Actually I will refer to a concrete set of recursive equations, which appealed to me as a possibly coming from an eigensystem.
Let us start with the Golden Ratio, which is the number $\varphi \approx 1.6180339887\cdots$, and it can be defined in several ways, one of them is through a recurrent process involving the Fibonacci numbers
$$\varphi=\lim_{n\rightarrow\infty} \frac{F_{n+1}}{F_{n}}
$$
where
$$
(1) F_{n} = F_{n-1}+F_{n-2}
$$
with initial values $F_0 =1,\,F_1 =1$. Actually, one can start with other initial values, say, $F_0 =1,\,F_1 =3$ and still converge to the same value of $\varphi$ when the proportion between predecessor and sucessor is taken as $n\rightarrow\infty$. The sequence of numbers is known as the *Lucas sequence*, and the component numbers as *Lucas numbers*. In fact, it shouldn't be hard to prove that the initial values imposed in (1) only shift individual members of the Lucas sequence, but again they will converge to $\varphi$.
So, how can one obtain a convergence to a value other than $\varphi$ ? It turns out that one could add more members to the recurrence in (1), for example
$$
(2) L_{n} = L_{n-1}+L_{n-2}+\cdots +L_{n-k},
k\in \mathbb{+Z}
$$
without lose of generality, the initial values to these Lucas sequences can be stated as
$$
(3) L_0 =1,\,L_1 =1\cdots L_{k-1} =1
$$
And let us define $\varphi^{k}$ the value to which our generalised $\varphi$ convergences as
$$
(4) \varphi^{k}=\lim_{n\rightarrow\infty} \frac{L^{k}_{n+1}}{L^{k}_{n}}
$$
Although it is, of course, possible to define a infinite number of recursions like those of (2), one could form a finite set of recurrent equations upon imposing the restriction that $\varphi^{k}-\varphi^{k+1}>0$, so it turns out that there are only 9 which are real, positive, and forming a totally-ordered set. Exceptions to this rule begin, of course, when $k>10$. So the list of those nine numbers is
$$
0.618034\cdots
$$
$$
0.543689\cdots
$$
$$
0.51879\cdots
$$
$$
0.508659\cdots
$$
$$
0.504145\cdots
$$
$$
0.502013\cdots
$$
$$
0.501017\cdots
$$
$$
0.50055\cdots
$$
$$
0.500108\cdots
$$
Each of those numbers is a solution of a recurrence relation. Now, let us imagine for a moment that the collection of those 9 numbers could be eigenvalues which might be obtained from an eigensystem, and that the recurrence relations are related with the eigenvectors of that system.
Let us imagine for a moment that the we already have the putative eigenvectors and eigenvalues, one way I imagined to approach the problem may be to start backwards: --apply Cayley-Hamilton's theorem, where the matrix of an eigensystem must satisfy its own characteristic polynomial. Let $\lambda$ and $v$ be any of the eigenvalues and eigenvectors of the system, so according to Cayley-Hamilton theorem (5)
$$
(5) p(A)\cdot v = A^n \cdot v+ c_{n-1}A^{n-1} \cdot v+\cdots+ c_{1}A^n \cdot v+c_{0}I_{n}\cdot v
= \lambda^n \cdot v+ c_{n-1}\lambda^{n-1} \cdot v+\cdots+ c_{1}\lambda^n \cdot v+c_{0}I_{n}\cdot v
$$
Since (5) is valid for any of the eigenvectors with its corresponding eigenvalue, in principle it would be possible to attempt to construct a system to solve the coefficientes of the characteristic polynomial from. Actuallly the correct thing doesn't seems to be to put the recurrence relations as the eigenvectors; rather it would be the function, which would be the solution to each one of them. Unfortunately a linear system is not obtained, that can be solved, but rather a collection of ten polynomials, each one of them (it seems) of degree 10 (6)
$$
(6)
\left[ (\varphi_0)^{10}+ (\varphi_0)^{9}c_9+(\varphi_0)^{8}c_8+\cdots+(\varphi_0)c_1 + c_0\right] v_0 = 0
$$
$$
\left[ (\varphi_1)^{10}+ (\varphi_1)^{9}c_9+(\varphi_1)^{8}c_8+\cdots+(\varphi_1)c_1 + c_0\right] v_1 = 0
$$
$$
\left[ (\varphi_2)^{10}+ (\varphi_2)^{9}c_9+(\varphi_2)^{8}c_8+\cdots+(\varphi_2)c_1 + c_0\right] v_2 = 0
$$
$$
\vdots = \vdots
$$
$$
\left[ (\varphi_9)^{10}+ (\varphi_9)^{9}c_9+(\varphi_9)^{8}c_8+\cdots+(\varphi_9)c_1 + c_0\right] v_9 = 0
$$
Where $\varphi_0 = 1$ is the eigenvalue corresponding to the trivial recurrence $L_n = L_n-1$.
Admitting the possibility of the existence of a matrix from which these eigenvectors and eigenvalues may be obtained, what would it be its form? what would it be its coefficients? what are the theorem(s) which justify the creation of an eigensystem either from those recurrence relations, or relating with them?
In summary: Given this finite set of linear recursions and their solutions (the set of nine numbers), how to construct a square matrix from them, which would be akin to creating an eigensystem, with the solution of the linear recursions as their eigenvalues (and possibly the linear recursions as their eigenvectors), if at all possible?
Here is one approach you might consider: consider just the Fibonacci example: write down a vector $u_n = (F_n, F_{n-1})$ (but make it a column vector) and multiply it by the matrix $A= \begin{\pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}$. This will give you the vector $u_{n+1}$. So yes, you can now use eigenvectors to find the growth rate of the sequence (and indeed to find the formula for $F_n$). This works very similarly for higher degree recurrences.
Hope this helps,
Neil
I do not know about eigenvalues for this system. But such equation is considered in book 'Combinatorics' by Vilenkin. I do not know if it is translated into English. But you can search. The main idea is to assume f(n)=r^n . The rest is similar to solving of linear differential equation with constant coeficients
To continue Calkin's approach, let $A$ be the companion matrix to an n-th degree polynomial (for simplicity, with real roots. One can then get an n-term recursion by taking the vector $u$ to be the most recent n values and $u_{next}=Au$. The ratio wil converge to the largest root (in absolute value) if that is unique --- for almost all starting values. One can have convergence to another root (even for the Fibonacci recursion) if one starts with an eigenfunction for that root, i.e., if one starts with F_1=1 and F_2=[other root] of the quadratic $x^2-x-1$.
I will add to my answer http://www.amazon.com/Combinatorics-N-Ya-Vilenkin/dp/1483248011 Combinatorics Paperback by N. Ya. Vilenkin, page 138
The companion matrix may need to have complex roots: indeed, for most recursions this will be the case. There are, however, some very interesting recursions where all the roots are real: for example, counting configurations of non-attacking kings on a kxn chessboard, as n goes to infinity: then there is a natural F_k x F_k (appropriately indexed Fibonacci numbers) matrix A_k whose eigenvalues and eigenvectors give the number of configurations. And since A_k is real and symmetric, all its eigenvalues are real.
It is clear that one gets real ratios at each step since, assuming the polynomial in question has real coefficients, the matrix is real and all steps involve only real numbers. However, this does not contradict my comment since either the largest root is real (so one can get the right real limit) or is complex (in which case it cannot be the unique maximizer of the absolute value: its conjugate is also a root).
Indeed, Thomas --- sorry, I didn't mean to be appearing to contradict you.
The Perron Frobenius theorem is also often useful in these settings. If you can show your recurrence satisfies it, then it will tell you something useful about the largest eigenvalue.