Euler-Binet Formula/Proof 2

Theorem
The Fibonacci numbers have a closed-form solution:
 * $F \left({n}\right) = \dfrac {\phi^n - \left({1 - \phi}\right)^n} {\sqrt 5} = \dfrac {\phi^n - \left({-1 / \phi}\right)^n} {\sqrt 5}$

where $\phi$ is the golden mean.

Putting $\hat \phi = 1 - \phi = -\dfrac 1 \phi$ this can be written:
 * $F \left({n}\right) = \dfrac {\phi^n - \hat \phi^n} {\sqrt 5}$

First we will prove the following

Lemma
Let $A$ be a square matrix which is neither nilpotent nor idempotent. Let $\lambda$ be an eigenvalue of $A$ and $\mathbf v$ be the corresponding eigenvector. Then for each positive integer $n$ the following equation holds:

Proof of lemma
The proof proceeds by induction. Clearly, the statement holds for $n=1$. Induction hypothesis: Suppose that $A^n {\mathbf v} =\lambda^n {\mathbf v}$ holds for some positive integer $n$. Then:

Proof
Let $A = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}$.

Let $I$ be the $2 \times 2$ identity matrix.

From Cassini's Identity, for each positive integer $n$ we have:

$A$ has the eigenvalues $\phi$ and $\hat \phi$.

Now we have that

By the preceding lemma we get for a positive integer $n$:

Let:
 * $\displaystyle B = \begin{pmatrix} \phi & {\hat \phi} \\ 1 & 1 \end{pmatrix}$

Let $B^*$ be the adjugate of $B$.

Next we will calculate the inverse matrix of $B$.

Then:

It follows that:

We obtain from \ref{prod_with_adj}:

So we find the inverse of $B$ to be:

By \ref{Ev_A} and the lemma, we find that:

Since multiplication of square matrices is associative, we have:

Thus:

So we get:

Hence the result.

It is also known as Binet's Formula.