linear-algebra

[knotty raven]

or echelon form whatever

but instead of producing that parameterization you start with it and want to see what matrix you can make

[wintry steppe]

if A is positive definite will A -aI still be postivie definite?

should be, right?

[native rampart]

No,Take A as identity and a=2

@wintry steppe

[wintry steppe]

ahh yes true, thx

@native rampart what if A is symmetric, positive definite and a is negative?

[native rampart]

Should be

[wintry steppe]

yeah that's what I thought as well

[native rampart]

X(t)AX+aX(t)X, a is positive

The left term is positive, because condition and right term is positive because sum of squares multiplied by a positive number

[fathom flax]

Any good websites you suggest to learn advanced linear algebra?

[prisma pier]

Quick question: is this equation a typo (the order should be switched, right?)

[fathom flax]

There are a lot of topics that could count as advanced linear algebra. What exactly are you trying to learn
@warm briar I don't really mind.

I've already covered the basic / intermediate topics, just want to go further.

[errant wyvern]

If we have a subspace S of space M4 (polynomials with a maximum degree of 4) such S={p(x) is an element of M4 | p(3)=p(0)=0 }; how would a basis of S look like?

[native rampart]

x(x-3)(ax+bx^2+c) will be the general element

{${x^2(x-3), x(x-3),x^3(x-3)}$} will be the basis

[stoic python]

DrunkenDrake:

[errant wyvern]

ahh i see, i must be missing some key thing for understading this, based on what do we go from p(3)=0 to (x-3) ? i assume there is some sort of relation between those two statments?

[native rampart]

p(3)=0 p(x)=(x-3)q(x) for some q

Remainder theorem

[errant wyvern]

only one things got me confused, the space is m4, so wouldnt the general element be x(x-3)(ax+bx^2+c)

as it is a space of polynomials with a degree of equal or lesser then 4?

[native rampart]

My bad,You are correct

[stoic python]

DrunkenDrake:

[errant wyvern]

ty for helping out

[pastel kettle]

Can u guys help me woth this??

Is it all false?

[median forum]

why is the first not a explicit set?

but ye, none of these should be

[pastel kettle]

wait

{0,2}

@median forum that's the first option

is it still all false

[median forum]

ah ok

either way you want to solve

[pastel kettle]

yeah solve for eigen value time (x,y) = (x, 2x)

but i don't see anyyy

[median forum]

$\lambda \begin{pmatrix} x \ y \end{pmatrix} = \begin{pmatrix} x \ 2x \end{pmatrix}$

from there you can find the spectrum of the matrix and its eigenvectors

[stoic python]

Fractal:

[median forum]

not sure if -2 or 2

cant read from your paper

[pastel kettle]

i wrote down x and 2x

[median forum]

so by def, an eigenvalue would be such that it solves that, yes

for some vector

so $\lambda x = x \land \lambda y = 2x$

[stoic python]

Fractal:

[median forum]

you can infer both the spectrum and the set of eigenvectors

[pastel kettle]

oh

so

empty set?

[half storm]

@median forum Is your semester over lol?

[median forum]

not at all

note reall, molee

that is solvable

[pastel kettle]

$\lambda x = x \land \lambda y = 2x$

[stoic python]

im_molee:

[pastel kettle]

hmmm

[median forum]

@half storm just started func anal, diff top, ring theory, qm, numerical PDEs and comp graphics

last week

[half storm]

Bruh how are you in here and doing all that

Lol

[median forum]

that and my thesis

I did the same number last sem and it went pretty weell

[gray dust]

qm let's go

[median forum]

yeye

[half storm]

How come you're taking qm?

Just cause?

[median forum]

just learned about poisson brackets

some chad version of lie brackets

uhh

its qm for math

[half storm]

Ill prog read something on it eventually

Cool. So you're just takimg the class cause its interesting?

[median forum]

and for the pure degree

but its p cool

[half storm]

Ah cool

[lucid cedar]

Hello again #linear-algebra,

Im trying to answer this question 6b. and 6c. are easy assuming 6a. is correct. I can confirm that their answer is correct but I do not understand how to generate such an answer. My gut reaction and initial attempt was that the only polynomials that would satisfy P(x) = P(y) was the constants, but then I remember that a quadratic has 2 roots and there for P(x) can equal P(y). This was my process so far with the question but I am still unsure how to generate a basis.

I had an idea right as I clicked post

I think it should work if I start with the polynomial (x-2)(x-5) and then multiply x on repeatedly for each successive term

and then yea thats why the x term is left out because it can not have 2 different roots

answered my own question. Sorry! Should I delete?

[prisma drift]

May someone check my work for this problem?

[gaunt field]

I just wanna confirm.. it's not possible to do LU decomposition on non-square matrices right?

[quiet crane]

does this mean i have to do REF operations to find out a?

[wintry steppe]

I think for square matrices u can check if the determinant is not zero then there is unique solution if i recall correctly

[half storm]

There are ways of doing this. You could perform the necessary elementary row operations in order to determine what a has to be so that the augmented matrix is can be transformed into an identity matrix with one column of zeroes.

[drowsy mango]

1 0 3 1
0 1 -1 0

How would I see that this is reduced completely

is it the 1 0 0 1 on the left

[drowsy mango]

https://gyazo.com/bafdf6baeb9adc5a4b63f00c89bace9a

I dont understand how z can be anything

its it because technically there is a 3rd row that is 0 0 0 0

[limber sierra]

if you name any value of z

i can name values of x, y that solve that system

[drowsy mango]

how would i be able to recognize that in a matrix form

because i understand what you are saying

[limber sierra]

looking for columns without a pivot

each column without a pivot corresponds to a free variable

the matrix of that system (consolidated) would look like $\begin{pmatrix}1&0&3&1\1&1&2&1\end{pmatrix}$

[stoic python]

Namington:

[limber sierra]

row reduction (symbolized in your image by the fancy arrow) gives us $\begin{pmatrix}1&0&3&0\0&1&-1&0\end{pmatrix}$

[stoic python]

Namington:

[limber sierra]

these are our pivots

so the columns without pivots correspond to free variables

ie the red column here

(the column on the right is the solution column, so it doesnt "count" - ie it doesnt correspond to any variable)

if you dont know what i mean by "pivot"

a "pivot" is the leftmost nonzero entry of a given row; if one pivot is below another pivot, it must also be to the right of it

[drowsy mango]

ok a lot to digest

but ty

so could that 3rd row have a pivot at all?

[limber sierra]

nope

this is a theorem of linear algebra: your pivots will always end up "in the same place" (if you put all the 0 rows on the bottom)

you might be more familiar with the equivalent statement, "every matrix has a unique fully-reduced (RREF) form"

(and all REFs have pivots in the same place as the RREF)

also, there is no 3rd row in that matrix

i'd assume you mean 3rd column

[drowsy mango]

yeah my b

you mean like if i swapped row 1 and 3

the pivots will be in the same place

1 and 2

my bad

[limber sierra]

if you swap row 1 and 2, the pivots are now wrong

this is because one requirement to be a pivot is to be the right of all pivots above it

but if you swap rows 1 and 2

then the pivot of the new second row will be to the left of the pivot of the top row

[drowsy mango]

okokok

[limber sierra]

in practice this just means

"you need to swap your rows to make a 'staircase'-ish pattern"

[drowsy mango]

this is a theorem of linear algebra: your pivots will always end up "in the same place" (if you put all the 0 rows on the bottom) Could you visualize this for me

[limber sierra]

its kind of hard to do that in a convenient way

but you might have for example

uh

[drowsy mango]

that's alright lol

[limber sierra]

$\begin{pmatrix}1&0&3&0\0&1&0&0\0&0&0&1\0&0&0&0\0&0&0&0\end{pmatrix}$

[stoic python]

Namington:

[limber sierra]

this is one row echelon form for given matrix

(indeed it's fully reduced, RREF)

the pivots are here

$\begin{pmatrix}-2&5&0&-6\0&1&0&1\0&0&0&17\0&0&0&0\0&0&0&0\end{pmatrix}$

bleh

[stoic python]

Namington:

[limber sierra]

there we go

thats another possible REF form

the pivots are in the same place

the point is

there's multiple different ways to row reduce a matrix

but as long as you "end up in" a form where:

• all rows, except 0 rows, have a pivot

• the pivot rows are in the right "order"

• the 0 rows are at the bottom

the pivots will be in the same place

[drowsy mango]

I understand

[limber sierra]

the important thing for our purposes is that the free variables will be the same

the coefficients might be different

but the variables that "matter" and the ones we can pick "freely"

wont be different

[though in that big example i gave, that system isnt solvable in the first place... but you should be able to get the idea]

[wintry steppe]

I am trying to solve for k but im a bit stuck

[dusky epoch]

what's your goal here

[wintry steppe]

Use gauss Jordan elimination to determine any value of k for which this set is linearly dependent

Thats the question

[wintry steppe]

<@&286206848099549185>

[versed topaz]

What value of k makes the rows of that last matrix linearly dependent?

[wintry steppe]

if k=1

[versed topaz]

yes

Does that work for your original three vectors?

[wintry steppe]

i think so?

so is that all the working i need to do?

[versed topaz]

yes

[devout void]

can someone help me logically what a curve is ?

[limber sierra]

in what context

generally in higher mathematics we take a curve to be the image of a continuous function from a subinterval of R onto a topological space

(roughly speaking, and definitions vary; the term is used a bit more specifically in diff geo)

but this is #linear-algebra so i'd assume thats not what you mean

"curves" arent exactly an object of study in linear algebra since, well, they're not linear

i mean obviously polynomial functions represent certain curves in R^n and linear algebra studies polynomial vector spaces

but the connection there isnt very direct

and linear algebra would generally studies the polynomial space in its own right, not its relation geometrically in R^n

can you be more precise about the context you're asking this question in? what do you mean "logically"? @devout void

do you mean, like, in terms of polynomial curve fitting?

[dusky epoch]

👻

[devout void]

okay so

@limber sierra hey

my book sucks at explaining something but we are studying curves

so if we want to draw a circle on a coordinative system we do x^2 + y^2 = r^2

that would be a circle with the origin (0,0)

ok ik how that is working

now my book says

"earlier we defined a line from a point and a vector basically (x-x0)=a*t where x0 is the first coordinate of the point and a is the scalar on i of vector

and so on for the other coordinates y and z

book also says we can do the same thing for curves

[limber sierra]

that grammar is hard to parse

but uh, sure, you can involve vectors if you want; vectors are effective ways to store a bunch of variables to manipulate

[dusky epoch]

i think you might be talking about parameterizations of curves?

[devout void]

@dusky epoch Yes

[limber sierra]

thats how i interpreted it

[dusky epoch]

yeah what about 'em

[limber sierra]

and theres no reason a parameterization cant involve a vector

[devout void]

the parametric equation of curve

wait wait

book says if x(t) and y(t) are 2 functions with 1 variable defined on a segment then we can say that x=x(t) and y=y(t), t belongs to the segment

so this sentence makes no logical sense to me

idk what the book is getting at

and we are on the plane

so curves on a plane

[dusky epoch]

either you're twisting what the book says or the book is shitty

[devout void]

so i totally understand how vectorial and algebraic equations of curves work

no totally how it is written

it literally says "segment"

[dusky epoch]

i can give you a "real world" interpretation of what's going on

[devout void]

yes pls

[dusky epoch]

imagine the variable t represents time

[devout void]

oh

ik that

[dusky epoch]

and imagine a particle moving on the plane

[devout void]

but how does it exactly apply here

we did that on lines

the only way im imagining it for a circle would be

with cos and sin ?

i guess

and for a certain angle i get a point

[dusky epoch]

x(t) and y(t) are the functions giving the x and y coords of the particle at any point in time

[devout void]

and for all angles would get each point

[dusky epoch]

the domain on which they are both defined is the time interval you are considering

[devout void]

|x(t) and y(t) are the functions giving the x and y coords of the particle at any point in time| this is what i needed

thanks Ann

u the goat

oh and one more thing please

[dusky epoch]

for a circle, if you take the default parameterization (cos(t), sin(t)) you get a particle moving ccw at a constant speed starting from (1,0) and traversing the entire circle in 2pi seconds

(or whatever unit of time you want to measure t in)

[devout void]

so it mentioned that F(x(t),y(t))=0

damn this book is shitty fr

[dusky epoch]

ah yeah that's

right so the implicit equation thing

that's a different way of specifying a curve

instead of a particle moving along the curve, you see the curve as the set of all points satisfying a certain equation

[devout void]

i kinda lost it

F means a function right ?

[dusky epoch]

F here is a function of two variables

[devout void]

so when i read F(x(t),y(t)) i think after both these vars get into a function they end up 0

[dusky epoch]

for example, for a circle of radius 1 centered at the origin, you would take the function x^2 + y^2 - 1

[devout void]

how is that possible ?

[dusky epoch]

F is a function of two variables. it takes two inputs, and it distinguishes between the first input and the second input.

[devout void]

so bear with my ignorance xd

if we say

f(x) = y

that means we do something to x

and it ends up as y

ik how it works with the cartezian multiplication

[dusky epoch]

lowercase f is a function of one variable here

[devout void]

but the brief idea that i have

yes

i just need to compare

so i understand this new thing

[dusky epoch]

y = f(x) can be rewritten as f(x)-y = 0

[devout void]

mhm

[dusky epoch]

you can read F(x,y) as "some expression involving x and y"

[devout void]

so when we say F(x(t))= y(y)

?

[dusky epoch]

?

[devout void]

its the same as F(x(t),y(t))

[dusky epoch]

no

[devout void]

oh

okay

but

why F(x(t),y(t))= 0 ?

idk if im stupid

if we do some function with x and y

it might be equal to 0

but in the curve case

we always have a radius

what am i missing here?

[dusky epoch]

not all curves are circles. i was just using a circle as an example.

anyway F(x(t), y(t)) = 0 is somewhat misleading

it's more about the relationship between these two representations

[devout void]

alright i guess makes kinda sense

thanks 🙂

[old flame]

In the step of additivity, why is $S(p(x)+q(x))=p(T)+q(T)$ valid ? The definition of S only specified a polynomial

[stoic python]

Otoro:

[native rampart]

Consider p(x)+q(x) as a polynomial h(x)

And apply S on h(x)

@old flame

[old flame]

So it will become $S(h(x))=h(T)=p(T)+q(T)$ ? And it follows yeah ?

[stoic python]

Otoro:

[native rampart]

Yes

[old flame]

Thanks, I am having a problem understanding polynomials applied to operators though

So polynomials are functions that takes some input x and outputs it in a form of $a_0+...+a_mx^{m}$. In this situation, the input is just an operator T instead ? Am I missing something here

[stoic python]

Otoro:

[native rampart]

Actually we deal with formal polynomials

They don't take a number x and x is a dummy variable that can be manipulated

And when you apply p(T) you are saying,instead of this placeholder x,I put T

Whever x exists

The input of S is a polynomial and output is an operator

[old flame]

May I ask what is formal polynomials ?

[native rampart]

Things,where you don't get to plug in x

And x is a dummy variable

Like x+2

When you say "polynomial",you are actually talking about polynomial functions

[spiral star]

the formal polynomials are sequences, and the dummy variable is also a special sequence

[old flame]

So first things first we're moving away from the familiar polynomials that I'm used to right ?

[native rampart]

Yes

I mean,you could use formal polynomials to define your polynomials

[spiral star]

if you are reading axler, he doesnt introduce formal polynomials which imo makes dealing with this awkward, but he gets around introducing ring theory

[old flame]

Yes I'm using axler

so from using the previous example, we require an extra function to define what the polynomial has for the values of x? and also, it is called formal polynomial, since the functions are of the form of polynomials but arent the ones Im used to before ?

[spiral star]

uh i can give you the gist of formal polynomials if you want

[old flame]

but how could I understand the materials of this chapter if I don't understand formal polynomials ?

[spiral star]

i think axler builds on intuition

[old flame]

and also could you explain "polynomials are sequence" lol no idea what this is

[spiral star]

do you know what a sequence is?

[old flame]

$a_0,a_1,...$ ?

[stoic python]

Otoro:

[spiral star]

yea, well formally, a sequence is a function from the natural numbers into a set

you basically map an index to an element of a set

$f\colon \bN \to M$ is a sequence

[stoic python]

Flow:

[spiral star]

M is any non-empty set

or in your example, the function would be named $a\colon \bN \to M$

[stoic python]

Flow:

[old flame]

so basically a set that allows us to list out its elements right ?

[spiral star]

and then the convention is $a_n := a(n)$ for $n \in \bN$

[stoic python]

Flow:

[spiral star]

so the index notation is just fancy notation for evaluating the function

[old flame]

oh okay

[spiral star]

you basically assign some element of M to every natural number

and since natural numbers are ordered you can list the elements in order of the indices

[old flame]

here a question though, so is every countably infinite set a sequence ?

[spiral star]

no, a sequence is a function

every function from the natural numbers into a non-empty set is a sequence

that's the definition

[old flame]

oh alright

my mistake sorry

[spiral star]

basically, every index is mapped to some element

that's the essence of a sequence

anyway, back to polynomials

[old flame]

okok

[spiral star]

to construct polynomials we first need a ring

in this case, we want a commutative ring with 1

i guess for starters it's okay to use fields

[old flame]

Im sorry, but I havent learnt rings yet

[spiral star]

so when i say commutative ring with 1, i mean just use $\bC$ or $\bR$

[stoic python]

Flow:

[old flame]

ah okay

[spiral star]

the generalization isnt really that important for now

let's use $\bR$ as an example

[stoic python]

Flow:

[spiral star]

the polynomials over $\bR$ will be sequences $\bN \to \bR$ with a certain property

[stoic python]

Flow:

[spiral star]

the property is, that those sequences have a finite support

so basically, we look at $\qty{a\colon \bN \to \bR \mid \qty{i \in \bN \mid a_i \neq 0} \ \text{is finite}}$

[stoic python]

Flow:

[spiral star]

only finitely many terms of the sequence are non-zero

so there must be an index k such that a_i = 0 for all i > k

for example, after 10 terms in the sequence everything is 0

or more generally, after finitely many terms, everything is 0

[old flame]

but the terms that you're saying, is it the coefficients or the entire term $a_ix^{i}$

[stoic python]

Otoro:

[spiral star]

i will come to that soon

but does it makes sense so far?

we take all the sequences that are only non-zero for finitely many terms

[old flame]

I could try to summarise

[spiral star]

okay

[old flame]

So polynomials are a sequence that maps $\mathbb{N}$ to $\mathbb{R}$. It has the property of finite support, which is that theres finitely many non zero terms. Explaining why we could explicitly writing them out I guess

[stoic python]

Otoro:

[spiral star]

alright, so yea, a polynomial selects finitely many non-zero elements from the ring in specific indices

i wanna give that set a name as well

$\bR^{(\bN)} := \qty{a \colon \bN \to \bR \mid \exists n \in \bN ~ \forall i \in \bN\colon i > n \implies a(i) = 0 }$

[stoic python]

Flow:

[old flame]

The output is $\mathbb{R}$ because our solution could be any number in $\mathbb{R}$ ? like by solving the polynomial

[stoic python]

Otoro:

[spiral star]

the codomain is $\bR$ because the sequence selects real numbers

[stoic python]

Flow:

[spiral star]

for every index it gives you a real number

except the definition of $\bR^{(\bN)}$ guarantees that after some point every index gets mapped to 0

[stoic python]

Flow:

[old flame]

so for every natural number, theres just a real number that corresponds to it though

[spiral star]

well yea

[old flame]

so its very different from the definition of the $x^{n}$ thing

[stoic python]

Otoro:

[old flame]

okay I'm following......

[spiral star]

it will recover the definition more or less when we introduce the meaning of the dummy variable

[old flame]

ohhhh, so this is the abstract version or smth like that ?

[spiral star]

yea, once we define how to operate with those sequences

okay, well to give you some intuition

we can use an analogy of tuples for the sequences

so let's say a(0) = 1, a(1) = 10, a(2) = 20, ....

since the indices are ordered 0 < 1 < 2 < 3 <...

we can imagine this is a tuple

( a_0, a_1, a_2, .... )

which in our case is (1, 10, 20, ....)

[old flame]

ah okay

[spiral star]

and there will be some point at which all elements are 0

like (1, 10, 20, 0, 0, 0, 0, 0, ....)

[old flame]

yup

[spiral star]

but it's okay to have zeros in-between

[old flame]

got it

[spiral star]

(1, 0, 20, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, ...)

[old flame]

it doesnt matter, but it just stops to some length

[spiral star]

but there is a final non-zero term

at some point

[old flame]

alright

[spiral star]

and we also allow just zeros everywhere

(0, 0, 0, 0, ...)

okay now we define some operations on that set

we need this to become a ring, so we need to define addition and multiplication for polynomials

addition is just point-wise as you would expect

[old flame]

so just like lists I guess

[spiral star]

if a and b are sequences then a+b is a sequence such that (a+b)_n = a_n + b_n

we just add the terms with the same indices

(a+b)_0 = a_0 + b_0 etc.

[old flame]

alright

[spiral star]

and multiplication is a bit more difficult

because we sort of have to encode the distributive laws

[old flame]

Don't mind me for interrupting, but I'm extremely sorry, I need to sleep soon, do you mind typing everything here and I will read it tmr morning ?

[spiral star]

sure it's fine

we can talk another time

[old flame]

I ping you tmr ?

[spiral star]

sure

[old flame]

thank you very much, I really appreciate it, cheers

[onyx ember]

Hello, I need help getting started on this question: Use propositional logic to prove this argument is valid: (A∧B)→(A→B')'

sorry thought this was differnt channel

ill head to correct one

[half forge]

can someone help me with this one

am i suppose to find random integers to plugin?

[half storm]

You have to show that it sastisfies the axioms of a vector space.

There are 8 axioms.

[half forge]

how do i show them?

can we do one axiom together

[half storm]

Sure you show that they're satisfied, if any one of these fails, then ti's not a vector space.

So we cna check the first one - commutativity of addition.

So we need to show that for all x and y in V $ x + y = y + x$

[stoic python]

TheDon:

[half storm]

V in your case is the set of ordered pairs of real numbers , so we need to show that if we take any $(a,b) , (c,d) \in V = \mathbb{R}^2$ then $ (a,b) + (c,d) = (c,d) = (a,b)$.

[stoic python]

TheDon:

[half storm]

So the proof is this:
Let $(a,b) , (c,d) \in V = R^2$ Then $(a,b) + (c,d) = (a + c, b + d) = (c + a , d + b) = (c,d) + (a,b)$

Hence we've proven commutativity of additon

[half forge]

oh okay, so we just pick random arbritary integers to prove this?

[stoic python]

TheDon:

[half storm]

There are a couple of things wrong with this statement.

1.) No the vector space is the set of ordered pairs of real numbers - not the integers
2.)We pick arbitrary vectors.

What I mean by this is that we do not make sure the statement is just true for a specific ordered. We need to show that the statement is true for any set of ordered pairs that we choose.

This is why we do not specific $(a,b) = (1,0)$ or $(c,d) = (2.3,4.1)$. We are showing that the statement holds for any sets of ordered pairs that we pick that belong to $ V = R^2$.

[stoic python]

TheDon:

[half storm]

and so this is why we just say $Let (a,b) , (c,d) \in V$. The only thing that we specify is that the ordered paris are in the set i.e. that the coordinates are real numbers.

[stoic python]

TheDon:

[half storm]

Does that makes sense?

[half forge]

yes

how would you do the scalar multiplictaion?

[half storm]

There are multiple axioms that have to be shown relating to scalar multiplication; your question is not very specific.

[half forge]

right, i got confused at this part since it has a second equation

[half storm]

O.k., so there are couple of things that we have to show involve scalar multiplication. 5 - 8 involve scalar multiplication.

So lets say that i wanted to show 5. Then what I need to show is that $1(a,b) = (a,b)$

[stoic python]

TheDon:

[half storm]

This is not hard.

So let $(a,b) \in V$ $$1 (a,b) = (1a,b) = (a,b)$$

done

[stoic python]

TheDon:

[half storm]

All i have done is used the definition of scalar multiplication that they have given us.

and then used that to conclude what I wanted to show.

[half forge]

okay, so scalar multiplication starts from vs 5 - vs8?

[half storm]

You begin proving properties that invovled scalar multiplication yes.

[half forge]

yes

[half storm]

That's right.

[half forge]

we barely strated learning this

i get lost lol

[half storm]

It's cool.

[open pivot]

Hey guys I'm stuck on this one. My thinking so far was to somehow find the standard matrix for F and invert it. I already know F(1,0) but I'm not sure how to use the additional information F(3,2) to help me out??

[native rampart]

What have you tried?

[tacit marsh]

Is q-p a vector

P+v a point

V-1.5W a vector

would p+q be a vector???

[oblique rune]

Is there a general formula for inverse of any matrix?

[native rampart]

Probably

There are algorithms to find the inverse of a matrix,if it exists

[oblique rune]

Mmhm

Maybe

[brittle juniper]

there are matrices that aren't invertible, so no you won't find a formula for the inverse of any matrix

[oblique rune]

Oh yes

So is there a general formula for an inverse of an invertible matrix?

would p+q be a vector???
@tacit marsh

Yep, vector+vector=vector

[brittle juniper]

in that case, there's a formula involving the comatrix but since the comatrix is itself a bit difficult to calculate, it's probably not what you're looking for

[zinc lava]

Easiest way imho to find an inverse is an augmented matrix with the identity matrix and rref.

[smoky oak]

What is a "picture"? And what is the difference between a "picture" and an "image"? I'm having trouble finding a definition for picture, but wiki has one on image.

[dusky epoch]

in what context

[smoky oak]

I'm following the MIT Introduction to Linear algebra textbook by Gilbert Strang. An example is "what is the picture of all combinations of cu". I can guess it may just mean what is the resulting set of all combinations but.. was hoping to find a clear definition somewhere

Let me find a video timecode too because he seems to refer to the equations themselves at one point

er, expressions

https://youtu.be/ZK3O402wf1c?list=PL49CF3715CB9EF31D&t=110

So he mentions row picture and column picture there, and in the book it asks for the "picture of all combinations of cu"

[left lily]

can someone help me :<

how do you do this?

I don't get it hehe

[native rampart]

Guess the function b

Ofc,It need not be the one you guessed,but the people who set this question expect that you use that function

[left lily]

ohhh

thank youuuu

[old flame]

@spiral star Sorry wrong sub lol, may we continue here ?

[spiral star]

lol

uh

yea

[old flame]

As i recall, we were on to defining multiplication for the sequence

[spiral star]

ah yea, i actually just wrote down what i wanted to say yesterday

so i have a minimal summary for you

https://cdn.discordapp.com/attachments/440230496396378124/751456470897524886/unknown.png

its just a bunch of definitions

in the end everything should be really intuitive to work with, and it will tie in directly with what your book does even though it doesnt mention this directly

[old flame]

ah alright, thanks dude.

you mean intuitive as in following what you have written ?

[spiral star]

working with polynomials works pretty much like how you would expect

you can basically apply what you know about manipulation of polynomial functions from school

[native rampart]

Except the substitution part

[spiral star]

yea i guess...

[old flame]

alright, I will start reading it now, will ask if theres any questions

sorry could you give me the latex file ? I couldn't really see itt

[spiral star]

open original

its really large

[old flame]

oh okok thanks

[prisma cairn]

Since what i am reading is about LA i thought i'd post it here. I don't get how to conclude that there'd be infintie zeros/roots (what is underlined)
https://cdn.discordapp.com/attachments/422677475760275466/751412862144217098/unknown.png

[native rampart]

Subtract lhs from rhs and call that polynomial p(z)

P(z) is zero everywhere,except at $z=\lambda_1$

[stoic python]

DrunkenDrake:

[native rampart]

Therefore p(z) will be a polynomial with infinite roots, which is not possible

@prisma cairn

[prisma cairn]

how do you know p(z) is zero everywhere except at lamda_1

[native rampart]

If z is not $\lambda_1$,you can divide both sides by (z-$\lambda_1$) in the original equation

[dusky epoch]

lambda

[stoic python]

DrunkenDrake:

[native rampart]

Which means lhs=rhs and p(z)=0

Except at that one point

[prisma cairn]

oh fuck yeah this is so dumb ahahaha

thank you, makes sense

@native rampart thanks!

[dark gale]

Hey Isaac do you know any ways to get the equation of a plane?

This is how I would draw up the problem.

[wary lily]

apparently OP doesn't know how to deal with parametric equations

But I just don’t know what to do with the separate variables if I don’t have a second line or something

[half storm]

The inverse of T cannot exist yea?

[pallid rampart]

Well it’a a linear map from a space with higher dimension to a lower dimension so it can’t be injective

So inverse doesn’t exist

It’s probably the inverse image

[half storm]

That's what I'm thinking

[pallid rampart]

Idk

[half storm]

Yea

that must be it.

the preimage or something

that's shit notation.

[pallid rampart]

$T\inv({(1,11)})$

[stoic python]

Whoever:

[half storm]

mhmmm

[pallid rampart]

[wintry steppe]

imagine using curly set brackets when denoting the preimage of a point

[copper pulsar]

How can I simplify $\frac{\vec{r}\cdot \overrightarrow{OP}}{|\vec{r}|^2}\cdot \vec{r}+\cos(\varphi)\left(\overrightarrow{OP}-\frac{\vec{r}\cdot \overrightarrow{OP}}{|\vec{r}|^2}\cdot \vec{r}\right)+\sin(\varphi)\frac{1}{|\vec{r}|}(\vec{r}\times \overrightarrow{FP})$?

[stoic python]

Anton S.:

[wintry steppe]

[copper pulsar]

[pallid rampart]

$(1-\cos(\varphi))\frac{\vec{r}\cdot \overrightarrow{OP}}{|\vec{r}|^2}\cdot \vec{r}+\cos(\varphi)\overrightarrow{OP}+\sin(\varphi)\frac{1}{|\vec{r}|}(\vec{r}\times \overrightarrow{FP})$

[stoic python]

Whoever:

[open pivot]

Asked about this before but got busy sorry, I am given a linear transformation F along with F(1,0) and F(3,2) and need to find F inverse of (100,-120). This is my working out that I’ve gone through twice and I seem to be wrong both times

[rose coral]

The original matrix looks good

But I think your inverse matrix is off

It should be $\frac{1}{20}\begin{bmatrix}-4 & 4 \ 0 & -5\end{bmatrix}$ I think

[stoic python]

The_Vman:

[open pivot]

Yeah you're right I'll have a look

Yep thats all good a stupid mistake haha thanks

[rose coral]

No problem

[old flame]

@spiral star So far I have two questions. The first one being, why is the multiplication defined like this, so its a sequence like this $(a_0b_i + a_1b_{i-1}, a_0b_i + a_1b_{i-1}+a_2b_{i-2},....)$ ? I'm guessing that the multiplication is defined as the first term of $a_i$ multiplied to the last term $b_i$ of the second sequence right ? However, I couldn't resemble this as polynomial multiplication lol, since theres only expanding brackets that I know of

[stoic python]

Otoro:

[native rampart]

Try taking 2 normal polynomials and multiplying then normally

The result is modelled on that

[old flame]

The second question is that $X^{i}$ is the ith term $(0,...,0,1,0...)$, for $\sum_{i=0}^{n}a_iX^{i}$. then for the term $a_0$, $X^{0}$ is what ?

[stoic python]

Otoro:

[native rampart]

i+1th term

Not ith

[old flame]

oh is that a typo then

[native rampart]

Yes

Or they consider the first term to be '0'th term

[old flame]

so followingg the definition, $X^{0}$ would be this $(1,0,0,0....)$ right ?

[stoic python]

Otoro:

[native rampart]

Yes

[old flame]

If I take two polynomials for say $a_0+a_1x$ and $b_0+b_1x$ then multiplying them would result in $a_0b_0+a_0b_1x+a_1b_0+a_1b_1x^{2}$ so how can I relate this to the definition ?

[stoic python]

Otoro:

[native rampart]

You get $(a_0b_0+a_1b_0x)+(a_0b_1x+a_1b_1x^{2})$ which is also the result of the multiplication defined above

[stoic python]

DrunkenDrake:

[old flame]

but the definition is $\sum_{k=0}^{i}a_kb_{i-k}$ so since this itself is also a sequence, the first term for $i=1$ would be $a_0b_1+a_1b_0$ then the second term with $i=2$ would be $a_0b_2+a_1b_1+a_2b_0$ am I wrong ?

[stoic python]

Otoro:

[native rampart]

Yes

But b2 and a2 are both 0

[old flame]

so this results in $(a_0b_1+a_1b_0, a_1b_1)$ ?

[stoic python]

Otoro:

DrunkenDrake:

[old flame]

but Im not sure how from the summation, I could obtain $a_0b_0$

[stoic python]

Otoro:

[old flame]

yeah I know thats missing

[native rampart]

Put i=0

[old flame]

oh but sequence is $i \in \mathbb{N}$ so I didn't know whether he counted 0 in

[stoic python]

Otoro:

[native rampart]

He probably did

Didn't know The debate about whether N has 0 or not has such serious consequences

[old flame]

yeah thats why I was stuck for a while for now

lol interesting

so I guess each term in the sequence corresponds to each term of the polynomial yeah ?

[native rampart]

Yes

[old flame]

ok thank you, gonna continue

[simple depot]

19. For what values of a does the set {(a,1,0),(1,0,a),(1+a,1,-a)} consitute a base for R^3?

So, since the dimension is 3, all I need to do is find the values of a for which the set has linear independence

for this i can make a matrix and solve it

but im not sure what the correct way to construct that matrix is

a      1    0
1      0    a
1+a    1   -a```

i did like that

cuz ive gone through to whole workbook writing them this way and had no problem

well thanks anyways

[wintry steppe]

gl getting an answer, apologies if i caused any confusion

and yeah, you can ping the @ Helpers tag 15 mins after posting if you don't get an answer / help

[simple depot]

dw its fine! thanks a lot!

[native rampart]

Check if determinant is non zero

[simple depot]

yeah, the determinant in non zero for any value of a that isnt 0

[half storm]

Or rather solve for the values of a so that the determinant is nonzero.

[simple depot]

but my question is more about the matrix itself

[native rampart]

Or do the gaussian elimination and check if you get identity

[simple depot]

yes

but

[native rampart]

You can do it both ways

[simple depot]

i just want to know if the matrix i constructed is correct or if it should be that, transposed

[native rampart]

Both are correct

[simple depot]

oh

really?

[native rampart]

Yes

[simple depot]

awesome :D

why is the transposed of what i did correct as well?

[native rampart]

Both have same rank

[half storm]

^

[native rampart]

So,If 1 is invertible the other is too

[simple depot]

okay but when i want to check if something is linear independant i say that the matrix times a set of scalars is equal to 0,0,0

right?

[half storm]

That is only the case of square matrices though. A matrix and it's transpose always have the same rank, but if the matrix is non-square then this method does not work.

[native rampart]

Are you sure?

[simple depot]

so wouldnt the first column of my matrix need to have the values corresponding to the first of those scalars?

[half storm]

At least for determining whether or the matrix spans the entire space.

@simple depot I'm not sure what you're asking "a matrix times a set of scalars"

oh o.k. i see.

Yes you need to deduce that those scalars are all zero.

[simple depot]

sorry i didnt learn this topic in english ahahha

Yes you need to deduce that those scalars are all zero.
@half storm yes!

ill write this in mspaint and show you lol

[native rampart]

Well,A sure way is to write the vectors as rows of matrix and use gaussian elimination

[simple depot]

this is what i mean, if it makes sense

if not just ignore me, its 2am and im stupid

[half storm]

Yes, and you need to deduce that $ \begin{pmatrix} \alpha_{1} \ \alpha_{2} \ \alpha_{3} \end{pmatrix} $$ = \begin{pmatrix} 0 \ 0 \ 0\ \end{pmatrix}$

[stoic python]

TheDon:

[simple depot]

yes

i know that

but

is the matrix that i constructed good?

as in

is it equivalent to what i wrote below it?

[half storm]

Oh

[simple depot]

thats what i dont quite understand

[half storm]

not quite.

[simple depot]

okay good

thats what i wanto to know hahah

[half storm]

It should be $\alpha_1 \begin{pmatrix} a_1 \ a_2 \ a_3\ \end{pmatrix} + \alpha_2 \begin{pmatrix} b_1 \ b_2 \ b_3 \ \end{pmatrix} \dots $

[stoic python]

TheDon:

[half storm]

Does that make sense?

[simple depot]

yeah, so i should always try to write the vectors as columns, right?

[half storm]

You can if you want to. It doesn't really matter; row and column vectors are really the same thing

[simple depot]

ohhh okay

that makes sense i guess

[half storm]

They are vectors in $\mathbb{R}^n$. But, generally, people will write the vectors as columns because they are columns of the matrix.

[stoic python]

TheDon:

[simple depot]

Got it! Thanks a lot!

Is there anyway I can +rep you or something like that on this server?

[half storm]

lol no.

[simple depot]

oh

well TTerra and you have been suuuper helpful! <3

[wintry steppe]

dm each of the mods personally and say "TheDon is a great and helpful server member who should be made Honorable or Moderator."

[half storm]

Don't need it. Most of us just like helping people out. It's good practice for us too.

[native rampart]

Why do people use non square matrices? Can't we just add zeroes to make it a square matrix?

[wintry steppe]

lol i didn't really help, i just vomited out some ideas and realized i was too tired to put them into a coherent answer, so i deleted them and apologized for confusing

[limber sierra]

@native rampart matrices correspond to linear maps from R^m to R^n

or similar for a vector space V over a field F

you lose that entirely if you force matrices to be square

at which point, whats the purpose of matrices at all

just seems like a really clunky way to write linear automorphisms

[simple depot]

lol i didn't really help, i just vomited out some ideas and realized i was too tired to put them into a coherent answer, so i deleted them and apologized for confusing
@wintry steppe you let me know that i could ping helpers and were super kind and realized you maybe couldnt help and deleted your messages to not let my Q get lost

10/10 would interact again

[wintry steppe]

if i had nitro i would give this many more flushed-variant reacts

[simple depot]

jajajajaja are there any flushed gifs?

hahahhaa*

https://tenor.com/view/psyduck-pokemon-blush-blushing-gif-5455322

[wintry steppe]

there are and you probably do not want to see the ones i would post

[simple depot]

oh

[wintry steppe]

anyways to prevent this from getting off-topic i shall go; glad i could help in some way

[native rampart]

@native rampart matrices correspond to linear maps from R^m to R^n
What I meant to say was add (if n>m)(n-m) vectors to basis set of R^n and create a copy of R^m.

[limber sierra]

why?

[native rampart]

Just for the sake of making matrix of T a square matrix

[limber sierra]

why?

like if you want to do that you can i guess

[native rampart]

Does that not give me anything useful?

[limber sierra]

but you lose your ability to talk about non-automorphc maps entirely

in exchange for what?

the theorems about square matrices almost always become "well, duh" if we're not talking about automorphisms

inveertibly for example

"a square matrix with a 0 row cant be invertible"

okay cool but

a nonsquare matrix cant be invertible ever

so we already know that

we dont get any new information

and in fact we lose information in that case

[native rampart]

Like what?

[limber sierra]

since nonsquare matrices can have one-sided inverses

informed by their dimensionality

[native rampart]

Ok

[limber sierra]

but we lost that information when we make them square

this is also ignoring the fact that vectors in CS are very rarely the same size, and yet computer scientists talk about maps between these vectors all teh time

that's like, literally the fundamental operation of machine learning

massive-ass vectors of wildly varying (and unpredictable) sizes

related by very large and complicated maps (m*n matrices)

[native rampart]

Are vectors in programming same as vectors in cs? Because scalar multiplication is not supported in a few cases

[limber sierra]

theres obviously more heuristics and shit to it than that

uh, you can code up scalar multiplication if you want

[half storm]

Yea

I was thinking that.

[limber sierra]

just iterate through the vector

[half storm]

It shouldn't be hard.

Yea

exactly that

just for loop through the entries of an array.

[native rampart]

Nice, Didn't see that

[half storm]

Yea real vector spaces have a direct counter part in CS it seems.

[native rampart]

What is the operation exactly?

[half storm]

You would just construct an array, and then you would run it through a for loop.

I'm not sure how you write code in here.

[native rampart]

what does 2(a1,a2,a3) give me?

(a2,a3,a1)?

[half storm]

Let a CS function take in an array and a number and just mutliply each element in the array by that number

It gives you what you would normally define scalar multiplication to be if that's what you wanted.

[native rampart]

What if multiplication is not defined for the element. {a,b,c} is a vector in c++ sense

[half storm]

I'm not sure I follow. You would just make a function that does that or you could use polymorphism.

in order to make it so that it looks symboically like you would want it to

[old flame]

@native rampart the definition for polynomial ring, just describes how the "polynomials" satisfy being a ring right ? Furthermore, the variable X just a sequence of the form $(0,...,0,1,0,...)$, so variable is arbitrary, since we could have different sequences containing different numbers e.g. (1,4,8,2,...,6,0,0,0,0,0,...)? and also did he rename the ring from $\mathbb{R}^{\mathbb{N}}$ to $\mathbb{R}[X]$ ?

[stoic python]

Otoro:

[native rampart]

The variable X is not a sequence

Also, What renaming?

[old flame]

but it was said that $X=(0,1,...)$

[stoic python]

Otoro:

[native rampart]

It is part of a sequence X=(0,1,0,0)... X^2=(0,0,1,0..)

And so on

[old flame]

oh okay so back to the positioning

but how is that a variable ?

[native rampart]

We are mapping a thing X to the sequence

So,You could just call it a sequence

[old flame]

so $X^{i}$ is natural numbers input while $(0,...,1,..)$ is the real output ?

[native rampart]

Like $ a+bx \mapsto (a,b)$

[stoic python]

Otoro:

DrunkenDrake:

[native rampart]

*including zero
Yea

[old flame]

so if X is a variable, then do we only change "X" itself ? im sorry but I don't really get

[native rampart]

That will change X^2,X^3..

$a+bx+cx^2 \mapsto (a,b,c)$

[stoic python]

DrunkenDrake:

[native rampart]

And so on

[old flame]

so from the remark below, the difference between $t^{i}$ and $X^{i}$ are $(a_0,...,a_n)$ instead of the sequence of $X$ and $t$ themselves ? cause $X$ and $t$ should have the same sequence right ?

[stoic python]

Otoro:

[native rampart]

What is t^i?

[old flame]

well from the remark it was just $\sum_{i=0}^{n}a_it^{i} \in \mathbb{R}[t]$

[stoic python]

Otoro:

[native rampart]

I don't think there is a difference if you use X or t

The former would belong to R[X](Assuming the a_i are in R),while the latter will be in R[t]

You can just write t in place of X,and you will have the same sequence

[old flame]

so its just a placeholder symbol representing the sequence of $(0,1,0,..) , (0,0,1,0,)$ and so forth ?

[stoic python]

Otoro:

[native rampart]

Yes

[old flame]

so its implying that $X$ is a variable, as in the idea of changing $X$ to be whatever we want, but its still the sequence we know of ?

[stoic python]

Otoro:

[native rampart]

Do not call X a variable

[old flame]

my bad im sorry, then what should it be

[native rampart]

Indeterminate

[old flame]

X is not a variable, but in turn just a placeholder symbol representing the sequences ? this should be what it is instead right ?

[native rampart]

Yes

[old flame]

but could I ask why X is not a variable ? whats the significance of a variable instead

[spiral star]

Variable is fine

Works just as well as indeterminate

[native rampart]

If you use variable for both polynomial and polynomial functions,it might get confusing

[tame steeple]

hi

may i ask how would i relate solving for the infimum of a condition number given its F norm

im trying to relate it to SVD but im still a little confused

i am given the 2 norm and F norm

[old flame]

Also, What renaming?
@native rampart rather, whats the difference between $\mathbb{R}[X]$ and $\mathbb{R}^{\mathbb{N}}$

[stoic python]

Otoro:

[native rampart]

There is a bijection,so they are the same as sets

[old flame]

but what is $\mathbb{R}[X]$ in the first place ?

[stoic python]

Otoro:

[native rampart]

They are not same as rings

Multiplication is defined differently

R[X] is the ring of elements of form $a+bx+cx^2...$

[stoic python]

DrunkenDrake:

[native rampart]

@old flame

[cerulean quest]

Hello! Can someone please explain to me why the solution set of every homogeneous linear system can be written as a linear span? And why the solution set of non-homogeneous linear systems is not a subspace of R^n?

I don't understand when the closure properties of vectors are satisfied and when it's not.

[spiral star]

@native rampart you kinda explain stuff opposite of how i defined it yesterday lol

[native rampart]

Also,X^i is (i+1)th element of tuple

[spiral star]

my natural numbers start at 0

and R is a ring, not real numbers

[native rampart]

So,you have a 0th element?

[spiral star]

X^0 = 1 = (1, 0, 0, ....)

[native rampart]

Yea,Otoro got confused over that

[spiral star]

yea didnt read the conversation because it's really long

i can clear up a bunch of things if needed

[native rampart]

Did I imply I meant R to mean real?

[spiral star]

so the indeterminate, let's call it X, is an element of the polynomial ring R[X]. each element of that ring is a sequence of elements in R, the element X is the sequence (0, 1, 0, 0, ....) where the 1 that turns up in the sequence is the multiplicative unit of R

i just saw either otoro or you use real numbers

like, i scrolled up to a first message that was 3 hours ago

its too much for me to read through right now

yea maybe i saw this one. anyway, so X^0 = (1, 0, 0, ...) and then X^1 = (0, 1, 0, 0, ...) and X^2 = (0, 0, 1, 0, ....)

i think one other question was about the name of the indeterminate

it's basically just a name we give those specific sequences

the polynomial rings R[t] and R[X] are equal, we just decided to give the indeterminate another name

both describe the same set with the same operations

the [t] and [X] are just notational

we can embed elements of the original ring into the polynomial ring by mapping c from R to the polynomial (c, 0, 0, ....)

[stoic python]

DrunkenDrake:

[spiral star]

i didnt use that tho

i defined $R^{(\bN)}$ instead

[stoic python]

Flow:

[spiral star]

which is sequences in R with finite support

and equipped it with the corresponding operations to construct R[x]

[old flame]

oh I thought ur capital R is real number lol my bad

so none of it is right ?

Yes, so $R^{\mathbb{N}}$ this ring with the operations of addition and multiplication makes up $R[X]$ ?

[stoic python]

Otoro:

[native rampart]

Just read what he wrote

i didnt use that tho

[spiral star]

@old flame the set $R^{(\bN)}$ together with the operations makes a ring $(R^{(\bN)}, +, \cdot)$

[stoic python]

Flow:

[spiral star]

an algebraic structure like that is always a set equipped with operations that follow certain laws

[old flame]

oh so the ring of $R^{\mathbb{N}}$ and $R[X]$ is bijective ?

[stoic python]

Otoro:

[spiral star]

you can't use $R^{\bN}$ because that would be all sequences in R

[stoic python]

Flow:

[old flame]

Im just confused about the difference between the two

[native rampart]

The product operation is different

[spiral star]

$(R^{(\bN)}, +, \cdot) = R[X]$

[stoic python]

Flow:

[spiral star]

with the usual abuse of notation when we can identify the structure by its carrier as long as the operations are unambiguous

it's like the first 3 lines in my summary

[old flame]

ah so this is $R[X]$

[stoic python]

Otoro:

[spiral star]

yea we use that notation if we refer to the ring $(R^{(\bN)}, +, \cdot)$ and we choose $X$ as the name of the indeterminate (or variable)

[stoic python]

Flow:

[old flame]

all good thanks. So in short, we've created a ring $(R^({\mathbb{N}} ,+,\cdot)=R[X]$, where elements of this ring is of the form $\sum_{i=0}^{n}a_iX^{i}$.

[spiral star]

X^i but i get what you mean

[old flame]

whoops my bad

[spiral star]

yes, they look just like the polynomial functions you are used to

[stoic python]

Otoro:

[spiral star]

but note that $a_i X^i$ is a product of two polynomials

[stoic python]

Flow:

[spiral star]

$a_i = (a_i, 0, 0, \dots)$

[stoic python]

Flow:

[spiral star]

and X^i is (0, 0, ..., 0, 1, 0, 0, ....) where the 1 is at the i-th position

[old flame]

yes, since a_i and X^{i} are both sequences , a_i being a map from the reals and X^{i} being the sequence of zeroes and 1

[spiral star]

a_i is not a map from the reals

a_i is a polynomial over the ring R

and so is X^i

you multiply them as defined above

[old flame]

oh I guess I misread this notation, $R \to R^{\mathbb{N}$, $a \to (a,0,...)$

[stoic python]

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

[spiral star]

it maps an element from the ring R to a sequence with finite support

i.e. a ring element is mapped to a polynomial

[old flame]

forgive me, but when I copied your notes, I replaced all R with $\mathbb{R}$ so its going to be a bit problematic lol

[stoic python]

Otoro:

[spiral star]

lol

i only use $\bR$ for real numbers, and i explicitly wrote in the text "ring"

[stoic python]

Flow:

[spiral star]

R means ring

but in a lot of examples, the ring will be something like $\bC$ or $\bR$ but not exclusively

[stoic python]

Flow:

[spiral star]

for example to evaluate the polynomial over the ring of endomorphisms, this analogy doesnt work anymore

but that was also why i wrote that stuff down for you

since you didnt seem to know why plugging linear maps into a polynomial can be well-defined

[old flame]

may I ask, for $a \in (R,+,\cdot)$, what are a

[stoic python]

Otoro:

[spiral star]

elements of the ring

for example if your ring is the integers, then a is an integer

if your ring is the real numbers then a is a real number

[old flame]

so it doesnt matter what a is, it just matter that a maps to a polynomial yeah ?

since you didnt seem to know why plugging linear maps into a polynomial can be well-defined
@spiral star yeah cause I want to know why it works

wasn't expecting the polynomials I once knew, wasnt really what it was lol

mind blowing

[spiral star]

i mean, it should be intuitive

3 = 3 * x^0 + 0 * x^1 + 0 * x^2 + ....

you can embed the elements of R into polynomials in the obvious way

3 --> (3, 0, 0, ...)

[old flame]

but then all real numbers could be done the same way

[spiral star]

hm?

[old flame]

written like what you have done above

[spiral star]

but that's the point

you can find the original ring again in the polynomial ring

that's why we can identify the ring elements with polynomials

we identify 3 with the polynomial (3,0,0, ....)

its an injective map from R to R[X]

that preserves the ring structure

[old flame]

so R could be anything and R is the subring of the polynomial ring of R

right ?

[spiral star]

every unital ring R is isomorphic to a subring of R[X]

[old flame]

ah alright

[spiral star]

you can find the original ring R again if you just look at the polynomials that have the form (a, 0, 0, ...)

if you add and multiply those, then only the first element will ever be affected

(a, 0, 0, ...) + (b, 0, 0, ....) = (a+b, 0, 0, ...)

is essentially the same as a + b = a+b

[old flame]

so its still the same structure

[spiral star]

yea the original ring is still there, but in a sense, R[x] is bigger

you can find R in R[x] but there are things in R[x] that cant be mapped back to R in a useful way

[old flame]

O.o

[spiral star]

it's a bit like talking about integers and rationals or whatever

every integer is a rational number but not every rational number is an integer

we kinda did the same here with polynomials

[old flame]

oh okay, I guess I will get into more details in an algebra class

so a similar thought

[spiral star]

yea, specifically if chose the integers as our ring to construct polynomials, then we can say every integer is a polynomial but not every polynomial is an integer

if you feel more comfortable with polynomial functions, then you can say you identify the number 3 with the function f(x) = 3

so every number c can be made into a constant function f(x) = c

but functions are more general

adding the functions is the same as adding the numbers

if f(x) = 3 and g(x) = 4 then (f + g)(x) = 3 + 4

so adding f and g is the same as adding 3 and 4

same idea applies to our formal construction of polynomials

3 = (3, 0, 0, ....) and 4 = (4, 0, 0, ...)

then 3 + 4 = (3+4, 0, 0, ....) = 7 = (7, 0, 0, ...)

[old flame]

so it is a similar idea, while we could construct the idea of functions from integers, we could also construct polynomials from integers, however they just have different structure ?

[spiral star]

yea we can take $\bZ[X]$

[stoic python]

Flow:

[spiral star]

polynomials over integers

we can construct polynomials over any unital ring

basically every structure that defines some sort of addition and multiplication as you would expect

(i dont wanna go into the actual ring axioms)

[old flame]

because polynomials can do the same things right

[spiral star]

they can do more

you can find a subset of polynomials that are basically integers and with addition and multiplication behave exactly like the integers

like i showed you before with the 3 + 4 example

but polynomials can do much more, we can write many more things that werent possible with just integers

polynomials are very common for all kinds of constructions in algebra

but anyway, you get the gist, right?

you can write polynomials as you would expect

but keep in mind that $a_0 X^0 + a_1 X^1$ is a sum of polynomials

[stoic python]

Flow:

[spiral star]

and $a_0 X^0$ is a product of the polynomials $(a_0, 0, 0, \dots)$ and $(1, 0, 0, \dots)$

[stoic python]

Flow:

[spiral star]

and $a_1 X^1$ is a product of the polynomials $(a_1, 0, 0, \dots)$ and $(0, 1, 0, \dots)$

[stoic python]

Flow:

[spiral star]

whenever you see a ring element you idenfity it with the obvious polynomial

[old flame]

its a sum of a product of polynomials to be exact

[spiral star]

\begin{align*}
1+10X+20X^2 = \
(1, 0 \dots) &\cdot (1, 0 \dots) \
+ (10, 0, \dots) &\cdot (0,1, 0 \dots) \
+ (20, 0, \dots) &\cdot (0, 0, 1, \dots)
\end{align*}

[stoic python]

Flow:

[old flame]

For the part of evaluating polynomials, $\varphi(r)=\sum_{i=0}^{n}a_ir_i$, is $r_i$ similar with $X^{i}$, as in it is another polynomial ? Moreover, could you explain what is evaluation homomorphism ?

[stoic python]

Otoro:

[old flame]

ah yeah, understood thanks

[native rampart]

Have you studied ring theory or group theory?

[old flame]

not yet I'm sorry

[native rampart]

A homomorphism is a map from A to B f,satisfying some conditions.
For ring homomorphism,(considering ring with identity as a ring) the conditions are:
$ \forall a,b \in A$ $f(a+b)=f(a)+f(b)$
f$(ab)=f(a)f(b)$
$f(1)=1$

[stoic python]

DrunkenDrake:

[native rampart]

Sorry,We are mapping a formal polynomial to a normal polynomial,where we can substitute stuff

[old flame]

cause from what I see from the definition, its basically defining the result of "substituting" r (an element of the ring) into the polynomial and obtaining a polynomial of r. So the homomorphism just concludes that the result is still an element in the ring right ? obviously this needs to be proved, but I will take it for granted for now

[spiral star]

when you evaluate $\vphi$ in $r$, then you replace $X = (0, 1, 0, ...)$ by $(0, r, 0, ...)$

[stoic python]

Flow:

[spiral star]

and then sum up the elements of the sequence

that will give you an element of the original ring R

that's what evaluation means

[old flame]

so its describing something like $a_ir^{i}$ ?

[stoic python]

Otoro:

[spiral star]

$a_i r^i = (a_i, 0, 0, \dots) \cdot (0, 0, \dots, r, 0 \dots)$

[stoic python]

Flow:

[spiral star]

but then the sum

ah i guess i can retrofit that

[old flame]

that will give you an element of the original ring R
@spiral star so its kind of like a method of confirmation about whether the mapping is valid ?

[spiral star]

nvm i leave the incorrect form it takes too long in discord

[old flame]

no worries

[spiral star]

its what drake said, you substitute X by r essentially

you transform the sequence into a ring element by summing it up

since it has finite support you sum up finitely many terms

its exactly what would happen if you substitute r for X as you know it from school

[old flame]

so substitution is valid if the sum is in the original ring

[spiral star]

you can always evaluate a polynomial in R[X] by plugging in a value of R

[old flame]

Alright. So in simplier terms, for the final part, the set of endomorphisms is also a ring and therefore it satisfies being a polynomial ?