linear-algebra

[quaint heart]

If there is a basis which is fixed under the map

then think about what the map looks like

everything is fixed

so its the identity

the identity looks the same in every basis

[dusky epoch]

as does any scaling map

[quaint heart]

that too

those are the only maps that always look the same

[slow scroll]

what do you mean exactly by "fixed under the map." Kinda stuck on that part

[quaint heart]

f(x)=x

[slow scroll]

ohh i think i see what you mean

[proper crescent]

It may help to think in matrix terms if that version is less helpful. Scalar multiples of the identity are the only matrices which commute with all others

So if you conjugate a scalar multiple of the identity with some change of basis you'll get the same guy back

[quaint heart]

and if some matrix is conjugated to itself by every invertible matrix

then it commutes with every invertible matrix

[proper crescent]

But really try to understand the geometric picture the others are giving. If a map acts the same way on everyone it should have the same matrix representation in any basis

[slow scroll]

uh define conjugate?

[proper crescent]

PAP^{-1}

[quaint heart]

is what you get when you conjugate A by P

Its a natural action of the set of invertible nxn matrices on M_{nxn}

[slow scroll]

when the eigenvalues of a matrix are all 1, it seems like the matrix would always just be the identity.

[quaint heart]

if it has a basis of eigenvectors

[proper crescent]

Imagine a rotation about some axis in R^3

[slow scroll]

okay

[proper crescent]

The only vectors that get scaled at all are on the axis, and they are preserved, so the only eigenvalue is 1

But it's not the identity

[slow scroll]

thats just the only real eigenvalue tho.

[proper crescent]

Well, as a map R^3 -> R^3 you wouldn't think of it as having complex eigenvalues. The matrix is allowed to act on C^3 and then you can say well there are other eigenvalues.

But also there should be a complex example as well

So we want the char poly to be (t-1)^n so just give me any upper triangular matrix whose diagonal entries are all 1

Actually I dunno why I had to think of the char poly but polynomials are how I process all of linear algebra lmao

But yeah the only eigenvalue is 1 but it's not the identity

[dusky epoch]

c o m p l e x i f i c a t i o n

[quaint heart]

For some reason I always try to make linear algebra problems into topology problems

[proper crescent]

Wait how?

[slow scroll]

let A be a torus shaped matrix....

[quaint heart]

Through continuity of functions like the determinant and the nth eigenvalue function

[proper crescent]

Weird

I mean I guess for certain problems it sorta is the correct way but also... idk somehow unless it screams itself as a topology problem that's not my instinct

When I learned the analytic proof of Cayley-Hamilton I was so upset lmao

[quaint heart]

I once tried to show that an open set of n dimensional vector space contains n linearly independent vectors using the fact that open sets are codimension 0 submanifolds and an n manifold cant be contained in a manifold of lower dimension

Or something like that

[proper crescent]

Good lord

Or wait I misread

Didn't see the word "open subset"

[slow scroll]

wait so im thinking about diagonalizing nxn triangular matrices with 1's on the diagonal, and it seems like the dimension of the eigenspace here would not be equal to n...

[proper crescent]

Yeah, "1 is the only eigenvalue" doesn't mean n-dim eigenspace

It just means there are no other eigenvalues

There are two types of multiplicity for eigenvalues

Algebraic multiplicity and geometric multiplicity

[slow scroll]

i got that

algebraic = geometric for diagonalizabilityu

geometric <= algebraic always

[proper crescent]

Yup, so if 1 has geometric multiplicity n then trivially you're the diagonal, since ker(A-I) = k^n => A-I = 0

So the example I gave was one where 1 has algebraic multiplicity n, meaning no other eigenvalues could exist, but you're not the identity, so the geometric multiplicity has to be smaller

[slow scroll]

oh...

i think ive reached a better understanding than before. thank y'all for the help!

[astral junco]

Multiplying these matrices togather should give one matrix that scales by 2 in x and y, rotates 45 deg, and translates 2 in the x direction and -2 in the y, right?

I tried applying that transformation to the square on the on the left, but it seems that the y value is off because it wasn't moved down, right?

[dusky epoch]

reverse the multiplication order

[astral junco]

i multiplied them in matlab. if i type matrixScale * matrixRotation * matrixTranslate * columnVector in matlab does it know that it needs to do the multiplication from left to right?

[dusky epoch]

no i mean like

translate * rotation * scale * vector

that's what i meant

[astral junco]

but couldn't changing the order change the resulting vector? are you always supposed to transformations in that order? sry i'm a little confused

i know i'm supposed to apply them in this order. sry i don't think i was clear enough

if i apply them in that order is the resulting square i got correct?

i'm just having trouble seeing why y value is not changed

like when i try to imagine the transformation in my head i see the square moving down, but i'm incorrect?

[broken hawk]

rotation and scaling commute, where you do the translation does matter

(assuming you rotate and scale around the same point)

[dusky epoch]

what i'm saying is that you're applying the transformations in the wrong order

and i'm telling you the order in which you need to apply them to get what you want

[astral junco]

ah i gotcha. thanks for the help

[valid citrus]

eigenspace is just the span of eigenvectors yea?

[broken hawk]

each eigenspace is the span of all the eigenvectors of eigenvalue n

[dusky epoch]

n to denote an eigenvalue

[broken hawk]

(or just the set of all of those, span is unnecessary)

I blanked out sry :P λ, n, whatever

all the same

[valid citrus]

so if i have to parametrize the eigenspace of a matrix A given the eigenvalue lamda = 2

i just remember i have to find the char polynomial and after that im stuck

[dusky epoch]

you want to what?

find a basis for the eigenspace of A corresponding to the eigenvalue 2?

that eigenspace is just ker(A - 2I)

[valid citrus]

damnit, forgot that, was going about finding the characteristic polynomial, thanks man

[dusky epoch]

charpoly will give you the rest of the eigenvalues

you don't really care about that here

[valid citrus]

yeah

basis is the same as parametrizing it?

[dusky epoch]

one can arrive at a basis from a parameterization and vice versa

[slim pebble]

anyone know how i can show that A is diagonalizable?

[dusky epoch]

depends on what A is

[slim pebble]

http://prntscr.com/nhsvxd

sorry idk how to type out a matrix so that's A

[proper crescent]

Find the eigenbasis

[slim pebble]

eigenbasis is basically the eigenvectors from all the eigenvalues correct?

[quaint heart]

Yep

Just find 2 linearly independent eigenvectors

[slim pebble]

i just wanna refresh to find eigenvalues is it (A-(lambda)I)?

[quaint heart]

First what are the eigenvalues

Yep

[slim pebble]

ok imma do that now

[quaint heart]

Or lambda I - A

Doesn't matter

[slim pebble]

wait order doesn't matter?

[quaint heart]

Det(lambda I - A) = 0

Is the equation you're solving

[dusky epoch]

det(M) = (-1)^n det(-M)

[quaint heart]

So Det(lambda I -A) = 0 iff Det(A-lambda I)= 0

[dusky epoch]

yes

[quaint heart]

Because of what ann said

For consistencies sake we pick lambda I -A

But both are correct

[slim pebble]

Det(A-lambda I)= 0

so use that to find the eigenvalues first?

and then find the eigenvectors after and that'll give me my basis?

[quaint heart]

Yep

Assuming that it does in fact have a basis of eigenvectors

If you have 2 distinct eigenvalues your good though

[slim pebble]

yeah i got 2 and -1

now i do (A-2I)x=0 for the eigenvector for the first one correct?

[quaint heart]

Yep

Or phrased another way

Ax = 2x

[slim pebble]

do i gotta get the rref form of the matrix after i subtract?

[quaint heart]

Just write the vector as (a, b)^t

And see what the matrix does to it

Why would you need to get the rref form?

[slim pebble]

i got it mixed up with something else

but after i subtract i got

-6 -6
3 3

[quaint heart]

Ok

So
[-6 -6] (a). (-6a -6b) (0)
[3 3] (b) = (3a + 3b)= (0)

Solve for a and b

It's pretty easy

@slim pebble

[slim pebble]

Yes is (-1,1) correct?

[quaint heart]

Yep

Or (1, -1)

Doesn't matter which

[slim pebble]

Now just gotta find the eigenvector for the other eigenvalue and that’ll make my P?

[quaint heart]

Yep

Then you change the basis to be in terms of those eigenvectors

Which makes your transformation look like a diagonal matrix

[slim pebble]

so in order to prove that A is diagonalizable you just gotta find the D matrix?

[wanton spoke]

For instance

[quaint heart]

the D matrix is just the matrix of eigenvalues

[wanton spoke]

Or find the eigenvalues

I'll let liquid continue his explanation 👀

[quaint heart]

it is unique up to permutation of the eigenvalues

ie you can list the eigenvalues in whatever order you like

but there are no similar diagonal matrices that arent composed of those eigenvalues

[slim pebble]

ah ok i get what you're saying

[quaint heart]

you can show this is true by noticing that the roots of the characteristic polynomial are invariant under change of basis

Since $det(PAP^{-1} - \lambda I)= det(PAP^{-1}- P \lambda I P^{-1})=det(P(A-\lambda I )P^{-1})=$ $det(P)det(A-\lambda I )det(P^{-1})= det(A-\lambda I )$

[stoic python]

Liquid:

[slim pebble]

hey @quaint heart

you think you can help me with part b?

http://prntscr.com/nht77q

[quaint heart]

where does 1 get sent? where does t get sent? where does t^2 get sent?

thats all the data that the matrix keeps track of

first off what is p(1)?

@slim pebble

[slim pebble]

p(1)?

[quaint heart]

oh lol

T(1)

[slim pebble]

you just plug 1 to the equation right?

[quaint heart]

idk why I thought the transformation was called p

yep

[slim pebble]

do i plug it in the 3p(t)+tp(t)+t^2p(t) ?

[quaint heart]

yes

the polynomial in this case is just p(t)=1

[slim pebble]

wait how you get that?

[quaint heart]

its a constant polynomial

[slim pebble]

sorry i missed class when she was going over this

[quaint heart]

so T is a transformation which takes a polynomial p(t) to 3p(t)+tp(t)+t^2p(t)

[slim pebble]

so... p(t) is the 2-t-t^2?

[quaint heart]

no

oh it is

in problem a

but its just a polynomial in terms of t of degree <= 2

[slim pebble]

because the p(t) the highest degree is 2 right?

[quaint heart]

Thats the vector space p(t) belongs to

$\mathbb{P}2$

[stoic python]

Liquid:

[quaint heart]

the space of real polynomials of degree <= 2

[slim pebble]

so what would t be?

[quaint heart]

First lets let p(t)=1

and find T(1)

and then we'll talk about t

[slim pebble]

wait you think you can explain over voice?

or nah?

[quaint heart]

yeah sure

[slim pebble]

T(1)=3-t+t^2

T(t)=3t-t^2+t^3

T(t^2)=3t^2-t^3+t^4

[quaint heart]

yep

now write it in term of the basis

T(1) = 3(1) -1(t) +1 (t^2) +0(t^3) + 0 (t^4)

[slim pebble]

{1,t,t^2}

T(t^2)=3t^2-t^3+t^4

[quaint heart]

R^n -> R^m means a mxn matrix

[slim pebble]

3    0    0
-1    3    0
1    -1    3
0    1    -1
0    0    1

{1,t,t^2}

3-t+t^2

[quaint heart]

T(a+bt+ct^2)=aT(1)+bT(t)+cT(t^2)

3 0 0
-1 3 0 2
1 -1 3 -1
0 1 -1 -1
0 0 1

[dusky epoch]

no

the polynomial itself has degree 3; how can a polynomial of degree 3 have a root of multiplicity 4?

[earnest vessel]

How did you get that?

[dusky epoch]

^

[earnest vessel]

And that is not true either, since two roots of multiplicity 3 give a total multiplicity of 6

And as Ann pointed out, the polynomial has degree 3

So the multiplicities must add up to 3

[ember pilot]

[wintry steppe]

do what it asks

nah idk dis

[oak briar]

if you multiply a matrix with a contraction of a (rank4) tensor, how do you call that?

[sand oak]

can someone help me out

I got That far

[native lodge]

,rotate

[patent glacier]

correct me if i'm wrong, but couldn't you just use the standard basis matrices for R^(2x2)

[stoic python]

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

[patent glacier]

like that would be e_1,1

[quaint heart]

That would work

[sand oak]

I'm just trying to follow the professors example...

[half ice]

@sand oak
A1 has the top left
A2 - A1 has the top right
A3 - A1 has the bottom right
none of them have the bottom left
You can span it with A1, A2, A3

Notably
A4 = A3 + A2 - A1
A5 = A4 - A1 = A3 + A2 - 2A1
They don't add anything to the span

[rare barn]

literally none of these are orthogonal to each other... am i missing something?

[dusky epoch]

these = what

[rare barn]

the two vectors being added

like 4/5,7/5 and 6/5,8/5

[dusky epoch]

there actually is one answer option in which the two vectors being added are orthogonal

d

[rare barn]

oh :/

true

sorry

[earnest vessel]

@rare barn you could also have known it because in none of the other examples you have a vector in Span{u}

[astral junco]

the basis for ker(A) is that vector correct?

[jagged pendant]

the set containing that vector alone would be a basis

[astral junco]

ok i see what you're saying. thanks

[astral junco]

does anyone know of a good youtube video or something that can help me with this?

i tried solving it by doing this:

[quaint heart]

3b1b has a linear algebra series

[astral junco]

idk if that makes any sense or not

i'm kinda lost

oh so

[quaint heart]

so basically the first column of the matrix corresponds to the first basis element in the source space

and the second column corresponds to the second basis element

[astral junco]

so it looks correct

[quaint heart]

your doing rows instead of columns

[astral junco]

oh

so transpose it and it's good?

[quaint heart]

yep

[astral junco]

nice thanks a bunch 👌🏼

[warm sluice]

Quick question: Is it a Cross Product or a Dot Product that equals to |A||B| sin (theta)?

[slow scroll]

cross

[warm sluice]

Ok! Thanks

[slow scroll]

np

[broken hawk]

if you ever forget it again, you can figure that one out pretty directly if you understand the geometric meaning of both sine and ×

because A×B has magnitude of the area of the parallelogram spanned by A and B. If you see A as being the base, then the height is |B|sinθ

note also that A×B ≠ |A||B|sin(θ)

the lefthand side is a vector, the right is merely its magnitude

@warm sluice

[astral junco]

what is this process called? is it a linear map or something?

i'm just trying to find out what i can search for on youtube to learn more about this.

or maybe someone can walk me through it?

[broken hawk]

the linear map is the thing described here, the matrix is a representation of it

as for how to do it, I recommend just watching the 3blue1brown series on linalg, it’ll teach you that and much more

https://www.youtube.com/playlist?list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab here you go

episode 3 should explain your thing, but I recommend starting at the top

he explains it much better than I could, and imo I could explain it pretty well

[slow scroll]

its a linear transformation. map/transformation can kinda be used interchangeably i think. Usually a linear map is a mapping from one vector space to another, while a transformation maps a vector space onto the same vector space.
To say that e1 maps to -3e2 is to say $$T\left( \begin{pmatrix}1 \ 0 \end{pmatrix} \right) = \begin{pmatrix} 0 \ -3 \end{pmatrix}$$
and to say e2 maps to e1 - 2e2 is to say $$T\left(\begin{pmatrix}0 \ 1 \end{pmatrix}\right) = \begin{pmatrix}1 \ -2 \end{pmatrix}$$

[stoic python]

kxrider:

[slow scroll]

so the matrix transformation needs to satisfy those two things

The only one that does that is $$\begin{pmatrix} 0 & 1 \ -3 & -2 \end{pmatrix}$$

[stoic python]

kxrider:

[astral junco]

alright seems to make sense

[hexed swift]

how the hell am i supposed to figure this out

[slow scroll]

#prealg-and-algebra xd

[broken hawk]

simply solve a linear equation

set up the general form for the equation of a line

insert the values you have

solve for teh unknown

done

[hexed swift]

ok but how do i do that.

yall assume i know how to do this

[broken hawk]

I assume you take classes that are supposed to teach you this

and that you take notes

[hexed swift]

this class is online, i've been teaching myself all the material :/

[broken hawk]

anyway, we’ll ignore that #prealg-and-algebra is prolly the better channel for this cause whatever, it’s linear :P

so, first of all, do you know the equation of a line

y = (something with x)

[hexed swift]

y=mx+b, i think?

[broken hawk]

ye

[hexed swift]

okay yeah

[broken hawk]

so, that’s the general form

now, you know the slope

do you know what the slope is?

[hexed swift]

yep its m, -2/5

wait

[broken hawk]

apart from the bit where it clearly says -1/2

[hexed swift]

1/2 i forgot what was in the screenshot ack

[broken hawk]

that it’s m is the bit that matters :P

so now you have the equation y = -1/2 x + b

now, you do know one pair (x,y) which satisfies this equation

plug them in

[hexed swift]

yeah i was given that. but. it isnt the y intercept, which is what i believe b is supposed to be?

[broken hawk]

no, that would just be a special case

just plug in the values

well, the y intercept would be b

you have to find that now

you can’t plug in b, it’s the unknown

[hexed swift]

i think i only have to write the equation

[broken hawk]

but you do know what m*1 + b is

yes, for that you need to find b

so, what is m*1 + b (where m is -1/2, as already discussed)

[hexed swift]

wouldnt it just be -1/2 + B?

[broken hawk]

yes, but what does that equal?

[hexed swift]

i have no idea, to be honest with you lmao

[broken hawk]

at x = 1, what is the y-value of the line?

[hexed swift]

0?

[broken hawk]

I’m gonna assume you just made a wild guess

read the question carefully. draw it

[hexed swift]

i didnt

wouldnt x=1 just be a vertical line at the x-axis point of 1?

[broken hawk]

what, no, that’s not what I meant

at the position where x is 1

you know what, sorry, but I give up. I can’t teach basic algebra

[hexed swift]

sweet. i'm sorry, this is just really hard for me, lol.

[broken hawk]

(I mean, I could. but I’d have to start at the beginning. with these things, starting somewhere in the middle is painful)

[hexed swift]

no its fine. im just going to cheat for this part lmao.

[broken hawk]

…that’s not fine

do you have any intention of ever taking any math classes ever again?

or ever using math in any way, shape or form

if so, you must understand this

these things are the foundation of everything to come

and you’ll struggle way more

[dusky epoch]

... @hexed swift?

[hexed swift]

i don't have to take another math class.

i could learn this if I had more time, and I plan to come back and learn it, however today is the last day I have to finish work.

[slow scroll]

you will almost certainly have to take more math after this

[hexed swift]

well, i'll just grab a tutor next semester.

[broken hawk]

grab a tutor for this stuff, not for whatever you’re working on then

[hexed swift]

i have no time left, todays the last day of my semester

this is the only thing i've really not been able to figure out, is writing out the equation itself without the y-intercept given. the entire rest of this unit has been fine

[broken hawk]

you got a pair (x=1, y=3). plug that into the equation, then solve for b

[hexed swift]

right

[proper crescent]

For what it's worth this kinda thing should be in #prealg-and-algebra

[broken hawk]

yea we mentioned it but I decided to ignore it because i don’t like changing channels :P

[proper crescent]

I'd rather you change channels in the future

But you can continue on here if you've been discussing for a while

[broken hawk]

but then I have to unmute the channel

[rare barn]

little confused with what to do here....

not sure how to find the distance just from the magnitudes, dont i need the original vectors

[broken hawk]

they form an orthogonal basis, so you know a lot about them

in particular you can rather easily compute projections

also remember the pythagorean theorem, I think you might need it

[rare barn]

yea but dont i need the vector itself to find the orthogonal proj

wait, isit b?

nvm i got it

3u2-2u4

root181

[heady monolith]

really basic question, if P is a projection matrix and v,w vectors, is <v,Pw> <= <v,w>?

[jagged pendant]

seems reasonable

[heady monolith]

actually jk I don't think that's right

[jagged pendant]

seems right to me but i havent written anything down for it

oh actually

(2, 1)
(-2, -1)
v = (1,-1)^T
w = (0,1)^T

[heady monolith]

let v and w be orthogonal

[jagged pendant]

then

[heady monolith]

and P be any projection to a subspace that doesn't include v and w

[jagged pendant]

<v,w> = -1

<v, Pw> = 2

theres a counterexample

[heady monolith]

yup

[jagged pendant]

huh

[heady monolith]

oh well, time to prove this lemma in a different way I guess

[jagged pendant]

whats the lemma, out of curiosity

[heady monolith]

eh, it's a research thing

trying to bound this error term that pops up

[jagged pendant]

i see~

[heady monolith]

the term happens to look like v^TPw

well, <v,Pw>

I don't really do math, I'm in stat/ML

[jagged pendant]

OK

cool!

good luck ^^

[heady monolith]

being a mathematician sounds fun though

[astral junco]

trying to find if this matrix has an eigenbasis. so basically my logic is to try to see if it has any eigenvalues because if not then there can't be an eigenbasis. if it does then i need to find the eigenvectors that make up a basis for R^3. is any of this correct or am i completely wrong?

[jagged pendant]

YA IT DOES SEEM FUN

well

since its in R^3

it definitely has at least one eigenvalue

cuz its characteristic polynomial has odd degree

and odd degree polynomials cross the x axis at least once

but ya

thats the correct way to go

[astral junco]

ok much appreciated

just making sure i sort of understand what's going on

[jagged pendant]

^^

[coral sage]

is there any generalization of vector matrix multiplication to other operations?

I'm looking for some sort of formalization for $A\widehat{v}$ where $A : Mat$ and $\widehat{v} : \text{Vec of} \mathbb{R} \rightarrow \mathbb{R}$

[stoic python]

Tyler (Aearnus):

[quaint heart]

Do you know what matrix multiplication is?

[coral sage]

except it's not multiplication it's application

yes ofc

On mobile will explain what I want better in a sec

[keen nexus]

Are you asking for something along the lines of linear transformations? Because if that's the case, then there's quite a bit of generalizing we can do from there.

[coral sage]

I want to be able to express the following:

For some vector $\widehat{v} = \begin{bmatrix} v_1 \ v_2 \ \vdots \ v_n \end{bmatrix}$, some matrix $A = \left[A_1 \hdots A_n \right]$, and some associative and noncommutative operation $\cdot$, I want to be able to express $A_1 \cdot v_1 + \hdots + A_n \cdot v_n$

is there an unamiguous way to do this without defining it myself? or is my best bet going to be explaining that in my writing

[stoic python]

Tyler (Aearnus):

[coral sage]

(edited to more accurately reflect what i'm doing)

[keen nexus]

Assuming $ A_{i}, 1 \leq i \leq n $ are all column vectors of $A$, it almost sounds like you're asking for a generalization of $A \vec v$ as some sort of span.

[stoic python]

TiredPatzer:

[keen nexus]

At least, that's the way it reads at the moment. Just trying to help clarify things

[coral sage]

I'm not considering the span of A or anything like that

[keen nexus]

Also, to be clear, I mention the span because when we have some matrix, $A$ multipying $\vec v$, we can express this multiplication as the span $v_{1} A_{1} + \dots + v_{n} A_{n} $

[stoic python]

TiredPatzer:

[coral sage]

ahh yeah I see what you meant. in that case yeah, I want a generalization of the span

though the elements of v are not necessarily constants

so it's the span of A if you're looking at a non-real (non-numeric?) space

[keen nexus]

Well, generally speaking, it's more like the span of the range of $A$, where $A$ is the matrix representation of some linear transformation, $T$

[stoic python]

TiredPatzer:

[coral sage]

i'm not 100% sure we're on the same page here

[keen nexus]

I suppose that's better worded as the span of the range of $T$, then. Sorry if that's a little vague - I'm being a bit non-rigorous.

[stoic python]

TiredPatzer:

[keen nexus]

That's alright. I don't think I'm being clear myself. But I do think I know what you're referring to - in general, the elements of v don't have to be numeric. They just have to be the elements of some field (and therefore satisfy all of the requirements to be defined as such).

[coral sage]

i'm in the middle of typing my thoughts up, but I want to do this because I'm modelling soft body physics using a vector of "nodes" and a matrix of distances

and i realized that calculating new positions/velocities for each of the nodes will be as easy as doing this sort of generalized sum across the columns of C

and I wanted a way to write that out correctly

[keen nexus]

Ok, I see. So, you're definitely describing some very specific things here. Firstly, your matrix C is symmetric with a main diagonal of 0s. What it also sounds like, to me, is these "nodes" are elements of this vector space.

[coral sage]

yeah

I think the problem here is that the matrix and the node vectors are fundamentally in different vector spaces

so what I actually need to do is bring them into the same vector space, then it's possible that I could literally just do vector and matrix multiplication

without having to do fancy things with summing columns with supplied operators or w/e

[keen nexus]

Essentially yes - that's where you'll have to pick up on some stuff about linear transformations.

Because you can define matrices that function as linear transformations from one vector space to another

and said matrices (not the C one you mentioned), would be composed of some basis of the vector space that you want to bring the element you're acting on to

[coral sage]

let me write up a little bit more and see if I can get further with this

I definitely need to formalize what a node is before I move any further

[keen nexus]

Sounds good!

[coral sage]

thanks!

[keen nexus]

Yeah, that's a good start - right now the idea of the nodes is a little vague.

No problem!

[coral sage]

this is what I have currently

i'm at the edge of my intuition, I have to break out the memo pad haha

[half ice]

@coral sage
Sounds a bit like you want to look into FEA analysis

I see what you're up to. I can look at what I know

[coral sage]

I appreciate it!
it's nothing that serious, I couldn't shake this idea of using a matrix of connections & i needed to write it down

[half ice]

You're going to want to include the idea that each connection wants to retain its shape

[coral sage]

also, I know for a fact using euler integration on $\pi$ and $\delta$ is going to make a really shitty simulation, but until I have an actual process for modifying $\alpha$ I won't mess with that

[stoic python]

Tyler (Aearnus):

[half ice]

Euler integration?

I figure you can model each connection with a second order DE. If any connection deflects too far, it stays that way

[coral sage]

was considering that

but i currently have a C in differential equations so that'd probably be a lil bit over my head

though it's worth a shot

[half ice]

Just not positive how the distribution of forces should be carried out

[coral sage]

writing it out

you need to turn the matrix C into an "offset matrix"

which encodes what position each constraint ideally would want its respective node to move to

(that's the best way i've managed to formulate that idea and I've been sitting on it for 10 minutes now)

currently trying to work out a numerical example on paper, will send it when I figure it out

Sorry for the crappy handwriting

But this is what I'm looking at

[half ice]

Are these nodes in 2D or 3D?

[coral sage]

2d

using complex numbers to model it

[half ice]

Gotcha

[coral sage]

node 0 is at position $0+i$, node 1 is at position $0.5+i$

[stoic python]

Tyler (Aearnus):

[coral sage]

sorry it was hard to read

[half ice]

Why (0 + i, 0, 0, 1)?

[coral sage]

Arbitrary choice, just allowed me to calculate it all out easily

Position is 0+i, velocity is 0, acceleration is 0, mass is 1

[half ice]

Gotcha

[timid ivy]

Let T be a linear operator on a finite-dimensional real inner product space V. T is self-adjoint if and only if there exists an orthonormal basis for V consisting of eigenvectors of T.

My book goes on to say that this is used extensively in many areas of mathematics and statistics. Can someone elaborate on this more? Does this theorem have a name?

Its kinda a corollary of Schurs theorem, so maybe people refer to that instead?

[jagged pendant]

spectral theorem, i think

[broken hawk]

that does sound like the spectral theorem, yea

[timid ivy]

The spectral theorem in the book is just a bunch of things that follows from normality or self-adjunct

So yeah that makes sense

[wintry dirge]

Simple quick question for my understanding:

Do the vectors v1 = (1, 0, 1) and v2 = (0, 1, 0) span the whole R3?

Do I need the v3 = (0, 0, 1) or is it already in v1?

[dusky epoch]

no, the span of (1, 0, 1) and (0, 1, 0) is not all of R^3, and no, (0, 0, 1) is not in their span

[wintry dirge]

What third vector do I need then in order to span all of R^3?

[dusky epoch]

well (0, 0, 1) will do

but there's many that fit that bill

[wintry dirge]

Any other to v1 and v2 linearly independent vector I guess

Thank you a lot for your help, I really appreciate it 😃

[lilac fern]

So if I should compute a coordinate vector directly, how should I do that? I know how to compute a coordinate vector, but I don't understand the question for the same exercise where it says "Check by computing directly" or something similar. Any advice here? 😃

[slow scroll]

can u post an example of what ur talking about?

[lilac fern]

Sure just a second

This is what it says. I understand the a, b and c part of the exercise. The d part is a little weird for me

[slow scroll]

what does section 4.6 tell you to do?

[lilac fern]

The (12) equation?

[slow scroll]

yea

[lilac fern]

I used that for solving the C part of the exercise as it says

[flint flicker]

According to my linear algebra textbook, the second property of a sub space of R^n is that If x and y are in U, then x+y is also in U

Would that then mean that any subspace of R^n must have infinitely many vectors in it?

[slender yarrow]

what if you only have the zero vector?

[flint flicker]

Ok

What if that’s not the case tho?

Assume there are vectors other than the zero vector

[slender yarrow]

if we're talking of an R-vector space then sure

[flint flicker]

Okay

I’m confused by that

[slow scroll]

a 0D space is the zero vector. A 1D space is a line in R^n

the set of point that lie on a line are in the subspace

[flint flicker]

Because in tandem with the third property, it seems like every vector in R^n can be reached

And I thought subspaces were supposed to be a subset of R^n

Proper subset

[slow scroll]

well usually we are talking about proper subspaces.
span{(1,0,0)^T} is a proper subspace of R^3

[flint flicker]

Hmm

Does a subspace in R^3 represent what I intuitively think it does?

Aka, a region of space?

[slow scroll]

pretty much. For a subset of R^3 to be a subspace, it should either look like a point, a line, a plane, or a cuboid

[flint flicker]

Gotcha

Thank you, that’s helpful to know

[slow scroll]

thats just because when you take a point on a line, and add it to another point on a line, you get another point on the line, and similarly for the others

[broken hawk]

uh

a subspace of R3 cannot be a cuboid @slow scroll @flint flicker

the 3D subspace is just all of R3

and also, the subspace must always pass through the origin

[feral crow]

I got (-2, 3, 0) + t(-1, 2, -1) for a

For b i got the determinant = 0 so R was at the meeting

For c I got x y z = 2 -5 -5

Which doesn't plug back in to (-2, 3, 0) + t(-1, 2, -1)

t = -4 works for x and y

but i got t = -5 for z

Is how I got (2 -5 -5)

Nvm I guess right answer is (2 -5 4) I did my matrix multiplication wrong

Ignore ^

[dense holly]

how do steady state vectors in Markov chains relate to eigenvctors

[grave plank]

• Howard Anton and Chris Rorres, Elementary Linear Algebra, 11th edition (2014), Wiley.
• Jim Hefferon, Linear Algebra, 3rd edition (2017).
• Friedberg, Insel, Spence, Linear Algebra, Prentice Hall (emphasis on theory).
• David C. Lay et al., Linear Algebra and its Applications, Pearson (emphasis on calculations).
• Steven Leon, Linear Algebra with Applications, Prentice Hall (emphasis on theory).

Out of all of these textbooks, which is most worth studying from?

[slow scroll]

ive never heard of any of those, but I've self studied about a semester worth so far from "Linear Algebra Done Wrong"

[quaint heart]

Ladw is pretty great

Also Hoffman and kunze

[opaque blade]

im given these 4 vertices, trying to find angle where diagonals intersect

i thought id use dot product but i dont see how dot product works here considering u is from (0,0) to (5,3) so its not tail to tail with v vector (2,3) to (3,0)

[slow scroll]

if you project u onto v, you will get the angle supplementary to theta

[slow scroll]

err i mean, if you use dot product u.v = |u||v|cos(pi - theta)

[opaque blade]

wym pi we work in radians

i mean degree

[slow scroll]

well whatever u work in a;sldfkjasf

[opaque blade]

im not sure whether the answer refers to vector u as the whole line (from 0 0 to 5 3) or if its from the intersection point to 5 3

[slow scroll]

doesn't matter, the angle is the same

huh i wouldve thought the angle supplementary to theta was the angle from the dot product, but your answer is saying otherwise

[opaque blade]

well shit

if anyone else has an explanation itd help me

or ill go tmrw ask someone hopefully i hve time

[slow scroll]

Explain why each of the following is not an inner product on a given vector space:

im pretty sure none of these are positive definite, am i right?

[quaint heart]

the first two are definitely not positive definite

what polynomial gives you that the third is not?

[dusky epoch]

any constant, i'm p sure

[slow scroll]

oh yea, thats tru.

[coral sage]

hey y'all does this look right

(ignore C_{01} I know that one's right)

[coral sage]

ah, @half ice, I figured out how to construct that matrix from the other day

[wintry steppe]

Does an eigenvector have to be a column vector?

Here the 0 which the matrix * the eigenvector is is assumed to be a column vector itself [0, 0, 0] vertically, but could 0 not be seen as a row vector, [0, 0, 0] horizontally?~~ And if 0 was taken as a row vector, wouldn't the eigenvector have to be a row vector as it would have to have n x 1 dimensions, so the the product matrix ([0, 0, 0] horizontally) could have m x 1 dimensions~~ Actually nevermind you couldn't multiply a 3 x 3 matrix with a 1 * n matrix, and that's what a row vector would have to be (not n * 1 dimensions)

[coral sage]

https://aearnus.github.io/2019/04/30/a-matrix-based-model-of-soft-body-physics btw

[wintry steppe]

Actually, I'm confused again. Your eigenvector has to be column vector if it's to the right of the matrix A, but if it's to the left then it can be a row vector since the dimensions would work. So why can't the eigenvector be to the left of the matrix when multiplying, which would mean the eigenvector would have to be a row vector rather than a column vector?

[half ice]

@wintry steppe
Still need it?

[wintry steppe]

@half ice It? If by it you mean some help, yep

Or clarification rather than help per se

[half ice]

The eigenvectors are the null space of A - Iλ
They aren't usually just a column or row of the matrix

They are vectors that the transformation does not rotate

[wintry steppe]

@half ice "They aren't usually just a column or row of the matrix" - which matrix are you talking about here? I mean the eigenvectors themselves are column vectors, like the one in the image I posted

"The eigenvectors are the null space of A - Iλ" - I don't understand what this means and at this point I'm too afraid to ask.

[half ice]

Nullspace → The null space of a matrix A are the vectors that A maps to 0. That is, Av = 0

In the example above, you're looking for the vectors that your bottom matrix maps to zero

[wintry steppe]

ok I get that

[half ice]

This forces an equation:
4v1 - 2v2 - 2v3 = 0

[wintry steppe]

makes sense

[half ice]

You can solve that for any one, so do this:
2v1 - v2 = v3

This gives you the general form of the nullspace. All of the vectors that are mapped to zero look like this:
[ v1 ]
[ v2 ]
[2v1 - v2]

For any v1, v2 you want

[wintry steppe]

I get the solving part, the part I don't understand is you're mapping your matrix A to zero, with the understanding of zero being

[0]

[0]

[0]

Rather than [0, 0, 0]

Which you could do if your eigenvector preceded the matrix A (was on the left)

and was a row vector rather than column

[half ice]

Remember that, in order to multiply two matricies, the number of rows of the first must equal the number of columns of the second

For a matrix A
Av is a vertical vector
vA is a horizontal vector

Just to make sure that matrix multiplication thing is met

[wintry steppe]

Do you mean that the product of the multiuplication is a vertical vector or horizontal vector?

[half ice]

Both v, and the product Av

[wintry steppe]

It sounds like you;'re agreeing with me then

[half ice]

I always have this in my head when multiplying matricies. Let B be a column vector, it's clear why a product is then a column vector

[wintry steppe]

Yeah I get that. My question is just would it be valid to have your eigenvector in the equation as vA=lambdax rather than Av=lambdax, so your eigenvector is a row and not column

vector

Also, another question, how do I find two orthonormal eigenvectors for a given eigenvalue if I've already found? https://math.stackexchange.com/questions/1383725/what-is-the-difference-between-orthogonal-and-orthonormal-in-terms-of-vectors-an seems to imply orthonormality is orthogonal+normalised, so should I just find two orthogonal vectors and then normalise them?

[half ice]

Sure! But it looks kinda dumb. We avoid row vectors whenever possible

You should get the same vector as a row with that process

[wintry steppe]

Why avoid row vectors?

[half ice]

They're just not used in favor of column vectors

[wintry steppe]

ok

[half ice]

Column vectors use space better, that much is clear

[wintry steppe]

Is my method for finding two orthonormal eigenvectors the right idea?

[half ice]

If λ is of multiplicity 2, then the eigenvectors associated to it will describe a plane. It's possible to find an orthonormal basis to this

This is called the Graham-Schmidt process I believe

[wintry steppe]

How do I know if lambda is of multiplicity 2?

[half ice]

How many eigenvectors it generates or how many rows reduce to zero when finding the eigenvectors

[wintry steppe]

I think it only generates one eigenvector

Since I only found one when solving to find the eigenvector.

[half ice]

The example above has 2

[wintry steppe]

The matrix I'm using is the one I posted ages ago

[half ice]

All of the vectors that are mapped to zero look like this:
[ v1 ]
[ v2 ]
[2v1 - v2]

This can be seperated like so:
[1] [0]
v1[0] + v2[1]
[2] [-1]

That's the one from Khan

Can't read it. Maybe try a snipping tool?

[wintry steppe]

It's a symmetric matrix and has two eigenvalues, but I don't know if that helps in finding the miultiplicity of the specific eigenvalues

[half ice]

E2 = E3, that's an eigenvalue of multiplicity 2

[wintry steppe]

Just to check, an eigenvector can't be populated with all zeroes can it?

nvm question answered https://math.stackexchange.com/questions/990016/can-the-zero-vector-be-an-eigenvector-for-a-matrix

[oblique crag]

Any1 able to help?

[half ice]

What are you looking to prove?

Those are the definitions of one-to-one and onto

[jagged pendant]

perhaps they were given different definitions?

[gray glen]

[half ice]

@oblique crag
Then do you know your definitions of one-to-one and onto?

[oblique crag]

@half ice i honestly have no idea 😅

I had same thought

Looked at it and was like that is the definition...?

Maybe examples?

[jagged pendant]

@oblique crag a common definition for one-to-one is the following:
for any two v,u in the domain (which is R^n here), if T(v)=T(u), then v=u

which is a different setup than (a) but is equivalent.

as for onto, the usual definition is what's on there in (b)

[oblique crag]

ya got that but, wtf is this stupid book askin me to do LOL

should i just rewrite what they said 😂

[jagged pendant]

can you find somewhere in the same chapter, these definitions? @oblique crag

[oblique crag]

I have an ebook and bad signal@rn but can try to load it

[jagged pendant]

ok

[rare barn]

If A is a zero matrix and b-> ≠0, then the normal equations for Ax-> =b-> have no solutions.

is this true?

(-> denotes vector)

[quaint heart]

Yeah, you can show that very easily

Or you can view your matrices as functions, in which case the zero matrix is the function which maps every point to 0

From which what you want to show immediately follows

[slim pebble]

http://prntscr.com/njq7ix

is this asking me to find the PCP^-1 and the scalar?

[half ice]

There's some way to write:
[ 0.1 0.1] = [a 0] [cosθ -sinθ]
[-0.1 0.1] [0 b] [sinθ cosθ]

[slim pebble]

is that the matrix product?

[half ice]

Yus

[slim pebble]

is that the same as PCP^-1 ?

[half ice]

No

It hasn't mentioned conjugation

[slim pebble]

oh ok... also, when do you know to use the determinant or not for the (A-Lambda I)

[half ice]

Eigenvalues/Eigenvectors satisfy:
λv = Av

It follows that the eigenvalues satisfy
det|A - λI| = 0

That form is useful for finding λ

[slim pebble]

oh so the det λ is for the eigenvalues?

[half ice]

Yus, it's a form that makes them easy to find

[swift plinth]

Need to find a Basis of this vector space . I dunno how I transform this Matrix into line gradually form. Pls help

[slow scroll]

i don't see the vector space... what does the L mean?

[swift plinth]

it says I need to find the basis of this under vector space

[dusky epoch]

are you translating this from another language

[swift plinth]

yes I am from germany

translated some words via google translater

[broken hawk]

can you give the original

I can speak german

[empty copper]

You can also speak funny German

[broken hawk]

this is true

[swift plinth]

original

Ich hab versucht , das Gleichungssystem mit gauß zu lösen. Da bekomm ich nach einer Umformung eine nuller Zeile

[broken hawk]

was ist L(…)?

[swift plinth]

so stehts im skript

denk das ist einfach nur die bezeichnung für den Unterrraun

m

[broken hawk]

okay, so, for everyone, L is just the span

I have no idea why they use L

[swift plinth]

I will ask university linz 😃

[broken hawk]

so, what system of equations did you try to solve? really all you have to do is find out a subset of those vectors which is linearly independent

(which will be either one or two of them, since you’re in a 2D space)

[swift plinth]

but how do i find this subset? I thought I need to solve this system of equations

[shadow gyro]

Could someone instruct me in this problem?
Let E denote n-by-n unit matrix, v be a n-by-1 vector, v' its transpose
is it true that det(E - v*v') = 1 - v'*v ?

[broken hawk]

@swift plinth you misunderstood the question, I think. you have four vectors, these define some subspace of ℝ², either a line or a plane. you now need to figure out what that subspace is, then write down a basis

you don’t need to solve any equations

just figure out which vectors are redundant

(that is, among those four vectors, some can be written as linear combinations of the others, kick them out until you’re down to the smallest possible)

[swift plinth]

ahhh now I got it !!! thanks Sascha!

[shadow gyro]

I checked some random vectors and it seems to check out, don't know how to proceed with the determinant though

[dusky epoch]

it's very easy to see this by using an eigenvalue argument

[shadow gyro]

I noticed it has 1 as eigenvalue though, I definitely missed something

[dusky epoch]

consider $vv^T$ as a linear operator on $K^n$. clearly, its rank is $1$ (since its image is the span of ${v}$), so it has $0$ as an eigenvalue with multiplicity $n-1$. its last eigenvalue is $v^Tv$, with $v$ as its corresponding eigenvector

[stoic python]

Ann:

[dusky epoch]

so then, $vv^T - E$ has exactly those eigenvalues but shifted down by 1. so they're $-1$ with multiplicity $n-1$ and $v^Tv-1$ with multiplicity 1

[stoic python]

Ann:

[dusky epoch]

and then $E - vv^T$ has exactly those eigenvalues but with the sign flipped

[stoic python]

Ann:

[dusky epoch]

so $\det(E - vv^T) = 1^{n-1} (1 - v^Tv)^1 = 1 - v^Tv$

[stoic python]

Ann:

[shadow gyro]

Thanks for the help. Is it inconvenient to directly manipulate the determinant in this case?

[dusky epoch]

it would be incredibly inconvenient to try and do entry-manipulation

[shadow gyro]

I tried to take the last term out of the determinant but it didn't word out. Thanks Ann.

[dusky epoch]

wrong channel

this belongs in #geometry-and-trigonometry @jaunty fern

[jaunty fern]

Oops

kk

[wintry steppe]

Yo

What u have

[elfin schooner]

o hek there is one linear algebra section

[wintry steppe]

So, I'm stuck on 2 questions on my HW. I don't want the answers, but just the thought process behind it.

The first one asks

2.) Let V be the degree n or less polynomial over the reals. Show the linear transformation derivative is not diagonalizable by finding its eigenvalues and eigenvectors

[elfin schooner]

sure

My internet is shite, but I'll try

Okayy

Uhh

[wintry steppe]

Hmm

[elfin schooner]

If you differentiate a polynomial with degree n, you get a polynomial with degree n-1

[wintry steppe]

I have idea but idk if I'm right. I can't write to see if u make sense

So I know that the derivative knocks down the power by one, n-1

[elfin schooner]

Yeahhh

[wintry steppe]

And then I'm stuck from herr

[hoary nimbus]

Sorry irrelevant

[wintry steppe]

Wouldn't u have a missing eigenvalue when u take derivative?

Yes

[elfin schooner]

^

[wintry steppe]

That much I can figure out

So wouldn't that mean u couldn't diagonalize that shit?

Right

So make derivative transformation matrix, then show it has missing value

Then maybe do induction and show some shit

Like, base case, then inductive hypothesis, but idk. I'm just throwing shit out

I probably can pull off the induction part, but how would I do the derivative transformation matrix?

Hmm.. Idk. I'm thinking something with operators, but idk. Have u done operators in the class?

Maybe some. Sort of matrix multiplication to get the polynomial down 1 degree?

??

[elfin schooner]

Hint : derivative of ax^n is nax^(n-1)

[hoary nimbus]

@wintry steppe try first degree polynomial first, maybe it will be easier to find the matrix further

[wintry steppe]

The derivative of ax^1 is just a

[hoary nimbus]

I mean start with a_0 + a_1 x

[wintry steppe]

Oh xD

[hoary nimbus]

Guess your problem mainly is finding the matrix then

[wintry steppe]

Look in notes

[hoary nimbus]

Hmm maybe starting with 2nd order would make it more obvious

[wintry steppe]

Was it talked about at all?

[elfin schooner]

Some basic facts

T:P^n -> P^n
Basis vector - (1,x,x^2,...,x^n)

T(p)= p' = A•p

Take a general polynomial p, differentiate ita and solve for A ig

[wintry steppe]

That makes much more sense

I think I can figure it out from here

[elfin schooner]

hECK yEA

[hoary nimbus]

Nice

[elfin schooner]

Lemme know the matrix aight

I'm curious too

[wintry steppe]

I'll show it to you on Monday when I get the solutions

[elfin schooner]

AAAAAH okayy

[wintry steppe]

Nice bois

What is other problem

U said u had 2

[elfin schooner]

@wintry steppe @wintry steppe Can I ask a few questions on how to study linear algebra if you don't mind?

[wintry steppe]

😂 I'm not a great person to ask about studying, but I can try

[elfin schooner]

After Psycho is done with their questions of course

I study math in a horrible college

I got only two students in my class

And our teacher is plain horrible.

[wintry steppe]

@elfin schooner I'm the wrong person on how to ask for studying linear algebra ROFL

[magic merlin]

damn two students

[elfin schooner]

xd

[wintry steppe]

Oh, u seem to know linear well

U solved his stuff

[elfin schooner]

eh but I still am not satisfied you know

[wintry steppe]

I mean, being at such a small uni is gonna be hard period. No other students to work with

[elfin schooner]

I've been studying alone, and most of the time, I have no idea if I'm doing it right or wrong

[wintry steppe]

Less options for classes

[elfin schooner]

I badly need a good mentor

[magic merlin]

less students = more time with teachers

[hoary nimbus]

Well how about reading the textbook linearly

[wintry steppe]

Like, it's gonna be hard. Ur best bet is a prof

Maybe a grad student if u have them

[elfin schooner]

My batch is the first

[wintry steppe]

Small classes usually means easier to ask questions

[hoary nimbus]

^

[elfin schooner]

My teachers don't care usually

[hoary nimbus]

Or getting lost either

[wintry steppe]

O

[elfin schooner]

I've asked several questions, but they don't answer them

[wanton spoke]

Wtf

[wintry steppe]

Well, idk how to help then. Maybe find one who does

[elfin schooner]

"I'll tell you tomorrow" they say, and the tomorrow never comes

@wintry steppe From where?

[wintry steppe]

Like, linear, or any subject u do, a PhD can answer u

Did you want me to ask my other question

[elfin schooner]

sure

[hoary nimbus]

Hmm I can relate the pain of not having a proper mentor

[elfin schooner]

@wintry steppe you ask first, we talk later

[wintry steppe]

How many math profs have u asked ur question to? Every single one? Like, a PhD in math can answer ur questions.

Best yeah, ask ur question yo

3.) Let v1, v2 be subspaces of a vector space V. Prove V= V1(+)V2 iff V=v1+v2 and v1 (intersection) v2 = {0}. Let T: V -> V be linear such that T^2=I. Prove V=V (+) Vt where Vt = { v is in V | Tv=v} and V = {v is in V | Tv = -v}

The formatting is a bit messed up since I am typing on my phone. I can also send the screenshot of the problem if needed

[elfin schooner]

OKAY THE FIRST PART OF THE QUESTION AAHHHHHHHHHHH

I LOVE THISS

[wintry steppe]

[elfin schooner]

I mean I remember solving that problem

It was kinda satisfying

Lets hit V1(+)V2 implies V1+V2 and V1 cap V2={0} first

[wintry steppe]

Okay

[elfin schooner]

Since a direct sum HAS to be a sum, the first conclusion is satisfied

Comes from the definition

[wintry steppe]

Yes, that's the easy part

[elfin schooner]

Now let v belong to V1 cap V2

Since v belongs to V1, v can be expressed as linear combination of basis of V1
Let v=(sum) a_i • x_i a_i are elements of field and x_i are basis of V1

Since v belongs to V2, v can be expressed as linear combination of basis of v2

Let v=(sum) b_i • y_i where blah blah blah

Before I continue, I wanna know what your definition of a direct sum is

[wintry steppe]

Let me check my notes on that

[elfin schooner]

@tulip flower channel is busy

Heads up, I'm traveling, might reply late

[wintry steppe]

Okay

I'm meeting with prof rn. Sorry I cannot help, but wolf seems like a linear slayer anyways

[elfin schooner]

😂 😂

[wintry steppe]

So u donut me me

[elfin schooner]

byee

[wintry steppe]

Let V be a vector space and {v1, v2,..., vn} be a subspace of V. We say V is the direct sum of the vi's written (you direct sum the vi's from 1 to k). If v is in V, then there are vi in v1 such that V= v1+...+vk

Ummmm...I may have written a bit too much of my notes haha

[elfin schooner]

Gotcha

So we know that v-v=0

ie a_i • x_i - b_i • y_i =0

[wintry steppe]

Makes sense

OH direct sums are for subspaces!

[elfin schooner]

aixi belongs to v1, aiyi belongs to v2, I think you can proceed

Hek yeah fam

[wintry steppe]

I just now got that XD

[elfin schooner]

I can present a good intuition for direct sum if ya wish

Note that I figured this out by myself so it might be wrong...idk

[wintry steppe]

🤔 I mean, I will also work with my classmates on this last HW assignment tbh.

He does allow for that

[elfin schooner]

oh lmao

That's really cool

[wintry steppe]

I'm going to do this tonight, then do all of my remaining HW tomorrow.

But ty for the help!

[elfin schooner]

Anytime

But

fammm

I'd really like it if you were to listen to my intuition to the definition of a direct sum

[wintry steppe]

Okay

[elfin schooner]

And share some opinion you might have

Here's a nice lil example

Let V=R2

You know that the subspaces of R2 are

-R2
-Straight lines passing the origin
-{0} duh

R2 and {0} won't help much in direct sums, so lets skip that

Let's just say the subspaces V1 and V2 are straight lines passing through the origin

such that V1=/=V2

[wintry steppe]

Okay

[elfin schooner]

Now the definition of direct sum says that If v is in V, then there are vi in v1 such that V= v1+...+vk

[wintry steppe]

Yes

[elfin schooner]

Now, you can like intuitively see that if you choose a point in line V1, and a point in line V2, you get another point in the plane

That vector law of addition

,w plot y=2x, y=3x

[stoic python]

Display results online and refine query

Requested by SolitaryWolf#6406

[elfin schooner]

let 2x and 3x be the subspaces...for now

[wintry steppe]

Okay

[elfin schooner]

Now, you can like intuitively see that if you choose a point in line V1, and a point in line V2, you get another point in the plane

Does this make sense now?

The sum of subspaces spans the space

[wintry steppe]

OH

That took a min to get

[elfin schooner]

An easy case would be to take the subspaces x=0 and y=0 i guess

[wintry steppe]

See, you know linear algebra

I wish you well with finding a mentor though

[slender yarrow]

i have a much more easy approach for your direct sum thing tho

[wintry steppe]

I'm all ears for wanting to learn the material.

[slender yarrow]

so yeah take a vector v in V1 cap V2

V1 cap V2 is a subspace of V, so -v also belongs to V1 cap V2 (so it belongs to V2)

[elfin schooner]

mmhmm

[slender yarrow]

now we can write the zero vector in two different ways

0 = 0+0 = v + (-v)

0 belonging to V1 and V2, v belonging to V1, -v belonging to V2

[wintry steppe]

OH

[slender yarrow]

and since we have a direct sum the writing should be unique

hence v = -v = 0

[elfin schooner]

heck

That's pretty neat

[wintry steppe]

Just one more day of lin alg after today. I can get through it!

[slender yarrow]

(i literally took my class notes, couldn't remember it on the top of my head)

[elfin schooner]

gaaa

That's really neat fam

Meanwhile, my teacher skipped sums because why not?

[slender yarrow]

heck

[elfin schooner]

Thanks for the proof!

[slender yarrow]

[wintry steppe]

OK I looked back at this and I'm not sure how to find orthonormal vectors here. Wolfram's eigenvectors for the a-b eigenvalue aren't orthogonal

v1 and v2 give a dot product of 1

[wintry steppe]

Nvm got it, found another orthogonal eigenvector using the x1+x2+x3=0 condition

[delicate spire]

hi, how would I go about finding a basis for the orthogonal complement of W, which is the set of all vectors [x,y,x+y]?

[livid crow]

Would I be correct in saying that If W is a subspace of Rn and y is in Rn then projwy is always the closest vector in W to Y

or would it not always be the closest such as the case with 0 vector

[half ice]

ProjW(Y) is the projection of a vector Y onto W.

[livid crow]

Meaning it will always be the closest

Even if its the zero vector

as its a shadow

The length is different but its basically the same

[half ice]

Closest?

[livid crow]

Closest vector in W to y

[half ice]

No. A projection doesn't take a subspace, it takes a vector

[livid crow]

Right but in the book that I am reading

it says the following

So what Iam trying to get at

is this always the case

if W is a subspace of Rn and y is in Rn

is projwy always the closest vector in W to y

Or would there be vectors that break this logic

[half ice]

No, the projection operation takes two vectors and returns one.

This passage is stating that there always exists a specific vector in a subspace with those properties

[livid crow]

read from y^

[half ice]

y-hat is unique in W, because of its protection properties

[livid crow]

I understand

Is it always the closest to Y

y^ is projwy

[half ice]

I'm not sure what it means by "closest"

Hmm. I suppose that is the distance for when y - y^ is minimum

[livid crow]

like how far away it is

Im just gonna take it as thats always the case

[slow scroll]

by looking at the characteristic polynomial of the operator.

are you asking about geometric or algebraic? @robust swallow

Suppose $(t-1)(t+2)^2 (t-3)^4 = 0$ is the characteristic polynomial for a particular operator. Then the eigenvalues are $1, -2, 3$. 1 has multiplicity 1, -2 has multiplicity 2, 3 has multiplicity 4. This is algebraic multiplicty.

[stoic python]

kxrider:

[slow scroll]

Suppose $\lambda$ is an eigenvalue for an operator $A$. Then $\dim(\text{Ker}(A - \lambda I) )$ is the geometric multiplicity of $\lambda$.

[stoic python]

kxrider:

[slow scroll]

lets say you compute the kernel of A - lambda I and you get (1,0,0)a + (0,1,0)b for all scalars a, b as the kernel.

If you found the kernel using rref(A-lambda I), the vectors should form a basis for the kernel of A - lambda I.

So just count the number of vectors (1, 0, 0) and (0, 1, 0) thats two vectors, so the geometric multiplicity is 2.

[dusky epoch]

👀

[slow scroll]

on part b, I said that $x$ can be written as a linear combination of the basis vectors. In other words, $$ (x, v_k) = \sum_{j=1}^n \alpha_j (v_j, v_k) = 0 $$. Not too sure what to do next tho xd

[stoic python]

kxrider:

[dusky epoch]

how did you do (a)?

[slow scroll]

when v = x, (x,x) = 0 which is true if and only if x=0.

[dusky epoch]

try to use the same idea in (b) as you used in (a).

i.e. try to arrive at (x,x) = 0

[slow scroll]

hmm i have an idea

Alright got it. I said $$ x = \sum_{k=1}^n \alpha_k v_k.$$
Therefore, $$(x,x) = \left(x, \sum_{k=1}^n \alpha_k v_k \right) \ = \sum_{k=1}^n \bar{\alpha_k} (x, v_k).$$
$(x, v_k) = 0 \forall k$ which means the sum is zero, and $(x,x) = 0$.

[stoic python]

kxrider:

[dusky epoch]

oh you're doing this in the complex case huh

ok

[slow scroll]

well gotta be precise about things ya know

[dusky epoch]

wait does your definition of inner product space involve it being over C

[slow scroll]

My book defines it for both. The problem here didn't specify but i assumed complex.

[dusky epoch]

ok fair enough

you can do (c) now by reducing it to (b)

[slow scroll]

oh yea, took me too long to see, but if $(x, v_k) = (y, v_k) $ for all $k$, then
$(x-y, v_k) = 0$ for all k. Therefore $x-y=0 , \implies x=y$.

[stoic python]

kxrider:

[slow scroll]

@robust swallow If you have an eigenvector, $\lambda$ for an operator, $A$, then the eigenspace is $\text{Ker}(A - \lambda I)$. Basically, its a space of eigenvectors.

[stoic python]

kxrider:

[dusky epoch]

it's the same as between a basis and a space...

@slow scroll λ is an eigenvalue, not an eigenvector

@robust swallow an eigenbasis is a basis for an eigenspace, evidently

an eigenvector is a (nonzero) element of the eigenspace

is there a problem you're doing rn?

@robust swallow

[stoic python]

Vagorge:

[slow scroll]

OOF i called an eigenvalue an eigenvector 🤦 🤦

[dusky epoch]

okay

so in that case it means there is a basis for all of K^n consisting of eigenvectors of A

of which there are thus n

well

n LI ones

so A is diagonalizable, and there is no discrepancy between alg and geo mults for any eigenvalue

[onyx tundra]

algebraic multiplicity of lambda_1 is presumably the multiplicity of (lambda-lambda_1) in the characteristic polynomial

is the geometric multiplicity dim(Ker(M - lambda_1 I))?

[dusky epoch]

indeed

[onyx tundra]

and then M is diagonalizable if and only if algebraic multiplicity = geometric multiplicity for all eigenvalues?

[dusky epoch]

yup

[onyx tundra]

cool, nice coincidence that this was here when I looked

at school we have to horribly blackbox diagonalisation so I've been trying to understand it properly, think this is the last piece in the puzzle

all they do for it is like

M = U^-1 D U, where D is a diagonal matrix consisting of the eigenvalues of M, and U consists of some corresponding eigenvectors

and then M^n = U^-1 D^n U

[slender yarrow]

just wondering Jean Valjean, are you french?

[onyx tundra]

nope I'm british

[slender yarrow]

rip

cause i recently found out about a french prof (on Youtube) explaining those concepts freaking well

[onyx tundra]

is it in french?

[slender yarrow]

it is

[onyx tundra]

I'm not bad at understanding spoken french I guess so that would be cool to see

[slender yarrow]

https://www.youtube.com/watch?v=jJmXnpjmk_Y&list=PLE8WtfrsTAin8EGRBTDZWpa_6RVO3o2Ab there's 3 videos, you could skip to the last one if you want ^_^ enjoy