linear-algebra

[wintry steppe]

conj(z*w) = conj(z)*conj(w)

mmm

but look the dot product here

it is z^n * conj(z^m)

so it reminds the same way, right? it ends in z^(n+m)

no

F

it ends in z^n * conj(z)^m

and whats the conj of z?

if z=a+bi then conj(z) = a-bi

yeah but i think im working on R

on R conj(x)=x

yeah so

[wintry steppe]

am i right?

tbh i feel just like your discord name

i don't know what's happening either

It's a bit weird if you don't know what T is and whether you are working on a real or complex space

i said T is a L2 Hilbert space

but i supposed u have read me xD

https://gyazo.com/a32ef912b6c8f7c948766138ce06a539

i'm not very familiar with these notation, but if my search is correct

L²(T) means the space of square integrable L2 operators on T

ye

but yes, on R

we havent worked on C yet

from T to R?

No, that this is for real number, not for complex

okay, in that case conjugate would be just identity function

you only need to conjugate when its on C

so z^n+m, 🙂

z^(n+m) yea

So i end with 1/2*Pi * (z^n+m+1)/n+m+1

what can i do now to say if n=m then the dot product is 1, else 0

?

now this depends on what T is

Does it not tell you more information about T?

Consider the Hilbert Space L^2(T)

bla bla bla

L2 are all the successions with dot product the one above

from my internet search, seems like $\mathbb{T}$ usually means the unit circle

[stoic python]

Carla_

[wintry steppe]

no, i got it, but i dont know how to translate

i speak a bit spanish

https://gyazo.com/6a68e0a35eaf09fa528c8858ae9bb3e7?token=c5f4f87525a5460b8a439a08c7fd48e9

these one are L2

that's $\ell^2$ i think, not $L^2$

[stoic python]

Carla_

[wintry steppe]

well, yeah, but i cant type that L :c

sorry

in your question seems like you're working on L² not l² since dot product is an integral

well, the next question is proof that the function 1/z cant be aproximated by polynomios like a0 + a1 * z + a2 * z^2 + ... + an * z^n in this space. Teacher gives us a hint which sais: proving that for any polynomio https://gyazo.com/353cc7348e560f2c39356aa1fcf1e18d

ok you are almost surely working with complex functions

wouldn't need to put absolute value inside square if it was real

i know this extends to complexs, but we arent working with them

And i'm pretty sure that T is the unit circle on complex numbers

i think she just copy pastes the integral, but rlly, on classes we havent touched complex

she mentions this works for complexs as well, but we havent worked with them

so i think she just uses the general notation, valid for complexs too

i don't see how this makes sense if not complex. maybe someone else can give more insight.

mmmm, okey, lets say it is Complex. Still, how can i proof that set is orthonormal?

if you're on the unit circle, there's something special that happens

do you see what z^n conj(z^n) is in that case?

0

?

nno

yes

1+i * 1-i

those are not in unit circle

Do you know how absolute value is defined for complex?

oh true

yeah it is 1

yeah is the distance to the origin

right?

yea, and |z|² = z * conj(z)

okey

but z^n * conj(z^m)

does it say something?

[stable kindle]

uh

[wintry steppe]

oops

[stable kindle]

consider 2?

[wintry steppe]

i mean |z|² = z*conj(z)

[stable kindle]

that sounds better

[wintry steppe]

but here |z|=1

[stable kindle]

ok

[wintry steppe]

so conj(z)=1/z

what

ah yes

by multiplicativity of ^n

conj(z)^n = z^-n

okey, right

so i end with

z^n-m

yea

so basically i have almost the same but witha minus

1/2*Pi * (z^n-m+1)/n-m+1

no

F

there's a case when that's not true

if n=m then the exp is 1

sry im being stupid

is ok, dont worry

yeah, it is 1

z^n z^(-n) = z^0 = 1

https://gyazo.com/c0c1c6cb0b30e215d6b842247164c14e

but i have z^n * conj(z^m)

you are integrating over a circle tho, not an interval

so you have to parametrize it

if n=m then can already see that is gonna be 1

cuz arc length of circle is 2pi

okey and i have to see that if n!=m it is gona be 0, right?

yes

thanks

[wintry steppe]

@wintry steppe hi again. Sorry for bothering, but i am stuck XD

you're not bothering, just ask

the parametrization of the circle is r(t) = (cos t, sin t). So r'(t) = (-sin t, cos t). So | r'(t) | = 1

But now i need f(r(t))

my f is z^n-m

idk what is f(r(t))

[still garden]

Is the part I highlighted a typo?

it should be alpha j

[wintry steppe]

you need to work with complex numbers.

R² is same thing as C except C has a multiplication, so in C it makes sense

yes but i dont know what to substitute

(x,y) in R² corresponds to x+iy in C

yes

oh

z is a+bi?

adasdasd

[thorny hemlock]

looks like it

[wintry steppe]

parameterization of circle is therefore cos t + i sin t

so it would be (cos t)^n-m + i(sin t)^n-m???

ofc not

F

it would be

(cos t + isin t)^n-m

aaaaaaaaaa

ye ye, totally my bad

F for newton

I killed him

you can also use that cos t + i sin t = e^(ti)

to write less

and if you're like me and don't know trig identities, this is a way to remember them lol

okey so i have to integrate

e^it*(n-m)

this is embarrasing

so i did something wrong.

https://gyazo.com/6488ac591cc7703f8bf2f75a1d14b036

where did i fucked up?

nvm, i forgot an i

but still

😢

so i end with 1/2*Pi * (0/m-n)

🙂

AAAAAAH because if n=m i cant integrate this way

lajsdhlasf

okey yeah, it is orthonormal 🙂

sry i was afk lol, maybe it would be better to ask questions about integration in #calculus tho

ye ye, i know

but i derped again

i was getting a 0/0

and then i realised ot was because if n=m then i cant integrate like that

but yes, for n != m i get the dot product is 0

and for n=m it is 1

Gucci

[wintry steppe]

Hey guys I could use some help with these two questions from my study guide. I honestly don't understand them and would appreciate a walk-through. Feel free to @ me Thanks 😄

[quartz compass]

what have you tried?

[wintry steppe]

I'm working on a different one rn Im not sure tbh though im really bad at linear algebra. I'd appreciate even a place to start

I understand the concept of similar matricies

and I'm assuming the first one will utilize A = PDPinverse with a NS matrix somehow

[quartz compass]

ok good,

what's it mean for a matrix to be nonsingular?

in particular, what do you know about its determinant?

[wintry steppe]

I think, there is a more rigiourous definition , but when its determinant is 0?

[limber sierra]

nonsingular means its determinant is NOT 0

[wintry steppe]

oops

LOL

[limber sierra]

i.e. its invertible

now, how are the determinants of similar matrices related?

[wintry steppe]

Its homogenous when 0 right

[limber sierra]

uh, homogenous doesnt apply to individual matrices, its a statement about matrix equations

or systems of equations

[wintry steppe]

ah nvm then

[limber sierra]

0 determinant means singular

[wintry steppe]

Im not sure , I only see determinants for similar matricies related to eigienvalues in my notes

det( B - lambdaIn)

But I feel like A is similar to B based on P

so im trying to make the connection between the det and P

[limber sierra]

okay im not sure exactly what material your course has presented

have you seen how you can write row operations as matrix multiplication?

[wintry steppe]

Yeah

[limber sierra]

actually thats probably not the best way to phrase this

let me revise

first off, we have A = PBP^-1 where A is invertible

can you justify B being invertible?

[wintry steppe]

If A and B are similar, they have the same determinant

If A is non-singular, the fact that A and B are similar implies that B is non-singular too

So B^-1 exists

Could make it be PAP-1 = B

ik you can Im looking in my notes for it

[wintry steppe]

That makes sense, this really cleared it up

[wintry steppe]

Once you've justified B^-1 exists, you just have to take the inverse of this

[wintry steppe]

Look up the proof if you're not satisfied with taking this fact for granted 😉

A^-1 = ( PDP ^-1) ^-1 you mean

Yup

Wait

Where does D come from?

It should be B

oops good catch

sorry i write D sometimes for diagnolized

lol

Cause i just learned about the diagnolized matrix

Ok now I know i have to just manipulate this to get B-1 = (PAP^-1)^-1

right

and that was like the point of the whole problem to show that it was possible

No need

im close now: I got to ( P A ^-1 ) ^-1 = ( PB ) ^ -1 stuck here

oh

[wintry steppe]

From this

Just expand the expression

$A^{-1} = (PDP^{-1})^{-1}$

[stoic python]

Laïka

[arctic stone]

I think you could also just multiply the equality by inverses and things from the right and left

[wintry steppe]

Yeah I ended up with A ^-1 and B^-1 tho

[wintry steppe]

ah ok

[arctic stone]

I mean wouldn't that be enough to show that these matrices are similar?

[wintry steppe]

Oh no I mean for this B-1 = (PAP^-1)^-1 I had A ^-1 still

I feel like I forgot to multiply by a identity PP^1 or something

[arctic stone]

Ohh

[wintry steppe]

Im so bad at this

and math in general lol

[arctic stone]

I think you can also multiply both sides by P from the same side, then P and P^-1 will cancell

[wintry steppe]

Yeah thats what i got

thank you though

its close enough

LOL

I think I at least understand it better

[arctic stone]

and you can multiply both sides by A^-1 or B^-1 on the same side

and stuff

[wintry steppe]

Ok so for the 2nd one this is what I tried doing

it seemed very abstract but I feel like I was on the right track maybe

[nocturne jewel]

A=[]

[wintry steppe]

Shhh I was going to fill it in but got lazy

[nocturne jewel]

$A=[a_{ij}]$

[stoic python]

moshill1

[wintry steppe]

Oh I wouldv'e written something way to compilicated lol

a1.... an

dots everywhere

im trash

[nocturne jewel]

that's equivalent, but what i put is the notation ive seen some people use

[wintry steppe]

I like your notation better

less dots

also the Identity matrix same thing but , yeah I moved on lol

[nocturne jewel]

yeah just write I_n is the nxn identity matrix

[wintry steppe]

noted

I was confused tho

how you would get lambda + r = 0 does the question mean to not solve for lambda

Or did it mean lambda = lambda + r

Cause to me it looks like they mean lambda = lambda + r but im not sure how they got rid of the values of A... or accounted for an

[quartz compass]

sounds like you're thinking about it wrong

take A+rI and multiply by v on the right

the goal is to show it simplifies to (lambda+r)v

[nocturne jewel]

$Av=\lambda v$ by definition, then consider $(A+rI)v$

[stoic python]

moshill1

[wintry steppe]

OHHHHHHHHHH

im dumb

rup

rip

Tbh not like I know what I'm doing at this point , def need to retake the class

For some reason the Gram Shmidt process seemed easy tho

[nocturne jewel]

I mean algorithms are ideally easy

[wintry steppe]

True , its just near the end of the class so its "supposedly hard". I guess its like learning Taylor series at the end of basic calc and realizing its not that hard

[nocturne jewel]

$det(xI-A)=det(-(A-xI))=(-1)^ndet(A-xI)$

[stoic python]

moshill1

[nocturne jewel]

so the polynomials are the same up to a +/- 1, which means they have the same roots and thus same spectrum of A

[fervent gulch]

Construct an example of a 2×2 matrix, with one of its eigenvalues equal to −1, that is not diagonal or diagonalizable, but is invertible.

[nocturne jewel]

Pick an invertible matrix st the char poly is (x+1)^2 I think

since x = -1 will be an eigenvalue of algebraic multiplicity 2, and geometric multiplicity of 1 (doing this mentally, that might be wrong)

it fails the characterization of diagonalizability

[fervent gulch]

geometric multiplicity ?

[nocturne jewel]

dimension of the corresponding eigenspace is the geometric multiplicity of the eigenvalue

[wintry steppe]

"how much linearly independent eigenvectors you get for a given eigenvalue"

[nocturne jewel]

equivalent since it's the cardinality of the basis for the eigenspace

[wintry steppe]

Sure

But I like the intuitive explanation

[calm yoke]

Suppose this is homogeneous, will it ever be determined?

I think for a=b=c=0 the W would cancel out, and it would be a determined, but my answer sheet likes to think otherwise

[limber sierra]

is this an augmented matrix, or a matrix A in the system Ax = 0?

also what do you mean by "determined"; just "neither over nor underdetermined"?

[smoky patio]

hope I'm not interrupting

any hints for this guys?

[pseudo thicket]

If {u,v} are linearly independent eigenvectors, then they must correspond to distinct eigenvalues.

this statement is false, right?

because of the geometric multiplicity?

one eigen value can have multiple linearly independent vectors?

[native rampart]

Yes,it is false

[wintry steppe]

they just have the same roots; in general the actual polynomials themselves might differ in sign

[dusky epoch]

show the determinant of this matrix is not congruent to zero modulo something, maybe

,w det {{5,6,2},{6,7,8},{2,8,3}}

[stoic python]

Results provided by WolframAlpha

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

Requested by Ann#0413

[smoky patio]

hmm

well yeah but I wasn't sure how

[dusky epoch]

wasn't sure how what

[smoky patio]

show its not congruent

[dusky epoch]

i evaluated det(your matrix) mod 10 just there

and got something that is very clearly not a multiple of ten

[smoky patio]

wait im missing something

how is that my matrix?

[lavish jewel]

mod 10

that's how

[smoky patio]

facepalm

how did you handle the term with question marks o-o

[limber sierra]

by just looking at them mod 10

"modulo 10" means "last digit" (in base 10)

[smoky patio]

ah..

I really have to understand modulo better.. thanks

nice

[smoky patio]

thanks a bunch!

[limber sierra]

the idea is that, if the matrix you gave has 0 determinant, then its determinant modulo 10 should be 0 (mod 10) as well

so ann computed the determiannt modulo 10

but its not 0

uh oh

[smoky patio]

uh oh

hahah

[limber sierra]

so your matrix cant possibly have 0 determinant

[smoky patio]

makes sense

tysm !

appreciate being pedantic :'))

[novel hamlet]

What does it mean for something to be semi-definetive

Or defibitive

[dusky epoch]

do you mean positive-semidefinite and positive-definite?

we say that a symmetric matrix A is positive-definite if <Ax, x> is positive for all nonzero x.

positive-semidefinite is the same except replace positive with >=0.

does this answer your question? @novel hamlet

[novel hamlet]

Yes that thingy

Got some homework on it due tuesday but cant understand what im supposed to do

Gona be back later to ask how to deal with this, i think I trymyself first

Is that <ax,x> same as cross product or dot product

[cursive narwhal]

I assume that $a \in \bR$ and $x \in V$, where $V$ is your vector space. $\langle ax,x \rangle$ is the usual notation for a 'dot' product on $V$, though we usually call it a scalar product

[stoic python]

Abhijeet

[cursive narwhal]

@novel hamlet

[novel hamlet]

Ah scalar product

Okay

[cursive narwhal]

But obviously, you can say <x,y>, where x and y are arbitrary vectors. Just look at the definition you've been given and work with that

This is also called an inner product, in case your book or whatever calls it that

[novel hamlet]

Idk the books are not in english

[cursive narwhal]

Which language is it in?

[novel hamlet]

Finish

[cursive narwhal]

Ah you are from Finland 😄

Nice

[wintry steppe]

Finland is a Fine land

[novel hamlet]

Ah, the books says to see eilen values if eigenvalues > 0 positively definite and If >=0 semidefinite

Eigenvalues

Ffs autocorrect

Not sure how to determine this for complex numbers, is 1-i positive or negative?

[crude falcon]

hey, I'm trying to prove that a set V is a vector space, for that I'm trying to prove that the + operation is a conmutative group, and I'm trying to see if the identity element exits, can I do this?
$(x1,y1) + e = (x1,y1) => e = (x1,y1) - (x1,y1) => e = 0$

[stoic python]

Jackieto

[cursive narwhal]

Neither

[limber sierra]

this doesnt make sense; you introduce e before you prove that an identity element exists

its circular

[crude falcon]

i mean I'm just using the definition of identity

if the result of that doesnt belong to V, then I can conclude that e doenst exists

[cursive narwhal]

What is your original question? Like, what is V exactly?

[limber sierra]

uh, this would be a way to prove e doesnt exist, sure

that would be a proof by contradiction that V is not a vector space

[crude falcon]

• operation is (x1,y2) + (x2,y2) = (x1+x2, y1+y2)

[limber sierra]

but that doesnt help you prove that V is a vector space

[crude falcon]

in R^2

[limber sierra]

what do you think the additive identity should be?

dont give me abstract symbols like e or 0

give me an ordered pair (a, b)

[crude falcon]

(0,0)

[limber sierra]

alright, so lets test it:

(x, y) + (0, 0) = (x+0, y+0) = (x, y)

and similarly (0, 0) + (x, y) = (0 + x, 0 + y) = (x, y)

so (0, 0) is indeed an additive identity

that suffices.

99% of the time, the best way to prove an "exists" statement (such as "there exists an additive identity") is to come up with an example and prove it satisfies the desired properties

in this case, there only exists ONE example, but thankfully its pretty easy to find

(0, 0) should be your first guess

[crude falcon]

alright that cleared it, what I'm not sure either is what I did to solve the ecuation is right algebraic wise, like is okay to pass the (x1,y1) as negative to the other side?

[limber sierra]

if youve proven an additive inverse exists, then sure

but proving additive inverses exist generally requires you to know what your identity is

so

[crude falcon]

does it matter which set the scalar used in a vector space is in? like in the example of R^2 is it required that the scalar belong to R or can it be anything

[lavish jewel]

it depends on which field the vector space is over

[crude falcon]

in R^2 for example?

[lavish jewel]

then it's a real number

[wintry steppe]

you could have R² over Q too for example, but its a different vector space

[dire thunder]

R^2 over Q

[dusky epoch]

R^2 but you only allow yourself to scale by rational numbers

this makes {(1,0), (sqrt(2),0)} LI

[wintry steppe]

yea

[dire thunder]

[limber sierra]

its a very messy space though, not super useful

R over Q sees some use in like, irrationality proofs i think? something like that

its also a good source of examples

[wintry steppe]

R² is same thing as C, not sure if that makes it useful tho

[limber sierra]

i dont think C over Q is particularly useful either

but perhaps im unenlightened

i could see some random algebraic number theory using it, but nothing comes to mind

not polynomial-ey enough

[dense oasis]

Hi, I know that $\dfrac{x^TA x}{x^Tx}$ is the Rayleigh coefficient, but what is $\dfrac{x^TA^{-1 }x}{x^Tx}$? Is it somehow related to the rayleigh coefficient? Is there some sort of inequality like $\lambda_{min} \leq \dfrac{x^TA x}{x^Tx} \leq \lambda_{max}$ where the two lambdas are the minimal and maximal eigenvalues respectivly?

[stoic python]

ch3ss

[verbal vessel]

I have a question. Suppose $A$ and $B$ are m*n matrices in R, show that if $N(A)=N(B)$ then there exists an invertible C such that $A=CB$

[stoic python]

Alireza

[calm yoke]

No, that is a matrix, and it is homogeneous

[calm yoke]

Like it has only one solution

[nocturne jewel]

is N() the kernel?

[verbal vessel]

Yes

I know that from rank-nullity theorem, rank(A) = rank(B), because both of them are m*n

So both of them have k vectors as their base,
base of A = {v1,...,vk}
base of B = {w1,...,wk}

But I don't know how to continue

[wintry steppe]

What if A=CB is the same as A=D^-1 B D

then A and B are similar matrices

What are some properties of similar matrices

[verbal vessel]

Why is this true?

[slow fjord]

5 vectors in 3D forms a regular pentagon.
How do I find the center point vector?

[wintry steppe]

First you need to define what the center of a regular pentagon is

[slow fjord]

How?

I'm coding it on c++

[lavish jewel]

you could probably write a function that consists of the sum of squared distances from one point in space to those 5 points, and then minimize it

[slow fjord]

I have a feeling that this will work. just need a reassurance:

Define a function that finds the midpoint of 2 vectors,
use it to find the midpoint of the bottom 2 vectors,
and use the func again to find the midpoint between the new vector and the top vector

or am I tripping?

[lavish jewel]

doing what i said, you get that the midpoint is the average of the 5 position vectors

and the function i proposed is convex, so there's a unique solution

so just add up the 5 points and divide by 5

give it a shot and lemme know

[slow fjord]

aaaah

cool

lemme try it

[dire yarrow]

#10, i cant figure out what Im doing wrong, i got two different angles of rotation

[stable kindle]

if sin(theta) = 1/sqrt2 and 0 =< theta =< 360, theta = 45 or 135

[dire yarrow]

Of course, dang im stupid, thank you very much

[slow fjord]

Awesome! This looks like a perfect midpoint. Thanks a lot, Edd 😄

[lavish jewel]

nicely, the result i gave you there has the name "convex combination"

you can read up on that, or use multivar calc as i did

[slow fjord]

Cool cool, I appreciate it

[dense oasis]

Thank you very much! Helps a lot!

[slender hull]

It seems to me the thing I'm being asked to prove is not true. Am I wrong?

Let z_k = x_k for an even k, k = 2n. Then
z_k(a + v) = cos(pi * k * ((a + v)/v))
= cos(2pi * n * ((a + v)/v))
= cos(2pi * ((na + nv)/v))
= cos(2pi * (((na)/v) + ((nv)/v)) )
= cos(2pi * ((na)/v + n) )
= cos(2pi * ((na)/v) + 2pin)
= cos(2pi * ((na)/v))
= cos(pi * 2n * (a/v))
= cos(pi * k * (a/v))
= z_k(a).

When k is even don't all​ the vectors in the eigenspace have the property z_k(a) = z_k(a+v)? Since the argument works for x_k and y_k, and therefore any linear combination of them.

[smoky patio]

I forgot to ask, is there any justification needed for this reasoning? like a theorem maybe?

[dusky epoch]

maybe that addition and multiplication "behave nicely" with modulo

[smoky patio]

true

[dusky epoch]

as in, congruence mod m preserves both addition and multiplication

and since dets are just a bunch of additions and multiplications in a trenchcoat the same is true for them

so if you take det(A) mod 10 then you can first take each entry of A mod 10 then do the det

and then take mod 10 again

[smoky patio]

ooooh

nice

gotcha

[tame mural]

If S is a set, what is the meaning of [S]?

[native rampart]

Context?

[tame mural]

A definition for graphs

E for edges is a subset of [S^2]

Ah I think it means that {1, 2} = {2, 1}

[sand sparrow]

Hi. I was trying to help someone with Linear algebra but the way the linear transformations are written confuses me:

To me the dimensionality doesn't make sense because if T(f) is the f(1) f(2) column vector that means f(1) and f(2) belong to R^1 and it looks like a p^1 to R^2 transformation

I am completely confused

[wintry steppe]

which picture is concerning you

just the first? or the second as well

[limber sierra]

f is an arbitrary element of P^2 here

check how the student's class is defining P^2

it probably means real polys of degree at most 2 or similar

[sand sparrow]

both basically. If I can understand one of them, I think I will be good with both.

[sand sparrow]

Oh ok. I never thought that P^2 could have been a polynomial of degree 2. That makes a lot more sense in that case. Thank you very much

It all makes sense now. His class was defining P as a polynomial

[limber sierra]

well, im not sure on that, which is why im asking you to check the notation

[sand sparrow]

It did plug in a polynomial of degree 2 for f in the P^2 example, when I asked the student.

[dreamy iron]

Is there an equivalent formulation for finite dimensional vector spaces?

like "All finite dimensional vector spaces are isomorphic to F^n, for some finite integer n" ???

also,

I know that every matrix transformation is a linear transformation, but the converse is false.

Is it possible to say that every linear transformation across finite dimensional domains and codomains is a matrix transformation?

[dreamy iron]

MSE says yes: https://math.stackexchange.com/questions/2170576/v-finite-dimensional-vector-space-and-isomorphic-to-mathbbrn

[smoky patio]

Assume part (a) is true, I'm working on part (b)

[deep warren]

i may come a bit out of context but yes, every finite dimensional vecstor space is isomorphic to K^{n} for some n

In fact, that n is exactly the dimension of the vector space

[smoky patio]

does this show then that B must be either positive definite or -B positive definite?

actually it doesn't really show anything

does it o-o

[deep warren]

Hmm i think youre using the variable z for two different elements of V

got kinda confused with that tbh

[idle flame]

you should first fix an element z in your space with the property you want, and then you can fix the subspace defined by it's span

[smoky patio]

ok ok thanks, sorry for the confusing work guys

[dusky epoch]

\cap btw not \bigcap

[wintry steppe]

after doing a good amount of precalc and calc, I am thinking of starting introductory Linear Algebra through self study and I would greatly appreciate any book recommendation on introductory LA

ofc I will be using the typical online video and practice resources like Khan Academy, brilliant, etc

[wintry steppe]

<@&286206848099549185>

[limber sierra]

what are you learning LA for? i.e. is your focus computational? algorithmic? proof-based?

[wintry steppe]

i am planning to pursue Computer Science ultimately

so for that

[limber sierra]

might want to try asking in #book-recommendations then

with that context

[wintry steppe]

sure, thanks a lot 👍

[quartz compass]

yeah that would be correct

I'm assuming (r,g,b) is a column vector multiplying on the right

sure I guess

[marble lance]

Why are you specifically addressing the honorables? Lol

[broken oxide]

Could someone help me with this?

Im not sure if i started it off right, I took v1, ..., vn to be the standard basis and then it would mean that V = [T]s and which would mean T is one-to-one and on-to since rank([T]s) = n meaking it invertable

[lilac stratus]

what does "V = [T]s" mean ?

[broken oxide]

The matrix representation of the linear transformation T with the standard basis

so the coefficients of T

[lilac stratus]

oh right

so yeah, that works

another way of proving it is to show that T^-1 will be the application that associates v_i to T(v_i) and extend it linearly (you can do it since the T(v_i)s are a basis of R^n)

[cursive narwhal]

You cannot take v_1,......,v_n to be the standard basis

They told you to work with any linearly independent list

However, you can use the fact that v_1,.....,v_n forms a basis for R^n and T(v_1),......,T(v_n) is another basis for R^n. Now, just check that T is surjective and injective. That will prove bijectivity.

I'm also not sure it's such a good idea to use matrix representations of a linear map for such a problem. It's not like it's necessarily wrong. It's just pointless because there's a direct argument that you can make.

@broken oxide

[broken oxide]

We actually didnt cover surjectivity and injectivity, so im not sure if i can use this method

[cursive narwhal]

What do you mean? You're doing linear algebra and you haven't learnt about surjectivity/injectivity of a map?

[broken oxide]

nope

Ohhhh u mean one-to-one and onto????

[cursive narwhal]

Yes

[broken oxide]

we didnt learn it by those names

[cursive narwhal]

Those are the weird words used for the same notions

[broken oxide]

Wait, why cant we assume U is the standard basis if vi can be any vectors in Rn. And if we have a lemma saying T is one-to-one and onto if rank([T]s = n

So if we prove it with the standard basis, would that prove thet T is basicially invertable no matter the basis

[cursive narwhal]

It doesn't say that v_1,.....,v_n is any linearly independent set in R^n

It says that v_1,.....,v_n is a linearly independent set in R^n

[broken oxide]

So would my method work if the question said it were any n vectors what were linearly independent?

[cursive narwhal]

Sure, then you could just use the standard matrix of T

but this is rather pointless; there's no need to use the matrix representation when you can just work with the map directly

[tranquil tangle]

sorry if this question has already been answered, but are there any good intro to linear algebra books? i want to start from the beginning because i have no experience with linear algebra.

i saw the books in the #books-old section, are those ones good for complete beginners?

also, what would i need to learn before doing linear algebra?

[idle flame]

to start an introductory course in linear algebra, I'd say basic geometry and trigonometry would be useful, understanding the physical meaning of a vector if you want some intuition to start, and the basics of vector geometry in 2 dimensions wouldn't hurt

[tranquil tangle]

alright, i'd say i know a good amount of geo and some basic trig. i also have some physics/kinematics knowledge, does the gilbert strang book/course cover what vectors are and basic vector operations?

oh ok, any other good books/courses then? ive tried a couple but they all seem too rushed or dont explain the foundations

oh ok

ill check it out!

do you have the link to the youtube series btw?

thank you!

ok, thanks ill probably do strang's book or another course after i finish this course

[idle flame]

I don't really know any good books that could give you a good introduction to linear algebra, I could look for some though.

But what I absolutely recommend is 3Blue1Brown's series in linear algebra

it gives you a ton of the basic ideas and much of the very needed intuition

you can find it in youtube

[tranquil tangle]

ah ok, ill look at that as well. i tried watching those videos a few months ago, but found them a little confusing, but i can definitely try again.

thx for all the help

[idle flame]

yeah you should watch them while you study the topic, otherwise the new concepts may seem a little alien

[lavish jewel]

are you trying to convert one color into another?

in rgb space?

if you're just transforming points, it's a matrix vector product

just answer this: you give an rgb color and it is transformed into another color

?

the thing is what you call "input" is weird

you're trying to show how a 3x3 matrix transforms 3d space, yes?

i would say the image is wrong

$\begin{bmatrix} R' \\ G' \\ B' \end{bmatrix} = \begin{bmatrix} C_R & C_G & C_B \\ D_R & D_G & D_B \\ E_R & E_G & E_B \end{bmatrix} \begin{bmatrix} R \\ G \\ B \end{bmatrix}$

[stoic python]

Edd

[lavish jewel]

this is what you want

i just changed the letters so that there was no confusion

it doesn't, you just mix xzy with rgb in a way that might be confusing

i just wrote it in a way that makes it clear which elements of the matrix multiply which elements of the vector

[stoic python]

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

[lavish jewel]

$\begin{bmatrix} R' \\ G' \\ B' \end{bmatrix} = \begin{bmatrix} C_R & C_G & C_B \\ D_R & D_G & D_B \\ E_R & E_G & E_B \end{bmatrix} \begin{bmatrix} R \\ G \\ B \end{bmatrix} = \begin{bmatrix} C_R \cdot R + C_G \cdot G + C_B \cdot B \\ D_R \cdot R + D_G \cdot G + D_B \cdot B \\ E_R \cdot R + E_G \cdot G + E_B \cdot B \end{bmatrix}$

[stoic python]

Edd

[lavish jewel]

as star person said, if the matrix is an identity, it's easy to show the sum is equal to the original vector

so that the delta is 0, as you wanted

what exactly are you calling an srgb derivative

the wording makes no sense

regardless of that, do you mean derivative as in calculus or something else?

that's a variable

the rate of change of what

you want the pointwise gradient of the vector field w.r.t. time?

the prime just means "new value"

or anything you want, really. it depends on the context

in this context, it looks like "it's an RGB triple, but the values are in general different"

so like, "new value after transforming"

[lavish jewel]

is that the direct sum of M and N?

i think it should be true as long as M is the complement of N and the sum of subsets is a direct sum

but not in general

[gloomy nebula]

Isn't $\alpha\wedge\alpha$=0?

[stoic python]

bunkermush

[gloomy nebula]

since wedge product of itself is 0

[quartz compass]

I suppose the point of the exercise it work that out to see for yourself

the fact that you asked means you were in doubt enough about it that you really should carry that out

[calm yoke]

Can somebody explain this?

Plz

[gray dust]

@calm yoke what's each letter mean?

[pearl scaffold]

Whats the point of gram Schmidt orthogonalization. Whats the point of converting the basis in a matrix to orthogonal basis

[calm yoke]

donde means where

[tough solstice]

Does someone know what the line above the x,y,z,w is?

[nocturne jewel]

Orthogonal vectors are nicer typically, and they're normalized so their norm is 1

But gram schmidt is used in matrix factorizations like QR

[tough solstice]

:z

What does a line over a symbol usually mean? Does it have any actions or meanings

ah

forgot about conjugation

thanks!

Whats the definition of global min/max?

When speaking of 2 functions

[nocturne jewel]

more calc than LinAl, but global max is the highest point on a function, min the lowest

[tough solstice]

👍🏿

[gray dust]

@calm yoke my best interpretation: E is a basis of some vector space and P is presumably an invertible operator on that space (guaranteeing H is a basis too), as a result P has the identity matrix as its matrix representation wrt E & H

[hallow edge]

Could anyone help me with this problem? I've tried setting up as a sum of two vectors equal to a and b. Although I came to no concrete conclusion.

Let B={b1,b2} be a basis for R2 and let T be the linear transformation R2→R2 such that T(b1) = 2b1+b2 and T(b2) =b2. Find the matrix of T relative to the basis B.

[desert light]

hi guys, does anyone know how to solve this?

[calm yoke]

Put all vectors in the same origin

[desert light]

what does that mean ?

[pearl scaffold]

#4

Doesnt (V . W)^2 = V^2 W^2

[lavish jewel]

so tell me, what does squaring a vector mean

[pearl scaffold]

Dot product'ing itself

[lavish jewel]

that's wrong though, that's magnitude squared

[pearl scaffold]

There's a vector squared and magnitude squared

[lavish jewel]

remember the dot product of V and W is | V | | W | cos(theta)

if you square that, you get a cosine squared too

so no

[pearl scaffold]

Oh yahh

And || P x Q|| = ||P|| ||Q|| sin theta

Bleh the absolute value sign gets messed up

I get the left side is |V|^2 |W|^2

But it says it should equal V^2 W^2

Maybe that's a typo?

Or maybe the italics implies that it is a scalar magnitude on the right side

[lavish jewel]

those letters on the right are in italics, not bold

presumably they stand for the length of the vectors V and W

which should match what you got

squaring a vector is pretty cursed

[pearl scaffold]

Didn't know italics to indicate magnitude was common practice

[lavish jewel]

me neither, but it's not bold, so it's a scalar

the book must have some section where it explains its own notation

you should revise it

[tough solstice]

Can someone explain to me what linear independence is? I dont really get it when i read it in my book

[naive whale]

What do you understand about it?

[tough solstice]

Something with the constants x and y

[nocturne jewel]

So let's say we have a set of vectors ${v_1,v_2,...,v_n}$

[stoic python]

moshill1

[nocturne jewel]

we say the set of vectors is linearly dependent if there are scalars $c_1,...,c_n\in\mathbb{K}$ where at least one is non-zero, $c_i\neq 0$, such that $0=c_1v_1+...+c_nv_n$ is true

[stoic python]

moshill1

[nocturne jewel]

the set is linearly independent if the only way $0=c_1v_1+...+c_nv_n$ is true is when $c_1=...=c_n=0$

[stoic python]

moshill1

[tough solstice]

And what happens if its linearly independent

Whats the reason behind it

[nocturne jewel]

what do you mean?

[tough solstice]

So what happens if its 0 or 1, whats the meaning behind it

i dont grasp the concept

[nocturne jewel]

Then the linear combination of the vectors in the set can only reach the 0 vector if all the scalars are 0

dependent vectors can reach the 0 vector non-trivially

[tough solstice]

Okay thanks!

[nocturne jewel]

simple example ${v_1,2v_1}$

[stoic python]

moshill1

[nocturne jewel]

so one vector is a scalar multiple of the other

$0=c_1v_1+c_2(2v_1)$ let $c_1=-2,c_2=1$

[stoic python]

moshill1

[limber sierra]

the vectors dont have to be scalar multiples of each other

consider $\left{\begin{pmatrix}1\0\end{pmatrix}, \begin{pmatrix}0\1\end{pmatrix}, \begin{pmatrix}1\1\end{pmatrix}\right}$

[stoic python]

Namington

[limber sierra]

this is obviously linearly dependent but no vector is a scalar multiple of another

perhaps you mean a LINEAR COMBINATION of the others?

but if they are scalar multiples, then they are dependent?
yes; suppose we have a set {v, sv}

so sv is a scalar multiple of v

(by the scalar s)

then setting a = 1, b = -1/s gives av + b(sv) = v + (-s/s)v = v - v = 0

and hence we have linear dependence whenever one vector is a scalar multiple of another

moreover, if we have a linearly dependent set, then we can express a vector as a linear COMBINATION of the others

[nocturne jewel]

The general idea is that if the set contains a vector which can be expressed as a linear combination of a subset of the other vectors, then the set is dependent, since you just take the subset that forms the linear combination and use those scalars, take -1 of the vector that gets made with that combination, 0 else.

[limber sierra]

for example, let's say au + bv + cw = 0 for a, b, c nonzero

then we can do algebra to rewrite this as au = -bv - cw

and therefore (since a is nonzero) u = -(b/a)v - (c/a)w

and hence u is a linear combination of v and w

this can be done for any vector in any linearly dependent set as long as its associated scalar coefficient is nonzero

linear (in)dependence seems like a weird construction but it behaves pretty nicely once you get used to working with it

[nocturne jewel]

span of a set of vectors is just the set of linear combinations of the vectors

$span{v_1,v_2}=c_1v_1+c_2v_2$

[stoic python]

moshill1

[limber sierra]

in fact, this gives us another way to think about linear independence

a set is linearly independent if its "minimal", in that removing any of the vectors would change its span

in a linearly dependent set, its possible to remove one of the vectors without changing the span

(since its just a linear combination of the others, as i showed above)

so it has "extra information" in a way

this is a very important concept: as it turns out, if span(S) = span(T) for two linearly independent sets S, T, then S and T must have the same number of vectors

we call S, T "bases" and their number of vectors the "dimension" of the space they span

the term "dimension" here should match how youre used to thinking of dimension; for example, one basis of 3-dimensional space is the three standard basis vectors

$\left{\begin{pmatrix}1\0\0\end{pmatrix}, \begin{pmatrix}0\1\0\end{pmatrix}, \begin{pmatrix}0\0\1\end{pmatrix}\right}$

[stoic python]

Namington

[limber sierra]

these correspond to the "x", "y", and "z" coordinates of the space

indeed, if you take any other vector in R^3 and write it as a linear combination of these vectors

[tough solstice]

How many solutions can there be for a linear equation system?

[limber sierra]

the scalar coefficients ARE the x, y, z coordinates of our vector

[nocturne jewel]

infinitely many, 1, or 0

[limber sierra]

(respectively)

[limber sierra]

nitpick: "infinitely many" is only possible if your space has infinitely many vectors

[nocturne jewel]

I was answering flush

[limber sierra]

yes

im aware

my nitpick still holds

[tough solstice]

👍🏿

[nocturne jewel]

you can have infinite solutions in a 3 eqn 3 unknown system though?

and 3 < inf

[limber sierra]

if youre solving systems over (Z/3Z)^7, then this space only has finitely many vectors

3^7 in fact

so it certainly cant have infinitely many solutions

[nocturne jewel]

no clue what Z/3Z is

[limber sierra]

the finite field of size 3

ie the congruence classes {0, 1, 2} modulo 3

[nocturne jewel]

Oh we used Z_3 for that

integers mod p

[limber sierra]

right

different notation for the same thing

the finite fields have like 5000 different notations

you also see F_3 or GF_3

the former standing for "field", the latter for "galois field"

its... annoying

(this notation is unambiguous because finite fields of the same order are isomorphic)

anyway, i digress

my nitpick doesnt really apply if the student is only solving systems over, say, R or Q or whatever

but its worth mentioning just in case

[nocturne jewel]

Nah my course spent 2 lessons on Z_p stuff and that was around midway through last term

so I forgot it existed

[limber sierra]

well there honestly isnt much theory to it tbh (at least vector spaces over Z_p)

it just comes up sometimes in constructions used in various places in mathematics and computer science

linear algebra is so well-behaved that we turn to infinite spaces almost immediately if we want to say anything interesting

the main utility to introducing them is as a convenient "simple" example of nonstandard fields we can define vector spaces over lmao

also matrices over Z_p tend to be faster to row reduce than matrices over Q or R in an exam setting

since you can sidestep screwing around with fractions

so that can be convenient as well

[tough solstice]

?_?

👍🏿

When speaking of ranks, what is this symbol called?

Phi?

[limber sierra]

rho (lowercase)

$\rho$

[stoic python]

Namington

[tough solstice]

thats the symbol for rank i assume?

[limber sierra]

i havent seen it before personally, but maybe?

[tough solstice]

👍🏿

[zealous junco]

is normal matrices that are not hermitian "rare"?

seems hard to think of a large set of them

nvm i guess ther are alot

like unitary ones

[wintry steppe]

Normal matrices are much more "general" than hermitian and unitary matrices which are considered special

Given a unitary /hermitian matrix, one can say a lot about their eigenvalues and even eigenvectors

whereas for normal matrices, any complex number can be an eigenvalue!

Indeed, any diagonal matrix is normal but a diagonal matrix is (usually) neither hermitian nor unitary

[native rampart]

Actually,For normal operators eigenvalues are real

And the basis in which it is diagonal is orthonormal

[wintry steppe]

normal matrices are unitarily diagonalisable*

[wintry steppe]

So their eigenvectors form an ON basis of C^n

[native rampart]

yes

[zealous junco]

hm isnt that only for hermitian that eigenvalues are real

[native rampart]

No, It's true for normal operators in general

See spectral theorem

[tough solstice]

when is a system of equations linear and non linear?

[native rampart]

mb

[zealous junco]

ah

[zealous junco]

linear is just when u have any variable x,y, e.g. in R^n and any constant a so that f(ax+y) = af(x)+f(y)

so a system is linear when u have say m of those f's

and u want f1(x) = k1, f2(x) = k2,..., fm(x) = km where k1,...,km are again constants

then its linear, if it cannot be written in this form its not linear

[crude falcon]

I have this subset: $U = {(x,y,t,z) exists in R^4 / ax+by+cz+dt = 0, a,b,c,d belongs to R}$ and I'm checking U is closed under addition

[stoic python]

Jackieto

[crude falcon]

but I'm not sure how to set up the vectors for doing so

like if I want to add two vectors, u + v, how would it work?

[simple scaffold]

Looks like ro

[lavish jewel]

please don't use latex if you're gonna produce such a cursed thing

[gray dust]

let u be an arbitrary vector in U, use the definition of U

there exist $u_1,u_2,u_3,u_4\in\bR$ where $u=(u_1,u_2,u_3,u_4)$ and $$au_1+bu_2+cu_3+du_4=0$$

[stoic python]

RokabeJintaro

[crude falcon]

alright

[gray dust]

let v be another arbitrary vector in U and do a similar thing

[crude falcon]

and the addition is just component by component like normal vectors right?

[gray dust]

the usual rule of adding in R^4 is entrywise

[marble lance]

Wait

{(1,0), (0,1)} is a basis

Not its span

[lavish jewel]

there is no "the basis", there are infinitely many bases for such a space

but yeah, if you take the cross product of two vectors, the result is orthogonal to both

[zealous junco]

hey so if i got f(x) = 1/2 x^T A x , then I computed grad(F) = A^T x

and then grad(grad(f)) = A

but if I do J(grad(f)) don't I get A^T

[lavish jewel]

you missed a 2 in the gradient

[zealous junco]

right yea

but the point is that is true right

(i edited)

because then theres a confusion

if its true

[lavish jewel]

J(grad(f)), or the Hessian, should be 2A

grad grad is like vector laplacian?

[zealous junco]

isn't Jacobian defined as the transpose of gradient for vector valued

hm

[lavish jewel]

it really depends on what notation you are using, to be honest

there is no unique definition for them

[zealous junco]

yea, so here i think we define grad(f) = [grad(f1)...grad(fm)]

[lavish jewel]

depends on your book or whatever you're reading

[zealous junco]

ok so lets assume grad(f) where f is sclar valued is just column vector

and then grad(F) = [grad(F1) ... grad(Fm)] where F is from n to m

[lavish jewel]

what is F

[zealous junco]

F is vector valued function from R^n to R^m

[lavish jewel]

mhm

[zealous junco]

then say f(x) = 1/2 x^T A x

yea sort of

I computed this and got grad(f) = A^ T x

[lavish jewel]

i would say the gradient of F is of size R^{n x m x m}

and sure, grad (f) = A^T x

[zealous junco]

then grad(grad(f)) = A

but heres where we learned that hessian is symmetric

and this is not symmetric?

[lavish jewel]

A is not symmetric?

[zealous junco]

yea

[lavish jewel]

o then you gotta be careful

[zealous junco]

wait i mean if A is not symmetric its still a smooth function

and grad(f) = A^Tx still holds

[lavish jewel]

yeah but it's not a quadratic form

[zealous junco]

but i think in class the prof said hessian of any smooth function is symmetric

[lavish jewel]

i don't know if that's the gradient anymore

[zealous junco]

ah

[lavish jewel]

i don't think it is

or is it?

you'd have to express it as sums and check for yourself

[zealous junco]

yea i think i did but let me send something

hm i prob did it wrong

missed some middle terms

i mean in this problem A is symmetric but im just wondering in general

nabla^2 is just nabla applied to nabla again if the functions smooth

sufficiently cont diff

like C^2 here

usually smooth just means smooth enough for the problem XD

but i think the official def is C^infinity

ah yea i was suppose to add it in the beginning

error

also i see my mistake, missed terms and added wrong terms

would be correct if A was symmetric

[lavish jewel]

yea

[quartz compass]

well there's an entire 2D plane of vectors to choose from that's got all vectors orthogonal to your single vector

let's say that vector is v, you can now just pick some arbitrary vector, say u, and compute u cross v and this resulting vector is perpendicular to v, and so is in your plane

so that means w = u cross v is a vector in your plane, to make another basis vector that's orthogonal to that you can do w cross v to get a new vector

and now you're done

there are other ways of getting a basis though too

haha yeah

depends on what your starting vector is you're trying to make things orthogonal to

if your starting vector is just (0,0,1) then you can just pick (1, 0, 0) and (0, 1, 0) directly

you don't wanna waste time computing cross products unless it's some really crazy vector like (1/2, 3/7, -9/sqrt(2)) or something where it's not obvious

also you can do it by computing dot products too, you could have instead taken the random vector you picked and instead of taking the cross product, do a dot product

then you find the projection onto v, and subtract out that

so w = u - (projection of u onto v) is another way to pick it

well, do whatever is easiest and fastest for you

[tough ore]

if I have a question about hyperboloids would this be the right place to ask?

I have to solve this problem

and I don't know how to prove that the tanget plane cuts the surface after two lines

we aren't using derivatives

so 2d is the way, I think:D

[tough ore]

Thank you

[velvet moss]

why does the transpose of an orthonormal matrice equal its inverse?

[wintry steppe]

define "orthonormal matrix"

[velvet moss]

square matrix where every column can be represented as a unit vector that is orthogonal to every other column

[quartz compass]

orthogonal matrices don't change vector length so let's say you take some arbitrary vector v and transform it by A to make u=Av

then since lengths are equal, $u^Tu = v^Tv$

[stoic python]

Merosity

[quartz compass]

we can plug in u=Av here

$(Av)^T Av = v^Tv$

[stoic python]

Merosity

[quartz compass]

$v^T A^T Av = v^Tv$

[stoic python]

Merosity

[quartz compass]

now put an identity matrix in the RHS

$v^TA^TAv=v^TIv$

[stoic python]

Merosity

[quartz compass]

subtract it over

$v^T(A^TA-I)v = 0$

[stoic python]

Merosity

[quartz compass]

since this holds for all v, we have

$A^TA-I=0$

[stoic python]

Merosity

[quartz compass]

that's pretty much it

[velvet moss]

thats actually really simple

what??

ok well ty

[lavish jewel]

the alternative is that the matrix has columns a_i, and since they are orthogonal and unit norm, a_i^T a_i = 1, but a_i^T a_j, with i =/= j, is 0

[quartz compass]

yeah that's way better

[rose umbra]

if T is invertible linear transformation, does it means that T^-1 is also linear transformation?

[lavish jewel]

yeah

[rose umbra]

@lavish jewel does it also mean T = T ^-1 or nah?

[lavish jewel]

no

[rose umbra]

i mean T(x) = T^-1(x ) ?

[lavish jewel]

also no

[rose umbra]

kk thanks

[frozen compass]

can someone help me with this one?

[lavish jewel]

you can find an orthonormal basis for the plane using u1 and u2 and gram schmidt

then you can do an orthogonal projection of y onto the plane using that basis

y - y_p is the component of y orthogonal to the plane

the length of that component should be the distance from the plane to y, if my last 2 neurons are working

[novel hamlet]

Im a bit confused, is cholensky decomposition A = TT^t plausible only for positively definite matrices or does it work for positiveluy semidefinite?

[novel hamlet]

actually now that i think of this

I have following block matrix and i need to show it is positive definite if and only if A1 and a2 are positive definite

I know that Det(a) = det(a1)*det(a2)

and that det(a)=product of its eigenvalues

but im having trouble here since if det(a1) and det(a2) are both negative det(a) is positive

[novel hamlet]

nvm was just stupid since det(A)=det(a1)*det(a2) = product of eigen values of a1 x egenvalues of a2 => eigenvalue of a1 is eigenvalue of A

[spiral star]

you dont even have to use the determinant

[novel hamlet]

yeah, i was just being stupid with my initial aproach

but not sure on how to do it without determinant

[spiral star]

well for one direction, suppose A1 and A2 are positive definite, write x = (x1, x2) and then x^T A x = x1^T A1 x1 + x2^T A2 x2 > 0 whenever x != 0

and for the other direction, you just need to use the eigenvalues

A is symmetric and all its eigenvalues are strictly positive, so you can diagonalize A

but you have two invariant subspaces, one for A1 and one for A2

so you diagonalize A1 with positive eigenvalues and same for A2

which makes them also positive definite

[novel hamlet]

well yeah that is more straight forward way of doing this

said me after wriiting A4 full on my way the hard way

[spiral star]

:)

[novel hamlet]

im kiinda stuck on this one I have

and i want to do QR decompose and show that

i was wise guy and went to wolfram

and the R does not looks like it wants to multiply by itself

[lavish jewel]

how so

[wintry steppe]

Given this dot product, https://gyazo.com/d7985836b10a11aa9ac976abbb196f15 , with coeficients on C, how can i prove z^n is orthonormal?

[autumn haven]

Is it true that the adjugate of an n-tuple is the vector with the conjugate of each term

[wintry steppe]

u asking me?

[autumn haven]

No 😦 im just tryna learn

[wintry steppe]

what are the a_ns and b_ns? (also don't post the same question in multiple channels)

nvm

i think i found the answer. Like, it is so simple but idk how to formalize it

https://gyazo.com/c5e48add93e3d2113c4fb190354bab77

well, you're already being helped with it in another channel, so

a_n and b_n are the coeficients

so if n=m, then the inner product will be 1

cuz <z, z> = 1

cuz it is 1*1 + ...

and the ... is a product with at least one of them 0

and if n != m then it is 0

for the thing above

[lavish jewel]

what is m

[wintry steppe]

mmm an arbitrary number (?)

like, i am asked to prove z^n is an orthonormal system

[lavish jewel]

and why do we suddenly have a power series

[wintry steppe]

then <z^n, z^m> should be 0 if n != m and 1 if n == m

no?

that means they are orthogonal

the inner product between anyone of them is 0

[lavish jewel]

queres definir un producto interno sobre $\mathbb{C}^{N}$?

[stoic python]

Edd

[wintry steppe]

no, the inner product is already defined xD

i have to prove that with that inner product, z^n is an orthonormal system

[lavish jewel]

pick a basis 1, z, z^2, ..., z^n

[wintry steppe]

[lavish jewel]

that's about all you need i guess

[wintry steppe]

uh

i think im not explaining XD terra could u? pls

[lavish jewel]

it's midnight and i'm hypoglycemic, i probably misunderstood

[wintry steppe]

given n and m natural numbers, write out the taylor series expansion of z^n and z^m about 0, that is, find the taylor coefficients. then use them to compute the sum

wtf

granted

this is all i can gather since you haven't posted the full question

[lavish jewel]

that's what i would've thought, the dual space of polys is fun stuff with factorials and lots of derivatives, but yea

[wintry steppe]

asdflaghsdah my question is on spanish

[lavish jewel]

explicame el problema en espanol

[wintry steppe]

i am given a hardy space, https://gyazo.com/a184135b228f19d16f7ebdbffcb69505 , where D = z € C / |z| < 1

and then

teacher asks us to prove if this is an inner product

https://gyazo.com/b459f50b090d1a8e12a387218bc5ff4c

i did, and it is

and now, i have to prove that https://gyazo.com/25ff3fc0c950250ba2f7dbc97aae7423 forms and orthogonal system on this space

ok so then what i wrote is how you ought to proceed with showing that z^n forms an orthonormal set, just wanted to make sure

wanted to make sure the domain you were considering this inner product on wasn't something bad and terrible

so, why do i need taylor? i mean, isnt enough proving that if n == m then the inner is 1, else is 0?

<z, z^2> = 0 * 0 + 1 * 0 + 0 * 1 = 0

<z, z> = 0 * 0 + 1 * 1 = 1

i mean, if n != m, either a_n or b_n will be 0, so that term will be 0 cuz it is a product with 1 of each members 0

only when n == m, a_n and b_n will be != 0 for that n in particular, for the rest it is gonna be 0

[lavish jewel]

yes, but you start with f and g which are polynomials

and these operations are over the coefficients

if i understood terra correctly, and i probably didn't, the taylor stuff is to get the coefficients of each term

[wintry steppe]

yeah, but the inner product doesnt care about the polynomios. Only the coesficients

[lavish jewel]

to be formal about it

sure, but that is the same as taking a bunch of derivatives and dividing by the corresponding factorials

[wintry steppe]

the coeficients can be complex, sure, but on this case, p(z) = z^n

the coeficients are always Real

1

$$z^n = \sum_{k=0}^\infty a_kz^k$$ where $a_k$ is $1$ if $k = n$ and $0$ otherwise. use this to compute $$\langle z^n,z^m\rangle.$$

[stoic python]

Terra

[wintry steppe]

huh?

ah ye

only if n = m a_k and b_k will be 1 for the same n

sdfasdf isnt this what i said above? Q.Q

[lavish jewel]

that's the same as doing this

but yeah

[wintry steppe]

okey, ty

[acoustic zodiac]

Doe someone have a proof for:

Given two spaces $V_\lambda_1$ and $V_\lambda_2$ where $\lambda_1 \neq \lambda_2$, the spaces are orthogonal

[stoic python]

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

[wintry steppe]

define V_lambda

[acoustic zodiac]

an eigenspace, i suppose. i don't know how to translate it properly, sorry. but the space of eigenvectors for each eigenvalue

oh, and this is assuming an endomorphism