How to prove the characteristic equation based solution of recurrence relations? What is the proof for / where might I find the proof to:
Let $c_1, c_2,..., c_k$ be real numbers. Suppose that the characteristic equation
$$r^k-c_1 r^{k-1}-...-c_k=0$$
has $k$ distinct roots $r_1, r_2,..., r_k$. Then a sequence $\{a_n\}$ is a solution of the recurrence relation
$$a_n=c_1a_{n-1}+c_2a_{n-2}+...+c_ka_{n-k}$$
if and only if
$$a_n=\alpha_1r_1^n+\alpha_2r_2^n+...+\alpha_kr_k^n$$
for $n = 0,1,2...$, where $\alpha_1,\alpha_2,...,\alpha_k$ are constants.
 A: A nice proof is given by using generating functions. Say your recurrence is:
$\begin{align*}
c_k a_{n + k} + c_{k - 1} a_{n + k - 1} + \dotsb + c_0 a_n = 0
\end{align*}$
with given values for $a_0, \dotsc, a_{k - 1}$. Define the generating function $A(z) = \sum_{n \ge 1} a_n z^n$, multiply your recurrence by $z^n$ and add up over $n \ge 0$. Recognizing some sums you have:
$\begin{align*}
c_k \frac{A(z) - a_0 - a_1 z - \dotsb - a_{k - 1} z^{k - 1}}{z^k}
 + c_{k - 1} \frac{A(z) - a_0 - a_1 z - \dotsb - a_{k - 2} z^{k - 2}}{z^{k - 1}}
 + \dotsb
 + c_0 A(z)
 &= 0
\end{align*}$
Multiply through by $z^k$ and solve for $A(z)$ to get:
$\begin{align*}
A(z)
  &= \frac{p(z)}{c_k z^k + c_{k - 1} z^{k - 1} + \dotsb + c_0}
\end{align*}$
for some polynomial $p(z)$ of degree at most $k - 1$ that depends on the initial values. Note that the denominator is the characteristic equation of the recurrence.
From calculus you know that this can be written as partial fractions:
$\begin{align*}
A(z)
  &= \sum_r \frac{A_r}{(1 - \alpha_r z)^{m_r}}
\end{align*}$
where $1 / \alpha_r$ is a zero of the characteristic equation of multiplicity $m_r$. Now, by the extended binomial theorem:
$\begin{align*}
(1 - \alpha z)^{-m}
  &= \sum_{n \ge 0} (-1)^n \binom{-m}{n} (\alpha z)^n \\
  &= \sum_{n \ge 0} \binom{m + n - 1}{m - 1} \alpha^n z^n
\end{align*}$
But the binomial coefficient $\binom{m + n - 1}{m - 1}$ is just a polynomial in $n$ of degree $m - 1$. Thus the solution is a sum of terms of the form $\alpha^n$ multiplied by a polynomial in $n$ of degree at most $m - 1$ if $\alpha$ is a zero of multiplicity $m$.
If any zero turns out complex, there will be a pair of complex conjugate zeroes. Writing out the complex numbers in polar form $\alpha = \rho \exp(\omega i)$ you get $\alpha^n = \rho^n \exp(n \omega i)$, giving rise to terms of the form $\rho^n \cos(\omega n)$ (they will turn out in conjugate pairs, the imaginary parts cancel out).
A: This can be shown with matrices. Note that your relation can be expressed in matrix/vector form as a system of equations.
$$\overrightarrow v_{n+1}= \mathbf M \cdot \overrightarrow v_n$$
 We know the solution to this recurrence relation....
$$\overrightarrow v_n=\mathbf M^n \cdot \overrightarrow v_0$$
This can be evaluated using eigenvalues, hence this is why you have eigenvalues in the solutions.
