Understanding Linear Recurrence Theorem I have been trying to understand this theorem but I could not get past a point. The proof has been taken from Kenneth Rosen's Discrete Mathematics.

THEOREM 1: Let c1 and c2 be real numbers.
  Suppose that r2 − c1r − c2 = 0 has
  two distinct roots r1 and r2. Then the sequence
  {an} is a solution of the recurrence relation an
  = α1 an−1 + α2 an−2 if and only if an =α1 r1n
  +α2 r2n for n=0,1,2,...,where α1 and α2 are constants.
PROOF : We must do two things to prove the theorem. First, it must be shown that if r1 and r2 are the roots of the
  characteristic equation, and α1 and α2 are
  constants, then the sequence {an} with an =
  α1r1n +
  α2r2n is a solution of the recurrence
  relation.  Second, it must be shown that if the sequence
  {an} is a solution, then an =
  α1r1n +
  α2r2n for some constants
  α1 and α2.
Now we will show that if an =
  α1r1n +
  α2r2n, then the sequence
  {an} is a solution of the recurrence relation.Because
  r1 and r2 are roots of r2 −
  c1r − c2 = 0, it follows that
  r12 = c1r1 +c2,
  r22 =c1r2 +c2.

Now until this part what I have understood(if I am right) that they have just taken a quadratic equation and split it into its roots. 
Now the following algebraic part did not make any sense to me. I want to understand this part:

c1an−1 + c2an−2 =
  c1(α1r1n−1 +
  α2r2n−1) +
  c2(α1r1n−2 +
  α2r2n−2) 
= α1r1n−2(c1r1 + c2) + α2r2n−2(c1r2 +
  c2)
= α1r1n−2r12 + α2r2n−2r22
= α1 r1n + α2 r2n
= an.
This shows that the sequence {an} with an =
  α1r1n +
  α2r2n is a solution of the recurrence
  relation.

Now I have omitted rest of the proof because I could no get past this point. If someone requires it I can edit it. or If someone can explain further part in simple english I would be very grateful for that. But for now I just want to understand the above part.
 A: Solutions to linear recurrence relations started making a lot more sense to me once I found a source that tied it to linear algebra.  I won't spell out all of the details, but "the interested reader" can verify most of it.

Consider $\mathbb R^\mathbb N$, the linear space of real sequences with termwise addition and termwise scalar multiplication.  The key idea is that the set of all real sequences that satisfy a linear recurrence relation is a subspace of $\mathbb R^\mathbb N$.  You can check for yourself that if $(a_n)$ and $(b_n)$ satisfy $f_n=\alpha f_{n-1}+\beta f_{n-2}$ then so do $(c:n\mapsto a_n+b_n)$ and $(d:n\mapsto ka_n)$.
$$c_n=a_n+b_n=(\alpha a_{n-1}+\beta a_{n-2})+(\alpha b_{n-1}+\beta b_{n-2})\\=\alpha (a_{n-1}+b_{n-1})+\beta(a_{n-2}+b_{n-2})=(\alpha c_{n-1}+\beta c_{n-2})$$
$$d_n=ka_n=k(\alpha a_{n-1}+\beta a_{n-2})=\alpha ka_{n-1}+\beta ka_{n-2}=(\alpha d_{n-1}+\beta d_{n-2})$$
So, since being a solution to a linear recurrence relation is preserved by addition and scalar multiplication, the set of all solutions is a subspace of all real sequences.  Since an $n$'th degree linear recurrence relation is defined by assigning the first $n$ terms of the sequence, it stands to reason that the dimension of the subspace is $n$.  Now we are just left with finding a basis for the subspace.

So assume that $(n\mapsto r^n)$ for some real non-zero $r$ satisfies the recurrence relation.  Then $$r^n=\alpha r^{n-1}+\beta r^{n-2}\\r^2=\alpha r+\beta\\ r^2-\alpha r-\beta=0\\$$
Long story short, for a two-term linear recurrence relation whose characteristic polynomial has distinct roots, a basis for the subspace of solutions $(n\mapsto r^n_1)$ and $(n\mapsto r^n_2)$, where $r_1$ and $r_2$ are the roots of the characteristic polynomial.  Therefore, the set of all solutions is the linear combinations of those two basis solutions.
