2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Can you prove that $1,\,x,\,x^2,\,x^3,\ldots$ is a basis for the vector space of polynomials? Or that the span of $1,\,(x-1),\,(x-1)^2$ is the same as of $1,\,x,\,x^2$ $\endgroup$ – Milo Brandt Jan 15 at 22:14
  • 1
    $\begingroup$ $1,\,x,\,x^2,\,x^3,\ldots$ is clearly spanning, but I'm not sure how one proves the independence of an infinite set. Is it enough to say that any finite subset is independent, as it has every element of the field (which is of characteristic $0$, if I might add) as a root while having a finite degree, therefore the entire set is independent? $\endgroup$ – V.Ch. Jan 15 at 22:17
  • 1
    $\begingroup$ It does suffice to show that every finite subset is independent (indeed, that usually how independence is defined for infinite sets), though your reasoning for showing that fact isn't convincing. $\endgroup$ – Milo Brandt Jan 15 at 22:19
  • $\begingroup$ Try writing out the change of basis matrix (up to any finite $n$) and note that it is triangular with nonzero diagonal entries. $\endgroup$ – Jair Taylor Jan 15 at 22:29
6
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Thank you for the succinct answer. To restate the question I had asked someone else below, supposing that $\{1, X-1, (X-1)^2,\dots\}$ is independent, would it then suffice to show that it has the same cardinality as $\{1, X, X^2,\dots\}$ to prove that it is a basis? $\endgroup$ – V.Ch. Jan 16 at 0:16
  • 2
    $\begingroup$ I've added a counterexample to my answer. $\endgroup$ – Bernard Jan 16 at 9:30
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.