that issue

[wintry steppe]

the something else is actually $\operatorname{rk}(A-\lambda E)^2$

[stoic python]

Xaositect:

[wintry steppe]

And in the case of one eigenvalue, you have $\begin{bmatrix} \lambda_1 & & \ & \lambda_1 & \ & & \lambda_1 \end{bmatrix}$, $\begin{bmatrix} \lambda_1 & 1 & \ & \lambda_1 & \ & & \lambda_1 \end{bmatrix}$ and $\begin{bmatrix} \lambda_1 & 1 & \ & \lambda_1 & 1\ & & \lambda_1 \end{bmatrix}$

[stoic python]

Xaositect:

[wintry steppe]

which one it is can be told by the number of lin.indep. eigenvectors (rk(A - lambda E))

[wintry steppe]

What is the hard thing with linear algebra?

Since for me it is just about transformations preserving linearity

(Prob. which is group homomorphism)

[wintry steppe]

Can someone ELI5 what is Explained variance score in https://scikit-learn.org/stable/modules/generated/sklearn.metrics.explained_variance_score.html?

[slow scroll]

https://cdn.discordapp.com/attachments/540211747613704221/550887627956682754/unknown.png

Bumping my question again. still no clue what to do

ehhh maybe ill just ask MSE

[half ice]

@slow scroll
Still looking for it?

[fierce rampart]

Just do the original proof backwards

[slow scroll]

Yeah I am. I’m aware this is pretty well known, but I’m looking for a fairly low level proof of this. Any hints?

[dusky epoch]

try going a bit more abstract and consider a linear transformation instead of a matrix

or perhaps view the matrix as a linear transformation. it's a bit of a moot point anyway

say, if $A$ is an $m \times n$ matrix, you can view it instead as a transformation $A \colon K^n \to K^m$

[stoic python]

Ann:

[dusky epoch]

(where K is the field this is all happening over)

@slow scroll can you take it from here?

[slow scroll]

an mxn matrix would be represented by the transformation A: K^n -> K^m. Since it has unique solutions, its matrix has pivots in every column, therefore n<=m and A is an injection. I suppose at this point, you would define a transformation B that maps solutions to A back to itself, but I also know that uniqueness does not imply existence, so I don't know how you would handle inputs of A that don't have a solution.

[dusky epoch]

mmmm

well remember that you don't need a two-sided inverse

all you need is a map $B \colon K^m \to K^n$ such that $BA = \mathrm{id}_{K^n}$

[stoic python]

Ann:

[dusky epoch]

ok wait, let me just make sure i don't get myself confused over this

[slow scroll]

I was thinking you had $A: K^n \to K^m$ and $B: K^m \to K^n$ but $B$ doesn't actually map every element of $K^n$, just the elements of $K^n$ in the range of $A$. I guess that doesn't actually matter though since for all $\mathbf x$ such that $A$ is consistent, $B(A(\mathbf x)) = \mathbf x$?

[stoic python]

bigboy77584:

[slow scroll]

maybe it would help if i tried to think about how I can't use injectivity to show that A has a right inverse

[brittle juniper]

Maybe you could be interested in the restriction of A in the preimage of its range

Because that application is bijective

[slow scroll]

idk what "the restriction of A in the preimage of its range" means. My description of B does sound bijective, but in order to be an inverse, B has to at least be onto (so solutions of Ax get mapped back to x), but without any kind of restriction, not all of the elements in the domain of B necessarily get mapped. I'm not even sure what that means. I guess that's what inconsistency is?

if you have a function $A: B\setminus x \to C$ then $A(x)$ would be inconsistent?

[stoic python]

bigboy77584:

[brittle juniper]

What I meant is that $A:K^n\to K^m$ may not be bijective, \ but $A:K^n\to\operatorname{Im}A$ is

[stoic python]

Tuong:

[wintry steppe]

you know what the inverse has to be from set theory

[slow scroll]

image is just the range, right?

[brittle juniper]

So you can find a reciprocal bijection for that application, and from it, build your B

[wintry steppe]

no

image of $f: A\to B$ is ${f(x)|x\in A}$

[stoic python]

neolithic:

[wintry steppe]

range is B right

[slow scroll]

i thought that was the codomain

[wintry steppe]

I take codomain = range = B

[brittle juniper]

Im $A={y\in K^n\ |\ \exists x\in K^m\ /\ A(x)=y} $

[stoic python]

Tuong:

[brittle juniper]

It has got a structure of vector space

[slow scroll]

i think i will just think more about this and come back to it later
anyway, just so whoever is curious can see, this is more what I was looking for to begin with:

Always open to other methods of proof tho

[brittle juniper]

writing everything in term of linear applications instead of matrices,

Working with $A:\mathbb K^n\to\mathbb K^m$.\ \
The equation $A(\mathbf x)=\mathbf 0$ has a unique solution and $A(\bf 0)=\mathbf 0$,\
so $\Ker A={\mathbf 0}$ and $A$ is injective.\
~\quad(it's a known characterisation of injectivity)\ \
This means $A$, as an application of $\mathbb K^n\to\im A$, is bijective.\
Let's call $B$ the inverse bijection.\
$B$ is an application of $\im A\to\mathbb K^n$ and $\forall\mathbf x\in\mathbb K^n,\ B(A(\mathbf x)) =\mathbf x.$
\ \
Then you can define the application $C:\mathbb K^m\to\mathbb K^n$ to be such that\
$\forall\mathbf y\in\mathbb K^m,\ C(\mathbf y)=\begin{cases}
B(\mathbf y)&\text{if }\mathbf y\in\im A\
\mathbf 0&\text{otherwise}\end{cases}$\ \
$\forall\mathbf x\in\mathbb K^n,\ A(x)\in\im A$,\
so $\forall\mathbf x\in\mathbb K^n,\ C(A(\mathbf x))=B(A(\mathbf x))=\mathbf x$.

[dusky epoch]

btw

@brittle juniper

[brittle juniper]

O crap

[stoic python]

Tuong:

[brittle juniper]

fixed

thanks

[dusky epoch]

you gotta be a bit careful tho. your C may fail to be linear

it's better to decompose K^m as the direct sum of Im A and some other subspace, and have C vanish on that instead

[brittle juniper]

O yeah mhmm

tried to do a projection without using the word projection so it's indeed kinda meh

B circ projection is what I'm actually looking for I guess

v2 incoming soon then!

[slow scroll]

ill take a close look at this stuff when i get home from work in a few hours lol

[brittle juniper]

Working with $A:\mathbb K^n\to\mathbb K^m$.\ \
The equation $A(\mathbf x)=\mathbf 0$ has a unique solution and $A(\bf 0)=\mathbf 0$,\
so $\Ker A={\mathbf 0}$ and $A$ is injective.\
~\quad(it's a known characterisation of injectivity)\ \
This means $A$, as an application of $\mathbb K^n\to\im A$, is bijective.\
Let's call $B$ the inverse bijection. $B$ is linear.\
$\forall\mathbf x\in\mathbb K^n,\ B(A(\mathbf x)) =\mathbf x.$
\ \
Let $E$ be a supplementary space to $\im A$ in $\mathbb K^m$ such that $E+\im A=\mathbb K^m.\
(there exists some for sure because of finite dimension)\
Let $p:\mathbb K^m\to\mathbb K^n$ be the projection such that $\Ker p=E$ and $\im p=\im A$.\ \
$\forall\mathbf x\in\mathbb K^n, A(\mathbf x)\in\im A$ so $p(A(\mathbf x))=A(\mathbf x)$.\
So $\forall\mathbf x\in\mathbb K^n,\ B(p(A(\mathbf x)))=B(A(\mathbf x))=\mathbf x$.\
$p$ is linear.\ \
$B\circ p$ is a linear application of $\mathbb K^m\to\mathbb K^n$ and works as a right inverse for $A$.

[stoic python]

Tuong:

[brittle juniper]

that would be v2

imo working with the linear applications themselves instead of matrices is more confortable

actually I've developed some kind of phobia towards matrices in the past few months

[wintry steppe]

Matrix is just a model of linear transformation which fits in cartesian space thus allowing easy calculation.
Change my mind.

[proper crescent]

Yeah I prefer proving things for linear transformations without resorting to coordinate computations if possible

[wintry steppe]

Matrices are for computers.

[brittle juniper]

People who share my opinion \o/

[proper crescent]

I often say matrix or linear map loosely, it's more like, idk I more often use characteristic/minimal polynomials than Jordan form

That's just how I was "raised" mathematically

[half ice]

I figure most people who do linear algebra are taught to think of things in terms of matrix algebra, and not as linear operations on vector spaces.

Yes, mathematicians hold the superior view, but not the common one

[proper crescent]

I mean really I think it's less how you're taught and more what you're doing.

Obviously you're taught to approach things in certain ways but I think it's less that all tasks are nicely amenable to both so much as, certain types of things beg for matrices and others beg for linearity

[restive palm]

If I may interject what resources would you recommend for learning Jordan normal form?

My lecturer just loves making us calculate stuff

But I don't feel like I know what's going on

[proper crescent]

Uh, I learned it in algebra actually because of the proof using module theory over PIDs. I think Hoffman-Kunze has a chapter on it?

[restive palm]

Ty Dami 👍

[sour garden]

Hey!

I learned it in Algebra 2 yesterday!

Although we skipped the part of the proof where you prove uniquness of cyclic submodules

[real wedge]

Our prof did not teach us this... pls help

[slow scroll]

@real wedge a lower triangular matrix is a matrix with all zeros except at and below the diagonal. so for a 2x2 matrix how many different entries would that be?

[lime cypress]

Good night guys. Need help with something. let me copy paste the message i already sent.

im given a complete rank column matrix (X), a n sized vector (Y) and h(z) = (Y-XZ)^t(Y-XZ). I need to find the z vector that makes the derivative of h(z) a null vector. Any hint? i tried doing the (Y-XZ)^t(Y-XZ) part with generic values but didnt found anything
oh h is R^p -> R and X is nxp
thus z is a p sized vector

Can the quadratic form be unhomogeneous? Working with h(z) i get Y^t*Y - Y*(XZ)^t +(XZ)^t*Y+(XZ)^t*(XZ) from the Y^t*Y - i can tell its not homogeneous but i doubt this how this exercise is supposed to be

[proper crescent]

@sour garden haven't actually seen that particular version, it was the one in D&F but in class the proof we did was basically that matrices over PIDs could be put into Smith Normal Form

[real wedge]

@slow scroll 3?

[slow scroll]

yes

[real wedge]

Btw our prof did not teach us vector space or any type of space that has matrices or whatever... we were just taught “subspace”

And like column space and that stuff

Nothing about polynomials or matrix spaces

[slow scroll]

do you know the vector space axioms? you can check that certain sets of matrices and polynomials (and more stuff) can satisfy the conditions to be a vector space

[real wedge]

No

My prof did not teach us it

I only know what subspace means

[slow scroll]

hmm i think for the purposes of this exercise, you are safe taking vector space to mean subspace.

The dimension of a space depends on the number of vectors there are in its basis. If you have the set of 2x2 lower triangular matrices (so vectors with 3 independent entries), how many vectors would be in its basis?

and i gtg eat brb

[real wedge]

I dont really understand the connection you make here, “the set of 2x2 lower triangular matrices (so vectors with 3 independent entries)”

[slow scroll]

@real wedge Its like a vector in R^3, where we would have (1,0,0) (0,1,0) and (0,0,1) in its basis. A basis for lower triangular matrices would look something like

,$ \left{\begin{pmatrix}1 & 0 \ 0 & 0 \end{pmatrix}, \begin{pmatrix}0 & 0 \ 1 & 0 \end{pmatrix}, \begin{pmatrix}0 & 0 \ 0 & 1 \end{pmatrix}\right}

[stoic python]

bigboy77584:

[slow scroll]

you can check that this set is linearly independent and does indeed span the space of lower triangular matrices

[real wedge]

Ok so they are not actually vectors

[wintry steppe]

@lime cypress hmm, one thing: what is the exponentation here?

Also could you put formulae into latex

[real wedge]

Thats what i was confused about

[lime cypress]

exponentation?

sure give me a min, its been a while since i used latex

if you mean ^t its the matrix transpose

[real wedge]

So a vector space can be any collection of whatever things that satisfy the axioms?

[slow scroll]

Yes exactly!

[real wedge]

Doesnt have to necessarily be typical vectors?

Ok cool

[slow scroll]

ye, you'll learn later on that there is a notion of "sameness" between weird spaces like polynomials and matrices to the vectors you are used to

[real wedge]

Cool

[wintry steppe]

Oh, transpose.

I see, it was hard for me to know that it is transpose without latex

[real wedge]

Its weird tho cuz our textbook doesnt even cover vector spaces yet its asked about on webwork

[lime cypress]

do you still need the latex? im confused how to do it

[wintry steppe]

@lime cypress so calculating the derivative of ||y-Xz||^2. Right?

That's fine

[lime cypress]

why ||y-Xz||^2?

[slow scroll]

yea that is weird. I know khan academy talks about subspaces before vector spaces. I guess it just simplifies the number of things you have to think about. when you are talking about subspaces, you can assume things like scalar multiplication and vector addition exist already.

[wintry steppe]

I thought that is obvious?

(x^T)x = ||x||^2

[lime cypress]

is || || the norm?

[wintry steppe]

Yeah, the norm

[lime cypress]

i haven't used a matrix norm yet, just the vector norm. is this important for it?

[real wedge]

Yeah, thx for ur help

[slow scroll]

np

[wintry steppe]

No.

Vector norm is enough

[lime cypress]

yeah i see it now

[wintry steppe]

Iirc Matrix norm is just an extension of it

[lime cypress]

when you multiply it, you do row*column but as its the transpose, its like the norm of each row

[wintry steppe]

Hm. Anyway

You've heard a lot about y-Xz, right?

[lime cypress]

that looks like when you are looking for eigenvalues but dont see how its related to this.

[wintry steppe]

Eigenvalues? Hmm

[lime cypress]

its the only thing i have heard it from. do you mean something else?

[wintry steppe]

Yep

[lime cypress]

what is it?

[wintry steppe]

I don't think eigenvalues aren't that related

[lime cypress]

i know, its just that its the only think i find similar with it

[wintry steppe]

Yep

So, one thing I'd like to say: how much do you know about null space and column space?

[lime cypress]

null space is the space that makes it null, but not sure about the column space, let me look

oh the column space is the range

[wintry steppe]

Ye

Then, what about linear regression?

[lime cypress]

not much. thats what you meant with y-Xz?

[wintry steppe]

Yep

[lime cypress]

@wintry steppe so y-Xz is the error term?

[wintry steppe]

Yep

Hmm, I think you should learn about these first before differential property

[lime cypress]

wow i couldnt have found that by myself. but i still feel a bit lost

[wintry steppe]

Because it will give intuition and hints to how to approach things

[lime cypress]

@wintry steppe anything else you can tell me about it? or reading about those is enough?

i have been at this for hours and wanna finish it 😦

[wintry steppe]

Yeah, first read those

But before that, I'm curious how you got there to differentiation before learning about linear regression and stuffs

[lime cypress]

well thats how the course is planned (linear algebra 2). I have missed many classes too but they didnt tell us about linear regression. Actually yes in other course but it was very mediocre, i have been thinking of studying it again by myself.

[wintry steppe]

Hmm

It should have been introduced in linear algebra 1

Sth like Least square methods.

[lime cypress]

we only saw matrices, determinants, SLE, vector spaces and euclidean spaces

[wintry steppe]

SLE?

[lime cypress]

well and a little about linear transformations, but i dont think we should count it

systems of linear equations

[wintry steppe]

I see.

Anyway, you know about null space

What about, column/rank space?

[lime cypress]

not much tbh, i only know that the rank is how many li columns the matrix has. In this problem i know that rank(x) = p and so all its columns are li, i guess it helps us in the regression part as it represent the independent variables

[wintry steppe]

Yep

But.. that is not enough to analyze this case

Then what about this?

How much do you know about
z^T X z

[lime cypress]

as whole or each?

well i guess it each as you cant do that. about z and z^t i dont know much just that its 1xp and about X just the rank part. Let me read more about this

@wintry steppe we can also tell that the null space of X is empty

[wintry steppe]

Uh, wait

Ah right, it should be empty for this to make sense

And I mispelled

I meant y^T X z

My memory is hazy on this, but I think it is related

[lime cypress]

from y i dont know much either :L

[wintry steppe]

Welp

So what kind of thing do you know of

Sorry if I was rude

I couldn't focus on this, cause I'm busy

I do think you need to learn how to deal with linear transformations.

One more thing, I recalled that this is definitely related with properties of X^T X

[lime cypress]

its ok man. i will check that

[lime cypress]

@wintry steppe kept reading about it and finally found it. This is the normal equation and basically what we are doing its find the minimal point possible to get the lowest residual. Definitively gonna keep reading about it, even i was able to solve it

Im sad thought i couldn't find out how to do just with algebra

[twilit yoke]

Hello.

I'm trying to do a review for my test on monday

College Algebra

Anyone here?

[faint lintel]

Linear algebra is not college algebra

ask in one of the question channels or in #prealg-and-algebra

[twilit yoke]

ok

Thank you

[wintry steppe]

Yup, that's right SunriseM

You could do this in pure algebraic way if you analyze the differentials

[lime cypress]

yes i know that, i guess i need to get better at algebra

[winter reef]

yo can someone tell me what is Characteristic polynomial for?

it helps find eigenvectors and eigenvalues

but like dont quite know whats next

[dusky epoch]

wdym "what's next"?

[winter reef]

like what are application of them

matrixes were pretty useful and I could tell when to use them

but like how are these useful maybe in other areas of mathematics

[sage mauve]

very

very very very

oh you meant about characteristic polynomials

[winter reef]

ye

[wintry steppe]

Isn't it a tool to understand the linear transformation

[sage mauve]

its useful in graph theory apparently

though im not at all versed in graph theory

[winter reef]

abastro ive had a lot of linear transformation before I started em

[sage mauve]

well like you said it helps with eigenvalues and eigenvectors

those two have a lot of applications

[winter reef]

like where

I havent had them introduced that m,uch yet

[wintry steppe]

Everywhere

There's not so much way to analyze a linear transformation

Eigen is an important one of them

[winter reef]

oh ok

[sage mauve]

theyre used to solve to obtain solutions to schrodingers equation

people use it to study elasticity

[half ice]

There's a bijection between a transformation, and the eigenvalue/eigenvectors accompanying it. That is, if you have the eigens, you can get the transformation back

[broken hawk]

the characteristic poly tells you some stuff about the linear transformation (e.g. you can read off the trace and determinant right away, its roots are eigenvalues) and thanks to cayley-hamilton you can use it to compute higher powers or inverses easily

[winter reef]

linear algebra is like cool af and aboslutely not cool at the same time, you know what I mean?

[wintry steppe]

I don't get what cool means, sorry

[winter reef]

cold

[half ice]

Until now, the best way to state a transformation was to say "this is what you can do to the basis vectors lol" and that isn't that great. It's not unique, and doesn't give much depth. Eigens are unique, and do give some depth