linear-algebra

[graceful osprey]

So just row reduction in that case?

[dusky epoch]

p much...

[rocky hill]

So I guess where I'm getting a little confused is which norm in some vector space induces a norm on T?

And to clarify, we're saying that some norm in X, or in the domain of T is the one that induces the norm on T? Or is it the norm of the codomain? Or is it both?

lol... 😭

[dusky epoch]

it's both

it's always wrt two norms

but if your operator acts from a space to itself

both norms are usually just whatever norm you're considering on your space

ie one and the same

[graceful osprey]

Alright, thanks Ann!

[dusky epoch]

@rocky hill it'd be more correct that the norms on X and Y induce a norm on L(X,Y)

[rocky hill]

okay lemme try an example:

Say we have T: X → Y where we're... concerned with say the 2-norm in X and Y... then this tells us... that the 2-norm of T is determined by...? okay uh I think I'm spinning my wheels here...

okay..

[dusky epoch]

L(X,Y) is the space of linear operators from X to Y, a vector space in its own right

the norm on L(X,Y) induced by the 2-norms is the operator norm as described earlier

defined by $$|T|{2,2} := \sup{\substack{x \in X \ x \neq 0}} \frac{|Tx|_2}{|x|_2}$$

[stoic python]

Ann:

[dusky epoch]

this... i think is the least notation-heavy that i can go

@rocky hill this make sense?

[rocky hill]

could you possibly give a concrete example? I feel like it's all there I'm just struggling to connect the pieces in my head in a cohesive way.

[dusky epoch]

hurgh

ok so let's say X = Y = R^2 both with the usual euclidean norm

and T is... let's say given by the matrix [ 1 1; 0 1 ]

my claim is that $|T|_{2,2} = \sqrt{2}$

[stoic python]

Ann:

[dusky epoch]

though i admit i am not 100% certain on that

[rocky hill]

ooof

[rocky hill]

if I'm taking the 2-norm of a matrix, and some of my eigenvalues are complex and want to find the maximum eigenvalue I can just use the modulus of the complex eigenvalues, yes?

(for spectral radius)

for reference this is the formula I have for $||A||_2$

[stoic python]

kickpuncher:

[rocky hill]

$$||A||_2 = \sqrt{\rho (A^T A)}$$

[stoic python]

kickpuncher:

[rocky hill]

where rho I'm under the impression is the maximum eigenvalue of A

what I don't understand is that A^T A is a matrix, not a scalar

so is rho actually the maximum eigenvalue of A^T A instead?

this text my teacher is using is absolute garbage

[dusky epoch]

if $A$ is $m \times n$ then $A^TA$ is $n \times n$

[stoic python]

Ann:

[dusky epoch]

$\rho(M)$ is spectral radius aka $\max{|\lambda| : \lambda \in \mathrm{spec}(M) }$

[stoic python]

Ann:

[dusky epoch]

and yes that applies to complex eigenvalues too

@rocky hill

[lone quail]

How do i find the equation of the plane which passes through a line r {x=3, y=t,z=t} and is parallel to: {x=2, y+2x=-1}?
https://cdn.discordapp.com/attachments/359052604149465088/632295929998737429/unknown.png
line r is formed when h=-1

(sorry guys could anyone help please? i think its the 4th time im posting this question...)

[steady fiber]

I do not understand what it means for a plane which passes through a line

[sonic osprey]

Probably the line lies inside the plane?

[steady fiber]

also what does it mean for it to be parallel to {x=2, y+2x=-1}

x=2, y+2x=-1? so y = -5?

x=2, y=-5 is a line

how can a plane be parallel to one line and pass through another

I can't help on this problem since I just don't understand it

[wind yacht]

A plane can be parallel to a line if they never cross

Think of as if you had a infinite pencil as the line and infinite paper for the plane

How can you orient those so they never cross

[steady fiber]

I get what that means

but it's kinda weird

to say something in a higher dimension is parallel to something in a lower dimension

you wouldn't say a line parallel to a point

it's just weird and somewhat confusing

[wind yacht]

I would say that it is weird but these are things to consider when studying higher dimensions

[brittle glacier]

Can anyone help me understand how to tackle a couple lin alg problems?

[steady fiber]

[undone garnet]

for A is nilpotent matrix

prove that A(A+B) is nilpotent too

any idea?

A, B is commutative

or is this problem wrong?

this idea is from this problem

prove that
det(A^2 + AB + B^2) = det(B^2)

and my idea is prove A(A+B) is nilpotent so that det(A^2+AB+B^2) = det( A(A+B) + B^2) = det(B^2)

[half ice]

@undone garnet
A, B is commutative? Then you have the power law:
(AB)^n = A^n B^n

Need a bigger hint then that?

[undone garnet]

hm...

I mean this may be wrong

because let B = I

(A(A+B))^k = A^k + .... + (A+B)^k

so we need (A+B)^k is = 0

(A+B)^k = A^k + ... + B^k

if B = I

so (A+B)^k can't be nilpotent

[half ice]

You know that A and B commute. Can you prove that A and (A + B) commute?

[undone garnet]

yeah

it's ez

A(A+B) = A^2 + AB = A^2 + BA = (A+B)A

[half ice]

So then you have the power law:
(A(A + B))^k
= A^k (A + B)^k
= 0 (A + B)^k
= 0

[undone garnet]

😮

damn

thank you 😄

[half ice]

We actually say that the nilpotent elements form an ideal. They are closed under addition, and "swallow" elements under multiplication

[undone garnet]

😄 nice

thanks a lot

[half ice]

Np. Now, the other part is wrong.
det(A + B) != det(A) + det(B)

[undone garnet]

no

det(A+B) = det(B)

if A is nilpotent matrix

[half ice]

Oh is that true? Lit

[undone garnet]

I can easily prove it by eigenvalue

[half ice]

Now I know

[carmine terrace]

https://i.imgur.com/mxQ4dbz.png

can anyone help me on calculating the probability of these?

I thought I knew how to do it, but I'm pretty lost now

[wintry steppe]

let sunday be time 0

the initial distribution is [1, 0, 0]

[cunning zealot]

hey guys

I have a weird question

I assume it's possible to solve the Brachistochrome problem in reverse

eg; The longest travel time for an object

right

[quartz compass]

yes, as long as you have the right constraints, might end up maximizing to infinity otherwise, good question to try to work out

[cunning zealot]

yeah i assume the near horizontal path is the answer

but i wonder if there's another way

or if there are more solutions to that problem

[quartz compass]

well do you know how to solve for the brachistochrone curve normally with calculus of variations?

[dusky epoch]

why is this in here

[quartz compass]

if you haven't learned how to derive the Euler-Lagrange equation yet, go do that first, cause you're in the wrong section lol

[cunning zealot]

Where is this supposed to be

We've done that in college but i don't really remember

I remember it being a stationary value where you derive for y'

I'd have to get some books and look it up :/

[quartz compass]

I'm about to head out but #multivariable-calculus is probably better place

[cunning zealot]

Which category is this supposed to be in btw? Which books would you guys recommend i could take a look at

Something with answers so that i can check my results

[honest citrus]

Could someone share some prerequisites for understanding quaternions?

Aka some research papers or good tutorials to really understand the concept/topic

[lone quail]

@steady fiber it intersects with a line which is formed by the three equations x=3, y=t, z=t, and is parallel to the line made of x=2, y+2x=-1

anyways anyone know how to help? i have no idea

[steady fiber]

if it's "parallel" to a line, the normal of the plane must be orthogonal to the line, so the normal vector of the plane dot product with a vector parallel to the line should be 0.
normal vector (a,b,c)
vector parallel to line (0,0,1)
(a,b,c)*(0,0,1) = c
so c must be 0

i'm just gonna assume "intersect" means the line is within the plane, so a vector parallel to that line dot product with the normal vector must also be 0, so
(a,b,0)*(0,1,1) = b
so b must be 0

so the normal vector is just (a,0,0), which implies that the plane is parallel to the the yz plane, so we need find a plane like that which contains the line x=3,y=t,z=t. Obviously, we can just pick the plane x=3.

you could also have noticed that you have two vectors "parallel" to the plane, and so you can find the cross product of those two vectors to find the normal vector instead

which will give you (-1, 0, 0) as the normal vector

which is the same answer, for when a=-1

[wintry steppe]

Can I define a linear subspace of V by going:
Linear subspace := U | (U⊆V) ^({0}⊆U) ^ ((u+w)∈U) ^ ((a*u)∈U)

[slow scroll]

well yea thats just the definition of subspace (missing the quantifiers you need for a, u, w)

[wintry steppe]

alright cool thanks

[dusk ruin]

Im doing an accelerated algebra course that follows no specific textbook and the lecture notes aren't very good

ive worked with vectors and matrices before

whats a good textbook that i can look at to gain more insight?

In my case right now we've looked at dimensions and kernels and stuff and hand ins require us to prove things about them.

ex: If a set of 5 vectors spans R^4 and we apply an invertible transformation A: show (accurately) that the images of the vectors will still span R^4

[wintry steppe]

Given: a family of vectors vi (infinite i) in V
Im supposed to define span(vi) with linear combinations w/o using span(vj) (j being finite)

How tf do I do this

[dusky epoch]

{0}⊆U

...

why not 0 ∈ U

[half ice]

Do they span a common space?

[wintry steppe]

@dusky epoch damn yeah thats easier XD

bc we like sets Ig?

[dusky epoch]

idk like

why would you even do this

why not write it out in words

[wintry steppe]

bc Im more afraid of doing it wrong in words than I am doing it in this way

anyways do they span a common space? vi and vj? yeah they are "the same"

Idk man Im stupid I just started linear algebra today

Im glad I made it through the set theory section of the script

[half ice]

Very nice, knowing what a span is on day one

[wintry steppe]

I have a newbie understanding of it

and its probably wrong

but the problem is I gotta hand in this sheet tomorrow

[half ice]

Is that exactly the question?

[wintry steppe]

if I dont I wont be able to write the finals

[half ice]

Maybe a picture of it might clarify

[wintry steppe]

well its German

but maybe

[half ice]

Ah yeah that doesn't help me

[wintry steppe]

haha

well maybe it will

[half ice]

People here do speak the language

[wintry steppe]

its (e)

the rest I pretty much got

ohne means dont use that

familie = family

the words are similar

raum = space

Menge = set

I miss A1 (e) the whole of A2 and A3 and I got A4 completely solved on my own

[wintry steppe]

😦

Im screeewed XD

[lone quail]

@steady fiber thanks a lot

[clear otter]

I'm trying to understand direct sums and products of modules. With the direct sum, my textbook says that the direct sum of a family of modules are families $(x_i)_{i \in I}$ such that $x_i \in M_i$ for each $i \in I$ and almost all $x_i$ are 0.

Why are almost all x_i zero?

I'm not even sure what that means to be honest. For example, if I have the direct sum of 5 modules, they are the 5-tuples right?

[steady fiber]

single dollar signs

for inline

[stoic python]

𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫:

[clear otter]

Thanks for that!

So is that to say if I have modules, $M_1, M_2, M_3, M_4, M_5$ that $M_1$ = {$(x, 0, 0 ,0 ,0) | x \in M_1$} or something along those lines?

And do that for each M_i?

[stoic python]

𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫𒐫:

[clever cedar]

lets say I have a matrix A and I simplify it to triangular matrix so i can easily find the determinant of that matrix

can i use that simplified matrix to find cofactors of matrix A?

or do i have to use the original matrix

[sonic osprey]

@clear otter Look at what almost all means

[clear otter]

@sonic osprey It means finitely many

I'm not sure why that restriction is needed though.

[clever cedar]

how do i decode that nonsense

all i see are diamonds

[sonic osprey]

@clear otter Think of a situation in which it would matter

[gloomy arrow]

How early/far in the course is eigenvalues/eigenvectors covered

[steady fiber]

was about a month and a half into a 3.5 month course

for me

I guess

[royal star]

totally dumb question but when sb says "f is a polynomial" does it necessarily have a finite degree?

[sonic osprey]

yes

[royal star]

ok thanks @sonic osprey

[glass patio]

does anyone know how to do piecewise graphing i'm very confused.

[wind yacht]

Break up the piecewise function into its functions with limited domains

[glass patio]

can you show me what you mean by that

[steady fiber]

that does not seem like linear algebra

maybe just #prealg-and-algebra

[glass patio]

sorry i'm new to the server

i didn't know where to go

[gray dust]

tbh trying to move the convo to a new channel right now would be a bit of a disservice even though this is the wrong channel

[steady fiber]

this is one of the more misused channels

[gray dust]

"linear ALGEBRA means i should ask my algebra q's here, right?" no but whatever, let em keep talking here if it helps

[half ice]

ABx means to apply B onto x, then A onto whatever that is

Namely, you have the associative property
A(Bx) = (AB)x

@flat sphinx

[paper egret]

find a basis for the space spanned by the following vectors

anyone here got any ideas?

wtf is it even asking lmao

[sonic osprey]

What's a basis?

[paper egret]

no freakin clue

im so lost in my lin alg class

vector spaces is fucking me hard in the ass

[sonic osprey]

Look at the definition

And slowly work through what it means

Both rigorously and intuitively

[paper egret]

you have a definition for me to start with

i cant find one in my book, or a simple one

[sonic osprey]

The one in your book is as simple as it'll get

[paper egret]

all i got

The pivot columns of a matrix A form a basis for Col A

[wintry steppe]

Show that: if (vi)i∈K is a linear independent family of vectors in a vectorspace V and 0 is not an element of K and v0 ∈ V −span(vi)i∈K , then (vj )j∈K∪{0}
is also a linear independent family.

Struggling with this proof

[sonic osprey]

@paper egret what book are you using

[wintry steppe]

wait does that belong to proofs or linear algebra now?

[paper egret]

linear agebra and its applications 5th edition

by lay

[sonic osprey]

Depends

The latter is nice because it ensures that what I'm thinking about matches up with what their book is talking about

The former option should really just be replaced with having me type out the formal definition and then explain it intuitively

You need to be able to work with rigorous definitions to work with vector spaces and so you need to struggle with them

[wintry steppe]

Should I post my problem into proofs?

I feel like the solution is just

claiming that 0 is part of the span

and that including it will therefore definitely give linear independence

but Idk

[sonic osprey]

@paper egret Look on page 150

of the book

[paper egret]

ok

wait is that ind eterminatnts

by lay

[sonic osprey]

what?

[paper egret]

if its on determinants

i cant look at that, we never covered determinants yet

[sonic osprey]

Section 2.8?

[paper egret]

oh

[sonic osprey]

Subspaces of R^n?

[paper egret]

oh

A basis for a subspace H of R n is a linearly independent set in H that spans H

ok

gotchu

so we take a set of vectors

if all of em are linearly independent

its a basis

for something

idk

[sonic osprey]

for H

[wintry steppe]

;_;

[paper egret]

ye

ok that makes sense

all i gotta do is find lin independent vectors

ez klap

[sonic osprey]

The idea is that

The set of vectors you were given

form a spanning set for your space right

[paper egret]

yes...

do you have a numerical example

we can run through

like which ones are a basis and which ones arent, for example

[half ice]

There's a common way to do this. Put the vectors into a matrix, and row reduce. Whatever column has a pivot represents the vectors you should keep to make a linearly independent set

[paper egret]

uhhh

hmmm

they are linearly independent

[half ice]

,w row reduce {{1,4,7},{2,5,8},{3,6,9}}

[stoic python]

Display results online and refine query

Requested by Kaynex#7520

[paper egret]

oh wait

that's what you meant?

[half ice]

Nope, those vectors are not independent

[paper egret]

ah ripperonis

[half ice]

That is, using the span of two of them, you can make the third

[paper egret]

yes

LIN ALG IS FUCKIN HARD DUDE

[half ice]

Nah you're learning. You'll get it

[paper egret]

bro im so lost in my class

😦

this makes me sad

[half ice]

Well, you know how to do that thing I just did

[paper egret]

row reduce, that was like 5 weeks ago stuff

vector spaces is ass

[half ice]

Row reduction is useful for a lot of stuff

[paper egret]

ok back to a basis

what makes a basis of a vector space

[half ice]

A few things important to know

Take a subset of a vector space, we'll call these vectors u, v, w.

span(u, v, w) = au + bv + cw
For any choice of scalars a, b, c

[paper egret]

when you say subset of a vector space

its the set of vectors

its a set of vectors*

under some vector space

[half ice]

Yup, just whatever vectors you want. No rules on how you choose them

[paper egret]

what really is a vector space

im not understanding what a vector space really is

like is there some analogy

[half ice]

It's a structure with two "things"

There are

• Vectors, that you can add/subtract together
• Scalars, which you can add/subtract/multiply/divide together

As well, you also get a scalar multiplication. You can multiply a scalar with a vector to get a new vector

[paper egret]

so how the cartesian is for functions

[half ice]

You're probably used to seeing vectors as arrows. You can always add two arrows together, and you can always scale them.

[paper egret]

vector spaces is for vector math

or cartesian is for points, where u can do shit with them

vector spaces is just the equivalent, but for vectors instead

is what i'm getting from this

[half ice]

Vector spaces are for a lot. The more general methods also apply to solving polynomials, differential equations, ect

[paper egret]

oh

but you get what i mean with my analogy?

idk if that's right at all

[half ice]

As a learner you should picture vectors as arrows. But be ready to leave that analogy, as they're useful for more than that

[paper egret]

ok

gotchu

is there a better analogy

?

[half ice]

Vectors are things you can add together
Scalars are "normal numbers"

You can multiply scalars and vectors together

[paper egret]

yee

[half ice]

To get abstract with it, note that polynomials fit the definition of vectors, where multiplying by a real number is the scalar

[paper egret]

wait

how

[half ice]

Let's look at quadratics. To get "handwavy" with it,

I can always add quadratics together. They are "like vectors".

I can always multiply a quadratic by a constant. Constants are "like scalars" and have a scalar multiplication

[paper egret]

yes

oh yes

es

yesyeyes

quadratics 3 components

the x^2, x, and constant

3 element vector

[half ice]

In the end, a lot of the theory that works on vector spaces help us find interesting things about quadratics

[paper egret]

oooo any examples?

[half ice]

Congrats, you just picked out the fact that this space has a basis. It's {1, x, x²}

{1, x, x²} are linearly independent because you can't make one as a linear span of the other two

And every quadratic can be made with the span of {1, x, x²}

Since span(1, x, x²) = a + bx + cx²

[paper egret]

yes

[half ice]

This is actually a pretty complicated example, so if it's actually sinking in, good for you

[paper egret]

ok uh back to vector spaces

i think im starting to get the hang of it

but i struggle a lot with proofs still

[half ice]

This is a vector space lol

But sure, let's talk about more traditional ones

[paper egret]

yes plz

lets just keep talking, this helps a lot LOL ngl

[half ice]

https://onlinemschool.com/math/library/vector/linear-independence/

This has some examples, let me know if you want to pick any of them out

[paper egret]

idk u can choose one

i dont want any bias for me

[half ice]

Well, the theorems here are important

span(u, v, w) = au + bv + cw
For any choice of scalars a, b, c

Is the set of vectors that u,v,w can make. This is a vector space inside your larger vector space

[paper egret]

ive already done most of this stuff already, i just need help with vector spaces and basis

yes..

i feel like i understand it a bit better, but thats far from fully mastering it

which i wanna try to do

[half ice]

u, v, w are said to be "linearly independent"
If au + bv + cw = 0 has no non-trivial solutions.

That is, a = b = c = 0 is always a solution, and we don't care about that one. Is there any other ones?

[paper egret]

nope not rly

[half ice]

Sometimes there are. If there are, the vectors are linearly dependent

[paper egret]

only the solution is the trivial

thus lin independent

otherwise lin depennt

that amkes sense

[half ice]

Another way of saying it
u, v, w are independent if u, v, w spans zero only once

So for example
[1, 2] and [3, 6]

Are linearly dependent because
3[1,2] - [3, 6] = [0, 0]

[wintry steppe]

waaait

can I use that in my proof

Show that: if (vi)i∈K is a linear independent family of vectors in a vectorspace V and 0 is not an element of K and v0 ∈ V −span(vi)i∈K , then (vj )j∈K∪{0}
is also a linear independent family.

[half ice]

Sure, that's a common definition!

[paper egret]

yes

that is tru

[half ice]

Often, finding the way to make 0 out of your vectors is the quick way to prove they're dependent

[paper egret]

yes

[wintry steppe]

So how do I prove my proposition?

[half ice]

So we get to the big one:
Take a set of vectors. If they are linearly independent, then the set is a basis for the space that they span

A basis is special. There's exactly one way to represent any vector in the space using the elements in the basis

[paper egret]

that's because the set of vectors in the basis are all linearly independent

[half ice]

Yup

It's a pretty easy proof to show there's no two ways to make a vector from a basis. Give it a shot?

Oops, phone corrected the hell out of me

Not that bad, there's only one

[wintry steppe]

soo what does my proof sketch look like now exactly

Ill claim that bc my set of vectors are linearly independent that that set is a basis for the space they span

jeez I feel so stupid Im sorry but I still dont get what Im exactly supposed to do like what are the ncessary steps

[half ice]

How large is the set vj?

How is v0 not in its own span?

[wintry steppe]

Idk 😦

[dense holly]

are two vectors linearly dependent if their determinant is zero when put in a matrix?

[gray dust]

if those vectors make up the columns of a square matrix A and det(A) = 0, then yes the vectors are LD

[wintry steppe]

a hermitian matrix is supposed to have n orthogonal eigenvectors
but first, why does it have any eigenvector?
wikipedia uses here the fundamental theorem of algebra on the characteristic polynomial
it does the job, but is there any more matrixy way, without using some weird polynomial?

[dusky epoch]

characteristic polynomial

weird polynomial

[glad pagoda]

lmao

[wintry steppe]

ok point taken

[dusky epoch]

i mean what'd you even expect

and why is the charpoly "weird"

it's perfectly natural

[stoic python]

tet:

[dusky epoch]

wdym

"over the reals"?

wdym "look like"

[wintry steppe]

what i mean is matrices used to be stuff really accessible to my limited human senses, like rotation
but something like the characteristic polynomial has absolutely no sensual interpretation for me

[dusky epoch]

"sensual"?

[wintry steppe]

without words

[dusky epoch]

...

[gentle knoll]

@wintry steppe to answer your question, the fundamental theorem of algebra will tell you that eigenvalues exist, assuming the field is algebraically closed (like the complex numbers). Eigenvectors must also exist because for a given eigenvalue x the matrix M-xI has determinant 0, meaning the corresponding transformation is not one to one, and so some nonzero vector gets mapped to 0. That vector is the eigenvector

[wintry steppe]

yeah i get that idea, but thanks for the description @gentle knoll

[leaden fiber]

how on earth does the first line go to the second :o

this is the case where lambada1=2

(ping me)

(for context it's the last example here http://tutorial.math.lamar.edu/Classes/DE/LA_Eigen.aspx)

[half ice]

@leaden fiber
The eigenvectors are of the form
[η η η]

That is, it's an eigenvector if all three of its values are the same.

Eigenvectors form a vector space, and therefore have a basis. A basis for this is [1, 1, 1] since you can multiply it by a constant and get every eigenvector

[2, 2, 2] would also have been acceptable

[leaden fiber]

if I get something that's not all the same

then is it not an eigenvector :o

[half ice]

Well, not an eigenvector for this specific eigenvalue

I assume there's other eigenvalues, and they each have their own eigenvectors

[leaden fiber]

So my list line here isn't an eigenvector, right?

[half ice]

This doesn't look like the same question?

[leaden fiber]

yeah not same question

trying to work my way around this question based on that one

[half ice]

Yours very well could be an eigenvector?

[leaden fiber]

how do I tell if something is an eigenvector?

[half ice]

If your work is correct, the steps look right

[leaden fiber]

dont' eigenvectors need to be the same for n1, n2 and n3?

or is it just for that case?

[half ice]

Just for that case. It happened to be true

[leaden fiber]

ohhh okay I get it!!

[half ice]

Eigenvectors are vectors that your matrix "retains the direction" of

[leaden fiber]

by the way is the format of my last line correct

or do I have to sub 1 for n2

[half ice]

No, that last line is very nice

Says what it needs to

[leaden fiber]

tysm!!

[half ice]

Np at all. Let me know if there's anything else

[leaden fiber]

is it an eigenvector if it is (0,0,0)?

[dusky epoch]

no

[leaden fiber]

ahhh okay!! so would the corresponding eigenvalue still be a valid eigenvalue?

also how do I check if it is diagonalizable?

[dusky epoch]

wdym "the corresponding eigenvalue"

if your only solution to Ax = λx is x=0 then λ isn't an eigenvalue

[leaden fiber]

ahh okay!

[spiral radish]

@leaden fiber 1853?

[ocean fern]

Hi. How would I approach this problem regarding vector spaces?

[half ice]

@ocean fern
Row reduce, and find whatever rows have pivots

[pallid rampart]

Let $A=\begin{pmatrix}1/7&3/7&3/7\ 3/7&1/7&3/7\ 3/7&3/7&1/7\end{pmatrix}$. What is $\lim_{n\to\infty}A^n$?

Oh jesus

[gray dust]

\\\\ is enough to properly separate the entries

[stoic python]

Whoever:

[pallid rampart]

Hell o

I wont do that

Thats just

Creating an empty line

[native lodge]

Diagonalization

Symmetric matrix is nice

Markov matrix too

Just take a look at eigenvalues

Any less than 1 will go to 0

1 will stay at 1

Anything greater than 1 will go to infinity

[dusky epoch]

this is 1/7 (3J - I) where J is the matrix of all 1s

uh

[brittle juniper]

[dusky epoch]

what

ok so spec(J) = {0,3}

[brittle juniper]

3-1=2

[dusky epoch]

spec(3J-I) = {-1,8}

spec(A) = {-1/7,8/7}

8/7 > 1

oof rip

[brittle juniper]

[dusky epoch]

so infinity then

,w [[1/7,3/7,3/7],[3/7,1/7,3/7],[3/7,3/7,1/7]]^1000

[stoic python]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Display results online and refine query

Requested by Ann#0413

[dusky epoch]

rig bip

,w [[1/7,3/7,3/7],[3/7,1/7,3/7],[3/7,3/7,1/7]]^100

[stoic python]

Display results online and refine query

Requested by Ann#0413

[dusky epoch]

oooffff

[pallid rampart]

Does it not exist

Wait no

There is a 1/huge number

[leaden ermine]

Hey, in an overdetermined system there are no solutions, but what does the null space look like? Is it empty as well?

[dusk ruin]

what can we infer if $x_i y_j = x_j y_i$ for some $i \neq j$

[stoic python]

aritmos:

[pallid rampart]

If A is a square matrix then it’s symmetric

But it isn’t

[dusk ruin]

therefore...

[sonic osprey]

Just make A a square matrix by getting rid of x_3, y_3

and see what your condition says about A

[dusk ruin]

but its a proof involving that matrix

i first need to prove rank nullity theorem

and show that in that specific case

that dim(ker(A))>=2 <=> rank <= 1

[wispy copper]

Hey, I’m 3 weeks into my Linear Algebra class this year and the first topic that we have almost finished so far has been about the Jordan Canonical/Normal Form.

I’m really struggling to find the motivations to learn the topic, feels like all that’s getting thrown at me are bunch of definitions and proofs and lengthy examples that don’t really feel clear to me about what is really going on.

Could anyone point me to some good resources I could get another perspective from? For what it’s worth, my class is almost fully focused on the theory with very few examples thrown in.

[cobalt tartan]

Is multiplication by a matrix distributive over a dot product?

[sonic osprey]

Depends what dot product

[cobalt tartan]

The standard one?

Like <x, y> = x_1y_1 + ... + x_ny_n

Because I have this question here,

Im' not really sure how to go about proving that it's true

I think that it's true

But an idea that I have is to let a be a vector in S, let s be a vector in S^perp but not in S^circ

And then show that $s \cdot a = 0 \iff A(s\cdot a) = 0 \iff (As) \cdot (Aa) = <s, a> = 0 $

[stoic python]

Liria ^(;,;)^:

[cobalt tartan]

Or something like that

Like to me it seems that inner products always go to R

Right?

@sonic osprey ?

[north sierra]

so what's the point of expressing a solution in parametric vector solution?

[urban bough]

Geometrically speaking, the matrix

1 0
1 0

Maps the x component of a vector onto the line x =y right?

[marsh cedar]

Yes, with respect to the standard basis.

[undone garnet]

any idea?

[pallid swallow]

hmm

N* is positive naturals?

and O_n is the zero matrix?

[dusky epoch]

yes

it's not entirely clear what the quantifiers are meant to be tho

[pallid swallow]

quantifiers?

2(AB)^k-AB=BA
hmm
A(2(BA)^(k-1)-I_n)B=BA ?

[hardy blaze]

http://prntscr.com/pl9j08

for these since u can write them out as linear combinations

then thats how W & H are subspaces right ..

[clever cedar]

is there a "best" row or column to preform cofactor expansion on?

for example 1st column

[hardy blaze]

ones with 0's

the most 0s

[clever cedar]

Excellent

in the case where all entries are non-zero

is it indifferent?

[hardy blaze]

yea

just try to do #s like 1 or 2

the smallest # so it doesnt get all ugly

[clever cedar]

oh that makes sense

[hardy blaze]

but if there is no 0

then it doesnt matter which 1 row/column u pick

just might take longer

[clever cedar]

awesome,

thank you for your time

[hardy blaze]

np

:^)

[wintry steppe]

having trouble understanding how i could actually solve this through back sub

if the matrix is banded wouldnt there be 2s+1 entries in the bottom row?

in which case back sub wouldnt even work

or would this matrix only be semibanded because its upper triangular?

ah i think i answered my own question

always helps to write things out i guess lol

[charred stirrup]

yo

for matrix A^4, is that equal to A^2 x A^2 ?

and consequently

matrix A^n for n that is wholly divisible by 2

it will be A^2 times itself n/2 times?

[half ice]

Yes, A⁴ is A×A×A×A

Or A²×A²

[charred stirrup]

how should I account for n where it is not wholly divisible by 2?

[half ice]

Like A³?

[charred stirrup]

find the base case?

[half ice]

A³ = A×A×A

[charred stirrup]

and then (A^3)^(n/3) ?

[half ice]

Wat

That's A^n

[charred stirrup]

holy shit

[half ice]

Wait no it's not

[charred stirrup]

you genius

it's not?

[half ice]

Wait yes it is

[charred stirrup]

i love u

gonna speed through this question now

[half ice]

K hol up

What's the question lol

[charred stirrup]

here is my solution so far

[half ice]

Oh, there's a pretty simple solution to this lol

[half ice]

Well, if you want it yourself, I'll let you have it. But try computing M^8

[charred stirrup]

= (M^2)^4

[half ice]

Yep, M^4 × M^4 should do it

[charred stirrup]

I guess your solution for odd number n is perfect then

should have thought about it better

[half ice]

I lied at you. There are three cube roots for any matrix

So that case gets weird

[charred stirrup]

worry not I will find M^2018 and just multiply it by M

[half ice]

That's a good idea, should work

Get M^8 yet? Should change your world

[charred stirrup]

working on M^3 :/ having trouble quickly multiplying the rows by the columns

always trail off and mess it up

i hope it is a get better the more you do kinda thing

[half ice]

I like to do this

It makes it pretty fast

[charred stirrup]

I like that picture

I dont have trouble finding the spot, it's just multiplying and adding the elements that make me mess up

[half ice]

Oh, well definitely have a calculator that does PEDMAS at the ready

[charred stirrup]

im scared lol i dont know if i get a calculator but if not they will likely give me kind numbers

wow

I did not expect M^8 to be the identity matrix

(identity matrix)^n for all real n is just itself yeah?

[half ice]

Yup

Aaaaaand so the question is now much easier

[charred stirrup]

by consequence of my first idea? (M^8)^n = M^8 for all n

[half ice]

M is a rotation matrix. It's a rotation by π/4. Doing this rotation 8 times brings any vector back to itself, and so M^8 = I

[charred stirrup]

but 2019 is not wholly divisible by 8

[half ice]

Nope. But you can get pretty close

[charred stirrup]

ohh

i see

[half ice]

Oh yes, M^(8n) = I for any positive n

[charred stirrup]

figure out how many 8s fit into 2019

[half ice]

Lel

[charred stirrup]

i goofed

i found 2018

[half ice]

Yup

And so it's clear why you needed to find M³

[charred stirrup]

ohhh my god

identity times m^3

you fucking genius

i was gonna do m^2 times m

im an idiot lol

[half ice]

Yup. The answer was right there the whole time

Top 10 anime betrayals

[feral mountain]

rotation of pi/4

[burnt heath]

any help is appreciated

[slow scroll]

@burnt heath Start by formally expressing what you are trying to show. (Think def of linear independence)

[charred stirrup]

is the solution some sort of trick with the identity matrix? or are the entries of matrix in terms of i and j

[slow scroll]

wat no. it basically tells you what to do: "start with the usual equation to check linear independence and try to use the assumption that BA = I_n"

[charred stirrup]

I wont lie Im very unsure of what the question is asking me

a 4x4 matrix that is equal to j-i

like ??

[slow scroll]

wait are you talking about a different question than KermitTheFrog's

[charred stirrup]

yeah

Aij = j-i

what is the question even asking?

[slow scroll]

so you have a matrix A = [a1 a2 a3 ... an] where a_i are the column of A. Since a matrix is defined by how it acts on a basis, in the basis e1, e2, ... , en we have that Ae1 = a1, Ae2 = a2, etc.

[charred stirrup]

we are definitely not on basis yet lol

[slow scroll]

The proof wants you to verify that the columns are linearly independent. In other words that the equation $$ \sum_{i=1}^n c_iAe_i = 0 $$ only when $c_1 = c_2 = \cdots = c_n = 0$. Then you want to use the assumption that $BA = I_n$ to help show this

[stoic python]

kxrider:

[leaden fiber]

how do I tell if a eigenvector is diagonalizable or not?

[quartz compass]

eigenvectors aren't diagonalizable

[leaden fiber]

oh right I'm blind looked at my notes wrongly

how do I tell if a matrix is diagaonlizable or not?

like what does it mean geometrically etc

[dusky epoch]

a matrix is diagonalizable iff the algebraic multiplicity and the geometric multiplicity are the same for each eigenvalue

or in other words

if dim(ker(A - λI)) is the same as the multiplicity of λ as a root of the charpoly of A

for all λ

[leaden fiber]

haven't heard of what dim, ker and charpoly is :oooo

what is algebraic multiplicity and geometric multiplicity lemme look that up :oooo

[dusky epoch]

uh

how tf have you not heard of DIMENSION

[leaden fiber]

ohhh

didn't know dim was short for that ;w;

[leaden fiber]

need help with this aahhh

so I know that the determinant of A is (lambda-4)(lambda-3)(lambda-2)(lambda-1)

what next >.>

don't fully understand the hint

[dusky epoch]

uh

wait

$|A - 5I| = \chi_A(5)$

[stoic python]

Ann:

[dusky epoch]

😛

$\chi_A$ denotes the charpoly of $A$ of course

[stoic python]

Ann:

[dusky epoch]

charpoly is short for characteristic polynomial

and you know $\chi_A(\lambda) = (1-\lambda)(2-\lambda)(3-\lambda)(4-\lambda)$

[stoic python]

Ann:

[dusky epoch]

it's not the determinant of A but of A-λI

@leaden fiber

(the hint's actually useless )

[leaden fiber]

wait what why

[dusky epoch]

why what

[leaden fiber]

oh

sorry brain wasn't working

[dusky epoch]

no like what are you confused at

[leaden fiber]

was dumb forgot that the determinant was determinant of A-lambda I

instead of just A

[dusky epoch]

det(A) is just the charpoly evaluated at 0 😜

[leaden fiber]

yeet

does that mean (1-lambda)(2-lamda)(3-lambda)(4-lambda)=0

or do I sub lambda as 0

[dusky epoch]

uh no

neither of these things

$\det(A - \lambda I) = \chi_A(\lambda) \ \det(A - 5I) = \chi_A(5)$

[stoic python]

Ann:

[dusky epoch]

= (1-5)(2-5)(3-5)(4-5)

[leaden fiber]

:ooo

yeet i figured it out tysm!!

[alpine lion]

Hey guys its probably a stupid question but we didnt go over it exactly and just want to make sure I get all points in my assignment. Topic is linear programming and we have to classify a few model into degenerate, unbounded etc. One of my models is ambigous (dunno if you call it that in english, it has multiple optimal solutions) and one point of the possible optimal solutions is also degenerate. So I should classify it as both or does one like "overrides" the other? (https://online-optimizer.appspot.com/?model=ms:vyJEvq7we6hwE76RR5nXR4ulnesrr7TN) This link shows a graph in Model overview

Oh solved it myself is only degenerate if theres only one optimal solution

[leaden fiber]

dunno how to do both parts A or B

does A have something to do with eigenvectors being linearly independent because they're distinct eigenvalues

or actually not because the question doesn't talk about the dimensions of A--the above statement would only be true if matrix A is 2x2 right

[quartz compass]

what have you tried?

[leaden fiber]

don't really know how to approach this question at all so haven't tried anything >.>

[quartz compass]

try something, anything

[leaden fiber]

can you guide me in a direction so I can try >.> I'm really stuck

I suppose I'm like, trying to learn the content through doing actual questions, so I don't really start out with a lot of knowledge

[dusky epoch]

i remember helping someone else do a problem which was worded EXACTLY like 4b...

like

it happened literally yesterday

is it 4a that you want help with, though

[leaden fiber]

could be a classmate :o

yeah

[dusky epoch]

ok

alright

so

do you want me to guide you through the solution of 4a step by step

[leaden fiber]

yes please!

[dusky epoch]

k

so

we have a matrix $A$

we have two eigenvalues of this matrix, $\lambda_1$ and $\lambda_2$, and we're given that $\lambda_1 \neq \lambda_2$

[stoic python]

Ann:

[dusky epoch]

and for these eigenvalues, we're given two eigenvectors: $v_1$ and $v_2$ respectively

[stoic python]

Ann:

[dusky epoch]

just stating the problem once again

[leaden fiber]

mmhm

[dusky epoch]

so can you write out, as an equation, the sentence "v_1 is an eigenvector of A with eigenvalue λ_1"?

[leaden fiber]

A-λ_1 I=v_1 A?

oh wait no

[dusky epoch]

no

[leaden fiber]

A-(lambda1 * I)=(v1 * lambda1)

right

[dusky epoch]

no

it's way, way simpler than that

and really if i were in a worse mood i'd make a terse remark like "well then clearly you don't know what an eigenvector is"

Av_1 = λ_1 v_1

[leaden fiber]

oh yeah you're right :o

whOOPS

[dusky epoch]

and likewise, Av_2 = λ_2 v_2

i hope that much doesn't need explaining

[leaden fiber]

yep

[dusky epoch]

and we're told to prove that v_1 + v_2 ISN'T an eigenvector of A

so how might we do that?

[leaden fiber]

try finding a lambda that would give A(v1+v2)=lambda * (v1+v2)?

and then there's probably a contradiction somewhere

[dusky epoch]

sure. i'd have called it μ instead of λ to distinguish it more clearly from λ_1 and λ_2 but yes exactly that's what i had in mind

assume it IS an eigenvector, then arrive at a contradiction

so now we ASSUME that there exists a scalar λ such that A(v_1 + v_2) = λ(v_1 + v_2)

what can be concluded now

or rather, how can this equation be rewritten in a more workable form

[leaden fiber]

which cannot be because A would have to = lambda

[dusky epoch]

no

wrong

A is a matrix

not a scalar

[leaden fiber]

uwu

[dusky epoch]

it makes no sense for a matrix to equal a scalar

[leaden fiber]

try equating A to lambda*I?

[dusky epoch]

no

you're overthinking it

again

[broken hawk]

well, I don’t think overthinking is the right word

more like

[leaden fiber]

being dumb?

:P

[dusky epoch]

overthinking is EXACTLY the right word

[broken hawk]

I think it’s more not having internalized that for matrices and vectors, Av = Bv does not have to mean A=B

this is true for numbers, of course

but not here

[dusky epoch]

texit seems to be back online

$A(v_1 + v_2) = λ(v_1 + v_2)$

oh ffs

ok fine don't really need texit here i guess

we have this equation

how can it be rewritten? how can we use what we know about v_1 and v_2 to make this more workable?

[leaden fiber]

are v1 and v2 linearly independent? or is that not important?

[dusky epoch]

v_1 and v_2 ARE linearly independent, but that's going to come in a bit later

[broken hawk]

(would be a good exercise to prove it, in fact)

[dusky epoch]

for now

look at the left hand side

A(v_1 + v_2)

[broken hawk]

wait, dependent? did I misread sth earlier on? I thought v₁ and v₂ were EV with different eigenvalues?

or was that a typo

[dusky epoch]

typo

anyway

A(v_1 + v_2)

how can you rewrite that

@leaden fiber

[broken hawk]

(so to clarify they’re independent but yea, not relevant here)

[leaden fiber]

Av1+Av2

or can you not do that with matrices :o

[broken hawk]

you can

[dusky epoch]

you can and that's exactly what i expected you to write

now simplify further

recalling what we know about this entire setup

[leaden fiber]

just a multiple of A right

bc they're eigenvectors?

[dusky epoch]

"a multiple of A"

what

no

Av_1 + Av_2

what can be done to simplify this further

knowing what v_1 and v_2 are

[leaden fiber]

multiplying matrices by their eigenvectors basically is transforming it along the same direction, right? so Av should be in the same direction as A? or is that V

[broken hawk]

A doesn’t have a direction

[dusky epoch]

no you're overthinking again

[broken hawk]

A is a matrix, not a vector

vectors have directions

matrices don’t

[dusky epoch]

look back at what i wrote

here

lemme copy and paste it i guess

$Av_1 = \lambda_1 v_1 \ Av_2 = \lambda_2v_2 \ \lambda_1 \neq \lambda_2$

[broken hawk]

you really only have to plug in definitions and apply simple computation rules

no thinking is required except for the setup

if you’re trying to visualize the setup you’re already overthinking it

[dusky epoch]

god fucking damnit why is texit working in bots but not here

[leaden fiber]

I'll go to bots gimme a sec

[broken hawk]

hang on I’ll type it out

[dusky epoch]

there we go

[broken hawk]

that works

[dusky epoch]

@leaden fiber

[leaden fiber]

lambda1v1+lambda2v2?

[dusky epoch]

ok alright there we go finally

so now we have this

A is out of the picture entirely and we may just as well forget about it, apart from the fact that v_1 and v_2 are LI by virtue of being eigenvectors of A with different eigenvalues

[atomic flint]

hey i was wondering where i would go to ask about vector spaces

[dusky epoch]

this channel is the right place but it's currently occupied

so you'd go to a questions channel

[atomic flint]

ahh ok thank you

[leaden fiber]

how does them being LI come in?

[dusky epoch]

well

it's not quite obvious yet, is it

[leaden fiber]

(by the way, aren't any two vectors that aren't on the same line linearly independent)

[broken hawk]

correct, if your vector space is ℝⁿ

but “lines” don’t make sense in all vector spaces

[dusky epoch]

but maybe we can rewrite this equation in a form where the linear independence of v_1 and v_2 can be but to use

also,

1. linear independence is a notion that applies to SETS of vectors, not just to pairs
2. you're overthinking again

but maybe we can rewrite this equation in a form where the linear independence of v_1 and v_2 can be but to use

[leaden fiber]

but “lines” don’t make sense in all vector spaces why :o

[broken hawk]

because “lines” are a geometrical thing

[leaden fiber]

also Ann the equation can't be further simplified, can it?

wdym spaces :o

[dusky epoch]

well do you know what linear independence means

like the actual definition instead of vague geometric mumbo-jumbo

[broken hawk]

storm, we can discuss this some other point for now you’re just avoiding the problem at hand

and I don’t wanna help you with that

(spaces is just short for vector spaces here)

[leaden fiber]

you can't make any of the vectors as a set as a combination of the other vectors?

[dusky epoch]

no

[broken hawk]

no, that’s spanning

[dusky epoch]

that's not quite "vague geometric mumbo-jumbo" but it's not the definition either

[leaden fiber]

((ahh I was referring to you as spaces sorry, how should I call you?
and okay! I'll stop the convo about that here :) ))

ehh it's not?

[broken hawk]

((oh in my nickname it refers to topological spaces, which have nothing to do with vector spaces. just unfortunate naming))

[dusky epoch]

you should have it in your notes somewhere

[leaden fiber]

ah yeah

this right?

[dusky epoch]

yeah that

ok great

so you see

that equation up there

has zero on its right-hand side

is there anything we can do to our equation to make that the case there too

here it is again for ease of reference

[leaden fiber]

minus (lambda1v1 + lambdav2) both sides...?

[dusky epoch]

yes

[leaden fiber]

actually should be minus lambdav1+lambdav2 both sides right

[dusky epoch]

yes

sorry, somehow the 1 slipped me

see that's why i said i would rather have called it μ instead of λ. that way the visual mixup wouldn't have happened

[leaden fiber]

so λv1+λv2-λv1-λv2=0

[dusky epoch]

anyway w/e that's beside the point

uh

what you've written is a tautology

[leaden fiber]

μv1+μv2-λ1v1-λ2v2?

[dusky epoch]

i mean you called it λ so let's continue with λ unless you want to retroactively rename every instance of λ (without subscript) to μ

[leaden fiber]

ah okay

[dusky epoch]

what you get, then, is this

do you see where to go from here

[leaden fiber]

do the Xes in the linear independence equation have to be scalars or vectors?

or does that not matter

[dusky epoch]

i'm going to ask you to look at that question again

and tell me

what do you think

what are the x_i

are they scalars

or are they vectors

especially given that the picture you posted makes a point to mark some objects in bold and others not

[leaden fiber]

uuh they're scalars?

[dusky epoch]

yes precisely

[leaden fiber]

yeet

yeah so that can't happen because lambda1-lambda will have to equal lambda2-lambda

but lambda1 and lambda2 are distinct

[dusky epoch]

linear independence has less to do with the coefficients being equal to one another and more with them all being equal to zero but w/e yes

you get that λ_1 = λ_2, which directly contradicts the distinctness of λ_1 and λ_2

and that's it

[leaden fiber]

ahhh okay I get it, tysm!!

sorry for being so dumb and annoying <3

[quartz compass]

don't be sorry, practice some matrix and vector multiplication problems so you're not doubting things like A(u+v) = Au+Av

[dusky epoch]

have i recommended 3b1b's essence of linear algebra

[broken hawk]

and also make sure you understand the difference between vectors, scalars and matrices intuitively

so you don’t even get the idea to utter things like “In the direction of A”

[quartz compass]

concrete computations with matrices with entries will help build an intuition of what's going on and what you can do, since in some sense they're just cute wrappers for those computations

[leaden fiber]

(( does u and v have to be scalars/vectors/matrices for A(u+v) = Au+Av to apply ))
((I know it does for scalars but how about vectors and matrices))

gonna check that out Ann!

[quartz compass]

do you know the criteria required to be able to multiply two matrices A and B together?

[leaden fiber]

yeah whooops was dumb and forgot that matrices don't have directions (or at least, I wasn't sure if matrices had directions)

yeah

[quartz compass]

do you know how to represent a vector as a matrix

[leaden fiber]

same number of columns of the first one with same number of rows for second one right

uhm, no? isn't a vector already a 1xn matrix

[quartz compass]

depends

you can have row vectors or column vectors

is a 1xn matrix a row or column

[leaden fiber]

(oh no i forgot runs to check)

[quartz compass]

this is good it's highlighting what you need to know

[broken hawk]

it’s probably pedagocially better to say vectors are column vectors here

[quartz compass]

these are the fundamentals,

[broken hawk]

and leave it at that

(obviously you can have a vector space of row vectors too)

[quartz compass]

there's nothing intimidating about a row vector

it's just a row

[broken hawk]

yea but the representation of linear maps as matrix left-multiplication doesn’t work with them

(you could represnet it as right-multiplication ofc)

[quartz compass]

only reason I bring it up is because a 1xn matrix is a row

and that's what they said

no need to get into any discussion about this other stuff

[atomic flint]

Hey i was wondering how do i go about doing this question

http://prntscr.com/plirg7

[dusky epoch]

do you know what "closed ball of radius 1 centered at (0,0)" means

[atomic flint]

not exactly

[dusky epoch]

then go back to your notes and find the definition of that there

[atomic flint]

i mean i have this in my notes

it is just a generalized statement about "closed ball"

[dusky epoch]

yes that's what a closed ball is

you want $\overline{B}_1\big( (0,0)\big)$

[stoic python]

Ann:

[dusky epoch]

the set of all points $(x_1, x_2) \in \bR^2$ such that $d\big((x_1,x_2), (0,0)\big) \leq 1$

[stoic python]

Ann:

[dusky epoch]

this is just spelling out the definition of a closed ball

[atomic flint]

so in the case of my example, of d(x,y) = 2|x1-y1| + 4|x2-y2| how would i go about that as the thing which you gave me has nothing about the y terms

[dusky epoch]

$d\big((x_1,x_2), (0,0)\big) \leq 1$

[stoic python]

Ann:

[atomic flint]

so is it just a circle about the point (0,0) with a radius of 1?

[dusky epoch]

not a circle in the euclidean sense, no.

d here is your metric

c'mon don't overthink it

[lone quail]

How to solve this question?

(I had to wrote it on paper cos i translated it from italian)

[steady fiber]

is that a 2 in from of z

[lone quail]

Yes

[steady fiber]

you can't find a single solution for it

if that's what you're looking for

you'll get infinite solutions

[lone quail]

Yeah but i need the general solution

I dont know how to find it

[steady fiber]

set up a matrix for it

$$
[
\left[
\begin{array}{ccc|c}
1 & 1 & 2 & -1\
1 & -1 & 0 & 1\
\end{array}
\right]
]
$$

[stoic python]

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

[steady fiber]

there you go

like this matrix

then row reduce and you'll get your general solution

[lone quail]

Yes but this isnt the general solution

Its asking to find the subspace

The question

This doesnt give the subspace

[steady fiber]

solve it

then find the line of points

which is the general solution

[lone quail]

Yes

[steady fiber]

and that line of points is the subspace

a line is a subspace of R^3

if it goes through 0 that is at least

0 is just the 0 vector

or a plane of solutions is possible too, idk

[lone quail]

Unfortunately you dont get the right answer this way

I tried

Anyways this is the solution, try help me understand it please:

The subspace required is the plane containing r and passing through the origin. The generic plane containing r has equation given by:

x+y+2z-1 + λ(x-y-1) = 0

Then it says that the origin is in this equation if and only if lambda=-1, then substituting we have y+z=0, which is the equation of the subspace

@steady fiber

So guys what the reasoning behind this approach?

[winter acorn]

Could someone just quickly refresh me as why there is as many pivot columns in A as there are in A transpose?

[wintry steppe]

how could i formally prove that x-x0 is not a scalar multiple of 1?