Finding a basis of the form $(1,a,a^2,a^3...)$

Question

If V is a subspace of $\Bbb{R}^\Bbb{N}$, which consists of all rows $(x_0,x_1,...)$ of the real numbers meeting the conditions $x_{n+2}=x_{n+1}+x_n$.

How could i prove that V has exactly 2 vectors of the form $(1,a,a^2,a^3...)$ and that those vectors also form the basis of V?

So far i only know that they the vectors need to be linearly independant and that every vector in V is a linear combination of these two vectors.

Answer 1

HINT

I would combine the form of the vector $(1,a,a^2,\ldots)$ with the recurrence relation you are given, which yields $a^{n+2} = a^{n+1} + a^n$ and solve for $a$...

Answer 2

$V$ is clearly of dimension $2$ because every sequence in it is uniquely determined once you choose $x_0$ and $x_1$.

Consider the left shift operator $S$ on $V$ given by $S(x_0,x_1,x_2,\dots)=(x_1,x_2,x_3,\dots)$. Then $S^2=S+I$ on $V$. The eigenvectors of $S$ are precisely the sequences of the form $b(1,a,a^2,a^3,\dots)$ where $a$ are the eigenvalues and so roots of $a^2=a+1$. Since this equation has simple roots, the eigenvectors are linearly independent and so form a basis for $V$.