linear-algebra

[weak needle]

Now multiply the two expressions we have

[dawn fractal]

becomes I

wait so i just multiply $(U^*A^*U)(U^*AU) = I$ and say that that's simply $D^*D$?

[stoic python]

!superficialsicko

[dawn fractal]

i overthought that.

ofc it's I only because A is unitary

so D*D = I

essentially meaning D* = D^-1

[weak needle]

Yep

[dawn fractal]

meaning the complex conjugate of each entry of D, i.e. every eigenvalue of A, is equal to its mult. inverse

[weak needle]

Ok you kinda overcomplicated things, DD*=I is good enough, no need to move back and forth

[dawn fractal]

right

let $\lambda = a+bi\in\C$ be an eigenvalue of $A$. then $$\overline{\lambda} = a-bi = \lambda ^{-1} = \frac{a-bi}{a^2+b^2}$$

[stoic python]

!superficialsicko

[dawn fractal]

therefore the modulus of any eigenvalue of A must be 1

nice

[lavish jewel]

yeah, it's enough to make D D* = I, which is pretty straightforward for diagonal mats

[zealous junco]

im not sure how this proves it but i think u can use inner product, its really fast that way

<Av, Av> = <A*Av, v> = <v,v> = 1

[radiant yarrow]

What does S to F here mean? how does a function go from a set to field?

[lavish jewel]

a field is also a set

[dusky epoch]

^

[lavish jewel]

with fancy properties sprinkled here and there

[weak needle]

I mean this just says that if <Av,Av>=1, then <v,v>=1 right?

[dusky epoch]

"a function from S to F" just means "a function with domain S and codomain F"

[radiant yarrow]

Okay but what does 'going from one set to another' mean?

Ohh

[lavish jewel]

it'S what every function does

[radiant yarrow]

co domain?

[lavish jewel]

take in elements from one set and spit out elements from another set

the input set is the domain, the codomain is the output set

[dusky epoch]

no space between co and domain

[lavish jewel]

thanks, i wasn't aware of that

[radiant yarrow]

Okay, won't the output set be called range?

[dusky epoch]

range is ambiguous

it could refer to codomain or image

[zealous junco]

yea, I guess I didn't elaborate but v is a normalized eigenvector of A

[radiant yarrow]

image?

But okay

codomain it is

[zealous junco]

so like ull get |lambda| = 1

[dusky epoch]

codomain is part of the specification of a function, i.e. what outputs it's allowed to produce

image is the set of all outputs it actually produces

[radiant yarrow]

Okayy

wait why isn't this specified before?

[weak needle]

Oh, nice that is way better

[radiant yarrow]

Like I heard you don't need more than HS algebra to learn this

[dusky epoch]

because the book assumes basic set theory as prereq i guess?

[radiant yarrow]

Basic set theory?

[dusky epoch]

yes

[native rampart]

Take a set S,let f do something,get a subset of F, which may be F itself

[radiant yarrow]

I've never heard about codomains even in apostol calculus

How they're different from image

[lavish jewel]

ann explained it just above

[radiant yarrow]

Yeah

[native rampart]

.

[lavish jewel]

for example, you can have a linear function from R^3 to R^2

R^3 is the domain, R^2 is the codomain

but the image may very well be just a line in R^2, not all of it

or just 0

[radiant yarrow]

okayy

I didn't completely get it

[lavish jewel]

imagine a matrix of all 0s

[radiant yarrow]

Yeah

O

[lavish jewel]

the image is the 0 vector

[radiant yarrow]

Oh

[lavish jewel]

but it's a [0,0]^T vector

which is part of R^2

[radiant yarrow]

Okayy

[lavish jewel]

specifying the domain and codomain is like giving the bare minimum info about a function. it doesn't tell you much about what it actually does

[radiant yarrow]

Okayy

[wintry steppe]

Is the sum of two invertible matrices necessarily invertible?

Answer is no:

Let A = I (Identity Matrix) and B = - I.

Both are invertible.

but the sum of them not.

so far so good. How can I prove it?

formally?

[limber sierra]

that suffices as a disproof

the way you prove such a statement false is by providing a counterexample

and showing it doesnt work

if you want to be really complete, i guess show the 0 matrix is not invertible

[wintry steppe]

Could I find a general way to this?

I mean

for example I proved that AB is invertible.

[limber sierra]

the way you prove statements will be different from how you disprove them

[wintry steppe]

hmm.

[limber sierra]

to prove a statement about a type of object, you do what you did - show that the statement follows from the definition/some manipulation/etc

to DISPROVE such a statement, you just give an example of it failing

[wintry steppe]

okay.

[limber sierra]

sometimes there will be many many counterexamples, sometimes therell only be 1

but it doesnt matter

you just need to give 1

(and show why it fails)

[stone valve]

in linear algebra, how does one interpret the search for a vector v so that Av = Bv' <-> v = A^-1 Bv' and wht does it mean to if a couple of vectors is a base of (R,R³,+,.)
is it just transformations using matrices? or how does one get their head around this
the main issue i'm having is to see what would happen if B in this case isn't an I3 matrix but a different one, or is that not possible?

[nocturne jewel]

The 1st thing you did just relies on the invertibility of A

$Ax=b$ iff $x=A^{-1}b$, assuming A is invertible

[stoic python]

Mosh

[stone valve]

Yes

wasn't the question but fair to mention it :))

[nocturne jewel]

well it was

maybe phrase the question better if it wasnt why Av=Bu iff v=A^(-1)Bu

the "search for v" is just solving the corresponding system

[stone valve]

but

how does one visualize the "base"

is it like a distorted version of the x,y,z axis?

[nocturne jewel]

basis is just a set of vectors which every vector can be written as a unique combination of them

in R2/R3 it's just a set of vectors which define the concept of the axes, as well as 1 "unit" in that direction

[stone valve]

so a bit like the parallellogram rule for vectors but with added properties? Like the sum of three vectors form one final one

ah

gotcha

[nocturne jewel]

B is a basis of V if B is indep. and spans V

that's all it is from the vector space perspective

[stone valve]

oki

thanks 😄

[wintry steppe]

guy.s

what is the dim(C^3)

3 or 6?

[torn hornet]

dimention over what field?

[wintry steppe]

hmm.

If it's over R , then it's 6.

if it's over C, then it's 3.

[torn hornet]

ye

[wintry steppe]

Am I right?

[twilit anvil]

yes

[dawn fractal]

if $A\in\C^{n\times n}$ is normal and unitary, \is it true that any eigenvalue of A is nonzero?

[stoic python]

!superficialsicko

[quartz compass]

what's the determinant of a unitary matrix?

@dawn fractal

[dawn fractal]

no idea

hmmm

well

for starters, $\det(A^*A) = 1$ since $A^*A = I_n$

[stoic python]

!superficialsicko

[dawn fractal]

but $\det(A^A) = \det(A^)\det(A) = \overline{\det(A)}\det(A)$

[stoic python]

!superficialsicko

[dawn fractal]

so det(A) is nonzero

i.e. determinant of unitary matrix is nonzero

[quartz compass]

yeah, but something more precise than this

|det(A)|^2 = 1 is what you showed

so det(A) = e^{i theta}

I think you had this as a problem the other day too, but yeah

at least good to see again, if it had a 0 eigenvalue the determinant would be 0, so that answers your question

[dawn fractal]

yeah, ive had these problems for the past week lmao

im finished now tho

[dawn fractal]

interesting

[quartz compass]

yeah

[hollow ruin]

Can anyone please help with where to start on this? I tried putting T in the functional but couldn't do it, do I substitute the v1, v2, v3, v4 into the integrals?

[zinc wasp]

Can anyone help me with this concept. the answer is -5 17 -52 but how did it get there

[hollow ruin]

Can anyone please help with where to start on this? I tried putting T in the functional but couldn't do it, do I substitute the v1, v2, v3, v4 into the integrals?

[wintry steppe]

Hello, is it true that if det(A)=1 and 1 isn't in the spectrum of A then there always is a solution for (A-Id)x=b?

[scenic fulcrum]

The condition det(A) = 1 is useless here

[zinc wasp]

<@&286206848099549185>

[scenic fulcrum]

$v_C = P_{C\to B} v_B$

[stoic python]

Boomer

[zinc wasp]

I don't follow the lingo though I need to see it written out to understand what goes where

[scenic fulcrum]

In fact, when you have a vector x written with the column matrix X in basis B and the column X' in the basis C and if P is the matrix for the change of basis from B to C you have : X = PX'

Here you have to compute P^-1 X to get the coordinates X' in the basis C

[wintry steppe]

so $1 \notin \sigma(A) \implies det(A-Id)!=0$?

[stoic python]

FeverDream

[torn hornet]

mhm, think back to what it means to be an eigenvalue and it will be obv why this is true

[wintry steppe]

ohh I see it now, thank you

[zinc wasp]

I just don't get it I need a problem to work and see the steps step by step. I'm sorry @scenic fulcrum

[scenic fulcrum]

Chet are you sure of the direction of the arrow in your matrix? Its weird

[flint jackal]

I believe what you do is, find the inverse of the 3 x 3 matrix then do matrix multiplication with the 3 x 1 matrix

[scenic fulcrum]

Yes but the inverse of the matrix is a burden to compute

[zinc wasp]

Ok thank you I shall try that and see

[rose umbra]

what is the fomula for this distance : ?

i already have the orthogonal basis

[zinc wasp]

@scenic fulcrum yeah it's all from zybooks and they throw in some odd stuff

I think it is a terrible teaching tool

[scenic fulcrum]

Scarlet you have to find the projection P(2x^3) of 2x^3 over the subsace and make the distance between both vectors

There is an important formula for the projection

[rose umbra]

do you mean proj(2x^3,1) +proj(2x^3,x) +proj(2x^3,x^2) ?

[scenic fulcrum]

No

If I note <,> the inner product

And e1 e2 e3 the orthonormal basis you found

<u,e1>e1 + <u,e2>e2 + <u,e3>e3

u=2x^3

[rose umbra]

thanks

[zinc wasp]

@scenic fulcrum that didn't work it gave me a different answer

[scenic fulcrum]

What is the right answer?

[zinc wasp]

-5. 17. 52

[scenic fulcrum]

Ok

[zinc wasp]

I don't get how it go there though

[scenic fulcrum]

Because there is a mistake in the statement, the direction of the arrow

I think it should be C -> B and not B -> C

In the case "C -> B" you just have to write PX, the product between P and v in basis B

which gives your answer

[zinc wasp]

Yeah that was the correct way for the next problem

[scenic fulcrum]

In fact, when you have P(B->C) to obtain v(B) from v(C) it's easy, you write v(B) = P(B->C)v(C)

On the opposite side, compute v(C) from v(B) is hard, you have to compute P(C->B) which the inverse of P(B->C) beforehand

and write v(C) = P(C->B)v(B)

[zinc wasp]

Can you help explain this part to me now please?

@boomer

@scenic fulcrum

[hallow lodge]

is the formula for finding the nullity of a transpose matrix the same as the original matrix, where nullity(A^T)+rank(A) = # of columns, or do we switch to # of rows?

[gritty swift]

A^T is a regular matrix

so if
nullity(A) + rank(A) = # cols
nullity(A^T) + rank(A^T) = # cols in A^T = # rows in A

also rank(A^T) = rank(A) is a common result in linear algebra

[hallow lodge]

ok so the nullity(A^T) = (# cols in A^T) - Rank(A)

yea

[gritty swift]

so yes nullity(A^T) + rank(A) = # rows in A

[gritty swift]

yes

[hallow lodge]

ok thx

[chrome obsidian]

Why did I get c and d wrong?

[gray badger]

dam, u guys are smart

[limber sierra]

perhaps they dont want the x=, y= part?

just at a glance, havent checked your math

but thats weird notation

[wintry steppe]

Guys, wanna ask, what is the geometric depiction of it?

or what would be the way of thinking of it?

[limber sierra]

im not sure trying to visualize this is a great approach

[wintry steppe]

What does f do?

Yeah, sure. But It's asked away.

have to solve it.

[limber sierra]

are they more specific than that?

because thats a bit vague

[wintry steppe]

nope.

They just said, calculate with the help of basis vectors.

[lavish jewel]

show the original in german

[limber sierra]

ah they want you to describe what it does to the space by looking at the image of the basis vectors

[lavish jewel]

hmm yeah, describe the transformation geometrically

[wintry steppe]

yeah.

Since it's 3D, it's not so clear to me.

[limber sierra]

you can view e_1, e_2, e_3 as representing the three "axes" of space

[lavish jewel]

it looks like a rotation of the xy plane to me

[limber sierra]

well, this might help:

x_3 is mapped to x_3

and not otherwise used or changed

so the third axis is held constant

ie you dont need to think about it

you can just look at x_1 and x_2 and imagine its distorting a 2d plane

[lavish jewel]

it's a 45° rotation, not sure in which direction off the top of my head

[limber sierra]

ccw

[lavish jewel]

so try taking the inputs (1,0,0), (1,1,0), (0,1,0), and their negatives

and see what happens

[limber sierra]

(1 0 0) is mapped into the first quadrant

[lavish jewel]

mhm, seems so

[wintry steppe]

i did.

let's visualize it by GeoGebra.

:D

[limber sierra]

well he just described what its doing

[wintry steppe]

z- axis remains still.

[lavish jewel]

45° ccw, as nami stated

it's a rotation on the xy plane

[wintry steppe]

It'll play with x and y axes right?

[lavish jewel]

mhm

well, i keep saying xy plane, i mean x1x2 plane according to the problem's notation

[wintry steppe]

It's a linear transformation right.

We could also expect scaling of the vector itself.

Am I right?

[lavish jewel]

this one seems to preserve length

[wintry steppe]

Yeah, it preserves I think according to the calculations with bases.

How about z?

Could I say these rotations are around the z axis?

[wintry steppe]

my understanding of it.

specifically 45 degree ccw rotations.

[lavish jewel]

yeah

you chose like the worst way of drawing that xD

[wintry steppe]

That's a gift from god himself.

my drawing is at its best.

[dusky epoch]

bold move, pinging a mod like this

[lavish jewel]

oh shit

that's a nice emoji

[zealous junco]

i should go back to exam revision instead of importing parrot gifs

[lilac stag]

why do we ever learn cramer's rule if it's O(n*n!)?

[limber sierra]

its not computationally useful but its handy for some proofs

see the second and third example here https://en.wikipedia.org/wiki/Cramer's_rule#Applications

i agree that its typically given too much weight in linear algebra courses, though.

should just be stated as a lemma/trick and otherwise not really focused on

but i think lecturers see it as a way to make students practice determinants AND solving systems at the same time

[wintry steppe]

i heard its also used for over finite fields

and gives a simple formula for inverse of 2x2 matrix

[quartz compass]

I'd also argue it's geometrically meaningful

[atomic hound]

i'm stuck on understanding the last inequality, how it goes from "less than or equal" to "less than". I think it has something to do with choice of z but i can't see it. any help would be appreciated

[scenic fulcrum]

$|a_m|z = |a_0|+...+|a_{m-1}|+|a_m|> |a_0|+...+|a_{m-1}|$

[stoic python]

Boomer

[scenic fulcrum]

And then you multiply the inequality by z^(m-1)

[atomic hound]

ah that's perfect, i see it now. thank you so much!

[scenic fulcrum]

👍

[olive tundra]

Does anyone have notes on linear algebra?

[nocturne jewel]

Yes I do, cause I took notes

[olive tundra]

Can you share them with me?

[nocturne jewel]

No

1. They're physical
2. Using someone else's notes is bad practice, as they are less effective than making your own

[olive tundra]

I haven't even started my uni... i just wanted to look at it on surface to have an idea what am I dealing with

[nocturne jewel]

Look at 3b1b's essence series then

[gritty swift]

you can also watch some of https://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2010/video-lectures/ after 3b1b's series

[hollow oar]

https://linear.axler.net/LinearAbridged.pdf

it's axler's linear algebra done right with all the proofs and exercises removed

good for skimming

axler is more advanced than most university's linear algebra courses though

even more advanced than the honors linear algebra course that my university teaches

[gritty swift]

^ its very abstract though, if you're doing a computational/applied class then it isn't that good to read

[prisma cairn]

It's applied in theoretical physics

(hopefully, that's why i read it)

[quartz compass]

determinants are good

[hollow oar]

i can agree sort of

we wasted almost an entire month of my linear algebra course painstakingly proving every single determinant result

didn't even get to cover inner product spaces as a result

we spent literally two weeks learning that linear functions exist

in linear algebra

but luckily i got to do gaussian elimination a billion times

[quartz compass]

lol sounds like you should blame your professor for that one, not determinants

[twilit anvil]

umm how do you spend two weeks learning linear functions exist

actually , dont answer that

:^)

[olive tundra]

Thanks

[olive tundra]

Thanks

[hollow oar]

it was bad

we also lost about a week or so because of the texas electrical infrastructure failing in february, but somehow (according to the syllabus) we didn't end up actually losing any time

well i guess we lost time, it's just if it hadn't failed we would have covered extra material

[lapis fern]

i am slowly learning how difficult axler actually is: linear algebra done right =/= linear algebra done easy.

[hollow oar]

yeah it was the first textbook i used after spending like a month or two learning how to write proofs. it was very challenging

[lapis fern]

same!

[hollow oar]

like 6 months after going through the first 3 chapters i returned and it was much easier. math gets easier the more you do

it's just suffering

[lapis fern]

LMAO. I don't like the sentiment, but I am forced to agree.

[hollow oar]

it's not all suffering to be fair it's very cool at the same time

[lapis fern]

when a proof comes out clean, and i totally get what's going on, it's a rush.

those moments are fleeting, lol.

[hollow oar]

yeah and just the learning more concepts is fun

[lapis fern]

im banging my head against chapter 3 right now. i get like 10% of it.

[hollow oar]

yeah that one was rough the first time i encountered it. very rough i think

the later half about the dual space was probably the hardest part

[lapis fern]

🙏

[twilit anvil]

youre banging your head now, but, you'll get used to it (:

[frosty vapor]

do friedberg linear algebra

[lapis fern]

As opposed to axler?

[frosty vapor]

idk

i like friedberg a lot

no idea how axler is

[crisp pilot]

Ah yes, his linear algebra series is a visual treat

[lapis fern]

I have a small QQ about $\mathcal{L}(\mathbb{R}^2, \mathbb{R}^2 )$: do the linear transformations in here have their own structure? Like they form a group??

[stoic python]

Videlicet

[lapis fern]

Surely that group is not Abelian cuz like a rotation composed with a flip, is not the same as a flip composed with a rotation.....

and the group is finite, no? There are only so many families of linear transformations from the plane to the plane...
Scaling
Flipping
Rotation
Sheering

[native rampart]

You can take any L(V,W) and the linear transformations will form a vector space wrt addition and scalar multiplication

[lapis fern]

But if the action is function composition ? Do we get a group structure?

[native rampart]

No,you never will(if you consider L(V,W))

[lapis fern]

sad

Okay.

[native rampart]

Because consider T(1,0)=(0,0) and T(0,1)=(1,0)

[native rampart]

That's why GL_n(R) is a thinh

[lapis fern]

what is that? I don’t have many skill points invested in proper abstract Algebra

[native rampart]

Set of matrices with nonzero determinant

They form a group wrt matrix multiplication

You can fix a basis,and take their linear transform analogues

[lapis fern]

Okay. Tyty

[quick oak]

anyone has any idea how to approach?

[lavish jewel]

seems like they want the matrix in the canonical basis for R3 and R2, so you can try converting the outputs of A, which are given as coordinates in the basis T, to the canonical basis of R2

[quick oak]

@lavish jewel Is the A is given to us in base-T? I should convert it so standart bases?

[lavish jewel]

it's given with the domain coords in the basis S and codomain coords in the basis T

if i'm not mistaken, it should be as simple as multiplying A by a matrix whose columns are the vectors in T

someone can correct me if i misunderstood

[quick oak]

lemme try

@lavish jewel the multiplying result is this

what now you think?

[lavish jewel]

i'm not sure what else to do tbh

[quick oak]

I found a question like this,

is it kind of the same with mine?

[lavish jewel]

yeah same idea

just like that

you don't need what they call here B_T, only the B_w

aside from that, they start in the standard basis

yours starts with the standard basis for R3, but the other basis T for R2

in other words, it's like one already multiplied that B_w^-1

so to "undo the effect" of B_w^-1, you multiply B_w (i.e. make a matrix whose columns are the vectors of T, and multiply it from the left)

so i do think what i proposed was correct

[wintry steppe]

so i have a question. usually, to get the linear application u set on the colums the images of ur input, right?

so for my case, i have to get the linear application that transforms e_2n into e_2n-1 and e_2n-1 into e_2n

basically $(1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, ...) -> (\frac{1}{2}, 1, \frac{1}{4}, \frac{1}{3}, ...)$

[stoic python]

equis de

basically $(1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, ...) -> (\frac{1}{2}, 1, \frac{1}{4}, \frac{1}{3}, ...)$

[wintry steppe]

whats the matrix of this application?

[lavish jewel]

a permutation matrix

you can get it by swapping some columns of an identity matrix

[wintry steppe]

how to make a matrix with this bot?

[native rampart]

1 && 2\\
2 && 3\end{pmatrix}```

[wintry steppe]

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

🙂

[native rampart]

Double \ not single

[wintry steppe]

ah

well, is this the matrix?

[stoic python]

equis de

[native rampart]

That will be part of your matrix

The first 4x4 square will be that

[wintry steppe]

ye okey

but it is a patron

basically it is

e2, e1, e4, e3, e6, e5

and how do i prove this is isometry?

i mean, it is obvious, only the position is changing

not the values

maybe... the sum of isometries is an insometry?

so i can do 2 for even and odd numbers

: $

[native rampart]

Show UU^T=I

[wintry steppe]

what?

isometry isnt ||Tx|| = ||x|| ?

wait, i can actually rearrange the elements on x

[native rampart]

Isn't isometry (Tx).(Ty)=x.y?

[wintry steppe]

i dont think so (?)

i mean, idk if that is the same

In mathematics, an isometry (or congruence, or congruent transformation) is a distance-preserving transformation between metric spaces

[wintry steppe]

So makes sense if it is defined as

[native rampart]

mb,so it should be ||Tx-Ty||^2=||x-y||^2

[wintry steppe]

yeah, norms are preserved

[wintry steppe]

so for x =

[native rampart]

Anyway you know T(0)=0 here

So this reduces to (Tx).(Tx)=(x).(x)

[wintry steppe]

sqrt(1+1/4+1/9+1/16+...)

yeah, i can rearrange elements, cant i?

i mean, for general case, not particular

to prove it is injective is made by proving T(0,0,0,0,0,0,...) = (0,0,0,0,0,0,...) only?

and to prove it is surjective, matrix has to have inverse?

[native rampart]

Well injective actually means surjective wrt matrices

[wintry steppe]

what is wrt

[native rampart]

With respect to

[wintry steppe]

oh, rlly?

[lavish jewel]

i don't think that's right?

you can have matrices be rank deficient for only the rows or only the columns

[wintry steppe]

what is the symbol * on matrixes?

A*A

is the product?

[lavish jewel]

you'll have to check the notation in the book or article you're reading

can be anything from product, to convolution, to complex conjugate, to complex conjugate transpose

or more

[wintry steppe]

from those options only the product fits xD

[lavish jewel]

what are you reading

A*A looks like a gramian matrix to me

[wintry steppe]

its an exercice

actually no

lel

it sais

prove that A*A = AA* = I

identity

[lavish jewel]

so yes, complex conjugate transpose

show that A is unitary

[wintry steppe]

ye

[lavish jewel]

• means complex conjugate transpose

not product

[wintry steppe]

So if A belong to L2

[lavish jewel]

the product is implicit

[wintry steppe]

it just the transpose?

[lavish jewel]

what's L2

[wintry steppe]

https://gyazo.com/a29a48423283a869283e781d1c7a4360

this L

[lavish jewel]

just post the problem

[wintry steppe]

it on spanish tho

this is a comment from the teacher

[lavish jewel]

that's my first language

just hit me up

[wintry steppe]

we have to prove this

https://gyazo.com/fa654e761028c8b8902972c30e82ad34

the i=0 is a mistake, it should be i=j

[lavish jewel]

aight

[wintry steppe]

yeah so first i think proving A = A*

and then A A = identity

[lavish jewel]

no

A need not be A*

[wintry steppe]

o.O

[lavish jewel]

you just need A* = A^-1

[wintry steppe]

well, okey

but in this case A = A* = A^-1

any unitary matrix is costant number on the diagonal, and conjutes of top half on the bot half, right?

[dusky epoch]

tfw no \langle \rangle

[lavish jewel]

as an example

you can see the main diagonal of A is not constant

[wintry steppe]

Can sm help

[lavish jewel]

this might be a better fit for either #prealg-and-algebra or one of the questions channels

[dusky epoch]

see pins

[wintry steppe]

Okay srry

[lavish jewel]

no problem 👍

[signal creek]

Hi,
is this the place to ask about linear/affine/projective Transformations?

[nocturne jewel]

yes, linear transformations are linear algebra

[wintry steppe]

How do you denote the set of lines through the origin in R^3 in set notation?

[signal creek]

ok thanks,
are some linear transformations in 3d, projective Transformations in 2d ?
for example a rotation around the z-axis x-axis

something like that

[lavish jewel]

sort of?

if you do the final step of setting z back to 1

you represent the xy plane as [x,y,1], transform it in 3d, and renormalize the resulting vector's z coord to project it back

i would say it's more a matter of choosing to represent it as such

[signal creek]

ok thanks!

[gray dust]

each such line is the span of a nonzero vector, so the set can be written as

$\brc{\Span\brc x:x\in\bR^3\sm\brc0}$

[stoic python]

RokabeJintaro

[wintry steppe]

Why?

I don't think this makes it clear that is the set of lines through the origin

Wouldn't this also be planes through the origin?

[gray dust]

why planes

[wintry steppe]

Why wouldn't it be?

[lavish jewel]

because you can't span a (nondegenerate) plane with a single vector?

[gray dust]

do you know what span{x} (x!=0) is?

[wintry steppe]

It's a scalar multiple of a single vector

in R^3

[gray dust]

the set of all scalar multiples of x

that's a line not a plane

[wintry steppe]

Okay

But it's not necessarily through the origin

For example, (6,6,6) is in your set?

Oh never mind it is through the origin

[wintry steppe]

Hmmm, I need help here

I tried to get the 2 new matrices subtracted by 1 and 2, then reduced echelon form, but it doesn't work

The answer is completely different

[scenic fulcrum]

You have no unicity of the basis

@wintry steppe

[heavy crown]

Hi, is this enough to say T is injective?

because in my notebook it says:
If T is injective and {u1,u2,...uk}⊆U linearly independant then {T(u1), T(u2),...,T(uk)}⊆W linearly independant too
so I wondered if I could use it the opposite way like I said

[gray dust]

no, counterexample: define a linear map T:R^2->R, T(x,y)=x. then {e_1} & {Te_1} are linearly independent but ker(T) is nontrivial ie T isn't injective

[heavy crown]

Oh I see, this makes things harder
I was trying to rely on to solve what I needed to prove

sorry for not adding it in the question

I thought saying if T is injective so KerT is 0 and then can prove it

but your counterexample proved me wrong hmm means I can't say it

[gray dust]

also generally the converse of a true statement need not be true, so you can't go about assuming the converse of stuff you learned in class to be true

[heavy crown]

Yes I definitely agree 😅

[gray dust]

@heavy crown for this exercise just use the given info

[wintry steppe]

How do I find a basis for U = {p in P_4(F) : p(2) = p(5) = p(6)}? One basis that I think I found is 1, (x-6)(x-5)(x-2), (x-6)(x-5)(x-2)x, but I think there isn't an easy way to verify this

[heavy crown]

Yes I figured out later thank you for your help:)

[verbal pivot]

why not?

[wintry steppe]

How do I verify it?

I have to show that it span of that list is a subset of U and U is a subset of that list

Showing that it's linearly independent is not as difficult

[verbal pivot]

mmm, maybe i'm missing something since I'm looking at this superficially, but it's clear that, since U is a subset of P4, and U is not equal to P4 (for example, the polynomial x is not in U), then the dimension of U must be less than 5, now, the three vectors you listed are linearly independent, because they are of different degrees, the condition p(2) = p(5) = p(6), cannot be met by any polynomial with degree less than 3 (excluding the constant polynomial of course), so, the only options left are third degree polynomials and fourth degree polynomials, if you were to add another polynomial to the list, then it must fall on one of those three classifications, but you already have those spaces covered

[wintry steppe]

I agree with all of that, but that's not very concrete

We need more algebra than words

I think if we can prove that

deg U ≠ 4, then we are done.

Since as you said, deg U ≥ 3, and deg U < 5.

[verbal pivot]

yeah, you're right, maybe you could suppose that there exist some polynomial q in U such that it's li to the others three, and arriving at a contradiction?

or you could try taking any vector of U, and showing that it descomposes in terms of the three vectors that you have chosen

maybe there's a easier way that I'm not seeing lol

[wintry steppe]

Yeah but that seems difficult?

Or very tedious

[verbal pivot]

well, is the fact that any polynomial of the form ax+b, or ax² +bx +c is not in U, enough for you to conclude that U is of dimension 3?

you could say that U, has at least, three basis vectors, but with showing that ax+b and ax² +bx +c are not in U, you are excluding 2 vectors from the whole 5 basis vectors of P4 ?

[heavy crown]

can anyone help me point out what I am missing or doing wrong?

[heavy crown]

<@&286206848099549185> please tag me if you answer 🙏

[scenic fulcrum]

MX = 0 <=> the columns of X are in Ker(M). There are n-r independant possibilities for each column. Then dim Ker T = n(n-r)

@heavy crown

[heavy crown]

Hi thanks for replying !
I agree there are n-r independant possibilities for each column, which is dimN(M)
but why dim Ker T = n(n-r)?

[scenic fulcrum]

Let e1,...en-r a basis of Ker M.

Each column of X in Ker T is written in the following form Xi = ai1.e1 + .... + ain-r en-r

So you have to choose n-r coefficients by column

And n(n-r) coefficients for the entire matrix

@heavy crown

[heavy crown]

that's good explanation thank you ! @scenic fulcrum

[wintry steppe]

how do i get the first element in a vector? i was asked to get $x_1+y_1$ as a result from a series of vector operations where $\vec x = \langle x_1, x_2\rangle$ and likewise $\vec y = \langle y_1, y_2\rangle$, i got to the point where $\vec z := \vec x + \vec y = \langle x_1+y_1, x_2+y_2\rangle$ but now i am stuck on getting the first element from $\vec z$.

[stoic python]

Edward.

[wintry steppe]

Ping me if you have any ideas.

[limber sierra]

is <_, _> meant to represent the contents of the vector here?

or an inner product?

[wintry steppe]

contents, yes

they are components not inner products (which i use \cdot for)

this is actually a multivariable question but there's no channel here for that class

[limber sierra]

well, then consider that x_1 + y_1 is a content of a vector whereas \vec{x}, \vec{y} are vectors

so you need an operator that takes in vectors but spits out contents

dot product comes to mind

||take the dot product of \vec z with <1, 0>||

[wintry steppe]

thank you.

also there is #multivariable-calculus i will ask there next time.

[sudden narwhal]

So, uh I have a Fibonacci sequence defined by :
$F_0 = 0, F_1 = 1$ and $F_{n+2} = F_{n+1} + F_n$ for $n \in\mathbb{N}$
They asked me to prove that $Fn^2 + F{n+1}^2 = F_{2n+1}$
Any Ideas ?

[stoic python]

Herels

[limber sierra]

induction

[sudden narwhal]

hmm

that's easy to say induction when we are working with natural numbers, but for real I don't know how to do the second part of the induction here

[limber sierra]

well, lets say youre doing your inductive step, and assuming your statement is true for all naturals $< k$ and you wish to prove it for $k$

[stoic python]

Namingtong

[limber sierra]

$F_{2k + 1} = F_{2k} + F_{2k - 1} = F_{2k} + F_{2(k-1) + 1}$

[stoic python]

Namingtong

[limber sierra]

do you see how I did that?

and now looking at $F_{2(k-1) + 1}$, since $k - 1$ is less than $k$, we can apply the inductive hypothesis

[stoic python]

Namingtong

[limber sierra]

try to go from there.

[sudden narwhal]

you switched all the n into 2k - 1 ?

[limber sierra]

if you mean the ns here, yes

and then 2k - 1 = 2(k-1) + 1 from basic algebra

if you dont understand how I knew to do that, I knew $F_{2(k-1) + 1}$ was a \textit{goal}, so to speak, since thats our induction hypothesis; and $2(k-1) + 1 = 2k - 1$ so I was looking for that

[stoic python]

Namingtong

[limber sierra]

and i found it when i applied the definition of F_n

[sudden narwhal]

I know all that

continue

[sudden narwhal]

I'm just not sure about this line

[limber sierra]

apply this fact

for n = k-1

this is what we assumed when we started the inductive step

i.e. that the statement is true for n < k

and k - 1 is < k

[sudden narwhal]

yes

[limber sierra]

(in hindsight, i believe you only needed to assume it for k-1)

(not for all n < k)

(but whatever, doesnt make a difference)

[sudden narwhal]

now we want to prove it for n = k and more

[limber sierra]

so what does this become when you replace F_(2(k-1) + 1) using the inductive hypothesis?

[sudden narwhal]

$F__(2k) + F__(2k+1) - F_(2k)$ ?

I hate those discord commands

[limber sierra]

no... you shoul have some squares

youre using this (with n = k-1) to replace F_{2(k-1) + 1}

[sudden narwhal]

oh my bad

[limber sierra]

anyway sorry, i have to go but hopefully you get this finished

someone else may be able to step in

[sudden narwhal]

😦

[limber sierra]

(although you may have better luck in #proofs-and-logic )

[misty storm]

technical question: if a set S forms a basis for the space V

then it can be said that "S spans the space V"?

[verbal pivot]

yes, that's one of the conditions for S to be a basis of a space

[wintry steppe]

what is naive gaussian elimination?

[nocturne jewel]

if S is a basis for V, it must span V

however not every spanning set is a basis, since you can have a dependent set which spans

I'd believe so.. but I'm not 100% sure

since $(x^3-x^2-3x-1)$ mod 2 has a solution of x=1

[stoic python]

Mosh

[hollow garnet]

isn't it -1? Not 1?

[coarse rain]

it's mod 2

and -1 = 1 (mod 2)

[hollow garnet]

I guess I should reword this, my question is why are we taking mod 2 here? Isn't finding eigenvalues mean that we are finding zeroes in the characteristic polynomial?

[zealous junco]

cuz the field is F2

[hollow garnet]

Ah

I see, I never worked with F2 so I guess I wouldn't know this

[wintry steppe]

Okay, I'm still stuck on this.

Just got back from helping grandma and passing out for an hour.

Okay, so I get two new matrices from subtracting 1, and 2 from the rows.

Then, I try reducing to echelon form.

But the answer I get is completely different from the key

[clear fjord]

Anyone here good with convex hulls? If so, please ping me (@)anyone knows how to calculate the effiency gap of elections?

[wintry steppe]

is the adjoint of a transpose of a matrix the same as the adjoint of the same matrix without the transpose?

[wintry steppe]

Suppose that U and W are subspaces of R^8 such that dim U = 3, dim W = 5, U + W = R^8. Prove that U directsum W = R^8

Can I just say that dim(U + W) = dim U + dim W - dim(U cap W), and so dim(U + W) = dim(R^8) = 8 = 3 + 5 - dim(U cap W)

So dim(U cap W) = 0 and therefore they only share the zero vector

[north anvil]

Why is the span of a vector v denoted as span{v} with curly brackets instead of parentheses?

[wintry steppe]

It isn't

It depends on the book you use

In my book it is not

[wicked palm]

I'd imagine it's because the span of a vector is a set

[wintry steppe]

@wicked palm That's not really relevant, because the output is a set

The input is a list

[wicked palm]

that's the one

[slate stratus]

Anybody have any idea how to do this?

[wicked palm]

couple of things I've noticed: the null space is defined by the basis [1, 1] so the null space is just the span of that vector, and since the basis of the null space is of dimension 1 you know that the dimension of the solution set must also be 1 be the rank nullity theorem

[slate stratus]

how do you graph it?

[wicked palm]

it's dimension 1 so it would just be a line

[north anvil]

For vectors, what is the difference between the following:

1. The span of 2 vectors, u and v

2. The linear combination of the same 2 vectors, u and v

[limber sierra]

the span is the set of all possible linear combinations

its like asking whats the difference between "ℤ" and "an integer"

a linear combination of $u$ and $v$ is a vector of the form[ au + bv]
where $a, b$ are fixed scalars

[stoic python]

Namington

[limber sierra]

the span of $u$ and $v$ is the set of ALL linear combinations; that is, ALL possible values of $au + bv$ given by picking ANY scalars $a, b$

[stoic python]

Namington

[north anvil]

Oh okay

Thank you

Linear combination -> vector

Span -> set

Rather, the set of all linear combinations, as you mentioned above

Appreciate that, @limber sierra

👍

If two vectors are not multiples of each other, then their span creates a 2D plane

Does that mean all vectors with a nonzero magnitude are IN said span?

What about a vector that has components that are only 0s?

[limber sierra]

no; it might be a 2d plane in 3 (or more) dimensional space

that said, the 0 vector is in any span

set your scalars to 0

[north anvil]

oh ok

[limber sierra]

0u + 0v = 0 the 0 vector

[north anvil]

👍

Thank you

So a vector can intersect a plane in a 3d space at only one point

and if this is the case, then the vector is not "in" the plane?

Assuming this plane is created by the span of 2 3-dimensional vectors?

[limber sierra]

im not sure what it means for a vector to intersect a plane.

[north anvil]

given that an arrow can represent a vector, can't a vector penetrate a plane at some point?

[wicked palm]

if the vector isn't in the span of a plane then it "sticks up" out of the plane is how I visualise it

[north anvil]

yeah yeah

except the vector's tail and tip are on opposite sides of the plane

[limber sierra]

do you not center your vectors at the origin? (or at least at some common point)

[north anvil]

Oh this is just arbitrary

[limber sierra]

if so, then the plane visualization doesnt necessarily make sense

[wicked palm]

in linear transforms your vectors are always centred at the origin, what you're describing is an affine transformation

but I still don't think it would be in the span

[north anvil]

A plane can be created by 2 3-dimensional vectors that isn't the x-y, y-z, or x-z plane

Oh alright

[wicked palm]

correct

[north anvil]

so a vector with its tail at the origin can still intersect said plane

but isn't in the span of the vectors which creates the plane

the span is the plane in this case, i believe

[wicked palm]

ok maybe a more algebraic approach would help here

[north anvil]

possibly

[wicked palm]

[x, y] is in the span {[a, b], [c, d]} iff there are scalers j k such that [x, y] = j*[a, b]+k*[c, d]

[north anvil]

👍

[wicked palm]

also don't think of vectors intersecting the plane like it's a short line

vectors aren't lines they're directions

[north anvil]

with some magnitude

but yeah

@wicked palm appreciate the help 👍

Thank you 🙂

[wicked palm]

np

[wintry steppe]

I don't know if this is the right channel but I'm not getting answers in the questions so. I'm trying to solve an overdetermined system (4 equations, 3 unknowns). afaik, there are 3 ways to solve: matrices, lesser squares, and newton's method. (correct me if I'm wrong). I was wondering how these methods would be ranked in terms of "complexity" of the math involved.

[limber sierra]

newtons method isnt used for linear sytems

it can be used but theres no reason to

least squares (along with newtons method) only approximates a solution, gaussian elimination produces an exact one

so one usually wouldnt say least squares "solves" a system either

anyway, gaussian elimination is very simple - its just a way of representing the "high school" method of elimination with matrices

least squares TECHNICALLY involves solving a very simple differential equation

so it technically requires more background

but its not much harder in practice - you dont actually use techniques of diff eqs

(its faster for computers, but again, only approximates)

newtons method is very very simple and also easy to visualize if you know what a derivative is

i guess if you want them:

ranked in terms of "complexity" of the math involved
itd be gaussian elim < newton < least squares? but really none are particularly difficult with a bit of practice

gaussian elim is literally just the high school method of elimination

its just that you write matrices instead of equations

@wintry steppe

[wintry steppe]

okay. I should've been more clear bc now I have no idea. I'm doing trilateration with four spheres (satellites) to determine a point on another sphere (earth). The radii (distance from the satellites to the unknown point) is known for each sphere, now I just have to find the intersection point. So I have 4 equations (distance formula) and 3 unknown variables (x,y,z of the unknown point). Which method would be best to use?

[limber sierra]

gaussian elimination if you want an exact solution

although you can just plug this into any CAS

and it should do it for you

like type the equations into wolframalpha or whatever

er wait hold on

these equations are nonlinear

that makes life... harder

and means you probably wanna use least squares or newtons method

id wager the former would be better but im not really intimately familiar with the pros and cons of each

you can still try plugging the equations into wolframalpha though, it might be able to figure em out

or excel

[lapis fern]

so Axler here is defining addition of two linear maps......

i'm actually not understanding what i shouldn't know and what i shouldn't be assuming.......

so things i know:

1. S and T are linear maps individually.
2. S+T is linear by definition.
3. \lamba T is linear by definition....

why does he then say "you should verify that S+T and \lambda T as defined above are indeed linear maps"?

[stable kindle]

don't want you to assume 2 and 3

he says they're linear maps

but he wants you to prove they're linear maps

it's not by definition, it's just that when you work it out they are in fact linear

[lapis fern]

well, i guess he does two things there:

a. he defines what it means to add T + S
b. he makes a claim that it's linear....?

[stable kindle]

yes

[lapis fern]

okay, okay. gotcha.

i have another question:

L(V,W) is claimed (without proof) to be a vector space.

i need to show that it is closed under this definition of addition.

[stable kindle]

isn't that just the notation representing a vector space?

[lapis fern]

which to me means:

S,T \in L(V,W) \implies S + T \in L(V,W)

[stable kindle]

wait

oh, no, i am very stupid

i blame sleep dep

[lapis fern]

no, i don't think so. it's the set of all linear maps from V to W, V is the domain, and W is the codomain.

[stable kindle]

yes ok

[stable kindle]

yep

[lapis fern]

okay, so this proof is saying something slightly different from S,T \in L(V,W) \implies S + T \in L(V,W)

it's saying in need u,v in V......

[stable kindle]

well S and T are functionals, so they need to take something in? just completely arbitrarily

so when you're manipulating them the arguments have to be in there somewhere?

wait, no

i was confused

so they skip through closed under addition and scalar multiplication in the first paragraph

the rest looks like just showing distributivity and associativity and whatnot

[lapis fern]

oh.....that makes more sense......i dont know why they didnt say that out right.

[stable kindle]

so it goes

[wintry steppe]

Okay, I am having a very hard time understanding.

This has an algebraic multiplicity of 2. How did I end up with two separate vectors?

I am so lost when it comes tot rying to solve this.

Why is it the vector gives x2+2x3 = 0????

[lavish jewel]

algebraic multiplicity is not the same as geometric multiplicity

[wintry steppe]

I am just lost

Like why is x1 a free vector?

[lavish jewel]

as a simple example, take an identity matrix of size nxn. it has eigenvalue 1 with algebraic multiplicity n, but it has n different eigenvectors

x1 is free because the equation says x2 + 2x3 = 0

it doesn't matter what x1 is, because it is multiplied by 0

[wintry steppe]

I must be lacking some sort of concept.

Oh, I think I get it now.

But how is the second vector x1 = 1?

[lavish jewel]

well, what are the eigenvalues of the matrix?

[wintry steppe]

2 (algebraic multiplicity of 2) and 3.

[lavish jewel]

that sounds wrong

[wintry steppe]

https://www.slader.com/textbook/9780201658590-introduction-to-linear-algebra-5th-edition/314/exercises/13/

[lavish jewel]

,w eigenvalues of {{0,1,2},{0,1,2},{0,0,0}}

[wintry steppe]

This is where I got it.

[stoic python]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Edd#5182

[wintry steppe]

These are both vectors from subtracting the values from the original one.

I know my understanding is really bad when I can't even understand the key.

[lavish jewel]

i think you're missing too many concepts for this task. you will want to review how to compute eigenvalues again, and when it is that you get eigenvalues that are 0

because just by looking at that matrix, you should immediately see that it is rank 1

you should take a step back and review linear (in)dependence

[wintry steppe]

I really should...

I have no idea how I have a B in this class. I literally know and remember nothing.

I am so lost...

I don't understand linear independence

[marble lance]

A set of vectors is linearly indepdent if you cannot write any of the vectors as a linear combination of the other vectors.

[wintry steppe]

Hello. I have a simple question, I think.
A solution states that the first two vectors in this matrix (in rref) are not proportional.

Isn't it wrong, since they have one free variable?

Thank you..

[lavish jewel]

what do you mean?

if you rref that, you will get something where the first vector is of the form [1,0,x,x,x]

the second will stay as it is

[0,1,2,0,0]

there is no way to multiply the first one by a scalar and get the second one

[wintry steppe]

I guess my question is, are they proportional?

[lavish jewel]

what exactly do you mean by proportional

if it is in the sense i just explained, then no

[wintry steppe]

Don't really know exactly. My solution says that it's not linearly independent, because they have 1 free variable, but it also says that the first two vectors are not proportional (don't really know what that means)

[lavish jewel]

can you show exactly what the text says?

[wintry steppe]

I have to translate

[lavish jewel]

what's the original language

[wintry steppe]

norwegian

[lavish jewel]

oof

[wintry steppe]

:p

question: What is the dimension of W = Span(s).

[lavish jewel]

at most 3

[wintry steppe]

This is the matrix from S

[lavish jewel]

what are the v_i?

[wintry steppe]

first

row

is v1

and so on

[lavish jewel]

ah

should be dim 2

[wintry steppe]

That's right, how did you come to that conclusion?

[lavish jewel]

we have 2 linearly independent vectors

v1 and v2

linear combinations of these vectors will at most yield another 2 lin indep vectors

i.e. any 2 lin indep vectors on a plane can span the plane

[wintry steppe]

Oh, okay, got it, thank you

[wintry steppe]

Does anyone know why linear independence defined in terms of finite sums? As in, why shouldn't infinite linear sums imply dependence?

[weak needle]

Well, for starters we don’t even really have an infinite sum of vectors. We have a sum of finitely many vectors only because addition is associative so the sum for two can be extended

[dusky epoch]

if you talk about infinite sums at all you will have to talk topology

which in linear algebra is something you do not want to mess with

and also like, convergence bullshit

[wintry steppe]

problem asked about set of all functions from $(0 ... 1)$ to $\mathbb{R}$ ${g=1/(1-x),, f_0=1,, f_1=x,, f_2=x^2,, \dots}$ where g is the sum of all other functions by geometric sum, yet it's independent

[stoic python]

Orangus

[dusky epoch]

yeah

because if all you have is the vector space structure, you do not know what it means to add up infinitely many vectors

[wintry steppe]

I see, thanks!

[cold echo]

hi can someone help me with this? thanks in advance

[nocturne jewel]

1. check independence
2. check they span uniquely

independence is obvious since the basis vectors are all of different degrees

[wispy thicket]

u_1 and u_2 are subspaces of V

I found this as a way to get a basis for u_1 first https://www.slader.com/discussion/question/find-a-basis-for-the-solution-space-of-the-homogeneous-linear-system-and-find-the-dimension-of-tha-4/

would it be the same as getting the row-basis which my friend claims to be?

[cold echo]

is this right??

[drowsy flower]

If I have 2 basis B, C for same vector space with same number of vectors, how can I show that there must exist at least be one non-zero vector x, from that vector space such that [x]_B = c * [x]_C?, where c is some complex scalar. And the vector space is C^n so complex numbers. And [x]_B here means coordinate of vector x relative to basis B and same idea for [x]_C

so what I had discussed with some people before is that we can write [x]_C = P [x]_B for some matrix P. P in this case would be the change of basis matrix

But I don't really know how to find P and what I can do after this step...

[nocturne jewel]

so... change of base formula?

[drowsy flower]

change of base formula is [x]_C = P [x]_B where P is the matrix from base B to C

but since B and C are arbritary basis, how do I represent P?

Do I just introduce new variables to it?

I know that P is n by n matrix

okay so what I was thinking that matrix P should be the same as matrix MN where M is the change of base matrix from B to standard basis S, and N is the change of base matrix from S to C.

which means N = [w_1 w_2 ... w_n] and M = [v_1 v_2 ... v_n]^(-1)

but what do i do next.......

any ideas <@&286206848099549185> ?

[verbal pivot]

are you trying to say that each coordinate of x in base C, is the corresponding coordinate of x in Base B multiplied by this complex number c?

[drowsy flower]

not each

there exist some vectors x which satisfies that

[verbal pivot]

yes, but it's enough for only one coordinate of x to hold that property?, or should it hold for all coordinates?

[drowsy flower]

isn't [x]_B unique?

wdym my all coordinates?

[nova yew]

here for the question image of L is there another way rather than checking suitable options?

[verbal pivot]

, okay I got confused lol, by coordinates I mean the numbers that accompany each vector of the basis in the representation of some other vector, let's say that the vectors $v_1, v_2, v_3$ form a basis of a space S, and let $x \in S$, since $v_1, v_2, v_3$ forms a basis, we can write $x = a_1 v_1 + a_2 v_2 + a_3 v_3$, in this case each $a_1, a_2, a_3$ would be the coordinates of x with respect to the basis $v_1, v_2, v_3$, if the choice of basis is clear, one may write x = $(a_1, a_2, a_3)$

[stoic python]

b2unit

[drowsy flower]

[x]_B = (a1, a2, a3)

but x = a1 v1 + a2 v2 + a3 v3

[lavish jewel]

is this an exam?

[nova yew]

exam analysis, exam is over.

[verbal pivot]

ooh, so, if [x]_C = (b1,b2,b3) (the representation of X in term of the basis C), you're asking if there exist a vector such that each b equals each a multiplied by some complex number right?

$\lambda b_i = a_i, \lambda \in \mathbb{C}$?

[stoic python]

b2unit

[drowsy flower]

yeah so [x]_B = c[x]_C which means if [x]_B = (a1, a2 ... an) and [x]_C = (d1, d2 ... dn) then [x]_B = c[x]_C means (a1, a2, ... an) = (cd1, cd2 ... cdn)

[verbal pivot]

oh okay, I get understand, let me think a little

[verbal pivot]

a really quick thought, maybe you are looking for eigenvalues of the change of basis matrix

[drowsy flower]

yeah I think so as well

So i want to find its characteristic polynomial

but what is the change of basis matrix though?

[nocturne jewel]

a matrix in which you can multiply a co-ordinate vector to get the co-ordinate vector in the other basis

[drowsy flower]

like I was trying to do this

[verbal pivot]

you want to compute the matrix?

[drowsy flower]

I mean I know what change of basis matrix is, how do I find it is my question

In order to find the characteristic polynomial, I need to know what the change of basis matrix is

[nocturne jewel]

https://www.youtube.com/watch?v=1j5WnqwMdCk

Just google "how to determine change of basis matrix"

[drowsy flower]

but we want to find change of basis matrix from basis B to basis C which are both arbritary?

all that video is showing is [x]_B = M * c where c is the coordinates

[nocturne jewel]

then you have an arbitary matrix

You cant get specifics in a general sense

[drowsy flower]

yeah but then how can I find characteristic polynomial for arbritary matrix?

[nocturne jewel]

who said anything about char poly?

[drowsy flower]

well we want to show that [x]_B = c * [x]_C

so I was thinking c is an eigenvalue here

[verbal pivot]

maybe there's no need to go to such lengths as computing the poly char, showing that an eigenvalue exists may be enough for what you are looking for

[drowsy flower]

how else would we find the eigenvalues?

[nocturne jewel]

Can you post the question you're trying to solve? This feels like an XY problem...

[drowsy flower]

sure

[twilit anvil]

isnt it a well known proof that every linear operator from C^n to itself has an eigenvalue? so in this case one would just cite that statement for an eigenvalue

oops, sorry , meant to reply to @verbal pivot

[drowsy flower]

Let $B = {v_1, v_2 \cdots v_n}$ and $C = {w_1, w_2 \cdots w_n}$ be bases for $\mathbb{C}^n$. Show that there must exist $\lambda \in \mathbb{C}$ and non zero complex vector $x \in \mathbb{C}^n$ such that $[x]_B = \lambda [x]_C$

[stoic python]

meguuuuu

[verbal pivot]

Yes, I'm aware of that, but I wasn't sure if meguu knew so I didn't want to instantly bring it up lol, and midway the conversation I realized that I have a messed up in my head the differences between eigenvalues and generalized eigenvalues lol, so I was a bit uneager to keep talking

[twilit anvil]

oh, sorry about that... i shouldve thought about that earlier , whoops

ill be more careful

[verbal pivot]

but yea, if they can use that result, then I think that they are finished?

[nocturne jewel]

$\exists P\in\mathbb{C}^{n\times n}$ such that $[x]_B=P[x]_C$ by change of basis

[stoic python]

Mosh

[drowsy flower]

yes

[wintry steppe]

you can show it with char polynomial

and fta

[nocturne jewel]

since P is in C^(nxn), then there must be at least 1 eigenvalue (I dont remember the proof for it)

then rest is trivial

[drowsy flower]

Yeah I was thinking about characteristic polynomial as well but then I would need to find P first right?

I don't know the 1 eigenvalue proof

[wintry steppe]

Let A be a in C^(n×n), n>0 Then its char polynomial has roots cuz FTA

[nocturne jewel]

Ok so think about the char poly of P. it will be a n degree polynomial with complex co-efficients

[wintry steppe]

so has eigen

[drowsy flower]

the hint tells me that "Let S be the standard basis for C^n. Try applying the fundamental theorem of algebra to the characteristic polynomial of S_P_B * C_P_S"

[nocturne jewel]

$det(P-xI)=\sum_{j=0}^nc_jx^j \ c_j\in\mathbb{C}\forall 0\leq j\leq n$

[stoic python]

Mosh

[nocturne jewel]

then that has a root by Fundamental Theorem of Algebra, so there will be x such that det(P-xI)=0

[drowsy flower]

okay I think I sorta get it, but I am stuck in this idea that we need to find P first before we show that it must have roots. Am I correct or incorrect to think this?

[nocturne jewel]

P is just the change of basis matrix b/w B and C, so the columns are the basis vectors of B (or C, one of them lol)

[drowsy flower]

Aren't the columns , basis vectors of B if we are changing coordinates from B to standard basis S?

[drowsy flower]

which is why earlier I said this

should I go with that?

[wispy thicket]

what is E(z) here? I am lost in how to start d

[nocturne jewel]

E(z) is the set of vectors z+h where z is in V and h is in H (which of course is also in V)

[wispy thicket]

and Q is a set of such sets for every v in V?

[nocturne jewel]

yes

so the vectors in Q are sets of vectors in V, over the scalar field R

[wintry steppe]

guys, what are the principal components of a matrix?

[sudden narwhal]

real or complex coefficients

[wintry steppe]

so

eigenvalues?

[sudden narwhal]

eigenvalues are roots of the characteristic polynomial

if im not wrong

[wintry steppe]

So, how can i get principal components?

[sudden narwhal]

I dont understand your question

[wintry steppe]

how can i get the principal components of a matrix

[dawn urchin]

This was a question on my last exam that I got really confused on. I got 1 and I know c/d are impossible. And E=B^T but I don’t know how to get B or F?

For clarification this is an old exam that’s already been graded

[lavish jewel]

b should be impossible, since v in nul(B) means Bv = 0, so it cannot be in col(B) unless v = 0

f is possible for a symmetric matrix

[weak needle]

No not true

[lavish jewel]

just make a matrix with all rows orthogonal to v, and take M^T M = F

[lavish jewel]

do correct me if i made a mistake

[weak needle]

For example if we make the collums {1,2,3}, {-1/2,-1,-3/2} and {0,0,0} that is a matrix for b right?

[hearty cipher]

heyy guys so I'm currently taking linear algebra at uni and am sort of struggling xD anyone have any good literature or website/yt channel recommendations to help reach a good understanding and get through the topics fairly quickly?

[lavish jewel]

absolutely, good catch

[weak needle]

Also you can take F to be the 0 matrix

[dawn urchin]

I was so close! darn. Well thank you!

[lavish jewel]

the 0 matrix is the simplest case of a symmetric matrix with all rows ortho. to v, indeed

[weak needle]

Ah yes, that is true

[lapis fern]

hi #linear-algebra

when we say a vector space is of the form V= {0,v}

what does that notation mean?

[torn hornet]

whats the context for the notation?

[lapis fern]

ive seen it mentioned in MSE, but ive not encountered it before in texts

[torn hornet]

can you show me the post?

could mean a number of things at first glance

[lapis fern]

https://math.stackexchange.com/questions/1258364/what-kind-of-vector-spaces-have-exactly-one-basis

the only answer.

[torn hornet]

oh so it seems it is literally just the underlying set of the space

it only contains elements 0 and V

v*

so basically 1d vector space over F_2 it seems

[lapis fern]

what does that mean, underlying set?

[torn hornet]

so like

a vector space is like a set with additional structure right

this one is just two elements

[lapis fern]

oh, so this just lists them out explicitly.

[torn hornet]

mhm

[lapis fern]

tyvm

[wintry steppe]

having eigenvalues and vectores, how can i get the principal components of a matrix?

[wintry steppe]

I need a bit of help

I have no idea what the notation means

[lavish jewel]

you already have A and S

just follow the formula

[wintry steppe]

What is A?

Is just equal to S?

[lavish jewel]

read what you just posted

[wintry steppe]

Ohhh