Prove or disprove that {$1,x,x^2,...$} is a basis of the space of complex sequences

Question

By $x^n$ I mean the sequence $(1,2^n,3^n,...)$. My attempt :

Take any sequence $(z_0,z_1,z_2,...)$. I need to find scalars $x_i$ such that $(z_0,z_1,z_2,...)=x_0(1,1,1,...)+x_1(1,2,3,...)+x_2(1,2^2,3^2,...)+...$

Next I considered the system of equations :

$x_0+x_1+x_2+...+x_n=z_0$ $x_0+2x_1+2^2x_2+...+2^nx_n=z_1$

$.$

$.$

$x_0+(n+1)x_1+(n+1)^2x_2+...+(n+1)^nx_n=z_n$

The coefficient matrix is a vandermonde matrix whose determinant is nonzero. So the above system has a unique solution. The uniqueness proves that {$1,x,...,x^n$} is linearly independent for each $n$, and hence, the set {$1,x,x^2,...$} is linearly independent. But I'm not quite sure how to show that it is a spanning set. This is all that I could do. Any help would be appreciated.

Answer 1

The sequence $(1,0,0,0,0\cdots)$ is not expressible as a linear combination of finitely many of your sequences. Therefore your sequences do not make up a basis.

Let’s look at this a little more abstractly. Your sequences $\{x_n\}$, defined more fully as $x_i(m)=m^i$, are just the ordinary, every-day monomials of the polynomial ring $\Bbb R[X]$, if you’re allowing real scalars, which we might as well do. Any linear combination of your sequences will give a polynomial function $\sum_{i=0}^Ma_iX^i$, finitely many terms.

But it’s a basic fact about real polynomials $\,f$ is that if $\,f$ has infinitely many zeros, then $\,f$ is the zero-polynomial. My counterexample has infinitely many zeros, but is not the zero-function. Thus it’s not a polynomial function, and so not expressible as a finite linear combination of your sequences.

Answer 2

You can't do that. Your sequeces are indeed linealy independent, but they don't span the space of all sequences. For instance, the sequence $(1,2^2,3^6,4^{24},\ldots,n^{n!},\ldots)$ cannot be written as a linear combination of $x_n$'s; it growths too fast for that.