linear-algebra

[dusky epoch]

i'm... constructing the matrix from its diagonalization

[opal osprey]

oh! okay

I like your solution

let me see if it yields the same matrix I have

hmm

I had this matrix before but then I "corrected it" to something wrong

I see

this was great help. thanks, @dusky epoch

there's also another approach with angles i'm not quite comfortable with yet

[graceful sluice]

Hey

I m working on projectors, matrix, and diagonalisation

For the exercice 2,i m looking for the matrix of projection on D1 in parallel to D2

I found a base : B={(1,1), (1,-2)}

[upbeat blaze]

Can anyone help me on how to approach this question? I am so lost.. Any help is greatly apprecaited

[gray dust]

A=PDP^-1. A^2=? A^3=? A^k=?

[dusky epoch]

@upbeat blaze ^

[upbeat blaze]

Thanks!

:D

[heady dagger]

Is it enough to define a linear map $T:\mathbb{F}^m \to \mathbb{F}^n$ given by $\vec{v} \mapsto \bold{A}\vec{v}$ and say that represents the transformation caused by the matrix $\bold{A}$, or is some kind of proof required to link the two ideas?

[stoic python]

Paddy:

[cursive narwhal]

It is certainly true that if $T: \mathbb{F}^n \to \mathbb{F}^m$ is a linear map, then there exists a unique matrix $A \in M(m \times n,\mathbb{F})$ with $T(x) = Ax$ for all $x \in \mathbb{F}^n$.

So, the transformation can be understood in terms of its matrix representation. This does require proof.

[stoic python]

Abhijeet Vats:

[heady dagger]

Thanks, I'll google around to find it

[cursive narwhal]

you're welcome

yea it should be a proof that can be found in most texts on linear algebra

[shy atlas]

excuse me wtf is this notation

$\mathbb{F}^{m \times n}$

[stoic python]

soαρ:

[cursive narwhal]

the set of all m by n matrices with entries from F

[shy atlas]

yeah shit notation

[cursive narwhal]

i mean, the F^{m x n} thing reminds me of cartesian products so i tend not to use that

M(m x n, F) is pretty clear

[gray mason]

Never seen $M(m \times n, \mathbb{F})$

[stoic python]

Stephen James:

[shy atlas]

me neither

looks like #multivariable-calculus

[wintry steppe]

:))

[cursive narwhal]

Never seen $M(m \times n, \mathbb{F})$
@gray mason It's used in klaus janich's textbook. I don't believe i've seen it elsewhere

urh maybe older books on linear algebra might have it? idk

[shy atlas]

well F^{m x n} is axler notation

so its superior by default

[wintry steppe]

1 + 1 = 3....

[shy atlas]

yes

[cursive narwhal]

ew axler

[shy atlas]

ew ur mom

[cursive narwhal]

rekt

[tame mural]

Mmm the difference between metric and norm is very subtle

[shy atlas]

oh i never thought about that

[tame mural]

What's an example of a very popular metric space that teachers might use to help students differentiate?

[shy atlas]

well you define norms on vector spaces but u can define metrics on any set

[tame mural]

So on vector spaces, the differences are a bit moot?

[shy atlas]

euclidean metric ig

[brittle juniper]

I don't see how students would confuse metric and norm in the first place

[shy atlas]

ye u need like an inner product space

and then norm is gonna be defined using the inner product

kinda similar ideas ig

[tame mural]

It seems almost more natural to go from a distance metric to a norm

d(x, 0) is the norm

[brittle juniper]

like a metric takes 2 points as input and a norm only 1, that's one giga fat difference

[tame mural]

Can you impose a metric on a vector space which doesn't end up serving as the norm as well?

[brittle juniper]

you can have a metric on a vector space that can't be turned into a norm

because of the homogeneity requirement in the definition of norm

[tame mural]

hmm I see

[hazy gull]

in this matrix can you explain why the span of s is not the zero vector? $$\begin{aligned}\begin{bmatrix} 1 & & 0 & & -1 \ 0 && 0 & & 0\ 0 && 0 && 0 \end{bmatrix}\ .\ s\begin{bmatrix} 0 \ 1 \ 0 \end{bmatrix}+t\begin{bmatrix} 1 \ 0 \ 1 \end{bmatrix}\end{aligned}$$

[shy atlas]

span of S ?

[hazy gull]

or x2 whatever you want to call it

[shy atlas]

no idea what you're talking about

second column ?

second row ?

[cursive narwhal]

what is S?

[shy atlas]

no clue

[hazy gull]

like why is it s[0,1,0] instead of s[0,0,0]

[stoic python]

gramcracker:

[cursive narwhal]

not really sure what you're going for here

[tame mural]

Maybe if you wrote out your problem more fully

[cursive narwhal]

the span of s?

s is a scalar from a field

it's a constant

[tame mural]

Most people talk about span for a set of vectors, or a matrix

[cursive narwhal]

When you're talking about span, you always need to refer to the span of a set of vectors

can you post the original question, gramcracker?

[hazy gull]

oh its not a question. im just paraphrasing something my professor wrote

[shy atlas]

When you're talking about span, you always need to refer to the span of a set of vectors
span$({ }) = { 0 }$

[stoic python]

soαρ:

[cursive narwhal]

show me what your professor wrote exactly

[shy atlas]

fite me irl

[cursive narwhal]

the empty set is still a set, soap

[shy atlas]

but its not a set of vectors is it

gottem

[tame mural]

gg no re

[cursive narwhal]

haizzzzzzz

[hazy gull]

ok so in terms of x1, x2, and x3: x1=x2 x2=0

[cursive narwhal]

send a picture of what your professor wrote

[tame mural]

It's too hard if you can't recover the full problem, IMO

[hazy gull]

this is not a problem to be solved

[tame mural]

or the full paragraph, description, etc.

[cursive narwhal]

oh its not a question. im just paraphrasing something my professor wrote
you're paraphrasing something your professor wrote

i want to see what it is so that i can tell you what's wrong with what you've said

[hazy gull]

im just wondering why you would write x2 as [0,1,0]

its a video

[cursive narwhal]

......................................

[hazy gull]

ok, its gunna take me a bit to find it

[cursive narwhal]

take your time

[hazy gull]

ok, so x1 has 1 and x3 has -1, but it still applies. I forgot the bottom row too, sorry

also not sure if you needed the aumented 0 vector, but I guess that would have made it more clear.

so here x1 = x3. but x2 = 0? I'd assume anything multiplied by x2 should be 0, so idk why its being represented as a 1.

[ruby locust]

can someone check #help-0

[hazy gull]

$$\begin{aligned}\begin{bmatrix} 1 & 0 & -1 & 0 \ 0 & 0 & 0 & 0 \ 0 & 0 & 0 & 0 \end{bmatrix}\ E_{0}=\left{ s\begin{bmatrix} 0 \ 1 \ 0 \end{bmatrix}+t\begin{bmatrix} 1 \ 0 \ 1 \end{bmatrix}\right} \end{aligned}$$

[stoic python]

gramcracker:

[cursive narwhal]

Basically, the solution set of this linear system is of the form:

$$S = {x \in \bR^3: Ax = b}$$

where $A$ is the coefficient matrix and $b$is some other vector. In other words:

$$x = (x_1,x_2,x_3)$$

where $x_1,x_2,x_3$ are the variables used in the linear system. Now, after doing row reduction, you've basically determined that $x_1 = x_3$ and the $x_2$ can be literally whatever you want it to be. So, define $x_1 = t$, where $t$ is a parameter. Define $x_2 = s$. Because $x_1 = x_3$, it follows that $x_3 = t$. Hence, your solution set is of the form:

$$S = {x \in \bR^3: x = (t,s,t) = t(1,0,1)+s(0,1,0)}$$

Now, you can clearly see that $t(1,0,1)+s(0,1,0)$ is just a linear combination of the two vectors $(1,0,1)$ and $(0,1,0)$. So, in fact, the set above is just the span of these two vectors.

[stoic python]

Abhijeet Vats:

[cursive narwhal]

ping me if you want me to clarify stuff. I might not answer immediately if i'm busy

[ruby locust]

@cursive narwhal

why equation 8 works

[cursive narwhal]

?? why me??

That's just matrix multiplication

[ruby locust]

you said ping me ( sorry if disturbed)

Why multiplying a row with another row's cofactor gives 0?

[cursive narwhal]

you didn't disturb me, i'm just a bit too tired to think so i'll let someone address your stuff

sorry

[ruby locust]

no problem

[cursive narwhal]

just leave it there and someone will assist you soon enough

[pine jasper]

@ruby locust The left of equation 8 gives the (2,1) element on the right of equation 7 which is 0.

[dry pulsar]

If K is a field, solving a homogeneous linear equations with coefficients in K, has exactly one solution

The statement is false, right?

[sonic osprey]

it can have one solution

[dry pulsar]

But not always has exactly one solution? O im wrong

[latent harbor]

Im trying to study for an exam tomorrow but there is no answer key, im struggling finding correct answers to the true and false section

can anyone help with any of these?

[radiant jasper]

can anyone help with any of these?
@latent harbor

TFFFF TFFTF

[latent harbor]

Thanks dude i guessed and definitely didnt get that

[pallid rampart]

1: is the equation of a plane, which is not a linear subspace of R^3 but I would say it’s a linear equation. So this one is just definition, it depends on how your professor defines linear equation

2: If detA=0 then AX=B might have no solution at all.

3: Consistent system only means there is a solution

4: The first nonzero entry in each row of reduced echelon form must be 1

5: For AB to be defined, the number of columns A has must be equal to the number of rows in B

6: A matrix is invertible (non singular) if and only if it can be row reduced to the identity matrix

7: Yes, trace is only defined for square matrices

8: Try it out, if (AB)^(-1)=A^{-1}*B^{-1} then the product of AB with A^{-1}*B^{-1} (in that order!!!) must be the identity

9: (I’m lazy so I’ll just say the answer, I don’t want to type the explanation) Yes it’s true

10: If the rows of a square matrix are linearly dependent, then the determinant is 0

@latent harbor

I gave you hints to each one

Ig if you want explanation

[radiant jasper]

Can someone help explain this proof?

why v is of form u+w?

[simple hornet]

hi im trying to prove something

I don't know how to search for the proof of something like this

i was watching the lectures for lin alg by gil strang

and they said that when you invert the elimination matrices you get a lower triangular matrix L who's entries are just the corresponding elimination matrices' elements

so im trying to prove that for a general case

and in order to do this i made a general case of an nxn matrix with dots between the entries and all that

but im struggling to mutiply the two, so if anyone could suggest a better method id greatly appreciate it

bcs im sure there's a more elegant way to do this

[simple hornet]

nevermind i solved it

i used the summation definition of multiplication

and said the only way you're not going to get zero is when you multiply by the diagonal elements so the only worthwhile thing to look at is not the summation but rather a_ij = e_il x e'_lj where l is a number between k and n in the summation of a matrix

And one of the e_il or e'_il are only 1s if e_il = e_ll or e'_il = e'll

can someone point any mistake in my reasoning?

sorry if it doesn't make much sense like this ooof

[dreamy iron]

Context: AXLER LADR, chapter 2.B

So I feel as if something spooky just happened.....I was hoping there’d be some reference to the axiom of choice.

Every vector space has a basis.....is a proof I’ve always heard required the axiom of choice.

(admittedly axler only speaks of finite dim vector spaces here, so I don’t know how that plays into it.)

[dusky epoch]

2.32 says every FINDIM vector space has a basis.

not all spaces are findim.

[dreamy iron]

Yes. So there’s no need to appeal to the axiom of choice?

(I’ve not encountered AOC yet, was really hoping this would be where that would happen, in a chapter about the basis of vector spaces).

[dusky epoch]

in the findim case yes you don't need choice

[dreamy iron]

le sad

[dreamy iron]

I’m trying to prove this:

This is my attempt.

The solution manual I’m looking at says I should saying about the linear combo in part 1 being equal to zero, and then those scalars being zero.

Which I DONT DO......

[dreamy iron]

This is the soln manual proof.

[cursive narwhal]

Which I DONT DO......
@dreamy iron yea so your first part just needs to be fixed so that you set that linear combination to 0 and deduce that lambda_1, lambda_2, lambda_3 and lambda_4 are all 0.

In part b), you’ve shown that the span of the list on the left is a subset of the span of the list on the right but not the other way. You can just do this using mutual containment. Your proof has the right idea.

[wintry steppe]

how are solved linear equations?

[cursive narwhal]

solved linear equations are doing fine, thanks for asking

[dreamy iron]

Lol. I thought i did the mutual containment bidirectionally cuz it’s a arbitrary vector \mu

aw man.

[wintry steppe]

:))

[cursive narwhal]

I mean, I guess that would count. Idk, I’d probably get marks deducted for not showing it explicitly in an exam

[cursive narwhal]

Go to #multivariable-calculus

[wintry steppe]

but is a liniear equation...

[cursive narwhal]

It’s a linear differential equation. Anyways, just go to the channel above like I told you to.

[wintry steppe]

:))))))))))

ok

[dreamy iron]

Okay, thanks @cursive narwhal i will be more direct.

[keen sparrow]

hello, could anybody provide some intuition on why this method to find the inverse matrix works?

[gray dust]

did you learn that a row op on A can be viewed as left multiplying A by the row op's corresponding elementary matrix?

[keen sparrow]

yes

So far I understand that the total effect of all the row operations is the same as multiplying each side of the augmented matrix by $$A^{-1}$$

[stoic python]

Racso:

[cursive narwhal]

That's not what Rokabe's talking about

[keen sparrow]

yes, I am aware... I am just adding the intuition I have so far

I also understand row operations have an equivalent elementary transformation matrix

[stoic python]

Abhijeet Vats:

[cursive narwhal]

@keen sparrow Is that clear or no?

[keen sparrow]

Yes, it explains why applying the elementary row operations results in the inverse $$A^{-1} = E_m \ldots E_2 E_1 I_n$$ thank you @cursive narwhal

[stoic python]

Racso:

[cursive narwhal]

you're welcome

[keen sparrow]

I remember elementary operation matrices from Strang lectures. I think I need a refresher on those 🙂

[cursive narwhal]

sure

[ruby locust]

Show by a picture how a rectangle with area x 1 y2 minus a rectangle with area x 2 y 1
produces the same area as our parallelogram.

Great solutions manual...

[wintry steppe]

n=?

[cursive narwhal]

??

What

[wintry steppe]

wrong channel, you might want to ask in, say, #prealg-and-algebra or #precalculus

unless there's a neat linear algebraic way to solve this

[limber sierra]

the linear algebraic way is "try a bunch of n till it works"

[wintry steppe]

lmao

[limber sierra]

for example, the first n i tried worked

[wintry steppe]

is 3 no?:))

[limber sierra]

yes.

[spice storm]

is this right?

[torn silo]

,rotate

[cursive narwhal]

Your closed under addition step could've been written better. So, you have:

$$a_2+a_1 = a_0$$

$$b_2+b_1 = b_0$$

Hence:

$$(a_2+b_2) + (a_1+b_1) = (a_2+a_1)+(b_2+b_1) = a_0 + b_0$$

Since the sum satisfies the condition required for it be a part of the set, the sum belongs to the set and the set is closed under addition.

[stoic python]

Abhijeet Vats:

[cursive narwhal]

Same thing with your closed under scalar multiplication step. You could've written it in a clearer way. As it stands, your argument mostly just consists of writing a bunch of equations without really explaning what's happening. You should write things down as you write your proof.

@spice storm

[spice storm]

Thank you but I have limit of room and prof hates scrap paper

[cursive narwhal]

Reduce your handwriting size

[spice storm]

Is the zero vector right?

[cursive narwhal]

Yes, the zero polynomial does belong to the set.

Like I said, reduce your handwriting size and write out the details. Surely your professor won't mind if you do that.

[spice storm]

Okay, So is just the scalar and mulptican needs improvement.

Alright, I will try

thank you @cursive narwhal

[cursive narwhal]

You're welcome

[spice storm]

before i right the scalar mulpitcaltion @cursive narwhal Is this right?

$a_2+a_1 = a_0$

$d_2+d_1 = d_0$

Then:
$C(a_2+a1) = C(a_0) C is d_0 $

[cursive narwhal]

[spice storm]

So is wrong

[cursive narwhal]

It's barely understandable. I don't really know what you're doing.

I see a_2+a_1 = a_0. That's somewhat recognizable. But everything else kind of just seems like random symbol salad

My dude, listen to me.

Please please please please please please please don't just write a proof only with symbols and without explanation for what you're doing and why. Please put in some sentences in there so it seems less like a symbol salad. If you're introducing new variables, define them explicitly and explain what they are.

[buoyant hemlock]

If A does not have independent columns, can A^T * A still be invertible? The "proof" I often see is that if A has independent columns, then A^T * A has to be invertible but they don't mention the other way around (if A^T * A is invertible, then A has to have to have independent columns).

[wintry steppe]

try using rank(A^TA) = rank(A)

[buoyant hemlock]

oh thats right, so it can't be invertible. thanks @wintry steppe

[wintry steppe]

(if A is a complex matrix then things go wrong and you have to take the conjugate transpose)

but thats being nitpicky

[buoyant hemlock]

Ok, I'll keep that in mind. Only dealt with real matrices thus far.

[wintry steppe]

a lot of theorems about real matrices translate into theorems about complex matrices when you replace transpose with conjugate transpose, this being one of them

just something good to be aware of

the reason being that inner products on complex spaces do not obey symmetry, but "conjugate symmetry"

[buoyant hemlock]

I'm guessing those are topics that are covered in intermediate/advanced linear algebra?

[wintry steppe]

i guess so

im not too sure what that consists of but i saw it in my second linear algebra class so i'll assume so

[gritty frigate]

What is happening on step 3 ?

[stoic python]

TTerra:

[gritty frigate]

Yep I undestand that

But after that ?

[stoic python]

TTerra:

[gritty frigate]

u-v = c

So c . c = cx*cx + cy+cy

[wintry steppe]

i assumed that "step 3" refers to the third equality sign

is that what you mean by "step 3"?

[gritty frigate]

Where the first arrow is lol

Sorry I should have adressed that

[wintry steppe]

oh, that just follows from the fact that any vector dot itself is its norm squared

[stoic python]

TTerra:

[gritty frigate]

Oh thank you !

But..

How do I know u-v x and y

[wintry steppe]

im not sure what you mean

that equality in my last message holds in general, i just assumed that c was a vector with two entries (going by your notation)

[gritty frigate]

|u-v| is a another vector that is the result of doing |u|-|v|

[wintry steppe]

the norm of a vector is not a vector

it is a scalar

[gritty frigate]

Fuck..

I can not believe it..

I confused normal with vector notation

Sorry lol

[wintry steppe]

haha, guess it was my bad for not putting vector arrows on my vectors

[gritty frigate]

I wasted your time and also mine hahahaha

Nononono not at all

It was my bad

Thank you

😄

I imagine what you mind was going trought when I was doing the quesitons

[wintry steppe]

dont worry about it too much, just think as everything without subscripts in my above messages as vectors (and those with subscripts as scalars - components of vectors)

im so used to not using vector arrows, so i can see how it could be confusing to some

[gritty frigate]

Now everything is clear bro 😄 I waste a lot of time trying to understand every proff

It kinda makes me happy..

I love proffs

This kinda things happen when you were used to work with other stuff lol

[simple hornet]

I have a bit of a problem

it's very basic but im not sure why this isn't working

So it's about row reducing

rref

And my answer key is giving one answer, symbolab another, and im getting another answer

this is the answer from the answer key

and this is what symbolab is giving

I'm getting something totally off

I got an identity matrix except with a 1 on the first row and 2nd column

[simple hornet]

okay i tried again and ive gotten the answer key's answer but still that means that there were more than one rrefs for one matrix which is weird

[quasi vale]

in your answer key, divide first row by 1-i

[simple hornet]

or row echelon* sorry

[quasi vale]

then you have 2/(1-i) as the first entry, and 1 as the other

[simple hornet]

yes i managed to get that fine

it's just that symbolab is giving one answer, this is giving another

if i try doing it any other way i don't end up at the original

the original being the answer key's answer

[quasi vale]

they're both the same, multiply divide 2/(1-i) with 1+i, and do the same with symbo labs answer, divide first row by -i

then you have (1-i)/-i, multiply divide with i, and you get the similar answer as the manuals

[wintry steppe]

symbolab is rubbish

[simple hornet]

one sec ill try

oh i see they really are equivalent

hm actually one other thing

when i do elimination i get something completely different, i get one entry in the first pivot and another in the second

while the rest are zeros

ill try to trace out my steps but idk the latex for matrices

$A = \begin{pmatrix} 1 - i & -i \ 2 & 1 - i \end{pmatrix}$

hopefully this works

crap

[wintry steppe]

$\begin{pmatrix}
1-i & -i\
2 & 1-i
\end{pmatrix}$

[stoic python]

val:

funnynumberatyahoo.com:

[simple hornet]

well ig u did it for me

bahah

okay so i begin by doing $R_2 \to R_2 - \frac{2}{1-i} * R_1$

[stoic python]

funnynumberatyahoo.com:

[simple hornet]

so i get

[wintry steppe]

$\to$

[stoic python]

val:

[simple hornet]

ohhh

[wintry steppe]

and don't use * in latex it just looks ew

[simple hornet]

oki

$\begin{pmatrix}
1 - i & -i \
2 - (1-i) \frac{2}{1-i} & 1 - i - (-i) \frac{2}{1-i}
\end{pmatrix}$

[stoic python]

funnynumberatyahoo.com:

[simple hornet]

Then you get

$\begin{pmatrix}
1 - i & i \
0 & -1
\end{pmatrix}$

[stoic python]

funnynumberatyahoo.com:

[simple hornet]

Then you multiply row 2 by i and then add to the top row

and get

In the same step i can also multiply -1 times the last one right

so ill do those two in one shot

$\begin{pmatrix}
1 - i & 0 \
0 & 1 \
\end{pmatrix}$

[stoic python]

funnynumberatyahoo.com:

[simple hornet]

and that's what im getting

wait a second wait a second i see my mistake

1 - i - (-i) (2/1-i) == -1 it's 0

lmAo

[wintry steppe]

how could one justify why a matrix is an orthogonal projection onto a line?

for ex (1 1 1
1 1 1
1 1 1)
why is that an orthogonal projection onto L

[hollow fern]

is that an orthogonal projection?

that will map everything in R^3 to a line, but i dont think that that is an orthogonal projection

i think they are supposed to have the property that P^2 = P, where P is your orthogonal projection

someone correct me if am wrong please

[wintry steppe]

yeah that is an orthogonal projection, i just don't know how to justify it

wdym by P^2=P @hollow fern

[hollow fern]

okay so if you take your matrix

multiply it with itself

if you get the same matrix back again, you have an orthogonal projection

if you do not, then it isnt an orthogonal projection

i think doing that for yours yields a matrix of all 3's

[wintry steppe]

wait so you're saying (1 1 1 (1 1 1
1 1 1 *
1 1 1
1 1 1 )
1 1 1)

should equal the 111 thingyu

[hollow fern]

yes

[wintry steppe]

ok lol idk how to use latex

ok that's really strange

[hollow fern]

haha no thats fair

yeah i dont think that is an orthogonal projection tho

[wintry steppe]

sighs

ok well it is so um

what does someone else think

what is your definition of orthogonal projection

the matrix is not equal to it's square so i would not call it an orthogonal projection, since it fails to even be a projection

i call a linear operator a projection if it is equal to it's square. that is obviously not the case for left-multiplication by your matrix

orthogonal projection if it's a projection whose range and kernel are orthogonal complements

[wintry steppe]

ok so this is the original problem maybe this will help

so B is the orthogonal projection and E is the reflection i just don't know how to explain it

okay, you shouldnt have left out the 1/3 on B

in that case, then B is indeed a projection

oh oops

i didn't know that would make a difference

well now you have a nice example where it does

so why? does it have something to do with parawiseparallelness

what

idk either

parawiseparallel

idk someone mentioned it but i don't see how that's a justification

what do you mean? i have never seen that term before

im not sure its even an english word

oh ok

so how would you go abt this

well, you can check its a projection (i did it in my head but it might be good to write out the calculation)

then you verify your definition of an orthogonal projection

you should calculate the kernel and image of B and see if they are orthogonal

hm

i see, thank you!

[normal quiver]

how do i prove "If u and v are objects in V, then u+v is also in V"

[cursive narwhal]

You're just showing closure under addition. So, just prove that u+v satisfies whatever condition that defines V

@normal quiver

Do you have a particular question you're trying to answer? I could be more specific depending on the problem.

[normal quiver]

the example just says prove whether it holds or not

[cursive narwhal]

Show me the example. Like, the original question.

[normal quiver]

@cursive narwhal

[cursive narwhal]

Uh M_22 is the notation for the set of 2 x 2 matrices with real entries, yes?

[normal quiver]

yes

[cursive narwhal]

Yea okay, so if you take any v and u in V, these are just 2 x 2 matrices with real entries. When you add them in the way that addition is defined for this set, you produce a 2 x 2 matrix with real entries as well and that has to be in the set. So, it follows that u+v is in V.

[normal quiver]

so for proving any problem for ax1, does the result just have to be a 2x2? when would this ax1 not be true?

is that when the resulting set is a 0 matrix?

[cursive narwhal]

the 0 matrix isn't a set

and when you're verifying closure under addition, it can be for any kind of set that contains any other mathematical object for which you can define addition in a meaningful way

for instance, V could've also been the set of continuous functions from [a,b] to R.

In that situation, it wouldn't make sense to ask "does the result just have to be a 2x2

[normal quiver]

@cursive narwhal ok im sortof confused, i thought you said "produces a 2x2 matrix with real entries as well and that has to be in the set"

[cursive narwhal]

We're talking about the specific example for $V$ in the book. So, in this case, $V$ is just $M(2 \times 2,\bR)$ (this is just my own preferred notation for $M_{22}$.

Now, we want to show that for any $u,v \in M(2 \times 2,\bR)$, $u+v \in M(2 \times 2,\bR)$. All that amounts to showing is two things:

\

1. $u+v$ is a $2 \times 2$ matrix

\

2. $u+v$ has real entries

\

With the way that addition is defined on the set, both conditions above are fulfilled so $u+v$ does belong $M(2 \times 2,\bR)$.

[stoic python]

Abhijeet Vats:

[normal quiver]

do i have to plug in numbers to show 1 & 2?

[cursive narwhal]

Nope. Since $u,v \in M(2 \times 2,\bR)$, they are both $2 \times 2$ matrices that have real entries. When they're added together in the fashion described in the book, the result is a $2 \times 2$ matrix with real entries so the sum is an element of $M(2 \times 2,\bR)$.

[stoic python]

Abhijeet Vats:

[normal quiver]

ooh okay

what about ax2, where $u+v = v+u$

[stoic python]

Blackwaterunsc:

[crystal oracle]

I've just figured out that I can approximate any integrable function on a closed interval [a, b] with a polynomial via orthogonal projection with the inner product ⟨p,q⟩=∫_a^b p(x)q(x)dx. And to understand this, all that is needed is the first half of a linear algebra textbook. So, why haven't I heard of this before? Do people actually use this for any problems?

[cursive narwhal]

@normal quiver This is something you can just check by evaluating u+v and v+u with the definition of addition on the given set.

You know, evaluate them and see if they're equal to each other or not

[normal quiver]

do i just flip the u+v inside the result matrix and check if theyre equal?
so for ex u_1 + v_1 -3 = v_1 + u_1 -3 but for all the elements in that matrix?

[cursive narwhal]

Yeap and notice that if you do that for all entries, you literally get the definition of v+u on the right-hand side.

so u+v = v+u

[normal quiver]

i see :D
Last question. The book says there are 10 axioms. Is there a few cases where lets say, if i wanted to prove ax7, i would also have to prove the previous 3 or 4?

ok better put is like if i had ax7, but lets say i find that ax1 does not hold. would that imply that ax7 also does not hold

[cursive narwhal]

You should just go down the list and prove each axiom one by one. But honestly, if you did want to jump axioms, in some cases I guess it could be useful? I mean, idk, no specific examples are coming to mind where it would be extremely useful to do it like that

I typically just check one by one

[normal quiver]

cause like ax7 is $k(u+v) = ku + kv$

[stoic python]

Blackwaterunsc:

[cursive narwhal]

ok better put is like if i had ax7, but lets say i find that ax1 does not hold. would that imply that ax7 also does not hold
Sure. So, for instance, if you verify that the set has no identity element under vector addition, then you can't talk about inverse elements with respect to addition. Inverses require the existence of an identity with respect to addition.

[normal quiver]

hmm got it

thank you so much!

i love u

:D

[cursive narwhal]

I've just figured out that I can approximate any integrable function on a closed interval [a, b] with a polynomial via orthogonal projection with the inner product ⟨p,q⟩=∫_a^b p(x)q(x)dx. And to understand this, all that is needed is the first half of a linear algebra textbook. So, why haven't I heard of this before? Do people actually use this for any problems?

@crystal oracle reposted your question for you since it got buried, sorry about that 😦

[humble shale]

So for B, I found that it doesn't span R^2, but I don't understand what it wants me to do as far as describing the subspace it does span? can anyone help me?

[wintry steppe]

do you know what it should look like?

[humble shale]

A line?

[wintry steppe]

more specifically, a line through the origin

but yeah

[humble shale]

right

So that's all it's asking for? lol

[wintry steppe]

how many points does it take to determine a line?

[humble shale]

2

[wintry steppe]

mmhm

and you know the origin is gonna be one of them, cause the zero linear combination is in the span

can you give me another point?

[humble shale]

besides the ones in S?

[wintry steppe]

any point in the span of S

[humble shale]

Ah ok

[wintry steppe]

your favourite one

[humble shale]

do I need to solve it for the equation of the line? or just state that its a spans a line through the origin?

[wintry steppe]

just "a line through the origin" isn't very specific, since there are a lot of those

[humble shale]

true

[wintry steppe]

gimme your favourite nonzero point in span(S)

then try finding the equation of the line in R^2 through (0,0) and that point

that, if you did it correctly, is the description of your line

(this isn't the best way to do this kind of thing btw, but it's good to work from intuition sometimes)

(and your intuition for this seems to be spot on, so thats good )

[humble shale]

so it spans the line y = -1/2x 🤔

[wintry steppe]

sounds about right!

for reference, the "proper" way to do this would probably be to use some kind of row reduction on the matrix whose columns are the vectors in S

but this is R^2 and things in R^2 are familiar and nice

[humble shale]

thanks!

[zinc tapir]

Does every linear operator have eignvalues if F=C?

I was working a matrix and it seems i only found eigenvalues if its R

also I worked a different matrix and only found complex eigenvalues

[wintry steppe]

look into the fundamental theorem of algebra

[zinc tapir]

ya im familiar

but i don't get what my prof means by this statement

so i took the first matrix, tried to find eigenvalues for F=R

and couldn't find any roots that are real

[wintry steppe]

ok, so it has no eigenvalues over R

[zinc tapir]

say eigenvalue= +-2i

then i took the other matrix

and i found eigenvalues 2, -1, -1 (algebraic multiplicity 2)

those are obviously real

so i wrote that the matrix doesn't have eigenvalues over C

is that right?

[wintry steppe]

you should say that its eigenvalues are all real

since 2, -1, -1 are still complex numbers

just with imaginary part 0

[zinc tapir]

could i say

Let F=C then B has the same eigenvalues/eigenvectors with roots of the form a+-bi and b=0

[zinc lava]

I dunno if this is new, but I thought I'd share ... how does one reconstruct a polynomial of Nth degree with access to f(0), f(1), f(2), f(3) ... as many as you want?

Furthermore can one reconstruct a two (or more) variable polynomial given any and all f(x, y) for any non-negative, integer choice of x and y?

A cubic example:

Suppose you know N = 3 or less.
f(x) = ax^3 + bx^2 + cx + d

f(0) = d
f(1) = a + b + c + d
f(2) = 8a + 4b + 2c + d
f(3) = 27a + 9b + 3c + d

Finding a, b, c, and d are easy to calculate using matrices and linear algebra.

A two-variable example:
f(x, y) = ax^2 + bxy + cy^2 + dx + ey + f

f(0,0), f(0, 1), f(1, 0), ... I think the idea is understood.

[half ice]

Same idea exactly. Linear algebra will solve the coefficients

Just, you'll need 6 equations

[ocean sequoia]

what does this notation mean i don't remember it being introduced

[wintry steppe]

@ocean sequoia its the vector space of quadratic polynomials

i.e elements of the vector space take the form ax^2+bx+c

Where a b and c are real numbers

[ocean sequoia]

so it includes first degree polynomialsthen as well?

[wintry steppe]

Yes

[ocean sequoia]

thanks

[wintry steppe]

in this context a linear function is a quadratic with a = 0 :P

[umbral smelt]

Yeah first degree polinomial is in second degree polinomial because of 0x^2+bx+c

[limber sierra]

in this context a linear function is a quadratic with a = 0

be a bit careful here

these arent linear in a linear algebra sense (unless c = 0), just a high school algebra sense

the fact that this word is reused and means slightly different things is very annoying.

[ocean sequoia]

yea for linear algebra the c must be 0

[zinc tapir]

What is the difference between the minimal polynomial of T and the characteristic polynomial of T

[limber sierra]

multiplicities of the roots.

[sonic osprey]

Do you know both the definitions of the two things?

[wintry steppe]

@zinc tapir say you have char poly (x-1)(x^2+1)^2 = 0

= (x-1)(x+i)^2 (x-i)^2 = 0

Then the minimal polynomial for this would be (x-1)(x+i)(x-i)=0

Thats just one example but i hope that illustrates it

[zinc tapir]

oh

Thanks that helps alot

I was pondering what namington said and this clears it

[ocean sequoia]

Im a bit confused as to what this means and i cant seem to find a straight forward answer on google... I end up seeing tensor or Cartesian Product

is it simply a way of denoting that if we are trying to do with two vector spaces that the elements in the set must belong to both?

So here is an example in the book but i dont understand why this is a basis wouldnt (x,(1,0)),(x^2,(0,1))) span the space?

[sonic osprey]

I think you should understand what the Cartesian product of sets is first

[ocean sequoia]

ok thanks I wasnt sure what else to look at

ive def heard that and read about it before i just cant remeber it ill go look thanks

[spice storm]

How do I get integers instead of Decimals when I compute RREF on MatLab?

It keeps showing decimals instead and not integers

[strange obsidian]

#computing-software

[ocean sequoia]

ok so a cartesian product is simply a way of denoting the combinations of sets

[cursive narwhal]

"denoting the combinations of sets" What do you mean by this?

[ocean sequoia]

a way of showing which x values give which y values

and all the combinations of those from your sets

[cursive narwhal]

If you have A and B as sets, then A x B is just the set of ordered pairs (x,y), where x is an element of A and y is an element of B.

When you say "which x values give which y values", it kind of implies that the cartesian product represents some sort of function? I would say that you should stick to the formal definition or come up with a better analogy.

[ocean sequoia]

i should stick to formal definitions

[cursive narwhal]

and all the combinations of those from your sets
This isn't all too bad tbh

[ocean sequoia]

ok so if we have set A as (1,2) and set B as (4,5) A X B is (1,4), (2,4) , (1,5) , (2,5)

[cursive narwhal]

No, see, that is a problem

A = {1,2} and B = {4,5}. We make a distinction between the notation used for sets and that used for ordered pairs

When you say, "we have a set A as (1,2)", you're talking about something different from the set {1,2}, which is what you were originally referring to.

[ocean sequoia]

oh sorry yea see thats not me not being precise

i mean

[cursive narwhal]

But you are correct in that {1,2} x {4,5} = {(1,4),(1,5),(2,4),(2,5)}

where we enclose the ordered pair in braces to signify that it's a set of ordered pairs

[ocean sequoia]

yea i shouldve been more careful

[stoic python]

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

[cursive narwhal]

Are you trying to say that ${1,2} \in A$?

[stoic python]

Abhijeet Vats:

[ocean sequoia]

yes

[cursive narwhal]

That is wrong

[ocean sequoia]

how did you get the brackets to show lol

oh

ok

[cursive narwhal]

The elements $1,2 \in A$ but the set ${1,2} \notin A$. There's a difference between talking about the elements 1 and 2 and the set containing the elements 1 and 2

[stoic python]

Abhijeet Vats:

[cursive narwhal]

how did you get the brackets to show lol
follow the code that i used

[ocean sequoia]

hm ok I mean the whole set is just 1,2

[cursive narwhal]

Yes, it just consists of the elements 1 and 2

[ocean sequoia]

so i guess it should be ${ 1,2 } = A$

i cant get the brackets to appear what in the world

[cursive narwhal]

{1,2} = A, you mean? Yes, that's how you defined the set A.

[stoic python]

brzig:

[cursive narwhal]

${BLAH}$

[stoic python]

Abhijeet Vats:

[ocean sequoia]

${BLAH}$

what

i copied it lmao

[umbral smelt]

add \ before the brackets

[cursive narwhal]

^Yes

[umbral smelt]

${a,b}$

[stoic python]

Wilston Lynx:

brzig:

[ocean sequoia]

cool thanks

didnt know i had to use the escape character

[cursive narwhal]

Brzig, you should definitely learn some set theory before learning linear algebra. Otherwise, most of it is going to be very messy.

[ocean sequoia]

yea honestly Ive learned a bit but this is the first time its hitched me up

hm ok

[cursive narwhal]

It'll be a bit more problematic later. You deal with sets almost all the time and when you start talking about functions in LA, things really can get confusing if you're not fluent with sets

Maybe try reading through a chunk of Naive Set Theory by Paul Halmos and then come back to LA after that? See if it works for you. It's a very short book and you don't need to read all of it to start LA but certainly everything up to a discussion of functions

[ocean sequoia]

damn i really wanted to get through the eigenvector stuff in LADR

ill look at the book

do you think i would be good getting through that without it?

I was actually excited about that stuff lol

[cursive narwhal]

I've not really read a significant chunk of LADR so I can't say what you will definitively require. It's one of the recommended texts for Linear Algebra I at my uni and that's a proof course. You probably do need at least a basic understanding of sets to get through it.

I mean, you can try getting through that stuff if you want. I think you should read what you enjoy. I can't guarantee you'll understand a whole bunch of it.

[ocean sequoia]

Ok ill try to get through this part and if i start getting hung up on the set stuff ill come back to it

i went through book of proof a bit ago

[spice storm]

Does anyone know how to compute this on MatLab or maple? Been trying for an hour

[foggy gazelle]

hi does anyone understand something about Linear systems of equations with parameters

[torn silo]

its very likely that someone does 🤔

[foggy gazelle]

can I just type out the equation here?

Its for a multiple choice question
-3x + 5y -2z = -12
2x - 4y +4z = 18
11x -19x+a*z = 54

I just forgot where to start

[umbral smelt]

Try changing the system into an augmented matrix first, then try applying row operations

[wintry steppe]

Well just by inspection, if u1 = (1,2,1,4) and u2 = (-1,2,-5,0) then u3 = (2,0,6,4) = u1-u2

And since its a 4x4 matrix, there must be at least one nonzero vector in R^4 that maps to the Nullspace of A

[sullen pollen]

what is a linear variety?

[stable urchin]

Question

let's take eigenvalue l

now suppose l has a geometric multiplicity of 1 and an algebraic multiplicity of 3

given this

and also given that there is one single eigenvector v associated with l

is the following true?

$$(A-lI)\vec{u_1}=\vec{v},(A-lI)\vec{u_2}=\vec{u_1}$$

[stoic python]

Bivariate:

[stable urchin]

u1 and u2 are eigenvectors

[stable urchin]

<@&286206848099549185>

[wintry steppe]

@stable urchin Yes, geometric multiplicity is the number of basis vectors needed to span the eigenspace

[stable urchin]

okie

is the latex stuff correct

i is identity matrix

l is eigenvalue

[wintry steppe]

If u1 and u2 are linearly independent eigenvectors, then ur eigenspace is span{u1, u2}

[stable urchin]

v is the eigenvector associated with l

[wintry steppe]

So (A-λI) * v = c1 * u1 + c2 * u2

[stable urchin]

ohhhh

i see

[wintry steppe]

Thats only if

U1 and u2 are lin indep

If not then ur eigenspace is span{u1} or span{u2} (doesnt matter which one :P)

[stable urchin]

sorry i havent done much linalg but what does span mean

[wintry steppe]

Span is the set of vectors that can be formed by some basis

[stable urchin]

ohh

ok

[wintry steppe]

Basis being what are the "units" of tje vector space

[stable urchin]

yes

[wintry steppe]

So span{(1,0), (0,1)} = R^2

For ex

[stable urchin]

am i able to solve for both u1 and u2 given that I know A l and v

[wintry steppe]

Yes u1 and u2 are the basis of ur eigenspace

I.e the basis of the nullspace of (A-λI)

[stable urchin]

so i can just plug it into So (A-λI) * v = c1 * u1 + c2 * u2

[wintry steppe]

Well not rlly bc c1 and c2 are arbitrary

And we dont even know whether u1 and u2 are lin independent

[stable urchin]

they are

its a given for the problem im doing

[wintry steppe]

Ahhhhhh

My bad

[stable urchin]

its ok

[wintry steppe]

Then yes

[stable urchin]

Wait is the latex i put above also correct or no

[wintry steppe]

We need to solve (A-λI) * v = 0

[stable urchin]

yeah i did that

[wintry steppe]

And u have v?

[stable urchin]

i have v, l and A

[wintry steppe]

Is v in R^2?

[stable urchin]

v is a vector with 3 dimensions

A is 3x3 matrix

[wintry steppe]

What is v = ?

because u can rewrite it as the sum of 3 lin independent vectors

And those span ur eigenspace

[stable urchin]

wait what

Oh yeah by the way this is from this problem

[wintry steppe]

If v = (5,6,7) then v = 5 * (1,0,0) + 6 * (0,1,0) + 7 * (0,0,1)

What im asking is

[wintry steppe]

What did u ger for the coordinates of v?

[stable urchin]

5 5 2

[wintry steppe]

Wait wasn't this one 5 5 1?

[stable urchin]

Ill show the work maybe i did sth wrong

[wintry steppe]

So if v = (v1,v2,v3) and 5v1 - v3 = 0 and 5v2 - v3 = 0

[stable urchin]

I gotta go have dinner I’ll be like 5 minutes

[wintry steppe]

👌

-5v1 + 5v2 = 0 => 5(v2-v1) = 0

=> v1 = v2

-2v1 + 5v3 = 0 => v1 = 5/2 * v3
-2v2 + 5v3 = 0 => v2 = 5/2 * v3

So if v1 = v2 = 5/2, then 1/2 = 1/2 * v3 => v3 = 1

So we have span{v3(5/2, 5/2, 1)} which is equivalent to span{v3(5,5,2)}

I think ur right assuming the work before that point is xorrect

[stable urchin]

alr

good

so now i need to find the other two lin indep eigenvalues u1 and u2

i found on a math stack exchange thread that you can do this

This was an answer

I was unsure whether this is correct hence me asking about it here

Anyhow

can I use the latex I put above to solve this?

[stable urchin]

@wintry steppe

sorry for ping

[wintry steppe]

Can u send the thread

[stable urchin]

sure

https://math.stackexchange.com/questions/289353/system-of-differential-equations-with-triple-eigenvalue?newreg=e3459f01d49944d89c8616c9a43f7942

[wintry steppe]

@stable urchin this doesnt apply to ur question

[stable urchin]

oh huh

[wintry steppe]

This question is in the context of DEs

[stable urchin]

yeah my question is in context of DEs

hence the prime

[wintry steppe]

Ohh i didnt realize that

[stable urchin]

'

All good

[wintry steppe]

Thought it meant the points after transformation or something

My bad

So yea that should work then

[stable urchin]

$$(A-lI)\vec{u_1}=\vec{v},(A-lI)\vec{u_2}=\vec{u_1}$$

[stoic python]

Bivariate:

[stable urchin]

does this work too

That's the formula I used for like 2x2 matrices

the only diff is that im multiplying the identity matrix term by the unknown vector instead of the known

[wintry steppe]

Why the second equation

[stable urchin]

second one for u2 first one for u1

[wintry steppe]

Yes but why does that work

The first one makes sense as v is the associated eigenvector of λ

[stable urchin]

i mean its the same as from the thread

[wintry steppe]

I think that should work then

[stable urchin]

Alr

thanks!

[wintry steppe]

Np :)

[stable urchin]

Here's what I got

[stoic python]

Bivariate:

[stable urchin]

assuming v to be any non zero eigenvector makes no sense

in context of the thread

like if my eigenvalue has an associated eigenvector v why make a new one

[wintry steppe]

Its because of geometric multiplicity which can be atmost the algebraic multiplicity

I.e geo mult can be 1,2 or 3

[stable urchin]

well in this case its 1

since it gives me a non trivial eigenvector

but it asks to make one up

then find 2 more

kinda wack

[wintry steppe]

Yah man so wack u already know g

[stable urchin]

huh

[wintry steppe]

What

[stable urchin]

ok

[tiny grove]

Can someone help me find P using matlab commands?

and the basis for the eigenspace?

i tried using the given command, but i think its not working for osme reasoin

¯_(ツ)_/¯

[forest trail]

If I have two vectors
Vector A is parallel to the x-axis
Vector B is perpendicular to Vector A
While Vector B be perpendicular to x-axis?

[torn silo]

why don't you draw an example?

maybe that'll help

[wintry steppe]

@forest trail I'm not sure i understand the question

If vector A lies on the x axis and vector B is perpindicular to vector A, then vector B is perpindicular to the x axis

[forest trail]

Yes, thanks! That's what I was looking for.

[opal osprey]

does anyone know of any good book on matrix-based geometry? (affine transformations, orthogonal transformations, etc.)

basically dealing with isometries, etc.

I'm having trouble with these particular two definitions

(sorry about the portuguese, I'll translate:)

the first one says that if f is a direct transformation, its general expression follows that matrix, and it will be a rotation with centre at the origin and angle theta;
the second one states mostly the same, except for inverse orientation, and also adds that the f is now a reflection on the line l sub 0, with an angle of theta/2 with the xx axis

[wintry steppe]

Well

[opal osprey]

so they're both coherent with one another, right?

wiki and the book's definition

[wintry steppe]

yes

[opal osprey]

a bit confusing

[wintry steppe]

A reflection by angle θ is equivalent to rotating 2θ

To make this a little more intuitive, lets say you shoot a light beam 90 degrees at a wall

The light beam will do have an angle 180 degrees relative to the wall after reflection

[opal osprey]

I see

[wintry steppe]

As for rotation

[opal osprey]

my biggest problem is the two definitions depending on different angles

[wintry steppe]

Why is that an issue

[opal osprey]

well

I don't really know; it seems a little messy

I'm asking this because of a particular exercise

[wintry steppe]

Can u send the exercise

[opal osprey]

of course - 1 sec

I'm being asked to determine this kind of isometry

I know it is inverse, as det of the matrix < 1

so it's a reflection, either a normal one as we just saw or a sliding one

but then the teacher assumed cos (2theta) = 3/5 and sin(2theta) = 4/5

whereas, using his definition, we know that would correspond to cos(theta) = 3/5 where the true angle is theta/2

[wintry steppe]

Thats for a rotation

Not reflection

[opal osprey]

what is?

[wintry steppe]

A reflection is 2 * theta because it has to rotate θ backwards once to get to 0 and then θ backwards again to -θ

And you said this is reflection since det < 0

[opal osprey]

yeah - do you agree?

[wintry steppe]

Yes

[opal osprey]

alright --

[wintry steppe]

So ur teacher using 2θ is justified

[opal osprey]

I don't really get that step

[wintry steppe]

Ok ill draw it out

[opal osprey]

oof. thanks

but isn't there a jump somewhere if you assume it is cos 2θ?

[opal osprey]

I see

but I don't think my confusion arises from that

[wintry steppe]

Then what is it that u find confusing

[opal osprey]

given his own definition, the reflection matrix is defined as a function of cos(θ)

wrt to an angle θ /2

[wintry steppe]

Thats not what the definition says

[opal osprey]

(the other one is wikipedia's)

the teacher uses the notes above

[wintry steppe]

Ahh

There's no difference then

θ/2 ln Wikipedia is what ur teacher refers to as θ

θ on Wikipedia is what ur teacher refers to as 2θ

Personally i like ur teachers labelling better

@opal osprey

[opal osprey]

so here's where it gets confusing

because he assumes, by matrix inspection, that on that particular exercise, cos(2θ) = 3/5

wouldn't it be a better fit if cos(θ) = 3/5 where the angle is θ / 2? it's a minor issue, but i don't get why the unnecessary change in notation

thanks for the help, though!

it's just this is still very messy

[tame mural]

For a matrix A, how do you easily note that you want the basis vectors from A?

I know the hat operator is used to mention the unit vectors

[dusky epoch]

wym by "the basis vectors from A"

[tame mural]

Like the columns

[dusky epoch]

wait what are you doing exactly

[tame mural]

I'm trying to find notation while writing my notes on matrices

[dusky epoch]

yea ok but

what's the notation meant to be for

[tame mural]

mostly linear algebra intro notes

[dusky epoch]

no i mean

[cursive narwhal]

what is the context

of the question

[dusky epoch]

i'm not sure what you want your notation to denote

bc i feel like you're misusing some terminology here

[tame mural]

Like if I want to say matrix application is the basis vectors of the matrix scaled by the values in the vector

[dusky epoch]

what do you mean by "the basis vectors of the matrix"????

[tame mural]

the columns, is that the wrong word?

[dusky epoch]

the cols

ok

[tame mural]

I'm assuming that the matrix represents a spanning set

[dusky epoch]

the matrix doesn't represent anything

you're overthinking it

a matrix is a matrix

[tame mural]

How do you efficiently note the columns of the matrix?

[dusky epoch]

mmh

idk i'd just put a note like "$A_i$ denotes the $i$'th col of matrix $A$"

[stoic python]

Ann:

[tame mural]

But what if you were already using that for rows ~_~

[dusky epoch]

oops

$A_{i*}$ for $i$'th row and $A_{*j}$ for $j$'th col

[stoic python]

Ann:

[tame mural]

ah I see, that works

thanks!

[tiny grove]

Can someone help me with matlab?

[dusky epoch]

dunno, but until you post your question that's gonna be a definite no

[tiny grove]

oof i posted far above but ill post again

i have trouble finding the basis for eigenspace A

and finding matrix p

[dusky epoch]

[X, L] = eig(A) will put the eigenvalues in L along the diagonal and the eigenvectors in X so that A == X * L * X^-1 iirc

[tiny grove]

oh i se

is there a command to calculate the maximum erorr?

[dusky epoch]

uhh

look in the matlab docs ig??

[spice storm]

Is this correct?

[spice storm]

<@&286206848099549185> also need help with 22

I have two answers but I’m not sure which is right

[cursive narwhal]

@spice storm I haven't really looked through your calculations because it's 5am and I'm not in the mood to look at matrices. But yea, the set seems to be linearly independent.

For 22, the idea is essentially the same. What you need to do is to show two things:

a) The set of polynomials span P_2. This shouldn't be too hard.

b) The given set of polynomials is linearly independent.

[spice storm]

Okay, thank you. @cursive narwhal

[stable urchin]

Row operations

From the first matrix the third row is replaced with -3 times the first row + 2 times the third row

Did they do this right?

Shouldn’t it be 19 for A_33

Darn

Ok thanks

[stable urchin]

Question: what are basic and free variables of a system?

[ripe hamlet]

A variable is a basic variable if it corresponds to a pivot column. Otherwise, the variable is known as a free variable. In order to determine which variables are basic and which are free, it is necessary to row reduce the augmented matrix to echelon form.

[stable urchin]

Nice

Thank

[wintry steppe]

@stable urchin another way of thinking about free variables is when you can parametrize your solution

If your solution set is spanned by a line, then there is only one free variable

If its spanned by a plane, then there are two

Etc

[stable urchin]

Ohhh

I see

Thank

[stable urchin]

Another question

In context of a system, is a coefficient matrix the same as an augmented matrix?

Does the coefficient matrix include the numbers after the line or not

Actually wait

Don’t answer that

I’m not smart

They are very different things

[slow scroll]

ye

Ill answer it anyway though: if you have Ax = b, the coefficient matrix is always A and the augmented matrix is always [A | b]

[stable urchin]

Mhm

Thanks

[tiny grove]

do dimensions of colmn space and rowspace equal to the rank of a given matrix A?

[slow scroll]

yes. dim Col = rank by definition. dim row = rank by a theorem

[tiny grove]

and uh... when a question asks you to find basis of each eigenspace of A, is it asking for the eigenvectors for each eigenvalue?

[slow scroll]

yea, sort of. It is asking for all of the linearly independent eigenvectors corresponding to each eigenvalue. When you have all of them, you get a basis for the eigenspace associated with that eigenvalue.

[tiny grove]

I see, thanks a lot

[slow scroll]

npnp

[stable urchin]

Shouldn’t the arrow under the 27 bolts be going the other direction

Idk how circuits work

What does the counterclockwise thingy in the middle mean

[limber sierra]

this uh, doesnt look like linear algebra

[stable urchin]

It’s in my linalg textbook

It’s an application of them

[limber sierra]

well okay

so the thing is, early electricians were wrong about how electricity worked, and we pay for that to this day

to clarify: when you see an arrow drawn on a circuit diagram (like the arrow under 27 volts), this is the direction a hypothetical "positive charge" moves

the problem is, the thing actually "moving" in a circuit (in a Newton's cradle style) is an electron - a negative charge

so the "actual direction of motion" (the current), in this case, is counterclockwise, as the arrow around I_1 indicates

in practice, this doesnt actually matter - the choice of sign is arbitrary (hence why early electricians "got it wrong" and wrote arrows indicating the direction of positive charge, not negative charge - it doesn't matter so they couldn't tell the difference)

at least, that's what i'm assuming the circle around the I_1 means

it's not standardized notation, at least from what i've seen (which is admittedly very little)

does your textbook not define it?

[stable urchin]

It doesn’t

[limber sierra]

its probably fairly safe to disregard then

in any case, the arrow under "27 volts" is certainly in the correct direction

[stable urchin]

Alright

Thanks

[limber sierra]

not that direction really matters here

again it's all equivalent

[stable urchin]

Okay

[tame mural]

I previously thought of vectors as lists of scalars, as they are often written F^n, but then I discover that vector spaces can use matrices as their vectors.

How does that work?

Don't vectors require their elements to be scalars?

[dusky epoch]

no

a vector need not have any "elements"

the whole point of messing around with the vector space axioms is that you want to abstract away from F^n and develop a theory that works for spaces that aren't necessarily F^n

[gray dust]

your phrasing is quite odd. at first you probably work with F^n, set of all n tuples with entries in F, and when F is R you get your classic geometric view of pointy arrows in a box. but then you later expand the defn of vector space to smth more general than a box of pointy arrows, the heart of the defn being a set of things you can add up & scale. you then look at other vector spaces that don't exactly feel like boxes of pointy arrows like certain sets of polynomials or functions. then you eventually come across that F^(m.n), the set of m by n matrices with entries in F, is itself a vector space over F, with vector addition & scalar multiplication defined the usual componentwise way

[tame mural]

ah I see

I'm surprised when one says that vectors need not have elements/entries though!

@_@

[gray dust]

have you seen vector spaces beside F^n & F^(m.n)?

[tame mural]

nope

[shy atlas]

[tame mural]

I mean, polynomials I guess

but they are described by scalar entries

[shy atlas]

[gray dust]

*coefficients in F

[tame mural]

what's the most popular introductory example

of a vector space that isn't like the examples above?

[gray dust]

the other one i mentioned in passing is certain sets of functions like the set of continuous real valued functions on interval [a,b]

[dusky epoch]

I'm surprised when one says that vectors need not have elements/entries though!
vectors are abstract objects

from the viewpoint of linear algebra, a vector is solely defined by membership in a vector space and has no properties beyond those that follow from it

[fickle tree]

if (vectors) a * b = ||a|| * ||b||, what can we say of the relative position of the vectors of (vectors) a and b?
I am not sure to understand can I get some lead?
like is it that they are unit vectors?

[shy atlas]

you mean $\vec{a} \cdot \vec{b} = \norm{\vec{a}} \norm{\vec{b}}$ ?

[stoic python]

soαρ:

[fickle tree]

yes

[shy atlas]

do you know the general formula for a dot b which involves the cosine of the angle between them

[fickle tree]

cos alpha = b/a?

[shy atlas]

[fickle tree]

sorry linear algebra is really confusing to me I loved calculus but this is not my cup of tea coffee

[shy atlas]

fuck tea coffee is the way to go

i was referring to this btw $\vec{a} \cdot \vec{b} = \norm{\vec{a}} \norm{\vec{b}} \cos(\theta)$

[stoic python]

soαρ:

[shy atlas]

where theta is the angle between them

[fickle tree]

would it be 0, so cos0 =1, which makes it that $\norm{\vec{a}} \norm{\vec{b}} $

[stoic python]

π = 3:

[fickle tree]

so it means it's on top of each other?

[shy atlas]

yes they're on top each other like i was on top of @cursive narwhal 's mom

[fickle tree]

ok I understand thank you

[cursive narwhal]

so it means it's on top of each other?
When you say, on top of each other, it implies something geometric in nature

[fickle tree]

I had a few other questions that might come in the next minutes

[cursive narwhal]

Just say that the angle between them is 0. That's all.

[fickle tree]

u=(0,3,5)
v=(-2,1,0)
w=(8,k,-2)``` 

so by solving the 3x3 determinant I get that k=2, that's it? I feel like it was too simple like I am missing something?

[muted holly]

-3(4)+5(-2k-8)
= -12 + -10k -40
= -10k -52

right?

Then the three are linearly independent as long as det(M) is non-zero

[fickle tree]

oh I made a mistake? I got -20-10k

AH yeah I kept the -8

so 52/10=k which is the only value the question asks?

like that's it?

[clear sparrow]

What is the distinction being made here?

[limber sierra]

read the next sentence

matrices are defined formally as a linear function in a vector space (and there are multiple equivalent definitions); it just so happens that this coincides with the "rectangular representation" you're probably more familiar with (the fact that they coincide is a theorem)

[clear sparrow]

What should I google to read up on this?
"Matrix rectangular representation" and the likes got me nowhere

[stable urchin]

@clear sparrow it means that the “matrix” above isn’t actually a matrix. However it’s helpful to display it as one, aka a coefficient matrix.

[wintry steppe]

Rectangular arrays r just one of many ways to represent a matrix

But the abstract idea of a matrix is what really defines it

[clear sparrow]

Ahh I see
Thanks!

[stable urchin]

Quick question, what makes a solution to a system unique

[wintry steppe]

If A is an invertible matrix, then Ax = b yields a unique solution to the equation b

@stable urchin

[stable urchin]

Ooh I see

[gray dust]

this doesn't at all cover cases where A^-1 isn't defined

[wintry steppe]

I guess i should be a bit clearer a unique solution isnt in the Nullspace of any matrix A

But for an invertible matrix the only element of its nullspace is the 0 vector

[zinc tapir]

can anyone give a clear definition of the nilpotent linear operator my prof just skipped over it and i can't find it in my book

i don't understand the wiki

[slow scroll]

T is nilpotent if there is n such that T^n is the 0 map

$A = \begin{pmatrix} 0&1\0&0\end{pmatrix}$ is an example since $A^2 = 0$.

[stoic python]

kxrider:

[zinc tapir]

is that for some positive integer n

[slow scroll]

yes. By T^n, I meant T composed with itself as a function n times

When you have a matrix, function composition becomes matrix multiplication

[zinc tapir]

so T^2 is just TT right

[slow scroll]

$T^2 = T \circ T$. If you have a matrix $A$, then $A$ composed with itself is $A$ times $A$.

[stoic python]

kxrider:

[stable urchin]

Super quick question

Is this consistent

[zinc tapir]

oh T composed with T

[stable urchin]

Sorry if I barged in

[slow scroll]

yes bivariate. If there were a pivot in the last column of the augmented matrix, it would not be consistent since you would have 0 = * != 0.