TITLE: Understanding why the roots of homogeneous difference equation must be eigenvalues
QUESTION [6 upvotes]: There is some obvious relationship between the root solutions to a homogeneous difference equation (as a recurrence relation) and eigenvalues which I'm trying to see. I have read over the wiki article 3.2, 3.4 and the eigenvalues ($\lambda$ ) are hinted at as the roots, but I'm still not sure why these must be eigenvalues of some matrix, say $A_0$, and what the meaning of $A_0$ may be. 
It seems that to solve a homogeneous linear difference equation we find the "characteristic polynomial" by simply factoring one difference equation. However, typically to find the "characteristic polynomial" I would solve the characteristic equation for some matrix, 
$A_0 = \begin{pmatrix} 1 & 0 & 0\\ 
0 & -2 & 0  \\ 
0 & 0 & 3  \\  
\end{pmatrix}$
$(A_0 - \lambda I)\mathbf x = \mathbf 0$, 
then solve for the determinant equal to $0$, and then solve for each $\lambda$ e.g.
$ \det(A_0 - \lambda I)  = 0$
$(1 - \lambda)(2 + \lambda)(3 - \lambda) = 0$
Now suppose this also happens to be a solution to some linear difference equation, and so here the characteristic polynomial is $\lambda^3 - 2\lambda^2 - 5\lambda + 6 = 0$, and the difference equation is. 
$y_{k+3} - 2y_{k+2} - 5y_{k+1} + 6y_k = 0 $. Then, for example, $\lambda = 3$ is a solution for all k. 
Now, given we have found this solution to this difference equation, how can we explain some special relationship to $A_0$, other than $\lambda = 3$ happens to be an eigenvalue of $A_0$? Is there any meaning to make of $A_0$?
(cf. 4.8,  Linear Algebra 4th, D. Lay)

REPLY [2 votes]: The roots are eigenvalues. What is the operator they are eigenvalues of? It is the shift operator. 
Consider the vector space $V$ of sequences $a_0, a_1, a_2, ...$ (say of complex numbers). Define the left shift operator
$$S(a)_i = a_{i+1}.$$
In other words, $S$ takes input a sequence $a_0, a_1, a_2, ...$ and returns the sequence $a_1, a_2, a_3, ...$. Now we need the following three fundamental observations.
Observation 1: The linear homogeneous recurrence with characteristic polynomial $p$ is precisely described by the equation $p(S) a = 0$. In other words, the sequences satisfying such a linear recurrence are precisely the sequences in the kernel of $p(S)$.
Observation 1.5: The shift operator $S$ acts on the space of solutions to any equation of the form $p(S)a = 0$. 
Observation 2: Over the complex numbers, we can factor $p(S) = \prod (S - \lambda_i)$. Thus if $(S - \lambda_i) a = 0$, or equivalently if $a$ is an eigenvector of $S$ with eigenvalue $\lambda_i$, then $p(S) a = 0$. 
Observation 3: $(S - \lambda_i) a = 0$ if and only if $a_n = c \lambda_i^n$ for some constant $c$.
$S$ resembles in some sense the differentiation operator, which it is sent to if one sends $V$ to the space of formal power series via 
$$(a_0, a_1, a_2, ...) \mapsto \sum_{n=0}^{\infty} \frac{a_n}{n!} x^n.$$
Thus everything above applies (at least formally) to linear homogeneous ODEs with constant coefficients and their characteristic polynomials as well. The corresponding eigenvectors are the functions $e^{\lambda_i x}$.