linear-algebra

[hollow cedar]

and the other?

[broken hawk]

one is:
-a linear transformation which acts as a rotation (rigid motion with positive orientation) on a two-dimensional subspace, and does not act on the other base vectors (given an orthonormal basis)
the other is:
-an arbitrary composition of “simple rotations”, where a simple rotation is what I just described above as a rotation

the definitions are equivalent for n≤3

but matter for n≥4

[hollow cedar]

I see

so symmetrical matrix aren't rotational, except for Identity matrix which have a -1 somewhere on the diagonal

[broken hawk]

(in 4D you can have “double rotations”: rotate one space first, then rotate the other space which is strictly orthogonal to it)

an even number of -1

specifically

[hollow cedar]

4D?

how do I even visualise

[broken hawk]

how is that relevant

things exist regardless of whether you can visualize them

[hollow cedar]

I mean

rotating one space, and another which is orthogonal to it is so weird

sounding

[broken hawk]

well yea

higher dimensions are highly unintuitive

[hollow cedar]

yeah, that's why I couldn't visualise

[broken hawk]

I’ve come to terms with the idea of two mutually orthogonal planes which only intersect in a single point

but I still can’t visualize it

a double rotation matrix would btw just look like this

$$\begin{bmatrix}
\cos(\theta) & -\sin(\theta) & 0 & 0 \
\sin(\theta) & cos(\theta) & 0 & 0 \
0 & 0 & \cos(\phi) & -\sin(\phi) \
0 & 0 & \sin(\phi) & \cos(\phi)
\end{bmatrix}$$

[stoic python]

Sascha Baer:

[broken hawk]

up to reordering of the rows and columns

(in such a way that the cosines remain on the diagonal)

[hollow cedar]

hmm

interesting

[broken hawk]

one subspace is rotated by θ, the other by φ

as you can see, you can’t fit this into a 3x3 matrix

[hollow cedar]

yeah, kinda fuzzy in my brain

[broken hawk]

(and there is a theorem, which isn’t that trivial, that the composition of any two 3D rotation matrices will be another rotation matrix)

(where rotation here means simple rotation)

[hollow cedar]

Alright

that's quite a lot to take in tbh

[broken hawk]

think more about four dimensions

^^

[hollow cedar]

😄

btw bout the question

if symmetric

wait

part ii is true

cuz

cuz we can't take any rotational matrix as a counter example here

yes?

[broken hawk]

I’ve since forgotten the question

[hollow cedar]

second

https://gyazo.com/21199ff78a0def902df5fac1b6c21cb6

[broken hawk]

uh I believe there’s a theorem that says that if a matrix is diagonalizable, then its eigenvalues will be equal to its svd

however!

sqrt(λ²) will always be positive, even if λ itself was negative

[hollow cedar]

it's not lambda^2 tho

it's lambda of M^2

[broken hawk]

yes, which happens to be the same

[hollow cedar]

oh right

[broken hawk]

see: my proof from before

[hollow cedar]

lambda ^ k

-.-

my bad yeah

[broken hawk]

so the first line breaks on λ(M) = svd(M), and the second breaks on λ(M) = √(λ(M²))

[hollow cedar]

what do you mean 'first line'?

[broken hawk]

problem 1

[hollow cedar]

oh

so i ain't true cuz we could pull up a rotational matrix

and ii is always true

[broken hawk]

uh

reread what I wrote just before

[hollow cedar]

uhmm

first line breaks on λ(M) = svd(M)
Cuz we can pull up a rotational matrix
the second breaks on λ(M) = √(λ(M²))
That's always true cuz

oh wait,

lhs might be negative, whereas rhs will always be positive

[broken hawk]

to make it explicit:

$$\lambda(\begin{bmatrix} -1 & 0 \ 0 & -1 \end{bmatrix}) = (-1, -1) \neq \sqrt{\lambda(\begin{bmatrix} -1 & 0 \ 0 & -1 \end{bmatrix}^2)} = (1,1)$$

[stoic python]

Sascha Baer:

[hollow cedar]

Right

[broken hawk]

as it turns out the essence of multiple choice is knowing a fuckload of counterexamples

[hollow cedar]

when will the SVD then, in general equal the eigvalDec?

[broken hawk]

I believe whenever the matrix is diagonalizable, but don’t quote me on that

I have very little intuition for svd

it might acutally be whenever the matrix is “orthogonally diagonalizable”

that is, diagonalizable by conjugation with orthogonal matrices

yea, that sounds more reasonable

since in SVD you always have PΣQ where P and Q are orthgonal

and I believe orthogonally diagonalizable over $\mathbb{C}$ is equivalent to the condition that $A\bar{A}^T = \bar{A}^T A$

[stoic python]

Sascha Baer:

[hollow cedar]

but that's over C

[broken hawk]

and over ℝ I believe it’s whenever the matrix is symmetric

[hollow cedar]

say if I wanna restrict this to R

[broken hawk]

but in both cases I’m not 100% sure

[hollow cedar]

what do you mean whenever the matrix is symmetric?

SVD(M)=EGVD(M) whenever M is symmetric? But no..?

[broken hawk]

I know for a fact that whenever the matrix is symmetric (A = Aᵀ), then the matrix can be orthogonally diagonalized

I’m however not entirely sure if that’s an equivalence

and I claim that whenever the matrix can be orthgonally diagonalized, then SVD = EVD

[hollow cedar]

Oh

so part ii woudl have been true

then

if

[broken hawk]

if the root bit wasn’t there

[hollow cedar]

there wasn't the square rooted term in there

[broken hawk]

yes

I believe so

[hollow cedar]

mhmm, what's the SVD of that example btw

that's equal to EGVD of the matrix

okay

so yeah

[broken hawk]

I’ve asked a friend what she thinks, will update you if she answers

[hollow cedar]

Sure, and thanks

like for real.

[broken hawk]

she's better than me at the whole remembering theorems bit

i'm more of the "know a fuckload of examples and counterexamples" guy

if i don't know any counterexamples it's true in my head ^^

(dangerous mindset, I know)

[broken hawk]

okay @hollow cedar while she also didn’t know about svd, the symmetric ⇔ orthogonally diagonalizable is an equivalence

proof:

symmetric ⇒ orthognally diagonalizable is the spectral theorem, not gonna prove that

[woeful hawk]

gu ys

i need help

a/3 = b/7 basically i have to find a and b and then i have to calculate (a+b)/(a+3b)

i mean the second step is easy, but i have to find a and b

[broken hawk]

could you please notice the fact that I’m in teh middle of explaining something

and not butt in?

[hollow cedar]

...

[woeful hawk]

and im in a hurry to solve my exercises

[broken hawk]

there’s a ton of question channels

go there

[woeful hawk]

ok

[broken hawk]

this isn’t even linear algebra

[woeful hawk]

then what is it lol

[hollow cedar]

I was gonna call Woog

[broken hawk]

assume A is orthogonally diagonalizable. then A = Q⁻¹ΛQ = QᵀΛQ = QᵀΛᵀQ = (QᵀΛQ)ᵀ = Aᵀ, using both the fact that Q is orthogonal and that Λ is diagonal and therefore Λ=Λᵀ

so A is symmetric

[hollow cedar]

Oh okay that makes sense

what were we gonna use this for?

establishing EGVD=SVD

for symmetrics, in R

right?

[broken hawk]

yea. and I believe that should actually show it based on the following reasoning:

the SVD is unique. if the diagonalization uses orthogonal matrices, it is an SVD. therefore it is the SVD

(up to reordering of the diagonal values)

[wintry steppe]

yo can anyone explain me what a kernel and image of an operator is? i understand it in a function way, but i don't understand how kernel/image can be basis and why if operator is nxn n=dimension of kernel+dimension of image

by operator i meant linear transformation matrix

[broken hawk]

I can try

(sorry, I said something that sounded right, but wasn’t)

okay, I think I have thought of a correct way to explain it that should give some intuition

let’s do this in a bit of a “proof” style, actually

okay, first of all, consider the easiest case: the matrix has full rank. that means, the image is an n-dimensional space

in other words, every basis vector will be sent to a linearly independent location

so if you take your n basis vectors… let’s say n=10

you look where they land, and you see that they still span a 10-dimensional space

they’re still a basis

but now, let’s change this up a bit

take that same transformation from before, but tweak the target of one of the basis vectors so that it lies inside the span of the other nine

so now the image is only 9-dimensional, right?

it’s gotten a bit “flatter”

but now the thing that’s happened is that when you flatten it, you’ll automatically end up with some line that gets squeezed into 0

that line is the kernel, and it’s now 1-dimensional

so 9+1 = 10 still

now, since that is hard to visualize, let’s go down a few dimensions and think of n=2

ie a plane

pick your favourite basis, and do a transformation to it. there’s three things that could happen

1. the two vectors stay independent. in this case, the image has dimension 2, and the kernel is only {0}, so dimension 0

2. the two vectors become parallel. think of how such a transformation would look. try to imagine how it squeezes the plane together into a line. if you have good imagination (or have watched 3blue1brown’s video about this, which I really recommend), you’ll notice that there should be some line of vectors that simply get squished down into 0. that line is the kernel. the easiest case is of course if one of the basis vectors themself goes to 0. either way, the image is 1-dimensional and the kernel too

3. both basis vectors go to 0. in this case the entire plane just shrinks down to 0. image is 0-dimensional and the kernel 2-dimensional

so you see that every time we lose a dimension in the image, we automatically gain one in the kernel

that’s the best I can do @wintry steppe

I recommend you watch the 3blue1brown video series, he does a fantastic job at this

[wintry steppe]

oh god, didn't see ur response 😄

thank you so much

[broken hawk]

read through it, come back with questions

but also like just watch 3b1b

[wintry steppe]

ait thanks 😃

[broken hawk]

it’s the second video in this playlist https://www.youtube.com/watch?v=fNk_zzaMoSs&list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab

but start at the beginning

and then don’t stop till the end :)

this’ll give you all the intuition you’ll ever need for linalg 1

[wintry steppe]

ill skip through it hahah i understand it for the most part 😛 just was wondering about how kernel/image

[broken hawk]

aight

[wintry steppe]

but thanks :))

[broken hawk]

in that case yea pretty much just think whta it means to lose a dimension in the image

how that could happen

you’ll see that if you require linearity, you must gain a dimension in the kernel

(also honestly this series is good to watch even if you understand stuff already)

[wintry steppe]

aye

could give another perspective

[broken hawk]

(I’ve watched it several times and have always taken something from it)

[wintry steppe]

watched one of his videos the other day, they're great

[broken hawk]

oh yea watch them all

[wintry spruce]

If something says

coordinates of aj with respect to the basis a1 ... am

where m<j

Does this just mean that there's some set of basis vectors a1 ... am which are
linearly independent and aj can be written as a summation of them?

such as

aj = x1*a1 + x2*a2 + ... + xm*am

For some reason the wording has confused me

[loud prawn]

Hello everyone!

I need to solve the equation:
$ A \vec{p} = 0 $

[wintry steppe]

imagine using 3b1b for linear algebra references

peak brainletism

[stoic python]

tarantino:

[loud prawn]

So, yes, how can I do it? If there were 1 instead of 0, I would use an inverse matrix

p.s. Ow, yes, I need to find the vector

[broken hawk]

since you didn’t put any constraints I’m just gonna give you the answer you don’t wanna hear, which is p=0

(in all seriousness: for there to be any nonzero solutions, A must not be invertible in the first place, and you need to find a vector in the kernel of A. for small matrices you can honestly usually just guess these)

for larger ones, well, this is just a system of linear equations

which should have at least one degree of freedom (else 0 will be the only solution), so you can set one of the variables to 0 or 1 or whatever is convenient and then just solve them

what will always work but is probably too much effort in most situations is computing the LU-decomposition of A and then solving Up=0 instead

which will then be rather trivial

[leaden ermine]

Hi, quick question. I know the answer but was hoping someone could explain it better. Like what's a kernel, maybe an easy example etc. Thx in advance!!

Let A be a d×d matrix with rank k. Consider the set SA := {x ∈ Rd|Ax = 0}. What is the dimension of SA?

Can post what I have if needed

[wintry steppe]

$d-k$

[stoic python]

TendentiousTorturousTopics:

[wintry steppe]

The rank-nullity theorem says that the rank + the nullity = d; the nullity is the dimension of the kernel.

[leaden ermine]

Hey again!

So what does Ax=0 exactly mean? I'm unfamiliar with this theorem and was just doing some googling

ah think I got it, so rank(A) + nullity(A) = n

where n is nubmer of columns

[wintry steppe]

Ax = 0 just means that the matrix-vector product is 0

Now, one can ask the dimension of the vector space containing all x that satisfy this equation for a given A

[leaden ermine]

so the dot product of the matrix and the vector is zero?

is that the same as orthogonal?

or am i mixing up terms

[half ice]

There's no dot product between matricies and vectors. That's a matrix multiplication

[leaden ermine]

ah dot has to be same sized lengths

[half ice]

No, there's no dot product between matrices and vectors

Orthogonal vectors are two vectors that don't line up. Matricies don't have the same idea

[leaden ermine]

ok so im confusing concepts

so when he refers to the matrix vector product its simply like a system of equations?

let me see if i can find an example

[half ice]

You can multiply two matricies together and they make a new matrix. That's a matrix multiplication

[leaden ermine]

yep

[half ice]

This is the same idea, think of a vector as a 1-column matrix

[leaden ermine]

So we have a vector times a matrix, and their sum must be zero? and that means Ax=0?

[half ice]

The Kernel of A is all vectors x such that Ax = 0

Or, the vectors that A "annihilates"

[leaden ermine]

ah ok so kernel basically means = 0

[half ice]

The kernel is a set of vectors, it depends on A

It tells you a lot about A

[leaden ermine]

So for my question, we have a matrix called A which is dxd (Which I suppose means same number of unknown columns times rows) with rank k (I looked up how to calculate rank)

So in order to find the dimensionality of A we need to find the nullity

[half ice]

Sorry, I take that back lol

[leaden ermine]

so we need to find the vector tht makes A zero

and that provides the nullity?

sorry if im all over the place

trying to wrap my head around this

[half ice]

Rank-nullity theorem:
Rank + Nullity = d

If you know the rank, and if you know d, the nullity comes right away

Or, you can find the vectors such that Ax = 0, they will make a subspace. The dimension of that space is the nullity

[leaden ermine]

got it 😃

[glass sluice]

🆙 | Shocks leveled up!

[leaden ermine]

ok second question

Assume Sv is a k dimensional subspace in Rd and v1,v2,··· ,vk form an orthonormal basis of Sv. Let w be an arbitrary vector in Rd. Find x∗ = argmin x∈Sv w−x2, where w −x2 is the Euclidean distance between w and x. Express x∗ as a function of v1, v2,...,vn and w.

eh it came out abd

bad

ill take a screenshot

am i understanding this correctly so far?

actually maybe what i typed at the end was basically the answer?

x star = sqrt ((w1-x1)^2 + (w2-x2)^2 + .... (wk-xk)^2)

[wintry steppe]

x* is a vector

You should be reading more carefully.

It's asking you to find the closest vector in S_v to w.

[leaden ermine]

hm, does the orthogonal piece come int oplay at all?

im going to try and draw it out so i understand what theyre asking better

[wintry steppe]

Yes. You need an orthonormal basis to compute the vector that you need.

[leaden ermine]

eh im a bit stuck on this one, so just from the beginning we have a vector w

and its orthonormal(orthogonal) vector drawn as well

and we want to find the shortest distance from the end of the w vector, to the end of its orthogonal vector?

[wintry steppe]

you have a linear subspace

envision a plane in R^3 that crosses the origin and think of that as your subspace

that's a 2-dimensional subspace

now what's the vector in that subspace that's closest to your vector?

[wintry steppe]

About time last time i asked for this channel was scorned ,

[sand hedge]

spam enough and woog will relent eventually

[wintry steppe]

You are well on your way it seems

[stoic python]

Sascha Baer:

[broken hawk]

cause to me l2 norm sounds like you either mean the norm in ℓ² (space of square-summable sequences) or the norm in L² (space of square integrable functions… ish)

however, if I am understanding you right
closest vector is whichever minimized the norm of the difference, yes (with regards to whichever norm you are using, i.e. the one induced by your inner product)
and yes, that difference would be √2 with the regular 2-norm

[broken hawk]

let me actually look at the question perhaps

okay so here’s the important bits that I think you missed somewhat

basically, S is some subspace and v is a vector which is not necessarily in that subspace. and you have to find the one vector closest to v inside the subspace

obviously if S was the entire space that would just be v itself

I don’t know what you mean with “so just from the beginning we have a vector w
and its orthonormal(orthogonal) vector drawn as well”

what do you mean with its orthogonal vector?

you don’t have an orthogonal vector given

at all

you completely misunderstood the question

the visual you should have is some vector w in ℝ³, completely arbitrary direction, and a plane through the origin, again oriented in some different arbitrary directions. and you want to find the vector nearest to w inside that plane

[hazy fiber]

Can someone tell me why that's not a vector subspace ?

[broken hawk]

what have you tried to show and what have youmanaged so far?

[hazy fiber]

I know that if a vector s exists on W, and a vector v exists on W, then s+v should exist in W

[broken hawk]

and is that the case?

what would you have to prove for that to be the case?

[hazy fiber]

That's W

[broken hawk]

the black ones you mean

including the lines, of course

[hazy fiber]

So I'm not sure if theres any operation with v and s, creates a vector out of this picture

Yes

[broken hawk]

well, from the image, it is obvious that scalar multiplication will keep you in the set

[hazy fiber]

Yes

[broken hawk]

but consider e.g. being in the black set and then going straight down

or even just like, diagonally down

[hazy fiber]

But something like v - s or something like that can create that vector that I'm looking for it

[broken hawk]

shocks please just reread what was written a few more times, it’s been said about three times what you are looking for

you’re looking for a vector in that plane

that is as close as possible to the given one (which is not necessarily in the plane)

[hazy fiber]

Oh GG

Here is it

So w is out from the subspace

[broken hawk]

yep that would work

think about projections

[west widget]

wow

[broken hawk]

what? how can any arbitrary vector have distance sqrt2?

i dont see what you are though

[kind stone]

linear algebra == vectorial algebra?

Is a stupid question but i want to know if it is different

[broken hawk]

linear algebra is the study of vector spaces and linearity. this encompasses a whole lot.
on the more practical side, it is stuff like working with matrices and vectors and studying the structure of, say, 3D-space
on the more abstract side, this stuff can be extended to much more generality and a lot of things you can say about “arrows in space” also works for e.g. functions

for example, just like you can add vectors together, you can add functions together, and the derivative of a function shares many properties with a rotation matrix (they’re both linear operators)

[kind stone]

@broken hawk @leaden ermine in conclusion are the same

oh i see

thanks

[sand hedge]

No one rly cares ngl, most you’ll get is “this doesn’t belong here”

[half ice]

Agreed, unless you are way off with your channel choice, people are pretty chill

A lot of people do calc in pre-calc, you have to call that out

[broken hawk]

or you disregard the fact that a convo is ongoing to ask a question instead of going to #❓how-to-get-help

[sand hedge]

you’re not allowed to agree with me, woog forbids it

[half ice]

That one is annoying and we're always trying to get people to stop doing that

[sand hedge]

Ppl are lazy, can’t rly combat it sadly lmao

[half ice]

Gotta love people who find the busiest channel because they think they'll have the best odds of getting answered there

[sand hedge]

Deter sure but some doinks will still do it

[half ice]

Then post a huge sideways picture

[sand hedge]

Kek Purs rotate is a god send

[half ice]

A++ bot command

Pur is a godsend, all of the bot commands are much better after the last few months

[sand hedge]

Would give a uwu but woog purged one of the most used emotes

[broken hawk]

does it rotate this tho

[half ice]

🤔

[broken hawk]

no not you

[sand hedge]

,rotate 180

[stoic python]

Couldn't find an attached image in the last 10 messages

[sand hedge]

emote support pls uwu

[placid trench]

#combinatorics

[charred stirrup]

for 3. my understand is that multiplying a vector by a scalar of '0' will simply make the vector 0

what is the significance of saying 0 is perpendicular to the vector? I am confused by why that statement is written there

[burnt gate]

if vector a dot vector b = 0, then the vectors are perpendicular, by definition, and the zero vector dot any vector = 0, so the zero vector is perpendicular to any vector

but i don't know the context of that

[charred stirrup]

in linear algebra, do you also say that a zero vector is also parallel to all vectors?

[burnt gate]

vectors are parallel if they're scalar multiples of each other, and the zero vector = 0 times any vector, so yes

[charred stirrup]

@burnt gate thank you, i really appreciate your help

[burnt gate]

np

[charred stirrup]

what is the (a subscript i [ai] ) mean? does the i stand for initial?

[slim maple]

it's just a label for something

i can have a list of numbers 5, 3, 6, 4, ...

and relabel them a_1, a_2, a_3

etc

i is just the index

[charred stirrup]

A trace of a certain linear equation would mean a set of all b for every variable x?

[burnt gate]

here, you can read it as "either a_1 != 0 or a_2 != 0 or a_3 != 0 ... or a_n != 0"

[wintry steppe]

https://www.pearson.com/us/higher-education/product/Lay-Linear-Algebra-and-Its-Applications-4th-Edition/9780321385178.html

[winter sage]

hey, does anyone know what this means?

i'm thinking it just produces vectors between the two original ones?

a and b are both vectors

arbitrary ones

[burnt gate]

i think it wants you to draw the region

[wintry steppe]

"convex" enclosure

it has a name

the link i posted, chapter 4 or 5

read the best books, on LA, thats the link i posted , best book on LA you will ever read

[burnt gate]

there's also the condition that s + t = 1

which should reduce the dimension by one

hang on, it's just a straight line, isn't it

[charred stirrup]

how does a trace differ from a hyper plane?

[broken hawk]

? I don't see any connection between the two?

a trace is a number

a hyperplane is a geometric object

[charred stirrup]

i think i might be severely mistaken in their meaning

a trace is not a set of all possible points that a linear equation can have?

i also thought a hyper plane is a set of points or solutions?

[broken hawk]

trace to me is the sum of the diagonal values of a matrix

and one of those concepts that still elude me

[charred stirrup]

what definition should I have for a hyper plane?

because that definition in the picture and geometric object one confuse me

somebody explained it as "imagine you have a cube, and then imagine you stack an a large amount of square sized pieces of paper to make a cube"

each sheet of paper is a hyperplane

[broken hawk]

hyperplane I would define as an n-1 dimensional subspace

[wintry steppe]

yee

[broken hawk]

well, or a shifted version of thay

[charred stirrup]

uhh... n-1 as n is the dimension?

[broken hawk]

yes

[charred stirrup]

oh shit

thank you so much lol

so like in R^4 a hyperplane is a 3d object??

[broken hawk]

I would say a hyperplane is the set of all points $$\lbrace \mathbf{a} + \mathbf{u}: \mathbf{u} \in U\rbrace$$ where U is a given n-1-dimensional subspace and a is a fixed vector

the +a allows you to have it not go through the origin

[charred stirrup]

i love u right now

[stoic python]

Sascha Baer:

[charred stirrup]

then the geometrical definition i described, it just so happens to be that the set of all points makes those shapes?

[broken hawk]

well, your hyperplanes were finite

in linalg they'd definitely be infinite planes

[charred stirrup]

each subset of points in 'U' is one of the many infinite sheets of paper?

[broken hawk]

huh? no

a subset of U could be anything

[charred stirrup]

sorry, not subset

[broken hawk]

but they'd all lie within a "plane"

[charred stirrup]

like, each "point" inside the set U

i dont know what to call it lol

[broken hawk]

basically
in 2D, a hyperplane is a line
in 3D it's a plane
in 4D it's an infinite volume

etc

the U is there to define that as a subspace, and the a to push it away from the origin, cause all subspaces go through the origin

and this makes sense cause for systems of linear equations, the solution space is
(solution space to the homogenous system) + (one particular solution to the inhomogeneous system)

and the solution space to the homogenous system is in general any subspace, but if you only have one equation it'll be n-1 dimensional

and so we get back to the definition you had there

[charred stirrup]

what is the difference between the two formulas when im trying to solve for roots of unity?

[broken hawk]

the first is specifically roots of 1
the second is roots of any number. if you plug the values for 1 in (r=1, θ=0) then you should get the same as above

[charred stirrup]

bless your soul dude

you take paypal?

[broken hawk]

well that’s a first
nah I’d feel bad about that. if I was actively tutoring you I’d demand money but I’m just doing this for fun here

[charred stirrup]

lmk them rates and ill hit you up in your free time

[broken hawk]

I don’t think I have time for tutoring, sorry

[charred stirrup]

for the first fomula, let's say z= a+ bi

then z^n = 1 ?

[broken hawk]

but I have some time rn for questions

[charred stirrup]

that's a shame lol

[broken hawk]

yea

school life’s busy and I’m already a teaching assistant

[charred stirrup]

for the second formula, i use it for z^n and then i will find the roots of Z?

[broken hawk]

first formula tells you all numbers ζ with ζⁿ = 1
second formula tells you all numbers ζ with ζⁿ = r e^(iθ)

for your given r and θ

[wintry steppe]

@broken hawk hey u here?

[broken hawk]

nope

[wintry steppe]

😦

i have a question regarding the linear transformation i asked about the other day, just had time to watch 3blue1brown video and stuff

so what u said if rows/columns of a matrice are independent (has a full rank) that means that the dimension of kernel will be 0?

haven't managed to find his video explaining kernel/image 😦

[fiery tree]

assuming that the dimension of your domain is equal to your rank, yes

https://en.wikipedia.org/wiki/Rank–nullity_theorem

[broken hawk]

yes, if the rank is full, the kernel is trivial

this one is actually easy to explain too: if the rank is full, that means all the columns are linearly independent, which means the only linear combination of the basis vectors that gets mapped to 0 is 0 itself

[main karma]

X possibilities?

[winter reef]

either thing in brackets equal to 1 or the thing in the power equal to 0

[remote sage]

@main karma The total possibilities for this problem is x = 5, x = 2, x = 7, and x = 6

These are the steps I did when working on the problem.

So you'll end up with four X solutions.

[main karma]

Ty

[remote sage]

Absolutely. 😃

[vestal ravine]

this feels so basic, but i just cant remember anymore -.- what was the calculation to figure out if this one is linear or not?

[half ice]

This is a linear system of two equations, if that's what you're asking

It's not linear if it has x², xy, or anything like that. You can't multiply unknowns together

[vestal ravine]

yeah but "they said" put it all to the left = 0 and then use whatever formula (midnight or pq) to solve it, if it doesnt "solve" its not linear and vise versa - but i have no clue atm. how i should actually do that with a x y and z (and yes, that sounds stupid, i knew it a few years back and could bite myself for actually not remembering)

[broken hawk]

but… midnight is for quadratic equations?

how does this at all apply

a linear equation is just one where all unknowns appear in the form ax, where a is some number and x is the unknown

and, optionally, there may also be a single constant term

(in which case it’s an inhomogenous equation)

(which are a bit more annoying)

[half ice]

You can't solve this as there's 3 unknowns, but only 2 equations. It is still a linear system of equations.

[broken hawk]

I mean you can solve it, it’s just not got a unique solution

the solution’ll be a line

[vestal ravine]

i think u did hit something in my brain my mentioning "inhomogenous equations" i might have to look into it, pretty sure i had some writings of it somewhere

[charred stirrup]

does a vector which has "length 1" mean its magnitude is equal to 1?

and, for this type of question, the only way i could solve it was by simply guess (x,y) to be some real number and solving for z to make it true where 31x+4y+108z = 0

is there a more "legitimate" way to solve it, or a certain procedure I should follow? since i am simply just finding one of the infinite amount of vectors which will be perpendicular to the given vector x

[cobalt bough]

I'm worried about my performance in linear algebra. I got well below the class average on the last midterm. Are there any good online resources I can use to solidify the basics?

Linear independence, Ax=0, etc

I feel like I'm not missing much I just need to fill my knowledge gaps and be more confident

Especially something like a theorem cheat sheet I could go over would be good

[wintry steppe]

Most of the theorems aren't something to memorize; everything tends to connect in linear algebra.

Try Khan Academy's linear algebra stuff, and read Linear Algebra Done Right; if you're trying to memorize things in linear algebra, it will be a rough time for you.

[plush fable]

Find the determinant of this matrix

[glass sluice]

🆙 | thedon leveled up!

[half ice]

Could make it easier by doing some row reduction first

R1 - 2R2 and R3 - 4R2:

[0 -2 -5]
[1 6 1]
[0 -25 -2]

Taking the determinant along the first column:
-1 × [-2 -5]
[-25 -2]
= 121

[wintry steppe]

meh if you're going to do row reduction, might as well just do it all the way to a triangular

that's like the best algorithm

taking a determinant against a column is cool though

[hushed prairie]

What’s an extraneous root ??

[wintry steppe]

Can I get a duh help for a math problem

[wintry steppe]

I have to write a system of equations

Disregard the minimum problem

I'm assuming the inflows have to equal the outflows?

[wintry steppe]

Yes

Very introductory

Idk if what I’m doing is right

Nevermind. I think I got it

[crimson oak]

hey guys, does anyone have any proof based questions that require you to use the replacement theorem, if so it would be greatly appreciated if you could share them with me, thanks!

[cobalt bough]

@wintry steppe that's what our professor has been saying and I agree, I just wanted to have a reference so I could make exercises around them and really internalize them

Your suggestions sound useful though!

Hopefully what will happen is as I go deeper into the course and more connections are made things will "click" more

[wintry steppe]

That's the way it was for me.

[cobalt bough]

That's what happened to me in Calc AB in high school. I literally had a C until at some point I realized it was all the same concepts coming up over and over again and it clicked

I then aced the final

It was fucking awesome

[broken hawk]

@charred stirrup yes, length 1 = magnitude 1
as for finding two orthogonal vectors: guessing the way you did is the fastest approach to find one in such a simple case. To find a third, you could compute the cross of the found vectors.
If you're given a larger set of orthogonal vectors and need to find one more, Gram-Schmidt procedure always works but it's tedious

[west widget]

wum

[snow snow]

"Through three of the points with the coordinates (2,-7), (8,35), (12,61), and (17,98) can you form a straight line, which ones?"

How do I do this one?

The coordinates are way too big to draw out on paper and I'm not going to learn anything from using Desmos.

I recall something about being able to use multiplication with this stuff but I'm not too familiar with it.

[ornate widget]

Hello I have been watching a lot of linear algebra videos and finally understand what eigen values and vector means. Now I want to actually find eigen values

I understand that Ax = lamda x

watching another videos says that to calculate the eigen values we need to do (A- lamda I ) X =0

and that det (A- lamda I) = 0 but why?

i watched another video and it turns out the det is a kind of ratio pointing to how much a cube was scaled after being transformed to another cube with the help of a matrix

[broken hawk]

@snow snow let’s do this abstractly. let’s say x, y and z are three points on a line (each of these would be a pair of coordinates). then there exists some vector, call it v with y = x + t*v and z = x + s*v, where t and s are numbers

so basically what you have to do is, you start with one of the points, take the vector connecting it to a second and see if you can get one of the remaining points by scaling that vector

in a simpler numerical example

let’s take the points (1,1), (2,3) and (5,9)

you can write (2,3) = (1,1) + 1·(1,2), right?

here y = (2,3), x=(1,1), t=1 and v=(1,2)

and then you can see also that (5,9) = (1,1) + 4·(1,2)

so they’re all on the same line

so basically

[snow snow]

you can write (2,3) = (1,1) + 1·(1,2), right?
No idea, I'm very shaky on this specific part of the book

[broken hawk]

well, test it

don’t just believe me blindly

work it out

all I did was subtract (1,1) from (2,3)

which gives me (2-1, 3-1) = (1,2)

[snow snow]

Sure

I get that

[broken hawk]

and then I wrote the 1· to bring it into the same form as I wrote above

but ofc it’s not necessary

here, lemme show a different example

lemme take three points, oh, idk, let’s say (1,1) again, then (2,4) and (3,5)

now let’s take a look at the differences

(2,4) - (1,1) is (1,3)

and (3,5) - (1,1) is (2,4)

now, if those three points are to lie on a line

these two differnces would have to be multiples of one another

(convince yourself of this!)

[snow snow]

Alright, but how do I check if that's true for much larger coordinates like (17,98) like I have?

[broken hawk]

well, anyway, is (2,4) a multiple of (1,3)?

[snow snow]

No

[broken hawk]

so they’re not on a line

alright, now, for you

pick one of your points as a “focus”

say the smallest one

subtract that from the other three

[snow snow]

This is actually starting to make a bit more sense

[broken hawk]

to get the directions the other three are away from it

then, if you can find a pair which is multiples of one another you’ve found your three points; if not then you have to check the last remaining triple

[snow snow]

So from what I've gathered, a line is an infinite series of scaling points, so every point on the line must be a multiple of any other point, right?

[broken hawk]

mmmh almost

that works if the line goes through 0

but otherwise you have to add a constant to it

to shift the line

[snow snow]

Yeah.

Is there any way to find the y = mx+b equation without writing the graph?

[broken hawk]

by subtracting one of the points we’re essentailly getting rid of that constant

[snow snow]

Actually, that doesn't have anything to do with my question.

Ignore that

[broken hawk]

uh yea, you can enter the knwon values for y and x

and then solve a sysstem of equations

you’ll get one equation for every point

[snow snow]

Let's go back to the multiples part

[broken hawk]

yes

ugh why do I not have my drawing tablet with me

I could so do with adding a picture to this

but touch-pad drawn stuff is garbage

^^

well whatever

imagine there was a pretty picture

the differences between the coordinates are vectors pointing from one point to the other

you want to see if any of them are multiples of each other

[snow snow]

Yes

[broken hawk]

since there are only four possible gruops of 3 points that shouldn’t be too much work to verify

I believe oyu should be able to do it now

[snow snow]

If I want to find if they are multiples of each other, couldn't I divide the bigger coordinates by the focused ones to see if they both divide into an integer?

[broken hawk]

but feel free to do it here

well, no, again because they’re not on a line going through 0

[snow snow]

I think that's where I kinda get stuck through this whole explanation.

[broken hawk]

what we’re doing with subtraction is essnetially moving that chosen point to 0

[snow snow]

The part of the line going through 0

[broken hawk]

and the whole plane with it

and then, yes, you can do just that

though it doesn’t have to be an integer

it just has to be the same value in both coordinates

[snow snow]

Oh yeah

[broken hawk]

so yea, I think you should be able to do it

[snow snow]

Hmm

I'll try

Thanks for the help!

[broken hawk]

np

[snow snow]

Alright, I solved it.

I'm at the last and hardest question of the chapter now.

But I'm going to have to ask my teacher about it tomorrow.

[half ice]

@ornate widget
Were you trying to ask a question?

[tidal nebula]

#help-9 ?

[versed mauve]

hi can someone help me with this question

i am not sure how to do it

i know that all v's and w's are orthogogonal to eachother and so v.w = 0 for all w,v

i am not sure how to use that. i have made the deduction that their union is all v1, v2, .... vn and w1, w2, ... wm since we know that v =/= w for all v,w due to the orthogonal thing

but idk... i require a hint

[charred stirrup]

@broken hawk thank you again, always got a solid and concise answer

[broken hawk]

what did I do this time

[charred stirrup]

last night lol, you answered my question and i woke up and got it

lin algebra midterm tomorrow, just been studying all day

[broken hawk]

good luck. go to sleep early

rest > studying

esp after already doing it all day

[stone drum]

@versed mauve What is the condition for linear independence?

[charred stirrup]

still got about 6 ish hours til bed

[broken hawk]

american time, right

[versed mauve]

all of the coefficients must be 0 if the linear combination is 0

[stone drum]

So that's where you need to end up.

[charred stirrup]

ya boy repping out canada

[versed mauve]

yes i was thinking about using contradiction

so assuming there exists some coefficient =/= 0

and then finding a contradiction

[stone drum]

Seems like a reasonable direction.

[versed mauve]

the thing is i really do not want express each vector as a coordinate

[stone drum]

Make sure to use the fact that the set of vs and ws are linearly independent separately.

[versed mauve]

cause so many symbols and it is so confusing

is it true that the cardinality of the union of V and W is the same as |V| + |W| because all vs being orthogonal to ws implies they are all unique?

[stone drum]

It is if the unified set is linearly independent.

[versed mauve]

yes but i need to use orthogonal somehow]

like the unified set has no duplicates

if some v = some w then it is eliminated in the unified set

so writing down the union is hard

i am thinking this is why they tell us they're orthogonal

[stone drum]

You're on the right track.

But the cardinality rule comes because the two sets are linearly independent, not the other way around.

[versed mauve]

so i am doing a1v1 + a2v2 + ... + anvn + b1w1 + b2w2 +... + bmwm = 0

[stone drum]

Right.

[versed mauve]

but i have no idea where to go from here

[stone drum]

What does it mean that the v set is linearly independent?

[versed mauve]

that if a1v1 + a2v2 + ... + anvn = 0 then all a's = 0

but it doesn't neccessarily = 0

[stone drum]

And the same is true for the ws and bs

[versed mauve]

yea

[stone drum]

So -if- the av + bw set is -not- linearly independent then what must be true?

[versed mauve]

it is linearly dependent XD

[stone drum]

Yes, but we're doing proof by contradiction.

How would you get a 0 vector from two separate sets that were both linearly independent?

Think x + (-x) ...

[versed mauve]

but av = -bw

[stone drum]

Right!

[versed mauve]

i did this before

it didn't seem to lead anywhere

[stone drum]

Now, what condition in the question haven't we used yet?

[versed mauve]

the orthogonality

[stone drum]

Right.

[versed mauve]

we know that some v . some w = 0

[stone drum]

Is it possible to get av = -bw if v and w are orthogonal?

Not just some v. some w = 0

Every v is orthogonal to every w.

[versed mauve]

so like the v's exist in another dimension to the w's

[stone drum]

Yes

[versed mauve]

i can picture it

like going in the negative direction doesn't enter another dimension

[stone drum]

Right.

[versed mauve]

but that is really informal

[stone drum]

True.

[versed mauve]

the only mathematical thing i know about the orthogonal condition is that v . w = 0 for all v, w

so somehow i need to get a dot product

[stone drum]

Yes.

[versed mauve]

i can see that (a1, a2, ... an) . (v1, v2, ... vn) = - (b1, b2, ... bm) . (w1, w2, ... wm)

i want to somehow merge the a's and the b's

i can't though because there are n items in the a's and m items in the b's

so i think this is not the correct way

[stone drum]

Try doing it with one 1 v and w.

[versed mauve]

oo

i can only seem to get a v and a w in the same set

this is so fking confusing

[stone drum]

https://yutsumura.com/orthogonal-nonzero-vectors-are-linearly-independent/

[broken hawk]

why does this site assume ℝⁿ as the vector space, that is like a completely unnecessary assumption

[versed mauve]

the v's and w's are only orthogonal to eachother, so the only orthogonal set that can be guaranteed is one where there is a single v and a single w

but then the number of v's and the number of w's aren't the same, so i cannot pair each v and w

i am looking into their method

i am confused because the union is not an orthogonal set

[charred stirrup]

how is this read? does the left side not read "magnitude of vectors v + u squared" ?

[versed mauve]

that's ambiguous because it's like v + (u squared) but it depends on how u say it...

[half ice]

Magnitude of the vector v + u, then squared

[versed mauve]

"the magnitude of v + u squared

[half ice]

Plural is weird, since v + u is one vector

[charred stirrup]

ahh i see, thank you

if i take the absolute value of a complex number, i will always get a monomial that is a real number?

[versed mauve]

|z|? you get the modulus which is an application of the pythagoras theorem above

it is real yes

[charred stirrup]

what is the point of this theorem? does the left side not say to just take the dot product of u and v?

[half ice]

Yes. The right side uses the modulus, so it's a way to equate the two

[charred stirrup]

comparatively, what significance does it hold to simply taking the dot product normally? Like if u = [1 2 3] and v= [1 0 2]

[inland rock]

@charred stirrup what do you mean

You mean what does the dot product mean?

It's the cos thingo

and tells you how much one line lies on the direction of the other line

Usually, you find it important in directional derivatives/proving things are perpendicular

[charred stirrup]

@inland rock for my given u and v, the dot product would be = (1 + 0 + 6)= 7

[inland rock]

yes

[charred stirrup]

but in the formula, i understand that using the right hand side will also produce 7

[inland rock]

oh like what is the intuitive understanding of the right side?

[charred stirrup]

what is the point that more tedious/difficult right hand side when i can just do the dot product the way it was initially taught

yeah

[inland rock]

Hmm I dunno

I can't think of a reason that'd pop up in Lin Alg

But in diffeq

multi* I could see it being useful when calculating with functions

within the dot product

[charred stirrup]

should i ignore that theorem i posed then?

my course is an introductory linear algebra course lol, i dont want to overload myself with things i wont need on my midterm tmrw

[proper crescent]

I've seen that theorem come up in proofs

For example, proving that linear isometries are orthogonal

[inland rock]

@proper crescent that's a smart nickname I guess

I'd vaguely know it

like as in what it is

[charred stirrup]

linear isometries?

[inland rock]

it's like 3 minutes of memorization

theformula

I wouldn't look into it

[charred stirrup]

thank you for your guys' time

appreciate it

let's say u is in the 2nd dimension and that I can say it's made up of two vectors which move along the two axes of the dimension. the orthogonal projection would be the components of u that are perpendicular to a certain axis?

i get the formula, but i guess im saying i dont exactly know what an "orthogonal project" is

[inland rock]

it's what I said before

It's the cos thing

Where we "project" the top line onto the botto mand see how far along the z it is

[charred stirrup]

in the picture, by "top line" you mean u? and im imagining swinging it down to the z axis?

[woven grove]

oh i just took an exam on this for my multivariable calc class

@inland rock is it always the top to the bottom? what if we project vector z onto the directon of u

[inland rock]

just look at it the opposite way

@charred stirrup yes

[charred stirrup]

how to read this? the line with def:

and the fsubcript a , how would that be read?

the description they used is slightly confusing

[broken hawk]

“the map eff a from R n to R” is how I would read that

I don’t see anything that is confusing though

that seems like a pretty straightforward definition

[charred stirrup]

i think it's exactly the "Rn to R" part that's weirding me out

the realm of real numbers in some dimension to the realm of real numbers?

[broken hawk]

ℝ is just ℝ¹

in the context of linalg, ℝ is a 1-dimensional vector space over ℝ

[charred stirrup]

ohhh

so from some dimension that's real all the way down to the first dimension

[broken hawk]

that’s bad terminology but yes

[charred stirrup]

lmao sorry

[broken hawk]

ℝ³ isn’t the third dimension, it’s a space with three dimensions

[charred stirrup]

dont want to give ur brain a headache lol

[broken hawk]

the “third dimension” would be like… idfk the z axis

doesn’t make sense to enumerate dimensions

[charred stirrup]

thank you again,

super nervous for test lol

[broken hawk]

don’t be
(you are now cured. please say “wow thanks I’m cured”)

[charred stirrup]

wow thanks I’m cured

t-15 hours

question is solved, but what does the x bar mean? does that mean take the conjugate?

the x bar also shows up here

[wintry steppe]

I'm guessing it's just vector notation?

People have strange ways of denoting vectors.

[charred stirrup]

thank u so much

prof whack

i hope it's not actually something important

[wintry steppe]

Well, it says that $\bar x \in \mathbb R^4$, so $\bar x$ has to be a vector.

[stoic python]

TendentiousTorturousTopics:

[charred stirrup]

given this, i can say that the dot product of a and x is going to be 0, but

from this, it says the solution is only valid if s-p is perpendicular to a, why is that? if c is 0, should it be that just 's' is perpendicular to a?

[polar sail]

Hi guys, what's the main method to find a basis for a vectorial subspace ?

[broken hawk]

any basis?

also, what kind of vector spaces are we talking?

ℝⁿ / ℂⁿ? arbitrary vector spaces?

for finding any basis of a k-dimensional subspace of ℝⁿ I would perosnally just pick k random vectors in the space, verify if they are independent, throw out ones that aren’t if they’re not and try out some more until I’ve managed. Putting lots of zeroes into the coordinates helps.

if they have to be orthonormal, you can often guess that too, but if you can’t just get as close as you can get on your own, and then use gram-schmidt

[polar sail]

yes sry for my bad-formulated question

i was talking of vector spaces in R^n . But you have to verify if they generate the space no ?

[broken hawk]

if you have a set of vectors which all lie in a k-dimensional subspace (where k < ∞), then if two of the following are true, so is the third:
-the set is linearly independent
-the set spans the subspace
-there are k vectors in the set

so if you have k vectors and they’re all linearly independent (and in the space), then they’ll automatically span

provided you know the dimension of the space is k, of course!

[polar sail]

ok thx a lot , i have to review my lesson

[versed mauve]

i have been stuck for so long with this i just dont know where to go

[wintry steppe]

First, observe that V and W do not contain the 0 vector because they are linearly independent.

lemme phrase that better lol

Consider any vector in the union of V and W.

It cannot be written as a linear combination of the rest of the vectors. Here's the proof. Assume for the sake of contradiction that $\mathbf u = \sum\limits_{i=1}^n c_i v_i + \sum\limits_{i=1}^m d_i w_i$.

[stoic python]

TendentiousTorturousTopics:

[wintry steppe]

We consider two cases: $\mathbf u \in V$ and $\mathbf u \in W$ will follow by a symmetric argument.

[stoic python]

TendentiousTorturousTopics:

[wintry steppe]

If $\mathbf u \in V$, then we see that all of the $d_i$ must be equal to 0; otherwise, orthogonality is violated.

[stoic python]

TendentiousTorturousTopics:

[wintry steppe]

but then that's saying that $\mathbf u = \mathbf v_k = \sum\limits_{i=1, i \neq k}^n c_i v_i$

that violates linear independence

[stoic python]

TendentiousTorturousTopics:

[wintry steppe]

thus we have reached a contradiction.

The same contradiction arises when $\mathbf u \in W$. $\mathbf u$ cannot be written as a linear combination of the other vectors.

[stoic python]

TendentiousTorturousTopics:

[wintry steppe]

therefore the union is linearly independent

[versed mauve]

thank you, this mostly makes sense to me, but could you explain the orthogonality being violated

[wintry steppe]

if $\mathbf u = \sum\limits_{i=1}^n c_i \mathbf v_i$, where the $c_i \neq 0$, then $\mathbf u$ cannot be orthogonal to any of them

[stoic python]

TendentiousTorturousTopics:

[wintry steppe]

by taking $\mathbf u \cdot \mathbf v_k = \sum\limits_{i=1}^n c_i \mathbf v_i \cdot \mathbf v_k \geq c_k \mathbf v_k \cdot \mathbf v_k > 0$

[stoic python]

TendentiousTorturousTopics:

[versed mauve]

i see because vk =/= 0 from the beginning ok

[tulip galleon]

hey guys \

how can i calculate a direction vector that is parallel to my z axis from 3D point

what i am trying to do is i have point in 3D space and i try to rotate my object about an axsis which should be parallel to my z axis and pass by my 3D vector

one approch i thought about is using translation matrix and pluging in my position

and i multiply this buy vector = (0, 0, 1)

so i expect the output to be parallel to Z axsis

like this

[wintry steppe]

parallel to z-axis essentially means the direction vector is (0, 0, -1) or (0, 0, 1)?

[jagged pendant]

that's perpendicular.

[tulip galleon]

@wintry steppe yes

woog i dont understand what you mean

[wintry steppe]

There's a difference between vectors and lines.

If you want a direction vector that is parallel to the z-axis, then it is (0, 0, -1) or (0, 0, 1). End of story.

If you want a line that passes through a point that is parallel to the z-axis, then you can parameterize it by (x, y, z) = (x_1, y_1, z_1) + (0, 0, t)

[tulip galleon]

I want to make rotation about the z axsis of a local space but i think its done by translating it back to the origin first then preforming the rotation the translate it back

But now i dont know how i will rotate it vertically about another point scine my rotation vertically is not parallel to any axsis

[spring wolf]

1 channel at a time my dude

and the other one you posted in is busy

but i'll help anyway: how do you know to find the equation of a line?

[tropic shell]

true soz

y=mx+c

[spring wolf]

Looks like someone's already explaining it in the other channel

[broken hawk]

this is not linear algebra btw, even if there are lines and algebra involved

[spring wolf]

also a good point #precalculus is better I think (although might as well let it be answered as it is being in #prealg-and-algebra )

[broken hawk]

or, you know

#❓how-to-get-help-greekletter

where this sorta stuff should actually go

[ember pilot]

or here @pastel harbor

[pastel harbor]

pruned algebra instead

[tropic shell]

@spring wolf SOZ FORGET THAT ques

[spring wolf]

consider it forgotten

[viscid basin]

can some1 help me through this ^

[tidal nebula]

Definition of nilpotent.

[viscid basin]

T^n =0

[tidal nebula]

For some n > 1.

[viscid basin]

And for this case T^2=0

[tidal nebula]

Yes.

[viscid basin]

I don’t get what the question means by there is a basis of V in which T is given by the that matrix

[tidal nebula]

So the minimal polynomial is x^2 and the eigenvalues are 0.

Similar matrices ?

[viscid basin]

Why is the min poly x^2

[tidal nebula]

It could be x if T=0.

Work with the characteristics polynomial then.

Or by case.

[viscid basin]

Could you reformulate what this question is asking in another way

[tidal nebula]

If T^2 = 0 then there is a in F and P such that T=P^-1[0 a \ 0 0]P.

For T=0, a = 0, P is the identity.

@viscid basin ?

[viscid basin]

Why T of that form ,wasn’t it the given matrix

[tidal nebula]

T = [a b \ c d] in general.

@viscid basin ?

[viscid basin]

Idk man nothing’s clicking in my head

[tidal nebula]

You have the question and T in general. Find a and P.

[oak briar]

Greetings, is there any kind of computational smart trick to multiply two symmetric matrixes?

[tidal nebula]

Try with a small example.

[serene zenith]

Would this yield the distance between two points squared https://i.imgur.com/vrLy1bZ.png

I'm slightly confused where the power would go

Would the power go before the square root like this https://i.imgur.com/0GlmZke.png

[broken hawk]

no, that‘s just different notation for “square root”

the number in front of the root tells you which kind of root it is

you usually leave the 2 away though

[serene zenith]

So my first equation is actually correct?

[broken hawk]

yes, though it would be nicer if you just cancelled the two

provided $a>0$ (which it is here), $\sqrt{a²} = \sqrt{a}² = a$

[stoic python]

Sascha Baer:

[unreal gust]

hey there guys so i have this uhm question

so i have this equation (i dont have specific numbers it was on a test and i just briefly remember a part of it)

so there was an equation

hold up i gotta find out how to do index for x

and i have to run this equation through a 4x3 matrix

how do i exactly do it i have no clue tbh

matrix has 3 rows 4 columnts

columns*

ill try to give an example

so lets say this is a matrix of a linear map and the equation is going through it how exactly do i do this

i think im supposed to get a vector in R3

[winter sage]

can anyone help me with this problem? ive looked up and watched videos on system of equations with infinite solutions but none of them have anything to do with this problem

[slow scroll]

^ Would like to see an answer to the question above

When you add all of the equations together, you get 0 = b1 + b2 + b3. Would that just mean that all possible values for b lie on the plane b1 + b2 + b3 = 0?

[half ice]

That's clever, and probably right.

You would normally proceed with row reduction

[slow scroll]

,$ \begin{pmatrix}
1 & 0 & -1 & b_1 \
0 & 1 & -1 & b_1 + b_2 \
0 & 0 & 0 & b_1 + b_2 + b_3
\end{pmatrix}

[stoic python]

bigboy77584:

[slow scroll]

so i guess with row reduction the correct line of reasoning would be that b1 + b2 + b3 must equal zero for its system to be consistent..?

I think that makes sense
Anyway I have my own question:

Not really sure how to go about this.

[half ice]

Know the determinant?

[slow scroll]

nope

[half ice]

In general, if two matricies are consistent, then their product is

Hmm. How to prove that though? Let me think

[slow scroll]

i can prove that the columns of A^n are linearly independent. I'm just tripped up about the rows

[half ice]

Same process
A^T is consistent if and only if A is

[slow scroll]

hmmm but why?

[unreal gust]

so uhm

does anyone know the answer to what i asked

[leaden ermine]

Hi can anyone assist me in graphing 3d equations in mplot3d? Im trying to visualize some intersections in linear equations and having issues setting up equations. For instance x + 2z + 2y = 3. My current code is ``%matplotlib inline
import numpy as np
import matplotlib.pyplot as plt

fig = plt.figure()
ax = plt.axes(projection='3d')``

[unreal gust]

😦

[tidal nebula]

No enough, you need some mesh, x, y, z coordinates.

[rancid bear]

from mpl_toolkits.mplot3d import Axes3D

you have to import it.

[wicked jacinth]

Anyone have a recommendation for a PRML alternative?

That book is not really helping me understand the linear algebra behind ML

[wintry steppe]

what is linear-algebra?

[broken hawk]

linear algebra studies vector spaces and linear transformations, which are incredibly important topics throughout math and physics. seriously, it crops up everything.
it also involves working with matrices, which are a representation of linear transformations.

[wintry steppe]

interesting

[broken hawk]

it has applications in like, everything

[quasi mesa]

what sort of applications of matrices are involved with computing?

[broken hawk]

machine learning for one
linear algebra is also basically the foundation of quantum mechanics, so it’s pretty darn important to physics

also anything that involves displaying 3D stuff on a flat monitor is full of Linalg

so anything graphics related, like video game engines

[quasi mesa]

how does it relate to ML

[broken hawk]

neural nets are basically a pile of linalg

[quasi mesa]

and linalg can be represented with matrices

[broken hawk]

you store stuff in vectors and matrices and do stuff I don’t quite understand to update the system

I‘ve not really worked with it

just read a few articles

also, any time you hear anything with eigen- in front of it, that’s linalg too

which seems to be about every second thing physicists care about

[quasi mesa]

looks a bit complex for me lmao, im trying to find an area to research more to use on my personal statement for computing

any ideas?

[broken hawk]

linalg is actually pretty elementary, it tends to be one of the first things encountered in a university

the reason it’s so universal is because it is (comparatively) easy

yet powerful

knowing linalg will definitely come in handy pretty much no matter what

[quasi mesa]

as far as we go in school atm is solving equations in 3d space

using matrices

anyways thanks for the insights @broken hawk

[broken hawk]

as a first intro to it, I strongly recommend watching 3blue1brown’s series on it

they’re extremely accessible, and while they won’t make you good at it on their own, they’ll give you plenty of intuition

[quasi mesa]

any books you recommend for after that series?

[broken hawk]

I’ve not really worked with any, myself. depends also on what you wanna focus on, I’m more of the pure maths guy so I like my linalg to be less focussed on matrices and more on abstractions

[quasi mesa]

makes sense, i read somewhere they can be used with distributions which seems interesting

[broken hawk]

for that I found the book we used in class pretty good, Linear Algebra by Friedberg, Insel and Spence

but there are tons of books on the topic

with different emphasis…es

[sharp lodge]

By Jordan Normal Form, modulo basis, a matrix is basically a tree of eigenvalues.
So if you choose n eigenvalues, and if you put the eigenvalues in a tree graph where the root is just some meaningless element. And every eigenvalue can only ascend to an same valued eigenvalue, then that defines a unique matrix modulo basis.
Does such a tree have a name?

I've never seen people draw this

[lean aspen]

@sharp lodge what do you mean by ascend

[sharp lodge]

@lean aspen A tree has a root node at height 0. A node ascends (directly) to another node if it's connected to it and the height increases by one. So basically if you travel up the tree.

For example a matrix can be defined mod basis as the matrix with eigenvalues (2,3,3,3,3,3) with structure:

Root
| | |
3 3 2
|
3
|
3

I forgot a 3 but whatever

Also i cant see who's ripping my graph on the tablet

[slate vigil]

what would be the minimal generating set of a group for a rectangle under rotation?

not sure if this is the right room for this, should this go in number theory?

[slow scroll]

abstract algebra?

[slate vigil]

yeah

ill put there now

thx

[slow scroll]

np

[slate vigil]

dunno why its part of a linear alg class

lumow

[ionic island]

Any fancy theorems like cayleigh hamilton for linear algebra?

[broken hawk]

cayley-hamilton is a linear algebra theorem

[ionic island]

duh

but I meant are there any more fancy theorems like it I could maybe teach people about

[broken hawk]

spectral theorem

(symmetric real matrices / normal complex matrices are orthogonally diagonalizable)

[rotund quail]

um helo people

,tex \left( 4r+t\right)^4

[broken hawk]

well with that equation you can be sure it’s not linear algebra

[rotund quail]

wait

[broken hawk]

also you want \left( and \right)

not \big

[rotund quail]

its linear algebra?

[broken hawk]

something with ^4 is probably not very linear

[stoic python]

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

[rotund quail]

it is

I do not understand

[broken hawk]

well, what is the question

[rotund quail]

,tex {4 \choose 0} 4r^4 +{4 \choose 1}4r^3 t+{4 \choose 2}4r^2 t^2+{4 \choose 3}4rt^3+{4 \choose 4}4t^4

wait me sec

[stoic python]

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

[rotund quail]

fOR rEaL?

let me use other renderer

[broken hawk]

tip: use \binom{a}{b} inseat

[slender yarrow]

^

[broken hawk]

it’s got much nicer syntax

and doesn’t break half the time

[rotund quail]

oh

alright

anyway,

i dont understand the process of solving these

could you help me, please?

[broken hawk]

what do you mean with solving these

[rotund quail]

like kick those combinatorics

[broken hawk]

again, this isn’t linear, so… not really linalg

[rotund quail]

its Algebra

right

[broken hawk]

but not linear algebra

[rotund quail]

Then where is Polynomials located then

[broken hawk]

depends what you’re doing with them. solving polynomial equations is uh, algebra. but not linear cause well, they’re not linear equations

[rotund quail]

oh

okay, thx

[broken hawk]

but if you’re studying the space of polynomials

that’s linear algebra again cause they do form a vector space

linalg isn’t much about solving equations

[rotund quail]

i will just go on with #❓how-to-get-help categories then 😶

[ionic island]

Any important things we can infer using determinants and or easy way to calculate determinants?

I know the product of matrices A, B will net you a matrix with determinant det(A) x det(B)

but there doens't seem to be a particularly easy way to compute determinants if you don't happen to have convenient matrix factorisations

[broken hawk]

you can, in principle, use gaussian elimination to get it to a triangular matrix, where the determinant is then just the product of the diagonal entries
however, you must be aware of how each transformation changes the determinant

(swapping changes sign, adding rows does nothing, multiplying a row by a scalar mutliplies the determinant by that scalar)

I just described how it does

[ionic island]

:p I didn't read up till that part