How to prove that ${1,(x-1),(x-1)^2,\dots}$ is basis of the polynomial vector space?

Question

For some reason I am struggling to prove that this constitutes a basis of the polynomial vector space. I can tell intuitively that it HAS to be, but I'm having trouble actually showing it on paper, so help is much appreciated.

Answer 1

You can easily prove it without induction, using the properties of $K[X]$ as a ring ($K$ is the base field).

The map $K[X]\longrightarrow K[X],\; X\longmapsto X-1$ defines a $K$-algebra endomorphism, which is by definition a $K$-linear map. This endomorphism is actually an automorphism since it has an inverse endomorphism: the map $X\longmapsto X+1$.

Thus, the map $X\longmapsto X-1$ is a $K$-vector space isomorphism, and as such, it maps a basis onto a basis. The image of the standard basis $\{1, X, X^2,\dots\}$ is precisely the set $\{1, X-1, (X-1)^2,\dots\}$.

Edit:

To answer your question in the below comment, no, it wouldn't be enough to show it has the same cardinality as the standard basis: such an argument is valid only for finite cardinalities.

Counterexample: the set $\{1,X^2,X^4,\dots,X^{2n},\dots\}$ has the same cardinality as the standard basis, yet it spans the set of polynomials with terms of even degree $K[X^2]$, so you can't obtain $X$, for instance.

Answer 2

A polynomial can be written in the form $ax^n +.....+c$

Therefore, $1,x...,x^n$ is a basis. Now we we want to show that $1,(x-1),...,$ is a basis (call this basis $B$).

We use induction:

The polynomials with degree zero, are just constants and $B$ clearly spans them. Polynomials of degree $1$ can also be accounted for as $a(x-1)+c(1)=ax-a+c$ and if we adjust $c$, it is obvious. Then assume that we want to show that we can span any polynomial of degree $n$ can be spanned and we know that $n-1$ degree polynomials can be spanned.

$d(x-1)^n +f(x-1)^{n-1}+...+r$ So we can adjust $d$ to cover all the possibilities for our leading coefficient and adjust the coefficients of the degree less than $n$ terms to account for all possibilities. This spans and has independent elements since each base has different degree.