Diagonalizing an endomorphism defined on a vector space of sequences. Let $E$ be the vector space of sequences $\{ u_n \}_{n \in N}$ such that $u_{n+3}=5u_{n+2} + u_{n+1} - u_{n}$.
I am told that the endomorphism $f: \{ u_n\}_k \rightarrow \{u_{n+1} \}_k$ is diagonalizable and that $\{ \lambda_n \}_k$ is the eigenvector of $\lambda.$
I am left with a few questions:
What does diagonalizable mean in this context? How could one find a basis for such a vector space or show this property?
 A: $\newcommand{\Sp}{\mathrm{Sp}\;}$
Let $\Sp g$ denote the set of eigenvalues of the endomorphism $g$.
Let us assume $E$ is a real-valued vector space.
In order to find an eigenvalue, you have to solve $f(u) = \lambda u$ where $u \in E$ for some $\lambda \in \mathbb{C}$, thus: $\forall n \in \mathbb{N}, u_{n + 1} = \lambda u_n$.
Let be $n \in \mathbb{N}, \lambda \in \mathbb{C}$ and $\{ u_n \}_n \in E$ and let assume $\lambda \in \Sp f$.
Thus: $u_n = u_0 \times \lambda^n$ and $u_{n + 3} = 5u_{n + 2} + u_{n + 1} - u_n$.
Then: $\lambda^{n + 3} = 5\lambda^{n + 2} + \lambda^{n + 1} - \lambda^n$.
Then: $\lambda^3 - 5\lambda^2 - \lambda + 1 = 0$.
Let be $\mathcal{S}$ the set of solutions of the previous polynomial.
Now, we know that $\Sp f \subset \mathcal{S}$ and an eigenvector for an eigenvalue $\lambda$ would be the sequence $\{\mu \lambda^n\}_{n \in \mathbb{N}}$ for some $\mu \in \mathbb{R}$ which could be rewritten as: the set of eigenvectors for the $\lambda$ eigenvalue is the vector space spanned by the vector $\{ \lambda^n \}_n$.
I'm not exactly sure whether we can speak of diagonalizability for $f$, as $\Sp f$ seems finite to me and $E$ is of course not the direct sum of its eigenvectors (which are all of dimension $1$ whence $\dim E = +\infty$).
