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.

Since $\operatorname{det} A =-1 \ne 0$, $A$ is invertible.

$A$ is not nilpotent.

Proof
Suppose that $A$ is nilpotent. Then there exists a smallest positive integer m such that $A^m$ is the zero matrix. But since $A$ is invertible, we have that:

contradicting the minimality of $m$. Therefore such an integer cannot exist.

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 $n+1 \gt 1$ and $n+2 \gt 1$, so
 * $\displaystyle F_\left({n+1}\right) +F_n =F_\left({n+2}\right)$,

and $\displaystyle F_n +F_\left({n-1}\right) =F_{n+1}$.

Thus:

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

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

Now we have that

By Eigenvalue of Matrix Powers we get for a positive integer $n$:

Let:

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

Let $B^*$ be the adjugate 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 Eigenvalue of Matrix Powers, 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.