linear-algebra

[zinc timber]

I had an extra |,|

[teal grotto]

Ax/|x| still lives in R^n tho

[zinc timber]

yes but the elements of the image have norm 1

oof

[teal grotto]

so its in S^{n-1}

[zinc timber]

F

what tf am I saying

[teal grotto]

lmao im trying to figure that out

[zinc timber]

nvm

let me rewrite it

$x \mapsto \frac{Ax}{\norm{Ax}}$

[zinc timber]

then it maps $\mbb{R}^3\setminus {0} \to S^2$

[teal grotto]

okay, so were just doing it in the case of a three by three matrix

and domain should be R^3 cut the kernel of A

[zinc timber]

I said for 3x3 matrices

[teal grotto]

...

[zinc timber]

didn't I?

[teal grotto]

?

[zinc timber]

[teal grotto]

eff

[zinc timber]

will extend to Rⁿ later if possible

[teal grotto]

alr

[zinc timber]

probably is

idk yet

[teal grotto]

alr, so we got a map from some open subset of R^3 to S^2

Ax/|Ax|

to use browers, we need a function from B^3 to itself

[zinc timber]

yes

[zinc timber]

we won't use whole S2

take the intersection of S² with (x≥0, y≥0, z≥0)

[teal grotto]

alr

makes sense for the positivity part

[zinc timber]

positive entries guarantees that we have a map from that set to itself

[teal grotto]

different from Ax/|Ax|, right?

no

just resricting it to S^2 intersect the positive octant

okay, got it

[zinc timber]

now it's homeomorphic to B² to brouwer applies

[teal grotto]

hmmmmm

so the domain of this map is R^3 cut the kernel of A

[zinc timber]

yes

[teal grotto]

were restricting it to S^2 in the positive octant intersect R^3 cut the kernel of A

like... what if there is a hole, or a slice through the domain now?

im failing to see how anything is immediately homeomorphic to B^2 here

like what if u have something like this

[zinc timber]

hole?

[teal grotto]

kernel of A could pass through S^2 in the positive octant

lemme rephrase. what is homeomorphic to B^2?

[zinc timber]

S² ∩ (x≥0, y≥0, z≥0) =:D

[teal grotto]

okay, so my point is, that this map, Ax/|Ax|

could actually never have a point in this part of R^3 whose sum is zero, nvm continue

[teal grotto]

this situation cant happen because all the entries in A are positive. cool

[zinc timber]

yes

[teal grotto]

alr, so we essentially get a map from B^2 to itself

theres a fixed point

Ax = |Ax|x

[zinc timber]

yes

and since |Ax|>0 that's out +ve eigen value

[teal grotto]

oh wait lol

okay, nice

a lot more topological than LA

im looking at the proof on wikipedia for this

[zinc timber]

[teal grotto]

and it started off how i stated off here

[zinc timber]

cool

[teal grotto]

um.

i dont want to read the rest of it now lol

[zinc timber]

probably does something on B³ ∩ (+++)

[teal grotto]

yea. i wonder if ur result generalizes

[zinc timber]

probably does

[teal grotto]

it should

[zinc timber]

also not my result lol

[teal grotto]

lol

[zinc timber]

it's Frobenius result I think

[teal grotto]

i have one for u if you want

thought this one was actually cool

[zinc timber]

sure

say $A$ is diagonalizable, find the necessary condition on $A$ such that $\m{A & A \ 0 & A}$ is also diagonalizable

[teal grotto]

a $C^2$ function $u:A\subseteq\mathbb{C}\to\mathbb{R}$ defined on an open subset $A$ is said to be harmonic if it saticies Laplace's partial differential equation: $$\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}=0$$

Given a harmonic function $u:A\subseteq\mathbb{C}\to\mathbb{R}$, show that there exists a harmonic function $v:A\to\mathbb{R}$ such that $f=u+iv$ is holomorphic on $A$

[stoic python]

c squared

[zinc timber]

$\log(|z|)$ is holomorphic on $\mbb{C}^$ but it doesn't have a harmonic conjugate on $\mbb{C}^$

[teal grotto]

is it harmonic tho?

[zinc timber]

yes

[teal grotto]

domain has to be simply connected

my b

[zinc timber]

u get this function when you solve Laplace eq on R²

[lavish jewel]

are you also sure it's v: A -> R?

[zinc timber]

also called the principal solution

[teal grotto]

yea, A to R or A to C, shouldnt matter i dont think

[zinc timber]

it'll also be false

[teal grotto]

dude nah thats cap

[zinc timber]

you get the HC as tan^-1(x/y)

if your domain contains y=0 points then it's not continuous

[lavish jewel]

i'm under the impression functions C -> R are not holomorphic in general

i might be mistaken, it's 6 am and i literally just woke up

[zinc timber]

I'm talking about the set |x-2|≤1

[teal grotto]

read that wrong

[zinc timber]

circle at (2,0) with radius 1

idk they confuse me as well

[teal grotto]

whats ur ce if domain isnt simply connected and open

[zinc timber]

first find the hc of log(|z|)

probably atan(y/x) or atan(x/y)

either case you take a open ball on x axis or y axis

[teal grotto]

god damnit, i meant is simply connected and open

[zinc timber]

there's a thread on SoF discussing this

[teal grotto]

already saw ce for not simply connected and open

dude. if theres a c.e. for when its simply connected and open, im gonna flip

[zinc timber]

ye but the ball is simply connected and open?

[teal grotto]

yea, and u just said, if im not mistaken, that log|z| has a harmonic conjugate on the ball centered at 2 of radius 1

[zinc timber]

no

i said if we choose the ball in such a way that the denominator y/x is invalid then it won't have a hc

[teal grotto]

oh

well, i have a proof that if the domain is simply connected and open (or as phrased in my class, diffeomorphic to a convex set), then any harmonic function has a harmonic conjugate on that set

"So any harmonic function always admits a conjugate function whenever its domain is simply connected, and in any case it admits a conjugate locally at any point of its domain."

thank you wikipedia for verifying

[zinc timber]

https://www.rgpvonline.com/answer/mathematics-2/21.html

not defined on y=0

"locally"

[teal grotto]

wdym solve it on R^2?

[zinc timber]

laplacina works on dim n

$\Delta = \pdv[2]{x_i}$

[wintry steppe]

lol

silly looking x

[zinc timber]

when you dim = 2 you get the fundamental solution as $-\frac{1}{2\pi} \log{|x|}$

[hardy inlet]

this the John Cina Laplacian transform?

[teal grotto]

alr, so ur saying if i solve it for n = 2, anything that satisfies LPE has to look like this?

[zinc timber]

no

it's a kernel

you get solutions by convoluting with it

kind of Green's function of Δ_2

[teal grotto]

these words are not registering in my brain. dont know much about diff eq

[zinc timber]

$\Delta \Phi(x) = \delta(x)$ is what I am saying

[zinc timber]

anywya I'm rusty on pdes

and typing

[teal grotto]

i feel like a baby who understands nothing. thank you for trying tho lmao

[zinc timber]

@teal grotto ask yr prof about log|z| because I also need an explanation

[teal grotto]

i will. what should i ask him specifically?

[zinc timber]

that in a ball centered at y=4 with radius 1, domain is simply connected and log|z| is harmonic, but harmonic conjugate is not defined on y=0

then how can it have a hc

something like this

idk

[hardy inlet]

can i try and finish this elementary problem of mine. I.e.

[zinc timber]

show that they have same dim?

[hardy inlet]

im pretty sure thats what my friend went about doing

this is where we were left off

[zinc timber]

ok then I'll let c² handle

[hardy inlet]

:( dimention sounds easier

[zinc timber]

otherwise you can show that for $T\in \mathcal{L}(\mbb{R}^4, V)$, map $\phi : \mathcal{L}(\mbb{R}^4, V) \to V^4$ by $\phi(T) = (T(e_1), T(e_2), T(e_3), T(e_4))$

[zinc timber]

show it's an isomorphism

[hardy inlet]

which is the direction c² was going right

[zinc timber]

I think so

[teal grotto]

dimension doesnt work*

[zinc timber]

wym doesn't work

[teal grotto]

*without assumption that V is finite dimensional

[zinc timber]

they are working on FDVS

V is finite dim is assumed most probably

[hardy inlet]

this is what they were trying but i dont think its right

cause i think c brought that up about not being FDVS a few hours ago

[zinc timber]

probably is FD

[teal grotto]

it can be shown in more generality is my point. question statement didnt have it and i was erring on the side of caution

[zinc timber]

true

I would recommend c^2's approach over dim argument

[hardy inlet]

ok

one of the other probs mentions a FDVS so i dont think he would have just forgotten

[zinc timber]

as a sidenote even if V is infinite dim, the dim argument still works

[teal grotto]

see, thats the part i wasnt sure about

like, 4 times infinity is infinity

but like

that feels wrong

[zinc timber]

entire mathematics is wrong

[hardy inlet]

so dim or iso

[teal grotto]

this is indeed the question

[hardy inlet]

isomorphic mapping

rn I just have what u types to modify my errors

showing its a bijection

[teal grotto]

this is the inverse

sry for the ping

[hardy inlet]

how tho

[teal grotto]

compose the and see if u get the identity

[hardy inlet]

u want me to do this?

[teal grotto]

yea

and then the other way

[hardy inlet]

i dont think i've done a composition of homomorphisms before

[teal grotto]

right now u havent even shown they're homomorphisms. they're just functions at this stage

[hardy inlet]

right cause i haven't show preservation under addition/scalar

do i need to show that before the composition

[teal grotto]

nope

order in which you show the properties doesnt matter

in general, sometimes showing that a function is a homomorphism might help you show that it is surjective or injective, or vice versa, but here it doesnt really matter

[hardy inlet]

idk what im doing

[teal grotto]

need to write (Phi o phi)(T) = Phi(T(e1), T(e2), T(e3), T(e4))

[hardy inlet]

should i rename these e's that ryu gave me

[teal grotto]

the e_i's are the standard basis elements in R^n. e_i has a 1 in the i-th component and zeros everywhere else

you dont have to rename them

[hardy inlet]

oh ok

interesti g

[teal grotto]

just show that when Phi eats (T(e1), T(e2), T(e3), T(e4)), it spits back T

[hardy inlet]

should i make a not like this

cause i didn't know that was the shorthand

[teal grotto]

yea

[hardy inlet]

yum

idk what im doing. do I need to simplify T(e_i) and if so how

[teal grotto]

so

ur gonna have to evaluate (\Phi o phi)(T)(x1,x2,x3,x4)

and show that that is equal to T(x1,x2,x3,x4)

for all (x1,x2,x3,x4) in R^4

[hardy inlet]

for all

[teal grotto]

and no, you dont need to simplify T(ei).

lol is okay. u have a formula for Phi(v1,v2,v3,v4)(x1,x2,x3,x4)

[hardy inlet]

do i need to add some x1-4s to my inverse definition

[teal grotto]

no

[hardy inlet]

muh x's are gone

[teal grotto]

put them at the right most side.

rn those things are not equal

u need the rhs to be Phi(T(e1), T(e2), T(e3), T(e4))(x1, x2, x3, x4)

anyway, Phi(T(e1), T(e2), T(e3), T(e4))(x1, x2, x3, x4) = x1 T(e1) + x2 T(e2) + x3 T(e3) + x4 T(e4)

now use linearity of T

[hardy inlet]

what does this even mean

why do i have an element in R thats supposed to be in v

[teal grotto]

heh?

ur evaluating (Phi o phi)(T) at the point (x1, x2, x3, x4)

[hardy inlet]

Phi goes to V

but here its T(R)

[teal grotto]

Phi goes from V^4 to L(R^4, V). Phi(v) goes from R^4 to V

[spiral osprey]

Can a flow equation result in variables being negative after solving using RREF?

[hardy inlet]

is there anything wrong with the 2nd or 3rd line

[teal grotto]

no. after the last line, you can write = T(x1, x2, x3, x4)

[hardy inlet]

'why the commas nowq

[teal grotto]

the commas are there

they are just hidden in the e_i's

e_1 = (1,0,0,0), e_2 = (0,1,0,0),...

[hardy inlet]

oh right x1's a scalar

[teal grotto]

ye

so that shows that (Phi o phi)(T) = T

now you need (phi o Phi)(v1, v2, v3, v4) = (v1, v2, v3, v4) for all (v1, v2, v3, v4) in V^4

[hardy inlet]

ok let me try that solo for a min

wait whats in the parenthesis this time

vs?

[teal grotto]

that was a vector in R^4

[teal grotto]

now these are vectors in V^4

like, if V = R^2, then an element of V^4 would look like ((a,b), (c,d), (e,f), (g,h))

[hardy inlet]

think i broke somethign

[teal grotto]

there’s no T this time

[hardy inlet]

had this prior

[teal grotto]

on the left hand side

[hardy inlet]

oh

[teal grotto]

should be (phi o Phi)(v1, v2, v3,v4) and then just work through the definitions

[hardy inlet]

notsure where my R4 is coming in/out

[teal grotto]

you dont need the xi's here

[hardy inlet]

but muh definition

[teal grotto]

phi(Phi(v))

work outside in this time

[zinc timber]

still on that problem?

[hardy inlet]

[hardy inlet]

current status

[hardy inlet]

idk what u mean by that. how can I apply phi if i dont know Phi

[teal grotto]

phi(Phi(v)) = (Phi(v)(e1), Phi(v)(e2), Phi(v)(e3), Phi(v)(e4))

but you know what Phi(v)(ei) is.

[hardy inlet]

should i make some step to show that v = v1...v4

am i missing a step?

cause its 1v1 + 0v2 + 0v3 + 0v4, 0v1 + 1v2 + 0v3 + 0v4, ...

now we gotta show that homomorphism

"we"

I think i generally know how to do that but do I need to do it for both Phi and phi

isn't there a shortcut to show scalar and addition simultaneously

[teal grotto]

yes

[hardy inlet]

rn:

[teal grotto]

show that Phi(av+w) = aPhi(v) + Phi(w). or for little phi. whichever one u think is easier

and if a function is linear and has an inverse, its inverse will also be linear

so u only gotta check if one of Phi or phi is linear

[hardy inlet]

check one as in Phi OR phi, or scalar

(ax + ay) vs ax + y

[teal grotto]

checking ax + ay doesnt get you the whole story they should be equivalent

checking av + w or av + bw gets u linearity in one check

[hardy inlet]

ah i was thinking the av+bw

[teal grotto]

but yea, just check that av + bw works

for either phi or Phi

once you know linearity of one of the phi's, it will be automatic that the other phi will be linear

this is a good exercise, if a linear transformation is invertible, its inverse is linear

[hardy inlet]

is this the correct step?

[teal grotto]

Phi of that stuff

but yea

[hardy inlet]

did i break math laws

[teal grotto]

no. but now you gotta stick (x1, x2, x3, x4) to the end of each term in the first two lines

[hardy inlet]

here?

[teal grotto]

and on the first term

Phi(av + bw)(x1, x2, x3, x4)

[hardy inlet]

other than this issue...

(and the x1 typo)

[teal grotto]

all good

except

should be aPhi(v)(x1,x2,x3,x4) + bPhi(w)(x1,x2,x3,x4)

[hardy inlet]

facts

[teal grotto]

now you can conclude that Phi(av+bw) = aPhi(v) + bPhi(w) as functions, so your map Phi is linear

[hardy inlet]

showing either*

[teal grotto]

nice

[hardy inlet]

this is our total piece 💛

now im not gonna do this tonight; but this ones easier because the Fields/VectorSpaces are fixed, right?

same concept just 'hardcode' some iso

[teal grotto]

this time u can use dimension argument

[hardy inlet]

ooooh

[teal grotto]

finding a basis for each space is easy

[hardy inlet]

also is R^{2,2} common. it erks me

i'm used to M_{2,2}(R)

[teal grotto]

i prefer F^{m x n} or Mat_{m x n}(F)

[hardy inlet]

something like this?

unfortunately the teacher doesn't like $V \cong W$

[stoic python]

MattDog_222

[hardy inlet]

i hate words

Now rq can u use ur superpowerful brain to analyze if these "hints" are useful for...

[zinc timber]

[hardy inlet]

im scared why did ryu

[zinc timber]

characteristic polynomial

but I wonder if you are allowed to use it

[hardy inlet]

eigenvalues :P

probably not

but maybe

I know i vaguely covered them at the end of the prereq

[zinc timber]

otherwise $A=\m{a & b \ c & d}$ with $ad-bc \neq 0$ then show ...

[errant mist]

If the eigenvalues are non-degenerate is should follow that the eigenspace is one-dimensional and the algebraic multiplicity is equal to 1 right?

[zinc timber]

no

what's non-degenerate eigen value? ≠ 0?

[hardy inlet]

$\frac{\m{d & -b \ -c & a}{\det A } = A^{-1}$

[errant mist]

no, not repeated eigenvalues

[stoic python]

MattDog_222
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[zinc timber]

why are u ing me

[errant mist]

@zinc timber if lambda is a root to the characteristic equation then there are no repeated roots

[zinc timber]

then yes

[hardy inlet]

so is this useful

[grave garden]

Guys, is linear map unique ?

💀

[teal grotto]

i was also confused. bad emote?

[grave garden]

Dun scared me guys

[teal grotto]

non-linear map unique ?

[zinc timber]

no, just upper me for 2 consecutive messages

[grave garden]

I want to prove the last statement, but how to prove it is injective ?

From the construction of M, it is guarantee to be surjective, but i dunno about injective

[teal grotto]

what is M?

[grave garden]

( at the top )

[teal grotto]

yea. idk what that is

[grave garden]

Hmmm, i dunno what it is called but that M thing change the bases

Btw is bases and basis the same ? 💀

[zinc timber]

M is the matrix representation

[hardy inlet]

what does this do for me

(a+d)t*

[zinc timber]

let V be of the form

it's a vector space, not a matrix

[hardy inlet]

damnit i didn't even reread the question

so why did matrices be brought up

[teal grotto]

space of polynomials in A forms a vector space over the field

there is a particular bijective map on this space

[zinc timber]

this should give you $A^2 - (a+c) A + (ad-bc) = 0$

[hardy inlet]

wait so why is A a matrix

er wait no its not

its the variable

$A^2 - (a+d) A + (ad-bc) = 0$

[stoic python]

MattDog_222

[hardy inlet]

since i had a typo

[zinc timber]

it's not a variable

it's A

the A

[zinc timber]

this A

[hardy inlet]

ok so what do maps from 2D to 2D have to do with a 2x2 matrix

(everything ofc)

and i have 4 unknown variables

[zinc timber]

multiply both sides by A^-1 and rearrange

(A^-1 exists because it's assumed)

[hardy inlet]

(what does it mean)

simplifying a littlemore i have $A^{-1}=\frac{A-a-d}{\det A}$

[stoic python]

MattDog_222

[teal grotto]

should be - (a+d)I

cant subtract a number from a two by two matrix

[hardy inlet]

3am math

[grave garden]

Guys, in a vector space there can be more than 1 basis ?

[hardy inlet]

yes

[grave garden]

Hmmmm i thought it was unique

[hardy inlet]

e.x. in R2, [1,0] and [0,1] are bases

elements are determined by a unique linear combination of the basis

[hardy inlet]

then [2,0] and [0,1] are also bases

but to get the point <5,5> you need 2.5[2,0] + 5[0,1]

[hardy inlet]

$A^{-1}=\frac{A-(a-d)I_2}{ad-bc}$

[stoic python]

MattDog_222

[grave garden]

I see, thank you !

[hardy inlet]

https://tenor.com/view/oh-boy3am-gif-10356365

im gonna head off thanks for ur help c² and ryu. i'll be back on in like 8 hours

[cursive ginkgo]

Hi ! How can I find the distance between the elementary matrix E11 (one 1 on top left and then filled with n^2 - 1 zeros) to the vectorial subspace of matrix whose sum of diagonal terms is 0 ?

[dusky epoch]

in what metric?

[brittle gyro]

any ideas on how to show this for L > 1 given the steps of the Method of Conjugate Gradients?

B is symmetric positive definite

[spare widget]

I suggest looking into Saad's book

[zinc timber]

I wish I could recall these

[peak goblet]

i was trying to find a formula, anyone could help me pls

[peak goblet]

@wintry steppe

[spare widget]

the subtraction of which two numbers?

X,Y? P,Y?, X,P?

I am guessing X and Y

so (X+Y)Y = P and
(X-Y)X = Q

then solve the system, although those are not linear systems

I don't get what the purpose of the absolute value is there, X-Y is always non-negative since X>=Y

[wintry steppe]

@peak goblet

[peak goblet]

if x and y postive or negative when we open the module the sinal changes all expression

[spare widget]

X>=Y -> X-Y >=0

Either way, you get an polynomial equation of degree 4

This is still analytically solveable but it's painful

I would suggest a numeric root finder and then checking the ceil and floor integers of the found roots

[dusky epoch]

wrong channel

[zinc timber]

it's not that simple, since X, Y are 16bit signed integers, their products are also to be taken as 16bit signed integers, so we'll get wrap around if they are large enough. better way would be to find an algorithm, which was the op's original question

[spare widget]

Indeed the abs would make sense if the prescribed behaviour upon overflow is wraparound. I did the above assuming everything fits, as wraparound is not always the default: https://stackoverflow.com/questions/3679047/integer-overflow-in-c-standards-and-compilers

Anyways, this is off-topic as Ann noted.

So here's a linear algebra question

I have

[dense whale]

I also have a question

Small one

[spare widget]

$L = C - A(I-C)B, , C = diag(c) \implies \frac{dL}{dc} = I + AB^T$

[stoic python]

criver

[spare widget]

Is there an easy way to find $\frac{dL^{-1}}{dc}$

[stoic python]

criver

[spare widget]

If c was not a vector, but a scalar

Then I would have used:

$0 = \frac{d}{dc}(LL^{-1}) = \frac{dL}{dc}L^{-1} + L\frac{dL^{-1}}{dc}\implies \frac{dL^{-1}}{dc} = -L^{-1}\frac{dL}{dc}L^{-1}$

[spare widget]

Ask your question btw, don't mind me

[stoic python]

criver

[dense whale]

OK thank you

In the characteristic equation of an eigenvalue problem, which can be written like this:
$\sum_{m=0}^{n} c_m \omega^m =0$ , the $c_m$ s can be complex numbers right (I mean not completely real in the general case)?

[stoic python]

ElonMoist

[spare widget]

If the matrix is complex

You get it from det[A-wI] = 0

[dense whale]

Yes that's the equation we get it from

[spare widget]

So if A is complex, the coefficients will also be

[dense whale]

It would be all real if A has all real entries right?

[spare widget]

Yes

Think aboit thebdefinition ofbdeterminant

[dense whale]

Thanks for confirming

[spare widget]

There are only multiplication, sign flips and additions involved

[dense whale]

I was worried if I was missing something

[dense whale]

Yes

[spare widget]

And the reals are closed under those

If the determinant involved square roots you should be worried

The roots can be complex though

Even if the coefficients are real

[dense whale]

Yeah

Thank you

[spare widget]

A real symmetric matrix though will have only real eigenvalues

[zinc timber]

char poly is actually det(xI-A) not det(A-xI)

[spare widget]

does the sign matter?

[zinc timber]

if you are talking about char eq then no, if you are talking about the char poly then yes

it's taken to be the monic polynomial, det(A-xI) is not necessarily monic,

[dense whale]

If c is a scalar, all the diagonal entries of C world be c right?

[spare widget]

I meant if c is scalar that is true for any L that is invertible

Where each element is a function of c

I've probably defined the problem poorly

It probably should read:

[zinc timber]

what do you mean by C=diag(c)? like cI?

[spare widget]

$Lx = c \odot x + A((1-c) \odot B x)$

[stoic python]

criver

[spare widget]

c is a vector

nvm, I still have to work out the formulation

[zinc timber]

maybe edd can help

[grave garden]

Hiii guys

What would the statement be ?

[zinc timber]

just change N by T

[nocturne jewel]

$N=[T]_{\mathcal{B}}^{\mathcal{B}}$

[stoic python]

Mosh

[dawn stump]

would anyone know how to solve this. i cant find any notes about how to solve this

[bronze glacier]

I have a quick question, suppose I have the following subspace of R^n : span(v1,...vk) of vectors in R^n and we want to find a base for them, does it make a difference if I put the vectors as rows in a matrix and find a base or if I put them as columns and find a base? are these two methods equivalent?

[nocturne jewel]

Gauss Jordan on the Augmented matrix [A|0]

[hardy inlet]

aka RREF

[quiet wren]

Hey, I've got a quick question. If Q is a PSD matrix and $\lambda$ is its smallest eigenvalue then how does it hold that $v^TQv \geq \lambda | v|^2$ for all v?

[stoic python]

ciri

[dusky epoch]

are you sure you don't mean $v^TQv \geq \lambda \nrm{v}^2$

[stoic python]

Ann

[dusky epoch]

or. wait

oh yeah no nvm this should be it

[quiet wren]

yeah that's what I have written.

[dusky epoch]

can you show where you're getting the inequality without lambda

[quiet wren]

I see this is the case when Q = I?

[dusky epoch]

this is the case when Q = I and in general when λ ≥ 1, but you're not answering my question

[quiet wren]

I can't see any explanation for this

[dusky epoch]

...let me ask you again

where are you getting it from that $v^TQv \geq \nrm{v}^2$ for any pos-semidef matrix $Q$?

[stoic python]

Ann

[dusky epoch]

i can name you a matrix Q for which this most definitely doesnt hold.

[quiet wren]

Oh sorry I missed a lambda there

Just a typo

[dusky epoch]

bruh

okay, so what you REALLY meant to ask is why the inequality $v^T Q v \geq \lambda \nrm{v}^2$ was true

[stoic python]

Ann

[dusky epoch]

in which case, the answer is fairly simple: $v^TQv - \lambda \nrm{v}^2 = v^TQv - v^T (\lambda I)v = v^T(Q-\lambda I)v$, and $Q - \lambda I$ is pos-semidef by construction.

[stoic python]

Ann

[quiet wren]

All clear. Thanks !

[spare widget]

I have the following linear operator that depends on a vector c

$Lx = \sum_i A_icB_ix$

[stoic python]

criver

[spare widget]

Taking the vector derivative of the inner product below results in:

$\frac{d}{dc}\langle Lx, y\rangle = \sum_{i}A_i^*yB_ix$

[stoic python]

criver

[spare widget]

Is there anything that I can show for the the derivative of the inverse operator

$\frac{d}{dc}\langle L^{-1}x, y \rangle = ?$

[stoic python]

criver

[spare widget]

For instance I know that if L were to be a matrix that depends on a constant c, then

$0 = \frac{d}{dc}(LL^{-1}) = \frac{dL}{dc}L^{-1} + L\frac{dL^{-1}}{dc} \implies \frac{dL^{-1}}{dc} = -L^{-1}\frac{dL}{dc}L^{-1}$

[stoic python]

criver

[spare widget]

Can something similar be done for the above inner products involving a vectorial c?

I can do the following:

$0 = \frac{d}{dc}\langle L^{-1}Lx,y\rangle$

[stoic python]

criver

[spare widget]

But I am not sure how to translate the product rule, or whether it can be translated at all

[hardy inlet]

not sure whats really going on; but yesterday there was something about using a characteristic polynomial to find A^-1. so I solved for A²-(a+d)A + (ad-bc) = 0

[nocturne jewel]

Yeah since $A^2-\operatorname{Tr}(A)A+\operatorname{det}(A)I=0$

[stoic python]

Mosh

[hardy inlet]

yeah thats where i got the abcd from a $\begin{bmatrix} a & b \ c & d \end{bmatrix}$

[stoic python]

MattDog_222

[hardy inlet]

but idk what information that lends itself to

like we need $p(A) = \frac{A-(a-d)I_2}{ad-bc}$

[stoic python]

MattDog_222

[hardy inlet]

but thats not a polynomial

[lucid glacier]

Why not?

It's a polynomial in A

You have
$p(A)=\frac{1}{ad-bc}A^1 - \frac{a-d}{ad-bc}A^0$

Where by convention A^0=I

[stoic python]

RYC for mod

[hardy inlet]

so now wait how did we know A is a 2x2?

cause V is 2 dimentional?

where does this take me

is that just the answer?

[lucid glacier]

Well, that's one option, but you don't have to treat A as a matrix here

ad-bc isn't well defined for a linear transformation

Since you need to choose a basis

But det and trace are invariant of choice of basis

You can also do this by considering powers of A in L(V)

[hardy inlet]

this basis?

[lucid glacier]

Again

That doesn't mean anything

You need to. Choose a basis to talk about matrices of transformations in an abstract vector space

I just mean consider the set id, A, A^2, A^3, A^4 in the space L(V)

Think of what the dimension of L(V) is

If you have an arbitrary polynomial which A satisfies, then you can get a polynomial for the inverse. This is nice since you don't need to assume Cayley-Hamilton

[hardy inlet]

This was our 'hint'. U mean something like this?

[hardy inlet]

should i change approach to avoid the characteristic poly?

[nocturne jewel]

wdym

what is 32 suppose to mean...?

Can you post the question verbatim?

[half sleet]

#calculus

[hardy inlet]

ideas for...

would i just show its a subspace by T(av+bw) = aT(v) + bT(w) and

[subtle walrus]

and?

[hardy inlet]

i think thats it actually

[subtle walrus]

ok, how does that show it's a subspace?

[hardy inlet]

because its closed under addition and scalar multiplication

[subtle walrus]

i think you need to do some type checking

E is a subset of L(V, W)

take two elements, say T_1 and T_2

what type do they have? why is their sum in E?

[hardy inlet]

T_1(v) + T_2(v) = 0 + 0 = 0

[subtle walrus]

yes

i dont see how this is related to what you wrote originally

[hardy inlet]

its not really :/

[subtle walrus]

well ok, this is correct anyway

scalar multiplication follows the same way

i think you also need identity in E (or E nonempty) and then that should be it

[hardy inlet]

what would the identity be if everything goes to 0

[subtle walrus]

v_1 and v_2?

also what is T

[hardy inlet]

T is a transformation from V to W

[subtle walrus]

it shouldn't appear in this context

neither should v_1 and v_2

[hardy inlet]

then what do i need to make to show its closed

[subtle walrus]

what was wrong with this?

[hardy inlet]

v is the same right

[subtle walrus]

yes

[hardy inlet]

doesnt it need to be difference

[subtle walrus]

i mean technically you have to show (T_1 + T_2)(v) = 0

[hardy inlet]

ohhh

[subtle walrus]

E is the subspace of all linear maps that are 0 on v

so you have to show that the sum of two of those is 0 on v as well

[hardy inlet]

i think im missing up the order. I have
$(T_1 + T_2)v = T_1v + T_2v = 0 + 0 = 0$

[stoic python]

MattDog_222

[subtle walrus]

notation-wise it is now correct

[hardy inlet]

but wasn't I supposed to show that $(T_1 + T_2)v = T_1v + T_2v$ not assume it

[stoic python]

MattDog_222

[subtle walrus]

this is how the sum of two linear maps should be defined

i am sure you defined L(V, W) somewhere

i suggest to look at that

this will also answer your other question of what the identity should be

since this question asks you to show E is a subspace of L(V, W) you must have covered that

[hardy inlet]

so i need the identity for the subspace to hold?

[subtle walrus]

subspaces need to have an identity

[hardy inlet]

would 0 be the identity

since 0Tv = 0

[subtle walrus]

what is 0 here?

[hardy inlet]

a scalar on the left and an element of W on the right

[subtle walrus]

how is a scalar an element of L(V, W)

[hardy inlet]

its not

a transformation would be

but we already have a transformation

[nocturne jewel]

if your vectors are transformations, what is the 0 vector going to be?

[hardy inlet]

the zero transformation

[nocturne jewel]

yes

so show that the 0 mapping is in E

Hint: It's not hard, don't overthink it

[hardy inlet]

E is the zero mapping

[nocturne jewel]

No

E is a set

[hardy inlet]

the set of zero mappings

[nocturne jewel]

Yes, it ends up being that E={0}, so what?

you still need to show 0 in E

you've shown a linear combination of transformations in E is in E, however you still need to show E contains the 0 vector

[hardy inlet]

0 = T : Tv = 0

[nocturne jewel]

Sure... but write it explicitly

There's no reason to do pointless stuff like T=0

[hardy inlet]

0_E = 0Tv = 0_W?

[nocturne jewel]

$0v=0$

[stoic python]

Mosh

[nocturne jewel]

so $0\in E$

[stoic python]

Mosh

[nocturne jewel]

why do you have T here?

0 is the transformation you're considering

[hardy inlet]

cause vectors in E are transformations

[nocturne jewel]

Why do you have T

when you're doing a concrete example of the 0 mapping

This is like saying x=3 for 2x+1 means 2(3x)+1

[hardy inlet]

yea

[nocturne jewel]

yea what?

you think x=3 in 2x+1 means 2(3x)+1?

[hardy inlet]

no

so it wouldn't be 0T it would just be 0

[nocturne jewel]

yes

cause your mapping is the 0 mapping

denoted 0

[hardy inlet]

yeah T = 0, not 0T

[lucid glacier]

E is generally not 0

[nocturne jewel]

Yeah but you don't need to explicitly write T=0

"Since 0v=0 for any v in V, 0 in E"

[nocturne jewel]

Yeah realized after

[lucid glacier]

Anyways it might be and that shouldn't matter, you're right

The point is to prove that E is nonempty, this is usually easiest to do by proving the 0 map is in it as mosh said

[nocturne jewel]

Cause stuff with 0 vectors work very simply

[hardy inlet]

so for 5B if v != 0, dimE should be the number of vectors in a basis for E. but wtf would a basis of E even look like. just the zero transformation?

[nocturne jewel]

Well consider a concrete example

for example v=[1,2] and V= R^2

then $E={T\in\mathcal{L}(V,W)|T[[1,2]^T]=0_W}$

[stoic python]

Mosh

[nocturne jewel]

E is simply the set of all transformations b/w V and W st [1,2] maps to 0

[hardy inlet]

so like 2x - y = 0 for all

[1,2]dot[2,-1] = 0

[lucid glacier]

Do you know how a basis for L(V,W) looks?

Try and draw inspiration from that

It's a similar construction

[hardy inlet]

i dont think I know what it looks like

i feel like dimE = dimV + dimW or dimV * dimW

[gray dust]

a hint to find a basis of L(V,W) is to recall that a linear map is uniquely determined on a basis of V

[hardy inlet]

is a basis the columns in the matrix representation?

[halcyon spindle]

For T in L(V,W), we have the matrix representation of T is [T(v_1), …, T(v_n)] where v_1,…,v_n would be a basis in V.

[hardy inlet]

Yeah sorry I went ahead and turned in what I had. He said something like dimV-1 * dimW i think

[wintry steppe]

to get the least squares beta, its just (XTX)-1 XTY right?

im trying to solve this

so im assuming x would just be a row vector of [-2 -1 1 2 4]

but when i do the XTX -1

i get a determinant of zero?

[wintry steppe]

Is it true that matrix representation of L T changes with change in bases ?

[wintry steppe]

<@&286206848099549185>

[zinc timber]

yes

[wintry steppe]

Okay ty

[trim merlin]

Hello

[wintry steppe]

hi, I would like to find the roots and multiplicities of a polynomial which is R[x] (R real-ring, ring of polynomials). However, when I factor the polynomial I get one factor as (x^2+1). My question is do I take x=+-i as a root or no?

[dusky epoch]

R as in $\bR$?

[stoic python]

Ann

[dusky epoch]

if so then no

if you were factoring it in $\bC[x]$ then you would split $(x^2+1)$ into $(x+i)(x-i)$

[stoic python]

Ann

[wintry steppe]

R as in $\bR$[x] (polynomial ring)

[stoic python]

dim99

[dusky epoch]

okay so that's a yes

[wintry steppe]

so I will not use the x=+-i as a root

[dusky epoch]

yes, x^2 + 1 is irreducible in R[x]

[wintry steppe]

oke thanks @dusky epoch I think i got now why +-i is not a root

[spare widget]

when we take a derivative of an inner product wrt a single variable, in the finite dimensional-case we get a Kronecker delta:

$\frac{d}{dx_i}\langle x, y\rangle = \langle \frac{dx}{dx_i}, y\rangle + \langle x, \frac{dy}{dx_i}\rangle = \langle \delta_i, y \rangle$

[stoic python]

criver

[spare widget]

How does this generalize to infinite-dimensional spaces? That is, the Kronecker delta will become a Dirac delta, but how is the derivative formulated? Is it:

$\frac{d}{d x_a}\langle x, y\rangle = \langle \delta_a, y \rangle = y(a)$

[stoic python]

criver

[spare widget]

That is, how would I define d/dx_a

Is it just

$\frac{dx}{dx_a} = \delta(x-a)$

[stoic python]

criver

[spare widget]

Is this formalized anywhere?

It's some kind of derivative with respect to a lower-dimensional object

Also when would this be equivalent:

$\left(\frac{d}{dx}\langle f(x), y\rangle\right)(a) = \langle \frac{df}{dx}\frac{dx}{dx_a}, y\rangle = \langle \frac{df}{dx}\delta_a, y\rangle$

[stoic python]

criver

[spare widget]

where the left-hand side is to be understood as a Gateaux of Frechet derivative, and the right-hand side is what I have above

clearly in the finite dimensional case the above is equivalent (y is not a function of x)

[wintry steppe]

pinging this again if anyone can help

https://cdn.discordapp.com/attachments/540211747613704221/944423692711108650/Screen_Shot_2022-02-18_at_8.42.26_PM.png
im trying to solve this
so im assuming x would just be a row vector of [-2 -1 1 2 4]
but when i do the XTX -1
i get a determinant of zero?

[spare widget]

$\beta_0 + x_i\beta_1 + r_i = y_i, \quad \beta^* \in\arg\min |r|^2$

[stoic python]

criver

[wintry steppe]

im not quite sure what that means :/

[spare widget]

you want to find a line

b0 + x * beta

[wintry steppe]

yea exactly

[spare widget]

that has the lowest distance to all points

the above is the equations that you have

beta0, beta1 are unknown

you have the points (x_i, y_i) from your problem

r_i is the residual

$r_i = y_i - (\beta_0 + x_i\beta_1)$

[stoic python]

criver

[wintry steppe]

yes agreed

[spare widget]

now if you were to write this in matrix form:

$r = y - \begin{bmatrix} 1 & x_1 \ 1 & x_2 \ 1 & \ldots \ 1 & x_n \end{bmatrix} \begin{bmatrix}\beta_0 \ \beta_1 \end{bmatrix}$

[wintry steppe]

why do you include a column of 1s?

[spare widget]

for beta0

[wintry steppe]

i guess like how did u get that from the problem?

[spare widget]

from here

[stoic python]

criver

[spare widget]

note that this is just the equations in matrix form

[wintry steppe]

hmm ok ok so since its the minimal distance. 1 is the lowest

since 0 means no distance

[spare widget]

Let's call this matrix X, and the column vector with the betas beta

then

$\beta^\in\arg\min|X\beta-y|^2 \implies \
X^TX\beta^ = X^Ty \implies \beta^* = (X^TX)^{-1}X^Ty$

you get the equation by taking the derivative wrt beta and setting to zero

to be sure the derivative is:

[stoic python]

criver

[spare widget]

$\frac{d}{d\beta}|X\beta-y|^2 = 2X^T(X\beta-y)$

[stoic python]

criver

[spare widget]

setting to zero you get the normal equations

$X^TX\beta^*=X^Ty$

[stoic python]

criver

[wintry steppe]

ah ok yea so i understand that Beta = (XTX)-1(XTy)

[spare widget]

afaik beta_0 is just \sum_i y / \sum_j x

though not sure

[wintry steppe]

but even then to get XTX -1

i did what you said where i set a column of 1s and then the x's

and the determinant is zero?

[spare widget]

typically you do not invert the matrix explicitly

typically you solve it with an iterative solver, especially for large systems

[spare widget]

X^TX's determinant is zero? I doubt it

maybe if you have linearly dependent stuff

but then that means that you have multiple solutions that would fit

an iterative solver like CG would find one of those based on the initial guess

[wintry steppe]

OH i think i was making the mistake of saying

X = row vectors instead of columns

thus xtx -1 = singular

[spare widget]

you need X like this

it's a n x 2 matrix

so X^TX is a 2x2 matrix

if you compute X^TX it's explicit to invert

[wintry steppe]

yea i got a 2 x 2 matrix for xtx

[spare widget]

if you don't compute X^TX then you use an iterative solver applying each to a vector separately

[wintry steppe]

\begin{pmatrix}\frac{55}{19}\ \frac{12}{19}\end{pmatrix}

[stoic python]

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

[wintry steppe]

and then i got this as my final

my idk if its right?

[spare widget]

they ask you to plot it to check

the line should pass close to the points

[wintry steppe]

so here to get the residual its y - that?

where y is a column vector of the y values?

[spare widget]

yes

that's the residual

but you want to minimize

$\min_{\beta}|r|^2 = \min_{\beta}|X\beta-y|^2$

[stoic python]

criver

[spare widget]

which yields you the normal equations

[wintry steppe]

ok idk why but im not getting the right answer

hmm

bc i did this in R already and i got the right answer

but on hand its not right

[spare widget]

don't compute this manually

it's stupid to

[wintry steppe]

i have to for exams tho ;/

im studying for an exam

on the actual submission i did it in R and i got feedback on it

which is why ik the answer is right

but im trying to learn how to do it on paper

[spare widget]

well you have to compute 3 dot products

dot(x,x), dot(x,1)=dot(1,x), dot(1,1)

your X^TX looks like this

$X^TX = \begin{bmatrix} x^T1 & x^T x \ 1^Tx & 1^T1 \end{bmatrix}$

[stoic python]

criver

[spare widget]

1^T1 is just the number of elements

x^T1 is just the sum

[spare widget]

wait I flipped them

you're right

it's

$X^TX = \begin{bmatrix} 1^T1 & 1^Tx \ x^T1 & x^Tx \end{bmatrix}$

[wintry steppe]

1, 4 4 26

[stoic python]

criver

[spare widget]

no

1+1+1+1+1 = 5

[wintry steppe]

oh shit ur rigt lmao

awk

[spare widget]

now you need X^Ty

[wintry steppe]

yea im doing that one rn

19 36

in a column

2 x 1

[spare widget]

you know the inverse of a 2x2 matrix I assume

[wintry steppe]

yes

2.6 -.5 -.5 .1

so ill dot them

31.4 -5.9

in a column

hmm

oh wait

nvm

\begin{pmatrix}314\ -59\end{pmatrix}

[stoic python]

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

[wintry steppe]

that still doesnt seem right?

[spare widget]

well this doesn't look right:
2.6 -.5 -.5 .1

[wintry steppe]

yea

that was supposed to be

:\begin{pmatrix}26&-5\ ::::-5&1\end{pmatrix}

instead

[stoic python]

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

[wintry steppe]

:\begin{pmatrix}26&-5\ ::::-5&1\end{pmatrix}\begin{pmatrix}17\ ::28\end{pmatrix}

[stoic python]

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

[spare widget]

where's the det

[wintry steppe]

is (xtx)-1(xty)

[spare widget]

1/det

[wintry steppe]

determinant is 1

26 - 25

[spare widget]

why is there a 1?

wasn't it 5

[wintry steppe]

here?

[spare widget]

why are there 5s

the matrix is 5, 4
4, 26

[wintry steppe]

oh ur right

[spare widget]

1^T1 = 5

1^T x = 4

x^T x = 26

[wintry steppe]

\begin{pmatrix}\frac{13}{57}&-\frac{2}{57}\ -\frac{2}{57}&\frac{5}{114}\end{pmatrix}

[stoic python]

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

[wintry steppe]

this is xtx -1 then

[spare widget]

take out the 1/114 outside instead

then perform the multiplication, then get it back inside

check your X^T y too

[wintry steppe]

my xty is \begin{pmatrix}17\ :::28\end{pmatrix}

[stoic python]

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

[spare widget]

either way these are algebra calculation issues at this point

[wintry steppe]

yes

\begin{pmatrix}\frac{13}{57}&-\frac{2}{57}\ -\frac{2}{57}&\frac{5}{114}\end{pmatrix}\begin{pmatrix}17\ 28\end{pmatrix}=\begin{pmatrix}\frac{55}{19}\ \frac{12}{19}\end{pmatrix}

[stoic python]

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

[wintry steppe]

i still dont think thats right tho :/

i just did it in R again

so confusing

X = cbind(1, x)
y = c(1, 3, 3, 5, 7)
beta_hat = solve(t(X) %*% X) %*% t(X) %*% y
beta_hat
y_hat = X %*% beta_hat
RSS = sum((y - y_hat)^2)
RSS```

[opaque solstice]

HI why log10(100) = 2 <=> so log2
(100) = 2 times 3.322, or 6.644

[spare widget]

wrong channel, but 10^2 = 100, so log10(100) = 2. log2(100) = x : 2^x = 100

[zinc timber]

#prealg-and-algebra

[spare widget]

6<x<7

[spare widget]

any idea regarding this @zinc timber ?

[wintry steppe]

https://cdn.discordapp.com/attachments/495064710278938662/944659441398210570/D249AE6A-E8AB-4904-9467-66A9E7A84A36.jpg

is this valid?

[spare widget]

what are you trying to do?

$\alpha = u^TAv = v^TA^Tu = (Av)^Tu = (u^TAv)^T = \alpha^T$

[wintry steppe]

show that u^TAv = v^TA^Tu

[stoic python]

criver

[spare widget]

alpha is a scalar

so alpha^T = alpha

so you get the equality from there

[wintry steppe]

so this isnt valid/

[spare widget]

I don't get what's happening there

[wintry steppe]

im taking the transpose of (u^TA)^T v

then that becomes (A^Tu)v

which is then

[zinc timber]

ngl having hard time understanding d/dx_a on infinite dim vs

[wintry steppe]

A^T(uv)

then A^T(uv)^T

[spare widget]

same thing, that's why I am asking, but at least there's the intuition from the finite-dimensional case

[wintry steppe]

A^Tv^Tu^T

[spare widget]

and I tested some examples where the Frechet/Gateaux derivative match with this derivative that pops out a Dirac delta

[spare widget]

that is, there must be some sufficient conditions for which this holds

[spare widget]

don't take the transpose of this

if you start with alpha = u^TAv

then take the transpose of alpha: alpha^T = (u^TAv)^T = v^TA^Tu = (Av)^T u

and since alpha = alpha^T the two agree

[wintry steppe]

how r u able to switch the u and v tho?

[spare widget]

(AB)^T = B^TA^T

[wintry steppe]

but theres a matrix in the middle right

[spare widget]

ok

(ABCDEFGH)^T = H^TG^TF^TE^TD^TC^TB^TA^T

it flips the order

same with inverse

[wintry steppe]

ah ok so even if u have something in the middle

it still flips

OH

nvm

im so dense

[spare widget]

(ABC)^T = C^T B^T A^T = (BC)^T A^T = C^T (AB)^T

[wintry steppe]

yea im so sorry

my mind is fried haha

couldnt you literally then just say

u^TAv = v^TA^Tu because

if you took the transpose of u^TAv

it would just flip and be

v^TA^T u

[spare widget]

(u^TAv)^T = v^TA^Tu yes

but you have to show that u^TAv = (u^TAv)^T before that

which holds because both are scalars, transposing a scalar does nothing

[wintry steppe]

ah ok

[spare widget]

if those were not scalars

e.g. if it was

Av != (Av)^T

simply because one is a column vector, and the other is a row vector

the only way those can agree is if Av is a scalar

i.e. A is a 1xn matrix and v is nx1

[wintry steppe]

ah ok ok that makes sense tysm

[spare widget]

I figured at least what it should intuitively mean

in the finite dimensional case the derivative wrt some coordinate is the directional derivative wrt e_i, equivalently wrt the Kronecker delta_i

Then in the infinite-dimensional setting this generalizes in the sense that we pick the direction to be the Dirac delta

i.e.

$(Df)(\delta_a)(x) = \lim_{\lambda\rightarrow 0}\frac{f(x+\lambda\delta_a)-f(x)}{\lambda}$

[stoic python]

criver

[flint nexus]

Hi is this explanation over?

I dont want to budge in if a question is still being answered

[spare widget]

just ask your question

[flint nexus]

Hi Im having difficulty with matrix transformations with the polynomial space

I am in struggling specifically with the Differentiation and then Integration of Matrix
In my book its state if you use 2 functions which are linear
If one provides integration and then the second one differentiation
If we choose v = 1 both transformations will have a result of 0 when combined

[zinc timber]

thing is in infinite dim, delta functions are not a basis (because there are unc many of them)

[flint nexus]

Its very basic i believe

but i dont get it

[zinc timber]

it's a generating set

[flint nexus]

SO T^-1T(1)=0

[spare widget]

so do I need a reproducing kernel hilbert space then?

[flint nexus]

whilst T is differentiation and T^-1 Integration

[spare widget]

what is this v = 1

[flint nexus]

its the the vector in the linear transformation for differentiation

Its how its written

so I would assume its a number without x

so differentiation would be 0

Im give an example

[spare widget]

found something relevant: https://mathoverflow.net/questions/138392/validity-of-functional-derivative-using-the-dirac-delta-function?rq=1

[flint nexus]

If you have V=1,x,x^2,x^3 and W=1,x,x^2

So this is for the Differentiation

you have then Av=w

if you solve for the core of (Av)

then you would do coreW=0

there you find out that the core is onedimensional

cuz 1 stays in W if you change all x= 0

Now you know the rule that dim W+ coreW = DIM V

so then you know dimV is 4

Now thats for differentiation

for integration you do the other way round

but for W --->V

So T(v) would differantiate the vectors and T^-1 would integrate them

Now if you do the Integral of the differentiation of 1 = 0

Thats where i get lost

i get that the matrices with another so

A*A^-1=(3x3) and A^-1A = (4X4) so the top row is 0

but i cant make the connection with

[spare widget]

is the point to show that differentiation doesn't have a trivial kernel?

[flint nexus]

T^-1T(1)=0

[spare widget]

because sure, dc/dt = 0, regardless of the constant C

to recover this with integration you'd need to set appropriate boundary conditions

[flint nexus]

i meant kernel when wrote core btw mb

[spare widget]

all constant functions get mapped to 0

[flint nexus]

mmmm

[spare widget]

vice versa, an integral is determined up to a constant

[flint nexus]

i get that

[spare widget]

I don't get what the question is