linear-algebra

[stoic python]

shadowplayer67
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[keen carbon]

It looks ugly but that's the gist of my troubles. I'm fully aware that PP^t should reduce to the identity matrix and I know I don't have to expand it like this (as alternative methods have been suggested above). Nonetheless, the multiplication method should work but it's not because there's no clear evaluation to the terms in the matrix

The subscript specifies which eigenvector, the superscript specifies which coordinate

[stoic python]

shadowplayer67
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[ocean sequoia]

wait why doesnt the above give the identity

maybe im looking at it wrong but doesnt it evaluate to [[0,1],[1,0]]

nvm read it wrong

ok lets use real numbers

sorry this notation is making it hard to follow

can you edit the message to be e3, e4?

[keen carbon]

I can but then you'd lose out on the distinction between whether we're talking about 1st or 2nd eigenvector vs 1st or 2nd coordinate

[ocean sequoia]

true honestly im assuming the arthimatic has to be wrong somewhere is what im looking for atm

[keen carbon]

It's weird tho because I multiply the same for PtP and PPt but PPt is just wrong

like why should the square of the first coordinate of the first vector plus square of first coordinate of second vector add to 1

thing is, using real examples in R2, they do add up to 1 haha but i still don't know why

[ocean sequoia]

if you try it with real numbers it works

have you tried doing a matrix with real numbers

try a non trival example

[keen carbon]

e_1 = [1/sqrt(2) 1/sqrt(2)] and e_2 = [-1/sqrt(2) 1/sqrt(2)]

in fact in R2, it works with any vector e_1 = [cos(theta) sin(theta)] and e_2 = [cos(theta+pi/2) sin(theta+pi/2)] = [-sin(x) cos(x)]

[ocean sequoia]

did it work

have you seen the generic proof @keen carbon

you know this should work right

[keen carbon]

No I haven't seen it

I know it should work but I don't get why the PPT terms evaluate the way they do

[ocean sequoia]

oh

ok

here

if Q^t Q = I then Q^t = Q^-1

so then QQ^t = Q Q^-1 = I

[hard drum]

or QQ^t = (Q^t Q)^T = I ig as Q square

because isn't Q^t the inverse of Q by definition if and only if Q^t Q = I = QQ^t anyway

[ocean sequoia]

for an orthogonal matrix it is

[hard drum]

Well that's the definition of inverses, with Q^t instead of a general matrix e.g. P

[ocean sequoia]

P here is an orthogonal matrix

[hard drum]

i know, lol

[ocean sequoia]

oh my bad then

sorry

[hard drum]

oh dw dw

[wicked hamlet]

i know monotonically decreasing functions can be differentiated anywhere, but is the differential always <0

i.e. f(x) is differentiable and f'(x) < 0

[hard drum]

-monotonically decreasing functions aren't necessarily differentiable everywhere, or even continuous (e.g. consider f:R -> R with f(x) = x for x >= 0 and f(x) = x -1 for x < 0)
-the derivative doesn't have to be < 0; consider for example f(x) = -x^3, which is (strictly) decreasing but has f'(0) = 0. It does hold that where f'(x) exists, f'(x) <= 0 though (consider what happens when you take the limit of the difference quotient)

[wicked hamlet]

oh lol my bad then, i was thinking that was a condition they had

[hard drum]

afaik monotonically decreasing functions are differentiable almost everywhere though tbf

[wicked hamlet]

are there any types of functions which always give f(x) is differentiable and f'(x) < 0

or do they all give it under one condition or another

[hard drum]

I'd honestly just say this lol, idk any other conditions that would give that

[wicked hamlet]

ok thanks you

[hard drum]

np

[wintry steppe]

i have a lin alg question

tryna find the vector consisting of orthogonal reflection of <1,b,b> b is a parameter, against the vector <1,1,1>
and another question being x^2 + y^2 = (z^2)/3 what would be the parametric equation of this? i thought x = t * cos(u), y = t* sin(u) and z = sqrt(3)*t?

but im not sure

tag me if u answer pls cause otherwise i cnt see it

[torn stag]

@wintry steppe Reflection of a vector $x$ across the line spanned by unit vector $u$ is given by the mapping $x \mapsto x - 2(x - (x, u)u)$

[stoic python]

IlIIllIIIlllIIIIllll

[torn stag]

where $(\cdot, \cdot)$ denotes the inner product on $\mathbb{R}^3$.

[stoic python]

IlIIllIIIlllIIIIllll

[torn stag]

@wintry steppe There are a lot of ways to parameterize that 2-d surface. The simplest one would be to take $\phi(x, y) = (x, y, \sqrt{3(x^2 + y^2)})$, which will give you the top half of the curve.

[stoic python]

IlIIllIIIlllIIIIllll

[wintry steppe]

thank you!!!!

[restive spindle]

(╯°□°）╯︵ ┻━┻

[subtle edge]

Apply the kkt conditions and see which ones through drawing out the cdns which ones are slack

ie find the feasible region that these conditions hold

[dusky epoch]

linear program
kkt conditions

[limber sierra]

They changed the value of P in the equation at the top of your screenshot

That's it

If you imagine it as y = mx + b, m is your slope (-4/3) which doesn't change, while b is your y-intercept (P/3) which does

Changing the y-intercept without changing the slope causes the line to move up and down, but not change direction

Hence why they're all parallel

Anyway, this is not linear algebra; please ask questions on this topic in #prealg-and-algebra

Or one of the 10 questions channels

No, to some values between 150 and 250 it looks like

Hard to tell exactly from the screenshot

[calm yoke]

Any recommended books for Jordan forms and Diagonalization?

[wintry steppe]

linear algebra done right talks about those (and more)

[limber sierra]

so does any other linear algebra book

and youll get 99% less determinant hate

[vestal spire]

Linear Algebra and its Applications by Lay et al. is a good one

[calm yoke]

I like that one, but it has little material on jordan forms

[solemn lotus]

if A is a subset of Rⁿ, and the linear span of A is LS(A),
if LS(A) = A, that must imply that A is a set containing only the zero vector, right?

i.e. for a specific field eg R³ or R², there is a single set A which can satisfy this, being {(0, 0, 0)} or {(0, 0)}, right?

[dire thunder]

is A subset or a vector

but anyway

[solemn lotus]

sorry, corrected

[gray dust]

they likely mean A={0}

[dire thunder]

LS(R^n)=??

[solemn lotus]

oh

Rⁿ

hmm right that makes sense, i didnt think about that

yeah ty that makes sense

[dusky epoch]

so here's something that just crossed my mind

not as a homework assignment or anything

given two constant matrices A and B, det(A + xB) is obviously a polynomial in x

can we say anything about deg(det(A+xB)) from any properties of B?

i have a suspicion that, barring some potential edge cases, it might be that deg(det(A+xB)) will be equal to rank(B). and it is certainly true for B = 0 or B = I

but i'm not exactly sure how one would go about establishing this in the general case.

A and B are square matrices of size n of course

[wintry steppe]

A
0 * *

• 1 1
• 1 1

B
1 0 0
0 0 0
0 0 0

where * arbitrary

[zealous junco]

I wanna prove something about the atomic norm:

Let $\mathcal{A} = {a(f)v_k^H \colon |v| = 1}$ be the atomic set and where $v_k \in \mathbb{C}^M$ and $a(f) = \frac {1}{\sqrt N} \begin{bmatrix} 1 & e^{2\pi j f} & \cdots & e^{2\pi j f (N-1)}\end{bmatrix}^T$

[stoic python]

Anticipation

[zealous junco]

ok, now let $|X|_{\mathcal{A}} = {\sum_k d_k \colon X = \sum_k d_k a(f_k)v_k^H}$ be the corresponding atomic norm

[stoic python]

Anticipation

[zealous junco]

now given a matrix $B = \begin{bmatrix} a(f_1) & \cdots & a(f_l) \end{bmatrix}$ vandermonde (with the a(f) as columns), and which is not full column rank such that $X = BV$

[stoic python]

Anticipation

[zealous junco]

how do you show that $\sum_l |V_l|$ where $V_l$ are the rows of $V$ is not the atomic norm of X, i.e. there is another decomposition $X = AU$ so that $\sum_k |U_k| \leq \sum_l | V_l|$

[stoic python]

Anticipation

[zealous junco]

I think this is true because B is not full column rank, i.e. theres redundancy in represneting X with the $a(f_l) v_l^H$

[stoic python]

Anticipation

[zealous junco]

note here the decomposition X = AU means A has to also contain a(f_k) as columns, but the frequencies f_k can be different

[zealous junco]

and here d_k > 0 is another restriction

[lavish jewel]

what's the rank of A

[zealous junco]

it does not matter, like I'm trying to show there is an A

whatever its rank is, most likely its full column rank

[lavish jewel]

how many values of f do you have

if they're fewer than N, B is full rank

[zealous junco]

yea I assume $B$ is not full column rank

[lavish jewel]

(the problem wouldn't make sense otherwise btw)

or well, you need V to be sparse

the solutions are in general not unique, as you pointed out

but atomic norm min allows you to pick one

i would find it weird in this case for B not to be full rank because it physically implies poor experimental design

like bad sampling

[zealous junco]

yea, i guess a better way to phrase my question is how to show the atomic norm decomposition of X must have a(f_1), ..., a(f_l) to be linearly independent

[lavish jewel]

in general that's just not true

you need extra knowledge about X

i.e. being sparse in B

[zealous junco]

ok, cuz if you go to page 7 here

https://arxiv.org/pdf/1405.6585.pdf

I don't know how they can justify the last equality

in (17) they didnt really say how much they r summing over, just "k"

but on the other hand in this part, it is assumed that A is full column rank, since the vandermonde decomposition requires it to be, it seems

[lavish jewel]

idk what anything there means, so

can't comment 😛

[zealous junco]

so your saying that the atomic decomposition doesnt have to have linear independent steering vectors in general

[lavish jewel]

it tells you that on page 2

yes

you just need X and the atomic set to satisfy some properties

like X being sparse enough in A that it is uniquely identifiable

that's the whole point

if you know a priori that you want a sparse solution and the set has some nice properties, then sparse enough vectors can be uniquely identified

even when the set is lin dep

[zealous junco]

wait what paper is this? doesnt seem like the one i sent

[lavish jewel]

oh oops, i was looking at another atomic norm paper lol

but anyway yes

[zealous junco]

hmm ok

[lavish jewel]

spectral matrices don't play nice if you just keep increasing the frequencies tho

because what you get is that at some point, a vector late in the set is exactly one of the first ones scaled by a complex weight

this makes the girth of the set 0

and arbitrary signals are unidentifiable

i'm pretty sure you have f < N

[zealous junco]

you mean increase the amount of different frequency components in the signal?

[lavish jewel]

it doesn't even make sense to do it otherwise

no, i mean one avoids aliasing when sampling

and also that going to 360° (representing DTFT in polar coords) is the same as 0°, so you don't include it

then stop a second to realize that placing several sensors in space is identical to taking a DTFT over time

[zealous junco]

wait.. when you say f < N, what does f refer to, since each frequency component is normalized in [0,1)

[lavish jewel]

i was talking about the discrete case. in the continuous case, this is the same as saying f doesn't go up to 1, which is exactly what you already wrote

in this case you can have steering vectors that are largely correlated, but not identical

[zealous junco]

oh, i feel like i have to assume number of frequency components is less than N actually

[lavish jewel]

that's not necessary, but what you have to avoid is going over 360°

if you were using something like ESPRIT or other subspace-based technique, yes

if you're using sparse recovery, no

[zealous junco]

interesting, thanks for the help

I feel like i may also need to use ESPRIT to do vandermonde decomposition of a toeplitz matrix later

or is there a different method

[lavish jewel]

i've never directly done those, so idk

esprit is pretty straightforward tho

but yeah, for linear models, something like X = AS can be solved with ESPRIT if A has a nice structure and fewer columns than rows

you could have A with more columns than rows and still solve it if S is sparse, but not with ESPRIT

or i guess still possibly with ESPRIT depending on the structure of A and S

but that's outside of the classical usage you see in most papers

akin to 0 padding the hell out of S and/or changing it to another basis

btw the paper ur reading talks about the spark in page 2 or 3, i forget

might wanna take a look at that

it's related to the recovery conditions for these problems

[zealous junco]

thank you

[tranquil steeple]

Where does your Toeplitz matrix come from?

[zealous junco]

from solving the sdp here

or something similar to this

[tranquil steeple]

So this construction of the Toeplitz matrix, is it by design that it has a lower rank than n? (I have not read the article)

(just curious, probably not helpful to your problem)

I would say most symmetric Toeplitz matrices have rank n, unless circulant, where rank is n-1 (or you could construct artificial cases)

I still don't get the point of this Vandermonde decomposition 😛

[zealous junco]

not necessarily,

[tranquil steeple]

you are probably right 🙂 I just have not encountered it 🙂

[zealous junco]

bruh i am just beginning to learn this. Anyways the vandermonde decomposition is used to prove thm3

[tranquil steeple]

If you get a practical example, then please share here, I would love to see it 🙂

btw Peter Stoica that was coauthor of the other paper you shared here is a really nice guy, he has his office one floor below me. Try mailing him 🙂 ? (I actually dunno if he is retired now, I haven't been at work for over a year)

[zealous junco]

yea im actually gonna code some stuff using this and another paper this week, but I think its not really interesting/pretty trivial. I can share it if u want

[zealous junco]

ah ok, are you doing phd?

[tranquil steeple]

assistant prof, www.2pi.se

[zealous junco]

o nice

im just starting to learn this haha, still undergrad

[lavish jewel]

oof you know petre stoica?

[tranquil steeple]

Well, I have not been to his place or anything, but we always say hi in the corridors. I think I took two courses with him as teacher.

[lavish jewel]

sweet

[tranquil steeple]

Was a pretty interesting set of teachers in a course about different transforms, Stoica, Svante Janson (Erdös number 1), and Bertil Gustafsson (one a of the classic numerical analysis people) 😛

[cobalt merlin]

how are these the same, help

[nocturne jewel]

Definition of norm

$\norm{[x_1,x_2,x_3]}=\sqrt{x_1^2+x_2^2+x_3^2}$

[stoic python]

Mosh

[cobalt merlin]

but how is | | x | |^2 = | | OP | |?

[nocturne jewel]

idk

you haven't given any other context

no clue what OP is

[nocturne jewel]

Yeah, notation is fucked one second

[cobalt merlin]

OP is literally x

which is just confusing

cause of course x is also being taken in respect to O

origin

[nocturne jewel]

yeah, that is very dumb to follow

but it's just pythagorean theorem twice

[cobalt merlin]

I have been trying to figure this out for an hour

and my brain is not getting it

[nocturne jewel]

distance b/w 0 and (x1,x2,0) is gotten by pythagorean and is sqrt(x1^2+x2^2)

so then OP is the hypotenuse in triangle OPP'

length of OP' is sqrt(x_1^2+x_2^2) and length of P'P is x_3

so $\norm{OP}^2=\norm{OP'}^2+\norm{P'P}^2=x_1^2+x_2^2+x_3^2$

[stoic python]

Mosh

[cobalt merlin]

could you join vc, I am struggling to understand with the message

[nocturne jewel]

I dont do vc

[cobalt merlin]

like the only thing which makes no sense is | | OP | |, it seems off

everything else like I get

[nocturne jewel]

yeah that should be $\norm{OP}^2$

[stoic python]

Mosh

[cobalt merlin]

ya

[nocturne jewel]

so idk why you're getting so worked up over a typo

[cobalt merlin]

thank you, but fuck these profs can be confusing

because they dont say its a typo

my prof did not fix it and this just is itching me

[nocturne jewel]

They likely dont realize they made a typo...

cause they're busy teaching the lecture

[cobalt merlin]

ya, I guess thank you

[nocturne jewel]

Or just.. use common sense and realize it's a typo and shrug it off

[cobalt merlin]

but linear algebra does not really use common sense. Shit gets way to confusing beyond common sense

[nocturne jewel]

Recognizing something as a typo... isnt LinAl

[cobalt merlin]

fair

[wintry steppe]

can anyone help with this please

[nocturne jewel]

#prealg-and-algebra

this isnt LinAl

[dreamy iron]

What does it mean “extending f_i linearly to all of V”?

[limber sierra]

The v_1, ... v_n form a basis so any linear function is completely determined by what it does to those v_j

So for example, let's say V = R^2 and v_1 = (1, 0)

Then we can determine f((2.5, 0)) by f((2.5, 0)) = f(2.5(1, 0)) = 2.5 * f(v_1)

So if f = f_1 this is 2.5, if f = f_2 it's 0

Now to extend that example, say v_2 = (0, 1)

Then f_1((2, 1)) = f_1(2(1, 0) + (0, 1)) = 2f_1(v_1) + f_1(v_2) = 2 + 0 = 2

Whereas f_2 of that would be 1

You get the idea?

Basically each f_i represents the coefficient of v_i when computing your input as a linear combination of basis elements

[shut marlin]

Can someone explain this to me? How can you have two different matrix representations of the same linear transformation depending on the choice of basis for the matrix? If every linear transformation has a unique matrix representation, how is it not flat out wrong to assert that we can produce a second matrix representation of a transformation that is not equal to the first. Sorry, I know I've just asked a lot of questions but I'm trying to understand how this works.

[sonic osprey]

It's not true that every linear transformation has a unique matrix representation?

If you pick a basis, then every linear transformation has a unique matrix representation with respect to that basis. But if you choose different bases you'll get different representations

@shut marlin

[shut marlin]

@sonic osprey Thanks for the help. I think the gap in my understanding is the idea of having a matrix with respect to a different basis than the standard basis.

If I multiplied a random vector in the domain of T by M(T), will I get the same output vector irrespective of the choice of basis for M(T)?

[sonic osprey]

Depends what exactly you mean. Just like matrices, you can also represent vectors in different ways depending on the basis chosen. If you represent the vector in the same basis that T is being represented in, then yes it will be the same

[shut marlin]

Oh, I understand now. Thanks for your help!

[sonic osprey]

It might help to explicitly see this by calculating a few examples

[shut marlin]

Thanks, I'll give that a go right now.

@sonic osprey I just did some practice calculations and it definitely feels intuitive. Thank you so much for helping me to understand this.

[vivid field]

Could someone let me know if I put this into parametric form correctly ?

[dreamy iron]

So the preposition defines what f_i does on a basis of V, and the “extension “ comes when we consider what f_i does on an arbitrary vector v in V, thus this discussion about coefficients which youve provided.

I have that right?

[limber sierra]

yes

i didnt quite prove it though

but you might be able to see from my examples why its true

[dreamy iron]

The text has a full proof of the proposition. And online has several resources also. Thank you.

[hard drum]

I have to show the map $T: \mathbb R^3 \to \mathbb R^3, \mathbf v \mapsto Av$, where $A =\begin{pmatrix}0 & -1 & 0 \ 1 & 0 & 0 \ 0 & 0 & 1 \end{pmatrix}$ has exactly one two dimensional invariant subspace.\ The only way I could show uniqueness was by the following argument: \suppose $W$ is a two dimensional invariant subspace of $T$. Let $v_1,v_2$ be an orthonormal basis of $W$ (wrt dot product) and extend this to an orthonormal basis $v_1,v_2,v_3$ of $\mathbb R^3$. Now $Tv_3$ is orthogonal to both $Tv_1$ and $Tv_2$ (as T is orthogonal) and hence to both $v_1$ and $v_2$ as $T(v_1),T(v_2)$ is also basis of W (easy to show). Hence $Tv_3 = \lambda v_3$ for some $\lambda \in \mathbb R$, and as the only eigenvector of T is $e_3$ and $|v_3|=1$, $v_3 = e_3$ and hence $v_2$,$v_1$ are linear combinations of $e_1,e_2$ by orthogonality i.e. $W = \text{Span}(e_1,e_2)$ as was to be proved.

I just wondered if there was a slightly more slick argument to this, potentially one that even avoids using orthogonality or something, or is that the key idea here?

[stoic python]

potato

[wintry steppe]

oops sorry

[worldly tartan]

Is there any 2x2 matrix with real entries which is symmetric and skew symmetric other than of the form kI where I is the identity matrix?

[hard drum]

If A is skew symmetric and symmetric then A^T = A and A^T = -A, what does that tell us about A?

[worldly tartan]

A=-A

[hard drum]

indeed

[worldly tartan]

Also, thnx for helping out preiously

[hard drum]

np

[nocturne jewel]

A has to be the 0 matrix I believe

[hard drum]

I was asking in a socratic way lol

[still lodge]

ive got this shitty system of equations right

i know that since it's not square, it has to either have infinitely many solutions or none at all

is there any way to make a conclusion about which it is without having to row reduce this thing

[lavish jewel]

if you have to ask, the answer is usually no

unless it is super evident to the naked eye, (r)ref is one of the easiest ways to check

[still lodge]

pain.

[still lodge]

wdym by this tho

[lavish jewel]

either you can immediately tell because the problem is super easy, or you calculate stuff

[lavish jewel]

what? don't ask if you'll complain about the answer

[alpine ferry]

is it correct to say that $|a \hat{i} + b \hat{j}| = \sqrt{a^2+b^2}$?

[stoic python]

gmod

[hard drum]

ye (under the standard norm at least lol)

[alpine ferry]

ok thanks

[timber willow]

So initially I thought C) since the vector space V could contain vectors which are scalar multiples of each other.

[wintry steppe]

i don't follow your reasoning, why does that give a linearly independent subset of V strictly containing B?

[shell kindle]

Does anyone know what this line stand for? Like what does p(A) mean? I know A is the matrix, but what is p?

[lavish jewel]

presumably rank? but you'll have to check your notes for the notation

[shell kindle]

Aight, thx

[timber willow]

Any hints on how to setup the problem? I calculated the vectors... But don't know how to proceed after taht

[dusky epoch]

you need not do any calculation here

the answer in fact depends only on whether or not your points are collinear, which they aren't

the possible locations of D are the vertices of a triangle whose sides are parallel to those of ABC and have A, B and C as midpoints

like this

red = A, B and C
purple = where D could be

[timber willow]

Ohhh

Yes! that makes sense!

Thanks!

[stoic python]

tushar

[rich garden]

just for reference we have this

is P(R) the set of functions from R to R? i've only seen the script P used for the powerset

[grim bridge]

that's weird notation indeed

it's definitely the space of differentiable functions from R to R but if it wasn't either introduced before it's kinda weird

because P(R) is {0,1}^R and not R^R

[dusky epoch]

@rich garden P(R) seems to denote the set of all polynomial functions on R

[rich garden]

ahh

[grim bridge]

weird to do so when he doesn't mention polynomials afterwards and mentions the generality of "wherever f and g are differentiable"

[rich garden]

wait there totally was a section on that

ohhh i just found it

[grim bridge]

also very weird to not use R[X] instead

[rich garden]

thank you so much

[grim bridge]

ah they call a type of function a polynomial they probably don't even care about notation

[limber sierra]

eh, this is fine for a linear algebra context

it makes formalization of some algebra tedious but you dont really run into this issue in linear algebra

and explaining the difference between polynomials as objects and their evaluation functions is just gonna confuse students

[timber willow]

Any clue on how to plug this?

[limber sierra]

those functions dont look linear 🤔

[coarse rain]

I'd just try and find the corners

[cold maple]

Hello everyone, I was doing my first online homework assignment for Linear Algebra (I have unlimited attempts on the hw), and I genuinely believe that my professor made a mistake when she chose the "correct" answers for (c) and (a). For context, in this assignment, you have to identify how many solutions each augmented matrix in reduced row echelon form has. I believe the answer for (c) should be infinitely many solutions since every article and documentation that I have read thus far regarding augmented matrices says that if there is a zero row, then the system has infinitely many solutions. For (a), I am not entirely sure why the answer for that is infinitely many solutions since neither my textbook or the lecture my professor provided has an example similar to it. I think it should be none of the above since it does not meet any of the conditions based on what I read online.

So yeah, I would like a second opinion on this. I have used matrices before so I am pretty confident that I did it correctly and the mistake is on my professor's end.

[limber sierra]

I believe you're mixing up zero rows and zero columns?

Or rather

Could you give me a link that backs up your claim regarding (c)

Since I believe you're missing something

[cold maple]

Sure. Give me a second.

[limber sierra]

(a) certainly has infinitely many solutions since the first column corresponds to a variable that immediately gets 0d out

So that variable could take on any value without affecting the solution space

[cold maple]

https://www.math.tamu.edu/~jlewis/Threetypesofsolsets.htm

[limber sierra]

That source does not support the claim you made.

A system has infinitely many solutions when it is consistent and the number of variables is more than the number of nonzero rows in the rref of the matrix.

(c) has two variables (columns to the right of the line) and two nonzero rows

So this doesn't apply

[cold maple]

Gimme a second rq

[limber sierra]

If it helps, convert these matrices to the linear systems they represent

It might help you understand their behaviour better

[cold maple]

I also used this video as well https://youtu.be/18_oG-cD9Ck?t=697

A lot of the videos on YT were pretty consistent in whole zero row thing.

as long as the constant matrix was also zero

[limber sierra]

The matrices in that video are square (ignoring the last column)

Yours are not

[cold maple]

Is there a difference between square matrices and augmented matrices?

idk my textbook called it augmented

[limber sierra]

No I mean

They have the same number of equations as they do variables

In (a) of your image, you have more variables than equations

And in (c), you have more equations than variables

So what that video says doesn't apply

[cold maple]

I see.

[limber sierra]

The math.tamu.edu link you gave seems to have correct info for the general case

I dislike that the video focuses on zero rows instead of zero columns

(or more accurately, nonpivot columns)

[cold maple]

So a square matrix is when the rows = cols

[limber sierra]

Since columns are what actually determine your "freedom" of variable selection

Yes

[cold maple]

I'm assuming there is also a term for when rows > col and when rows < col

[limber sierra]

Overdetermined and underdetermined

[cold maple]

And that essentially impacts the way in which you decide if it has infinite, unique, and no solution systems?

[limber sierra]

It means the "tricks" in the video don't work

The first length you gave gives general "rules"

Alternatively, you could just imagine it as the system it represents

For example, the matrix from (c) represents:

x + 0y = 7
0x + y = 2
0x + 0y = 0

The last row is obviously irrelevant, so this simplifies to:

x = 7
y = 2

Which like... Immediately and obviously gives us a single solution

(7, 2)

[cold maple]

I see.

[limber sierra]

You could do a similar thing for (a), which would give you that your first variable (say x) always zeroes out

Meaning it doesn't affect the solution space at all

Since we know it has a solution (you can check this), this means it must have infinitely many

Since we could set the first variable x to whatever

And it wouldn't make it not a solution

for example, both (0, -11, 14) and (69420, -11, 14) solve (a)

[cold maple]

Alright, I think I'm starting to understand it more.

Thanks

[opal mortar]

hi guys, i have a first assignment about elementary linear algebra. is my answer correct? Or does this have to be processed first? I did it just looking at the terms of the reduced row echelon and row echelon matrices. and this is my answer

[limber sierra]

seems you missed a requirement to be in REM/RREM

your rows have to be ordered a certain way

[opal mortar]

so,must be processed first ?

[limber sierra]

they need to have a "staircase shape" with 0 rows on the bottom

idk what "processed" means

[opal mortar]

so what should i do for matrices

[limber sierra]

let me give an example

this is in REM because its nonzero entries form a "staircase shape"

this is not, because theres a step thats "too large"

if it was in REM, one of those (either the 1 or the 3 i circled) would have to be a 0

this is also not, because there's a 0 row not at the bottom, which ruins the staircase.

if it helps, there is only 1 matrix there in RREM, and only 1 in REM that's not in RREM.

the other 4 are all "neither"

[opal mortar]

so, (a) is RREM,
(b) not both
(c) not both
(d) not both
(e) not both
(f) Rem

[limber sierra]

not quite

A doesnt have the "staircase shape"

[opal mortar]

but, why (e) not both ? it has 1 and in each column the one the entries are zero

[limber sierra]

you're right, (e) is in RREM

(a) is not.

you're right about the rest, though.

[opal mortar]

ooh i see

[opal mortar]

ooh ok, thank you ^^

now I understand the outline 👍

[random axle]

If these two planes are perpendicular

My teacher says the normal of each of planes must lie on the other plane

I get that the normal of each of the planes must be parallel to the other plane

But why must the normal lie on the other plane?

Although I haven't drawn the normal vector well, you can see that it clearly does not lie on the other plane right?

[limber sierra]

perhaps your course is taking it as convention that all vectors and planes are centred at the origin?

in which case the statement is true

otherwise, you can "place" the vector anywhere sure

position of vectors isnt really a... meaningful concept

[random axle]

Here is the full problem

[limber sierra]

linear algebraically

[random axle]

So the aim is to find the plane equation, given we have one plane perpendicular and two points on the plane we are trying to find

[lavish jewel]

perhaps it is more clear to say that the direction vector of the normal lies on the other plane if you apply a corresponding displacement

[random axle]

This makes sense. So then how can we cross product two vectors and get the correct normal vector for the other plane IF that normal doesn't actually lie on that plane. Since as you said it needs a corresponding displacement

[lavish jewel]

if you x product without the displacement you stilk get the right direction

as nami said, the position of vectors isn't a concern

[random axle]

Okay, I see. Thanks.

[olive tundra]

Can anyone help me here?
I don't understand how interchanging 2 rows is an elementary row operation but interchanging 3 rows is not

[dusky epoch]

mostly convention

rearranging 3 rows can be recreated through several steps of swapping two rows

if anything, while doing Gaussian elimination you can rearrange the rows however you like so long as you don't duplicate or lose any

[olive tundra]

Yeah makes sense

Thanks @dusky epoch

[limber sierra]

though theres also a bit of pedagogical merit by focusing on interchanging 2 rows

since it makes it clear what permutation matrices youll use

and therefore, how it affects the determinant

(multiplies it by -1)

[quick sundial]

hi guys i have simple linear algebra question

could any one give me some help?

[lavish jewel]

go ahead and ask, someone will (possibly) answer

[quick sundial]

the question says that i have 4 vector and i have to check Independence so in the solution first we check the rank of the matrix that we created by taking every vector and put it in the matrix as a row, after we check the rank of the matrix we go to write their linear combination so ut them into matrix by taking every vector as a column (in order to find the relation between them if they are dependent)
my question is why we are this sequence of matrix sorting(first taking the vectors as a rows then taking the vectors as a columns)

[lavish jewel]

you could check independence using column operations too if you wanted

it's just convention

[unkempt matrix]

heyheyhey, can someone help me with matrix theory?

[hard drum]

[solid leaf]

Is there an easy way to find the nth root of a 2x2 matrix? It's symmetrical if that helps. According to google, there's a thing called Jordan decomposition but i dont think we are supposed to use it if we havent learned it so

nvm lmao we good

[lavish jewel]

symmetric matrices are diagonalizable

in a nice way

[worldly bear]

,rotate

[worldly bear]

is this the only way to do this?

[tranquil steeple]

you can do an eigendecomposition

[worldly bear]

i’m not familiar with that

[tranquil steeple]

$B^2=QLQ^{-1}$ where $L=\begin{bmatrix}0 0\0 4\end{bmatrix}$ and then $B$ is either $\pm Q\begin{bmatrix}0 0\0 2\end{bmatrix}Q^{-1}$

yeah, that was not so nice...

[stoic python]

Sven-Erik

[tranquil steeple]

there we go 😛

[worldly bear]

am i expected to know that two weeks into a linear algebra class?

[tranquil steeple]

your solution is fine 🙂

[worldly bear]

okay, just didn’t know if i would lose marks or anything

[tranquil steeple]

I would not think so

[worldly bear]

not sure if it was the indented solution

[tranquil steeple]

perfectly fine to solve like that

[lavish jewel]

your solution indeed looks perfectly fine

as it turns out, there's many ways to define "square roots of matrices", so you'll probably revisit the problem later on and run into the method sven mentioned

[worldly bear]

i was going to try and parameterize the variables and do it that way

then i noticed that -1 and 1 work

not sure if they’re the only ones though, i can’t imagine they would be

[tranquil steeple]

Those are the only solutions (unless I am too tired to think :P)

[worldly bear]

this worksheet was kinda tricky

like we learned about these different properties of matrices but it’s having us find examples to fit certain criteria

it feels so wrong to just guess and check, id rather do them a little more rigorously

[lavish jewel]

intuition is also important

[worldly bear]

that’s what’s saved me so much time instead of mindlessly plugging in values

[noble swan]

I've been told the row vector rule using dot product is just another way of solving a system of equations, same as an augmented matrix, but I don't understand how?

I do the dot product and I get a matrix that's the size of however many rows there are in the first matrix given to me, but what do I do with the smaller matrix to solve the system of equations?

[lavish jewel]

what's row vector rule?

[noble swan]

Honestly just seems like a fancy way of just saying dot product

[lavish jewel]

that's not the solution of a system of equations tho

it's just matrix multiplication

[noble swan]

Oh

Guess I interpreted it wrong

Thanks!

[wintry steppe]

can someome hwlp

qu

question 8

[clever fjord]

I can also ask a question

[tranquil steeple]

That is not linear algebra

[wintry steppe]

its from linear algebra chapter

[clever fjord]

I was kinda thinking bout that too :D. (I have done barely any and none in 1,5years tho)

[wintry steppe]

I think you create a linear equation then solve for a

irish higher level maths is aids

[clever fjord]

I guess that could be represented as a system of equations which can be represented as some matrix fkery iirc?

[wintry steppe]

mine?

[clever fjord]

yes

Idk tbh, as said, I haven't done any linalg in 1,5 years lol

[wintry steppe]

ugh

[clever fjord]

Actually nvm everything I said.

Looking at the picture, (a-2) degrees can't be that much smaller angle than (a+6) degrees. That is, if I understood the question correctly anyway

[wintry steppe]

could you help

I'm trying to understand it not get the answer

I checked the answer is 51⁰

but I want to lesrn to solve it

[clever fjord]

1 sec I wanna check if I understood this at all 😄

I did, so

a-2 = angle1
2*a+3 = angle2
a+6 = angle3
3*a-4 = angle4
angle1+angle2+angle3+angle4 = 360

[wintry steppe]

yes ik

but how do you get 51⁰

[limpid vine]

solve for a

[wintry steppe]

ik

bt idk how

[limpid vine]

that’s basic algebra

can you show me where you’re stuck?

[wintry steppe]

I don't get how to get 51⁰

[limpid vine]

2x + 5 = 10

what’s the value of x there?

[wintry steppe]

yes ik

its 2.5

[limpid vine]

yes

the other question is no different

[wintry steppe]

question 8

[limpid vine]

yes

it’s no different

[limpid vine]

did you read this?

[wintry steppe]

yes

ik

it equals to 360

but you solve to get 51⁰

[limpid vine]

you solve for a in the final equation

[wintry steppe]

I got

a = 7

[limpid vine]

can i see your work?

that’s incorrect

[wintry steppe]

tryna figure it out rn

[limpid vine]

can you simplify that expression?

[wintry steppe]

1a-7

[limpid vine]

no

can you write out your steps?

so i can see where you’re going wrong

[noble swan]

Bruv XD

[wintry steppe]

add and subtract like terms

here's no = part in it

[limpid vine]

yes, but the expression above is not equal to a-7

[wintry steppe]

ik

[limpid vine]

then your simplification was wrong

[wintry steppe]

a = 51

[clever fjord]

Ey, it took me way too long to notice there's actually only one variable there, too 😄

[noble swan]

You sure this isn't just normal algebra?

[wintry steppe]

no

[limpid vine]

so i’m asking you to write out your steps

so i can see where you made a mistake

[wintry steppe]

"use your knowledge of angles to form an equation and solve it to find the value of a "

thsts the question

[noble swan]

My guy, you added wrong

[wintry steppe]

uhoh

[limpid vine]

@wintry steppe can you simplify this, step by step, and write it out?

[noble swan]

Write the steps for maximo

[clever fjord]

and looking at this, you can substitute the first 4 rows to the last equation, and you get an equation with only as and numbers which will lead to a=51

[wintry steppe]

so lemme try that

[limpid vine]

i’m assuming you’ll notice your mistake yourself and i won’t even need to explain

[wintry steppe]

alri so

a1 a2 a3 a4 = 360

a+6= 90⁰

[noble swan]

Wha

[limpid vine]

what?

[wintry steppe]

nvm

im confused af rn

I don't get this at all

[limpid vine]

@wintry steppe you’re overthinking this

[wintry steppe]

ik

[limpid vine]

how many angles are in a circle?

[wintry steppe]

360⁰

[noble swan]

Just add your a's and your constants

[limpid vine]

yes

so if you have 4 angles

that add up to make a circle

then the value of those angles together

must make 360 degrees

[wintry steppe]

but idk how to get the value of a

ik it's 51

but how do I get it

[limpid vine]

i’m trying to explain that to you

[noble swan]

Just treat a as a variable man

[wintry steppe]

ik

[limpid vine]

so now you know that a1+a2+a3+a4=360

simply add up the angle values shown in terms of a

[limpid vine]

note: this isn’t the variable “a”

this is just a way to say “angle1 + angle2+…”

[wintry steppe]

(a-2)(2a+3)(a+6)(3a-4)=360

thats the questio

[limpid vine]

why are you multiplying them?

we said add

[wintry steppe]

I'm not

ik

so

[limpid vine]

(a-2) + (2a+3) + (a+6) + (3a-4)=360

(a-2)(2a+3)(a+6)(3a-4)=360
is not the same thing

[wintry steppe]

ik

alri so add the like basic algebra

[limpid vine]

alright, then refrain from omitting the "+", it's confusing

[limpid vine]

yeah

[wintry steppe]

7a-7 is my answee

[limpid vine]

that still looks off

can you write your steps, and show me your work?

[wintry steppe]

I did it in my head

[limpid vine]

can you write your steps, and show me your work?

[wintry steppe]

I did the question 5 times I cant be bothered now

[limpid vine]

then you'll continue to get it wrong

[noble swan]

Then why are you asking for help?

[wintry steppe]

ik I will

[limpid vine]

because clearly "doing it in your head" isn't working

[wintry steppe]

I worked it out normally 3 times

[limpid vine]

yes

[wintry steppe]

I scribbled it out bc I got mad

[limpid vine]

and you keep getting it wrong

i need to see your work

[noble swan]

?????

[limpid vine]

so i can help you understand how to get it right

[wintry steppe]

lemme re do it

[limpid vine]

what's a + 2a?

nvm

what's -2 + 3?

[wintry steppe]

1

[limpid vine]

and 1 + 6?

[wintry steppe]

7

[limpid vine]

and 7 + 4?

[wintry steppe]

there's no 1 in the equation

11

[limpid vine]

so how did you get 7a - 11?

[wintry steppe]

-2+3+6+4

[limpid vine]

yes, and we just did that

-2 + 3 = 1

(-2 + 3) + 6 = 7

((-2 + 3) + 6) + 4 = 11

[wintry steppe]

yes

[limpid vine]

so how did you get -11?

[wintry steppe]

wait oops

[noble swan]

You copied a constant wrong btw

[wintry steppe]

its be 7a+11

[limpid vine]

we'll get to that lol, i wanna cover the arithmetic first

[noble swan]

Gotcha

[limpid vine]

alright, if these were the angles, then yes, it'd be 7a + 11

but check again what the values of the angles are

one of them is incorrect

[wintry steppe]

i have to go

[clever fjord]

Well, you got the answer. Just fix the sign error here. Literally swap one + to - and u get 51

[wintry steppe]

which is?

[limpid vine]

look at the diagram

and look at the values you wrote down

one of the values you wrote down is wrong

[wintry steppe]

3a-4

so

7a+3

[limpid vine]

yes

[wintry steppe]

and then

[limpid vine]

@clever fjord don't

@wintry steppe go and read back what we've discussed

see where you have to plug in 7a + 3

[wintry steppe]

I dont have the time

is it 360?

or 7

[limpid vine]

what does 7a + 3 represent?

[wintry steppe]

360⁰ + 3

[limpid vine]

?

[wintry steppe]

idk bei

bro

I got 1 min

[limpid vine]

how did you get 7a + 3?

[wintry steppe]

bro my time limit is abt to run out

screen time

[limpid vine]

alright i don't want to rush the explanation

so just lmk when it runs out and i'll explain

[wintry steppe]

bri

quick

help

bro it's ober

[limpid vine]

how did you get 7a + 3?

[wintry steppe]

I get it tmrw

by adding and subtracting the equation

bro I have less then 2 mind

and my phone I'd locked till tmrw

[limpid vine]

complaining slows us down

[wintry steppe]

ik

[limpid vine]

you got 7a + 3 by adding up the angles, yes?

[clever fjord]

Well, he did go offline

Would yall mind explaining this step (sth I took from a lecture). I don't really remember the rules of multiplying vectors and their transposes together. Don't really have material going over this stuff at hand, but this just looks counter-intuitive to me 😅 (this wasn't on a linalg course so ye, no materials)

basically teacher gave the formula and said this is how it is

It's actually part of linear algebraic representation of Mean Squared Error, but this is the crucial bit I don't understand)

Also, sry, I don't speak maths in english 😄

[wintry steppe]

Im back

i took my school ipad

Yes i got it by adding all angles

[clever fjord]

Yes

[wintry steppe]

7a+3 =-*0

360

7a+3=360

[limpid vine]

so if 7a + 3 = 360, how can you solve for a?

[wintry steppe]

so

lemme work it out

7a =357

a = 51

[limpid vine]

👍🏻

[wintry steppe]

when simplified by 7

omfg ty

[limpid vine]

cmiiw, but this seems to imply that $(X\overline{w})^T\overline{y} = \overline{y}^TX\overline{w}$

[stoic python]

maximo

[clever fjord]

The counter-intuitive part to me is how y remains tranposed but (Xw)^T just becomes Xw

[limpid vine]

i'm assuming all of y,w, and X are matrices?

[clever fjord]

They are

as for size/shape (which shouldn't really matter here unless it does:D )
:

[limpid vine]

i don't actually know a lot of linear algebra but the idea seems to be that $X^T\overline{w}^T\overline{y} = \overline{y}^TX\overline{w}$, so the matrix products and transposes are doing some magic to equate to the same thing

[stoic python]

maximo

[limpid vine]

sorry i can't be of more help

[clever fjord]

Cheers, mate, anyway.

Yeah, I bet there's a video on the topic somewhere and tbh I could have found one while you were helping gbro :D.

[noble swan]

I don't really understand this question, can I get some help?

[limpid vine]

hm maybe it's as simple as saying: let $X\overline{w} = Z, Z^T\overline{y} = \overline{y}^TZ$, which may or may not be a property of matrices

[stoic python]

maximo

[hallow bobcat]

@noble swan Try to solve the system Ax = b

[noble swan]

Into echelon form or reduced echelon form?

[limpid vine]

i'm learning the same content rn, and iirc you want to make it into echelon/reduced echelon and see if the system of equations is consistent

if the system is consistent then b is W

[noble swan]

I'm a little lost on what W is exactly

Where it calls it the set of all linear combinations

[limpid vine]

the span of {[2, 0, 6], [-1, 8, 5], [1, -2, 1]} *i just realized i did the rows instead of columns, mb.

have you learnt spans yet?

[noble swan]

Yeah

I just interpreted span as the range in which the solution of the system of equation lays in

[hard drum]

yeah so $W = \left{ \alpha \begin{pmatrix} 2 \ -1 \ 1 \end{pmatrix} + \beta \begin{pmatrix} 0 \ 8 \ - 2 \end{pmatrix} + \gamma \begin{pmatrix} 6 \ 5 \ 1 \end{pmatrix} : \alpha,\beta,\gamma \in \mathbb R\right} \= {Ax : x \in \mathbb R^3}$

[stoic python]

potato

[hard drum]

so yes, 'is b in W' is equivalent to 'is there x s.t. Ax = b' and so the answer is 'yes' if Ax=b has a solution and no otherwise

[noble swan]

Ah, gotcha

So W is the span of A

[limpid vine]

A itself is a matrix so idk how accurate that statement is

[hard drum]

I'd use the term column space of the matrix here as a span is usually for a set of vectors

but the column space of A is precisely the span of its columns

[noble swan]

Huh, okay

I'm still confused on how to do part b

[limpid vine]

since W is the column space of A, and can be written like this

then just let alpha = 0, beta = 0, and gamma = 1

that shows that [6,5,1] is in W

[hard drum]

b is by definition of W

[noble swan]

So are alpha, beta, & gamma just weights in which I can plug any value I need?

[hard drum]

the third column is a linear combination of the columns

[hard drum]

Yeah, set as alpha,beta,gamma range over all real numbers, can be anything

i'm curious, how are you defining the span of vectors?

[noble swan]

I don't really understand the theory & concept behind span

[limpid vine]

alpha beta and gamma are analogous to x_1, x_2, and x_3

[noble swan]

I just think of it as the range in which the solution to the vectors lay within

[hard drum]

wdym by 'solution to the vectors'?

[noble swan]

Well, I guess not solution, but the area created by the heads of the vectors?

[limpid vine]

i think of the span of a set of vectors as the set of points which you can reach by adding up any number of weighted vectors within that set

so if you have the vectors [1,0] and [0,1], then you can use those vectors to span R^2

or more correctly would be "the span of the vectors is R^2" i think

then if we ask if a vector b is in the span of a set of vectors, we're asking if a linear combination of the set of vectors will equal that vector b

[noble swan]

Yeah, I thought of it as, if you have two vectors, the area between the two vectors creates a sort of "plane" or piece of the pie in which you can find a 3rd vector within that area

[limpid vine]

yeah that's the idea i'd say

[hard drum]

yeah

And so a linear combination of a set of vectors ${v_1,\dots,v_n}$ is something of the form $\alpha_1 v_1 + \dots + \alpha_n v_n$ where $\alpha_1,\dots,\alpha_n$ are (assuming you have a real vector space) real numbers

[stoic python]

potato

[hard drum]

that just formalises the idea a bit more of the vectors 'creating a sort of plane' etc

just combinations

the span is then the set of all those linear combinations

[limpid vine]

in a mathematical sense, we're asking if a vector x = [x_1, x_2, x_3], for x_1,x_2,x_3 in R, solves the equation Ax = b

sorry for the row vector, i don't know the latex for matrices yet

[noble swan]

Lmao, I don't know any latex XD

[limpid vine]

$\begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix}$

[stoic python]

maximo

[limpid vine]

poggers

[wintry steppe]

Hey guys need some advice/help does this statement I wrote for this problem seem okay so far ?```md

Exercise 1.2.4
Let 𝐴 be a square matrix.
Prove that if 𝐴−1 exists, then there can be no nonzero vector 𝑦 for which 𝐴𝑦=0.
(Hint: Try a proof by contradiction.)

If 𝐴 is a square matrix 𝐴∈𝑅𝑛𝑥𝑛 where 𝐴−1=𝐵 , we can prove 𝐴−1 exists meaning there is no > 0 vector 𝑦 in which 𝐴𝑦=0 exists via contradiction

Proof by constradiction statment:

If 𝐴−1 exists , then there CAN be a nonzero vector 𝑦 y > 0 for which 𝐴𝑦=0

case 1:

case 2:

case 3...

Since case 𝑋 <- ( insert contradicting case here) seems to contradict our initial statement making it false, then the opposite must be true ```

[nocturne jewel]

You've said the contradiction and... that's it pretty much

[wintry steppe]

Right, i'm confused if I even wrote that correctly lmao

[nocturne jewel]

Yes, the contradiction is that "If A has an inverse, then there exists a non-zero y st Ay=0"

[wintry steppe]

Thank you for verifying

one more question

I seem to remember you sometimes need to have 1...n cases and if you don't choose good cases to start with you could go on forever. How do you know what your base case is again? Its usual when like i = 0... or something right

[nocturne jewel]

#calculus

[nocturne jewel]

Im assuming you mean strong induction, in which case yes the base case is "whatever makes sense to start at"

[wintry steppe]

Yeah thats what I meant lmfao, thank you

[nocturne jewel]

if you're doing induction on the size of a matrix, then you should usually start at n=2 (n=0 is just if the identity matrix makes it true, and n=1 is a scalar matrix)

[wintry steppe]

Ohh ok good point

Pretty easy, then you just stop as soon as you get a contradicting case

@nocturne jewel

Looks good to start?

[nocturne jewel]

You dont need induction.. but sure

[wintry steppe]

Sorry idk what you mean

[nocturne jewel]

Assume A has an inverse and that Ay=0 has a non-trivial solution
Ay=0 has a non-trivial solution means Ay=b has more than one solution for some b, so A isnt invertible
therefore A doesnt have an inverse
Contradiction, QED

The proof is trivial w/ fundamental theorem

[hard drum]

A^-1 exists so if Ay = 0 then 0 = A^-1 0 = A^-1 Ay = y

[nocturne jewel]

alternatively, assuming $A^{-1}$ exists, then $Ay=0 \ A^{-1}Ay=A^{-1}0 \ y=0$

[hard drum]

lol

nice

[stoic python]

Mosh

[nocturne jewel]

y has to be a matrix times the 0 vector.. so it has to be the 0 vector if A is invertible

[wintry steppe]

Huh, wasn't the point to try and find a y thats > 0 , not change the value of b

[nocturne jewel]

I was just using a different equivalence from FTIM

A is invertible iff Ax=b has a unique solution for all b

[wintry steppe]

Yeah idk thats too advanced for me lmao, im just trying to pass this class and find the dumbest way to do this

[nocturne jewel]

FTIM shouldnt be advanced

[wintry steppe]

I dont even know what FTIM is

[nocturne jewel]

fundamental theorem of invertible matrices

[wintry steppe]

Linear algebra slanggg

[nocturne jewel]

it's how you tell if the inverse matrix even exists, and is a list of ~20 equivalent statements of invertibility

imo one of the strongest theorems in 1st year math

[wintry steppe]

I'm just getting a math minor lol, I dont memorize the theorums

If I had to memorize all the theorums from calc 1 I would kms

[nocturne jewel]

https://mathworld.wolfram.com/InvertibleMatrixTheorem.html

[wintry steppe]

yeet

[nocturne jewel]

yes, that's FTIM 1, most texts build it up

all the stuff about invertibility and A as the co-efficient matrix of a system

[wintry steppe]

Looks better?

If 𝐴 is a square matrix 𝐴∈𝑅𝑛𝑥𝑛 where 𝐴−1=𝐵 , we can prove 𝐴−1 exists meaning there is no > 0 vector 𝑦 in which 𝐴𝑦=0 exists via contradiction

Proof by constradiction statment:

If  𝐴−1  exists , then there CAN be a nonzero vector  𝑦  > 0 for which  𝐴𝑦=0  

Case 1  𝐴^−1𝐴𝑦={0}𝐴^−1,..𝑦=𝐴^−1{0}=0 

Since case  1  cleary resulted in only 1 possible solution where  𝑦=0  this seems to contradict our inital statment making it false, then the opposite must be true

Damn that formats ugly

[wintry steppe]

I'm assuming this was a similar concept for this problem as well ```md
Exercise 1.2.5
Let 𝐴 be a square matrix. Prove that if 𝐴−1 exists, then det(𝐴)≠0. (Hint: Use the fact that det(𝐴𝐵)=det(𝐴)det(𝐵) for 𝑛×𝑛 matrices 𝐴 and 𝐵.)

If 𝐴 is a square matrix 𝐴∈𝑅𝑛𝑥𝑛 where 𝐴−1=𝐵 , we can prove that if 𝐴−1 exists then det(𝐴)!=0 via the fact det(𝐴𝐵)=𝑑𝑒𝑡(𝐴)𝑑𝑒𝑡(𝐵)

Let 𝐴−1 exist , then (𝐴^−1)det(𝐴𝐵) = (𝐴^−1)det(𝐴)det(𝐵)

[little crater]

Question, so normal when we are talking about spans of a space i guess we describe it in terms of a series of basis vectors

i was wonder if the same methods in linear algebra can be used to describe something like polar coordinates or spherical coordinates

and the span of that basis

like We would say the angle that theta that rotates around the z axis and are distance away from the origin r can express all of R^2

so i didn't know if linear algebra sticks with vectors or if there is something similar with the topics of spans of specific angles around distance away from the origin

like rho , phi, theta can be used as a "basis" if i am using that correctly to span all of R^3

I guess each definition you define for an axis you rotate around would have to be given in the form of a vector anyways but Idk if that is anywhere away we can describe spaces

Example say we have a angle that rotates around a vector <1,2,3> as an axis of rotation and a distance away from the origin called R what is the span of this space

Idk if what I am saying makes any sense

[hard drum]

just my two cents but the polar system of coordinates isn't linear for example so it certainly wouldn't work in the same way

[little crater]

Yeah it isn't linear

but like we can describe the same space using it

[hard drum]

you can get some stuff locally, like basis vectors at a point, but yeah

[little crater]

and i didn't know if you have some space r^4 you can describe it with 3 axis of rotation and a distance away from the origin like how you use how use 1 angle for r^2 and 2 angles for r^3

[hard drum]

that is actually a thing ye

[little crater]

and maybe if you don't define your axis of rotation specifically around

x,y,z

but some vector

in x,y,z.....

that it some how makes a different subspace or something like that

[hard drum]

this thing basically is a thing lol

have used it exactly once ever

[little crater]

Yeah i have seen that

[verbal pivot]

I don't really get what you are asking, are you asking if polar coordinates can 'span' R2 just as we say a LI set of two vectors span R2?

[little crater]

sure like i guess how do we show that

but i was more so interested if we define an axis of rotation that is "off axis" using a vector

and some distance away from the orgin

[verbal pivot]

mmm, well, the unit vectors in polar coordinates (I dont remember their names in english ) are always perpendincular to one another, so they in particular are LI and span R2?

[little crater]

some line in that space like (1+t,2+2t,3+t)

and then have some distance r

like the black line at the top

and I want to rotate on this new axis i defined

oh

[verbal pivot]

hee

[little crater]

i see my issue might be

How you define that angle

i see

yeah i think it now falling apart because you have to give some way to define that angle rotation

i guess it would fall back into the old angles we used and that these new ones would just be a new definition or something

[verbal pivot]

mmm,

maybe you could just take the vector defining the line, take two perpendicular vectors to the vector of the line (so that way you have a orthogonal list that spans R3), and apply the usual spherical mapping to this new basis?

[little crater]

i guess

yeah just didn't know if you make define it something like

and now it rotates differently but i guess it just becomes a parameterization of x,y,z anyways

[zenith junco]

HELP

DOES ANYONE HAVE A PDF OF THIS TEXTBOOK

[verbal pivot]

Perhaps you could avoid the topic of defining the angle (maybe this is what you were pointing to with the last image) if you draw a circle that intersect twice the given line (and it's center on the origin), for each point of this circle you now can draw a vector (which in this case would be the analogous to rho in polar) and a vector perpendicular to it (which in this case would be the analogous to theta in polar), obtaining two vectors that span a plane containing that line, of course you can always write this point of the circle as a point mapped by cosine and sine, and with that you can avoid making a "geometric" argument for where the angle is defined (but of course, this angle coincides with that of the image you draw, if I understood correctly),

[wintry steppe]

Feel free to roast me

[nocturne jewel]

3rd line... what?

[verbal pivot]

can you explain what you did on the third line?

keep in mind that, on the left you have the sum of two determinants, which are just real numbers, and on the right you have a matrix, so that equality is pretty nonsense

[wintry steppe]

I was trying to distribute A^-1, and I thought that det(A* A^-1 ) was just 1

[hard drum]

it's the line before that that's the issue

not sure how second line here would follow from the first

[wintry steppe]

I was trying to distribute A^-1

[nocturne jewel]

multiplication doesnt magically become addition

is the problem

[wintry steppe]

Dont you have to foil

[nocturne jewel]

...

[wintry steppe]

lmfao

[nocturne jewel]

please enlighten me

on where you see something remotely of the form (a+b)(c+d) in det(A)det(B)

since det(A) is a number, det(B) is a number

so it's just the product of 2 numbers

[wintry steppe]

like wouldn't 3( xy) = 3x + 3y

[nocturne jewel]

????

You need to brush up on some basic algebra if you're saying 3xy=3x+3y

cause you are saying multiplication is identically addition

[wintry steppe]

I dont know im confusing myself

[nocturne jewel]

Assume A^(-1) exists ok

what can you tell me about AA^(-1)?

[wintry steppe]

The identitiy matrix

[nocturne jewel]

good

so $AA^{-1}=I_n$

[stoic python]

Mosh

[nocturne jewel]

take det of both sides

$det(AA^{-1})=det(I_n)$

[stoic python]

Mosh

[wintry steppe]

=1

[nocturne jewel]

yes

so you have 2 numbers, det(A) and det(A^(-1)) being multiplied together and getting 1

therefore, neither can be 0

cause assume one of them were 0, then you'd have 1=0 contradiction

$det(A)det(A^{-1})=1\implies det(A)\neq 0 \land det(A^{-1})\neq 0$

[stoic python]

Mosh

[wintry steppe]

Yeah I kept getting 1 = 1 but I think my issue is trying to get it into the form det(A) != 0 which is what I think they want. Thats what generated my huge mess

[half ice]

Must be a real relief to see it completed then!

[wintry steppe]

ohh wait

[nocturne jewel]

basically you just do contradiction w/ the assumption det(A) IS 0

[wintry steppe]

ur saying its not det(AA^-1) but det(A)det(A^-1)

[nocturne jewel]

then you get 1=0, contradiction, det(A)!=0

[nocturne jewel]

yes, by using the hint provided

[wintry steppe]

I love linear algebra

[nocturne jewel]

But yeah, I'd love to say what you did is salvagable, but it isnt

[wintry steppe]

🤡

[nocturne jewel]

Proof is tricky, especially w/ LinAl.. but you'll get there eventually

[hard drum]

cold blooded dang

jk

and ye it just takes practice

[wintry steppe]

I don't even know what I did , I was trying to hack together something that looked like 1 = something lmao

[nocturne jewel]

If you say multiplication is addition then I will be blunt \j

[hard drum]

I like how the question assumes A^-1 exists by saying A^-1 = B and then says 'if A^-1 exists'

[nocturne jewel]

The question itself doesn't

coding introduced that definition

[wintry steppe]

Oh no thats my bad i copied pasted from the last question lmao

[hard drum]

isn't that the question itself on the original ting

o

[wintry steppe]

thanks for catching that

[nocturne jewel]

The original just used A inverse, the B came up in the general product rule for det (not sure if it has a proper name)

[wintry steppe]

It was a really similar question so I just plopped it in

I gotta fix that too

Guess im just lost on where the crap B goes

Like how does it just magically disappear into AA^-1

[nocturne jewel]

wdym

[wintry steppe]

Isnt the matrix B a completely different matrix

[nocturne jewel]

B is a matrix with same dimensions as A

|AB| =|A||B| is just the general property

The A in it isnt the A in the question

[wintry steppe]

So confused

[limber sierra]

ignore the first line then

[wintry steppe]

I need to catch up to this part on Kahn Academy lmao

[wintry steppe]

Lmao best way to learn

[limber sierra]

or say B = A^-1

[hard drum]

the point is we're saying 'for all B, ...'

[limber sierra]

the first line was giving the general rule you'll be using, the rest of the work was applying that rule.

[wintry steppe]

is this true cause then it makes sense

[limber sierra]

yes, youre applying the rule |AB| = |A||B| with A = A, B = A⁻¹

[wintry steppe]

OHHHHHHHHHHHH

WHY TF DIDNT IT JUST SAY THAT

[nocturne jewel]

Cause it shouldn't have to tbh

[wintry steppe]

Its a substitution and you don't declare what your substituting ? Seem pretty confusing to me

[nocturne jewel]

no

[hard drum]

it's not a substitution

it's like saying 'all men are mortal' and then realising i'm a man so i'm mortal

it shouldn't need to say that if you let 'me' be a man then i'm mortal

[nocturne jewel]

Like, you know $u\cdot v = \norm{u}\norm{v}\cos(\theta)$, you can take the dot product of any 2 vectors you want, they dont have to be called u and v for you to use the dot product

[stoic python]

Mosh

[wintry steppe]

Its just saying B is suppposed to be the opposite of A ?

[nocturne jewel]

inverse of A

and no

[wintry steppe]

This whole things just dumb imo, terrible notation

[nocturne jewel]

it's just saying B is another nxn matrix

[nocturne jewel]

You're anti-formulas...?

[hard drum]

what was the original question?

[wintry steppe]

Anti-linear algebra notation yes

In calc if they have some confusing concept they at least have a different notation for it than another variable

[hard drum]

it sounds like the issue here is simply saying a formula holds for all (something)

not sure how it's unique to linear algebra

[nocturne jewel]

[hard drum]

what was the original statement of the question out of interest?

[nocturne jewel]

.

[hard drum]

oh soz

[wintry steppe]

Like if you take the determinant you'd denote it ` . Here its just like oh B isn't another matrix B even though that's how you'd write it , here you're supposed to guess that B actually meant B = A^-1 lmao

[hard drum]

sure, 'Use the fact that det(𝐴𝐵)=det(𝐴)det(𝐵) for 𝑛×𝑛 matrices 𝐴 and 𝐵.', so we can take B as A^-1, as A^-1 is an n x n matrix

B didn't mean A^-1 necessarily, it's a general statement

[nocturne jewel]

Yeah... they're not going to tell you how to do the proof

Like the thing you're getting upset at sounds like the fact they didnt tell you explicitly how to do the proof

[wintry steppe]

Im not upset I just think the notation in linear algebra is sub par, I prefer calculus , no hate

[nocturne jewel]

...

They're the same

Calc has general notations too......

$(uv)'(x)=(vu'+uv')(x)$

[stoic python]

Mosh

[wintry steppe]

[limber sierra]

lmao