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:
 * $F \left({n}\right) = \dfrac {\phi^n - \hat \phi^n} {\sqrt 5}$

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

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

Next we will prove: For all positive integers $n$ we have:

Proof
We will prove the statement by mathematical induction. Since $F_2 =1$, we have:

So the statement is true for $n=1$.

Induction hypothesis: Suppose that the statement is true for $n \geq 1$. Then:

But $n \ge 1 \implies n+1 \> 1 \land n+2 > 1$, so
 * $F_\left({n+1}\right) +F_n =F_\left({n+2}\right)$,

and $F_n +F_\left({n-1}\right) =F_{n+1}$. So the statement is true for $n+1$. {{Qed|statement}

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

Now we have that

By Eigenvalue of Matrix Powers 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.