$U$ be the set of all sequences of real numbers $(x_n)_{n\in \mathbb{N} }$ with $x_n + x_{n+1} = x_{n+2}$ find $\dim(U)$ and find a basis

Question

Let $F = Abb(\mathbb{N}, \mathbb{R})$ be the $\mathbb{R}$-vector space of sequences of real numbers.

Let $U$ be the set of all sequences of real numbers $(x_n)_{n\in \mathbb{N} }$ with the property $x_n + x_{n+1} = x_{n+2}$.

1. Show that $U$ is an $\mathbb{R}$-subspace of $F$.

2. Determine the dimension of $U$ and find a basis.

So 1. should not be so complicated $U$ is not empty since the sequence $x_n=0$ is in $U$.

If we take two sequences $x_n$ and $y_n\in U$ we have

for $x_n+y_n=:c_N$ $x_n+y_n + x_{n+1}+y_{n+1} = x_{n+2}+y_{n+2} \Rightarrow c_n + c_{n+1} = c_{n+2}$ so $x_n+y_n \in U$

and for

$\alpha x_n=:g_n$ we have $\alpha x_n +\alpha x_{n+1} = \alpha x_{n+2}\Rightarrow g_n + g_{n+1} = g_{n+2}$ so $\alpha x_n \in U$

For 2. I am having some problems.

What I though is:

Every sequence should satisfy $x_n = x_{n+2}-x_{n+1}$

So we can see every sequence as a linear combination of $x_{n+2},x_{n+1}$ so we have that the $\dim(U)=2$ and a Basis of $U$ is $<x_{n+2},x_{n+1}>$

I think that what I have written for 2. is wrong but that's the only thing that came to my mind. Can someone help me?

Answer 1

Regarding the basis for $U$, if you look for basis elements of the form $\lambda^n$, you'll see that $\lambda^2-\lambda-1 = 0$. Knowing that $\dim (U) = 2$, you have that any $(u_n)\in U$ can be written as $$ c_1 \left( \frac{1+\sqrt{5}}{3}\right)^n+c_2\left( \frac{1-\sqrt{5}}{3}\right)^n. $$