#Basis proof
kind spruceBOT
plain briar
well you want to see if X is spanning and linearly independent
for the spanning part, I suggest you try showing that you can write 1, x, x^2, x^3, ... etc as lin combos of polynomials in X
meager path
I get that, but since the polynomial is n-dimensions I don't know how I can do it without saying "it follows clearly that"
meager path
for the spanning part, I suggest you try showing that you can write 1, x, x^2, x...
It works for that
plain briar
okay
meager path
oh I get u
plain briar
it's an idea of proof yes
if you manage to show that, you've shown all polys can be written as a lin combo of polys in X
@meager path
meager path
Sorry yes I am reading, I am just doing what you said to see if I can establish a pattern 🙂
There doesn't seem to be a pattern. The only other thing I can think of is the binomial formula
meager path
induction works nice here
I was half way through that before I posted because I realised that if I have 2 basis and I add them, it isn't necasserily a basis as 2 linearly inde sets added isn't necasserily linearly inde. I.e. it didn't seem to work
Ill post what I did
so far anyway
plain briar
my idea is to split spanning and lin indep, so that it's easier to work through
steady estuary
R[X] is infinite dimensional
plain briar
and then for spanning to split the reasoning for each monomial X^d
meager path
R[X] is infinite dimensional
yes, that is the problem 😅
plain briar
maybe it's doable your way but it sounds like a huge pain
meager path
I don't follow you
plain briar
you're trying to prove everything at once with your induction
meager path
indeed. I thought about doing them independently but Idk what would be different except I would do it twice. For n=1, it is obvious that the spanning set is lin indep
plain briar
but also i think daddy_pi is right, your induction wouldn't show the proposition at hand
the proposition you're showing is for all n, X_n is a basis of R_n[X]
meager path
ye it doesn't work
plain briar
it will only work for finite polynomial spaces
meager path
I am just lost for ideas...
plain briar
for the spanning part, I suggest you try showing that you can write 1, x, x^2, x...
let's come back to this idea, shall we?
meager path
ok
plain briar
for all n, the polynomial X^n is a linear combination of vectors of X
meager path
ok
meager path
Ye that's clear
plain briar
now let's do strong induction, let's suppose we can write X^k as linear combinations of X for all k<d, and we wanna show X^d is also a lin combo of X
(are you fine with strong induction ?)
meager path
Induction works to show the set is spanning
plain briar
yeah I'm only focusing on the spanning part right now
meager path
ok
plain briar
(are you fine with strong induction ?)
reiterating my question @meager path
plain briar
np
plain briar
@meager path you back ?