Proof of Linear Homogenous Recurrence Relations with constant coefficient and with two distinct roots

I was going through the book "Discrete Mathematics and its Application" by Kenneth Rosen where I came across the proof the following theorem. The backward proof is fine but I did not feel the forward proof of quite satisfactory.

Theorem: Let $$c_1$$ and $$c_2$$ be real numbers. Suppose that $$r^{2} − c_1r−c_2=0$$ has two distinct roots $$r_1$$ and $$r_2$$ . Then the sequence $$\{a_n\}$$ is a solution of the recurrence relation $$a_n = c_1 a_{n−1} + c_2 a_{n−2}$$ if and only if $$a_n =α_1r^{n} +α_2r^{n}$$ for $$n = 0, 1, 2,...,$$ where $$α_1$$ and $$α_2$$ are constants.

Proof: We must do two things to prove the theorem. First, it must be shown that if $$r_1$$ and $$r_2$$ are the roots of the characteristic equation, and $$α_1$$ and $$α_2$$ are constants, then the sequence $$\{a_n\}$$ with $$a_n =α_1r^{n} +α_2r^{n}$$ is a solution of the recurrence relation. Second, it must be shown that if the sequence $$\{a_n \}$$ is a solution, then $$a_n = c_1 a_{n−1} + c_2 a_{n−2}$$ for some constants $$α_1$$ and $$α_2$$.

The forward proof

Now we will show that if $$a_n =α_1r^{n} +α_2r^{n}$$ , then the sequence $$\{a_n\}$$ is a solution of the recurrence relation. Because $$r_1$$ and $$r_2$$ are roots of $$r^{2} − c_1r−c_2=0$$ ,it follows that $$r_1^{2}=c_1r_1+c_2$$,$$r_2^{2}=c_1r_2+c_2$$. From these equations, we see that

$$c_1a_{n−1} + c_2a_{n−2}$$

$$= c_1(α_1r_1^{n−1}+α_2r_2^{n−1} )+c_2(α_1r_1^{n−2}+α_2r_2^{n−2})$$ $$= α_1r_1^ {n−2}(c_1r_1+c_2)+α_2r_2 ^{n−2}(c_1r_2 + c_2)$$ $$= α_1r_1^{n−2}r_1^{2}+α_2r_2^{n−2}r_2^{2}$$ $$= α_1r_1^{n} +α_2r_2{n}$$ $$= a_n.$$

This shows that the sequence $$\{a_n\}$$ with $$a_n = α_1 r_1^{n}+α_2r_2^{n}$$ is a solution of the recurrence relation.

The backward proof

To show that every solution $$\{a_n\}$$ of the recurrence relation $$a_n = c_1 a_{n−1} + c_2 a_{n−2}$$ has $$a_n =α_1r^{n} +α_2r^{n}$$ for $$n = 0, 1, 2,...$$ , for some constants $$α_1$$ and $$α_2$$, suppose that $$\{a_n\}$$ is a solution of the recurrence relation, and the initial conditions $$a_0 = C_0$$ and $$a_1 = C_1$$ hold. It will be shown that there are constants $$α_1$$ and $$α_2$$ such that the sequence $$\{a_n\}$$ with $$a_n =α_1r^{n} +α_2r^{n}$$ satisfies these same initial conditions.

This requires that

$$a_0=C_0=α_1+α_2$$, $$a_1=C_1=α_1r_1+α_2r_2$$.

We can solve these two equations for α_1 and α_2 . From the first equation it follows that

$$α_2 = C_0 − α_1$$ .

Inserting this expression into the second equation gives

$$C_1=α_1r_1+(C_0−α_1)r_2$$.

Hence,

$$C_1=α_1(r_1 −r_2)+C_0r_2$$.

This shows that

$$α_1=\frac{ C_1 − C_0r_2}{r_1−r_2}$$

and

$$α_2=C_0−α_1=C_0− \frac{C_1 − C_0r_2} {r_1−r_2} = \frac{C_0r_1 − C_1}{ r1−r2}$$

, where these expressions for $$α_1$$ and $$α_2$$ depend on the fact that $$r_1 \neq r_2$$ . (When $$r_1 = r_2$$ , this theorem is not true.) Hence, with these values for $$α_1$$ and $$α_2$$ , the sequence $$\{a_n\}$$ with $$a_n =α_1r^{n} +α_2r^{n}$$ satisfies the two initial conditions. We know that $$\{a_n\}$$ and $$\{α_1r^{n} +α_2r^{n}\}$$ are both solutions of the recurrence relation $$a_n = c_1 a_{n−1} + c_2 a_{n−2}$$ and both satisfy the initial conditions when $$n = 0$$ and $$n = 1$$. Because there is a unique solution of a linear homogeneous recurrence relation of degree two with two initial conditions, it follows that the two solutions are the same, that is, $$a_n =α_1r^{n} +α_2r^{n}$$ for all nonnegative integers $$n$$.

Doubts

In the forward part we are supposed to prove something of the form,

if "$$\{a_n\}$$ is a solution of the recurrence relation" $$=>$$ "$$\{a_n\}$$ is same as $$\{α_1r^{n} +α_2r^{n}\}$$".

Here in the above forward proof, for the purpose below,

It will be shown that there are constants $$α_1$$ and $$α_2$$ such that the sequence $$\{a_n\}$$ with $$a_n =α_1r^{n} +α_2r^{n}$$ satisfies these same initial conditions.

we are sort of using the conclusion of the above implication to show the conclusion is true if the hypothesis is true.

We know that $$\{a_n\}$$ and $$\{α_1r^{n} +α_2r^{n}\}$$ are both solutions of the recurrence relation $$a_n = c_1 a_{n−1} + c_2 a_{n−2}$$

I hope they are using the proof of the first part (backward proof) to say that $$\{α_1r^{n} +α_2r^{n}\}$$ is a solution of the recurrence relation.

Because there is a unique solution of a linear homogeneous recurrence relation of degree two with two initial conditions

I hope the above is theorem which exists and it is not dealt with in the book.

The entire forward proof for seems a bit weird to me and it seems that was sort of forcely made to agree the facts of mathematics.

Alternative proof: Let $$U$$ be the set of sequences and $$A_m : U \mapsto \mathbb{R}$$ be defined as $$A_m((u_n)_n) := u_{m+2} -c_1 u_{m+1} - c_2 u_m.$$ Then it is easy to check that A_m is linear. One also checks that $$(r_i^n)_n$$ satisfies $$A_m((r_i^n)_n) = 0$$ for all $$m \ge 0$$ and $$i= 1,2$$. It follows that $$A_m((\alpha_1 r_1^n + \alpha_2 r_2^n)_n) = 0$$ for all $$m \ge 0$$.
One obtains uniqueness of solutions if $$u_0 = C_0$$ and $$u_1 = C_1$$ by contradiction: suppose there were two solutions $$(u_n)_n$$ and $$(v_n)_n$$. Let $$m$$ be the smallest number such that $$u_m \neq v_m$$. The recurrence relation leads to a contradiction.
Now suppose $$C_0$$ and $$C_1$$ are given. We must check that there are $$\alpha_1$$ and $$\alpha_2$$ such that $$\alpha_1 r_1^0 + \alpha_2 r_2^0 = C_0$$ and $$\alpha_1 r_1^1 + \alpha_2 r_2^1 = C_1$$. Such $$alpha_i$$ can be found if the determinant of the matrix of the system is nonzero. One computes the determinant: $$r_2 - r_1 \neq 0$$ since $$r_1 \neq r_2$$.