# Euler-Binet Formula/Proof 2

## Theorem

The Fibonacci numbers have a closed-form solution:

$F_n = \dfrac {\phi^n - \paren {1 - \phi}^n} {\sqrt 5} = \dfrac {\phi^n - \paren {-1 / \phi}^n} {\sqrt 5} = \dfrac {\phi^n - \paren {-1}^n \phi^{-n} } {\sqrt 5} = \dfrac {\phi^n - \paren {1 - \phi}^n} {\phi - \paren {1 - \phi}}$

where $\phi$ is the golden mean.

Putting $\hat \phi = 1 - \phi = -\dfrac 1 \phi$ this can be written:

$F_n = \dfrac {\phi^n - \hat \phi^n} {\sqrt 5}$

From Definition 2 of Golden Mean: $\phi = \dfrac {1 + \sqrt 5} 2$

Therefore, substituting $\sqrt 5 = 2\phi - 1 = \phi - \paren {1 - \phi} = \phi - \hat \phi$, the above can be written as:

$F_n = \dfrac {\phi^n - \hat \phi^n} {\paren {\phi - \hat \phi}}$

## Proof

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

First by the lemma to Cassini's Identity:

$(1): \quad \forall n \in \Z_{>1}: A^n = \begin{bmatrix} F_{n + 1} & F_n \\ F_n & F_{n - 1} \end{bmatrix}$

Recall from Eigenvalues of Real Square Matrix are Roots of Characteristic Equation, we can find the eigenvalues of $\mathbf A$ by solving the equation $\map \det {\mathbf A - \lambda \mathbf I} = 0$

Therefore, taking the determinant of the square matrix below:

$\mathbf A - \lambda \mathbf I = \begin{bmatrix} 1 - \lambda & 1 \\ 1 & -\lambda \end{bmatrix}$

We obtain the equation:

 $\ds 0$ $=$ $\ds -\lambda \paren {1 - \lambda} - 1$ Determinant of Matrix Product $\ds$ $=$ $\ds \lambda^2 - \lambda - 1$ $\ds$ $=$ $\ds \lambda^2 - \lambda + \paren {\lambda \phi - \lambda \phi} + \paren {\phi - \phi} - 1$ adding 0 $\ds$ $=$ $\ds \lambda^2 - \lambda + \paren {\lambda \phi - \lambda \phi} + \phi - \paren {\phi + 1}$ regrouping $\ds$ $=$ $\ds \lambda^2 - \lambda + \paren {\lambda \phi - \lambda \phi} + \phi - \phi^2$ Square of Golden Mean equals One plus Golden Mean $\ds$ $=$ $\ds \lambda^2 - \lambda \paren {1 - \phi} - \lambda \phi + \phi \paren {1 - \phi}$ $\ds$ $=$ $\ds \paren {\lambda - \phi} \paren {\lambda - \paren {1 - \phi} }$

Therefore, $A$ has the eigenvalues $\phi$ and $\paren {1 - \phi}$.

We will now determine the eigenvectors of $A$

From the Definition of Eigenvector of Real Square Matrix, we have:

A non-zero vector $\mathbf v \in \R^n$ is an eigenvector corresponding to $\lambda$ if and only if:

$\mathbf A \mathbf v = \lambda \mathbf v$

For $\lambda = \phi$, we have:

 $\ds \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} \times \begin{pmatrix} v_1 \\ v_2 \end{pmatrix}$ $=$ $\ds \begin{pmatrix} \phi v_1 \\ \phi v_2 \end{pmatrix}$ $\ds \leadsto \ \$ $\ds \begin{pmatrix} v_1 + v_2 \\ v_1 \end{pmatrix}$ $=$ $\ds \begin{pmatrix} \phi v_1 \\ \phi v_2 \end{pmatrix}$ Definition of Matrix Product (Conventional)

We see from the above, when $v_2 = 1$ then $v_1 = \phi$ and we have:

 $\ds \begin{pmatrix} \phi + 1 \\ \phi \end{pmatrix}$ $=$ $\ds \begin{pmatrix} \phi^2 \\ \phi \end{pmatrix}$ $\ds$ $=$ $\ds \phi \begin{pmatrix} \phi \\ 1 \end{pmatrix}$

For $\lambda = \paren {1 - \phi}$, we have:

 $\ds \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} \times \begin{pmatrix} v_1 \\ v_2 \end{pmatrix}$ $=$ $\ds \begin{pmatrix} \paren {1 - \phi} v_1 \\ \paren {1 - \phi} v_2 \end{pmatrix}$ $\ds \leadsto \ \$ $\ds \begin{pmatrix} v_1 + v_2 \\ v_1 \end{pmatrix}$ $=$ $\ds \begin{pmatrix} \paren {1 - \phi} v_1 \\ \paren {1 - \phi} v_2 \end{pmatrix}$ Definition of Matrix Product (Conventional)

We see from the above, when $v_2 = 1$ then $v_1 = \paren {1 - \phi}$ and we have:

 $\ds \begin{pmatrix} 2 - \phi \\ 1 - \phi \end{pmatrix}$ $=$ $\ds \begin{pmatrix} \paren {1 - \phi}^2 \\ \paren {1 - \phi } \end{pmatrix}$ $\ds$ $=$ $\ds \paren {1 - \phi} \begin{pmatrix} \paren {1 - \phi} \\ 1 \end{pmatrix}$

We have now demonstrated that:

$\begin{pmatrix} \phi \\ 1 \end{pmatrix}$ is an eigenvector of $A$ with eigenvalue $\phi$

and:

$\begin{pmatrix} \paren {1 - \phi} \\ 1 \end{pmatrix}$ is an eigenvector of $A$ with eigenvalue $\paren {1 - \phi}$.

We now observe that:

 $\ds A^n \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ $=$ $\ds \begin{bmatrix} F_{n + 1} & F_n \\ F_n & F_{n - 1} \end{bmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ from (1) above $\ds$ $=$ $\ds \begin{pmatrix} F_{n + 1} \\ F_n \end{pmatrix}$ Definition of Matrix Product (Conventional)

Then we notice that:

 $\text {(2)}: \quad$ $\ds \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ $=$ $\ds \dfrac 1 {\phi - \paren {1 - \phi} } \paren {\begin{pmatrix} \phi \\ 1 \end{pmatrix} - \begin{pmatrix} \paren {1 - \phi} \\ 1 \end{pmatrix} }$

From Eigenvalue of Matrix Powers for a positive integer $n$, we have:

 $\ds A^n \begin{pmatrix} \phi \\ 1 \end{pmatrix}$ $=$ $\ds \phi^n \begin{pmatrix} \phi \\ 1 \end{pmatrix}$ $A^n \mathbf v = \lambda^n \mathbf v$ $\ds A^n \begin{pmatrix} \paren {1 - \phi} \\ 1 \end{pmatrix}$ $=$ $\ds \paren {1 - \phi}^n \begin{pmatrix} \paren {1 - \phi} \\ 1 \end{pmatrix}$ $A^n \mathbf v = \lambda^n \mathbf v$

Putting all of the pieces together, we obtain:

 $\ds \begin{pmatrix} F_{n + 1} \\ F_n \end{pmatrix}$ $=$ $\ds A^n \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ $\ds$ $=$ $\ds \dfrac 1 {\phi - \paren {1 - \phi} } A^n \paren {\begin{pmatrix} \phi \\ 1 \end{pmatrix} - \begin{pmatrix} \paren {1 - \phi} \\ 1 \end{pmatrix} }$ from (2) above $\ds$ $=$ $\ds \dfrac 1 {\phi - \paren {1 - \phi} } \paren {A^n \begin{pmatrix} \phi \\ 1 \end{pmatrix} - A^n \begin{pmatrix} \paren {1 - \phi} \\ 1 \end{pmatrix} }$ $\ds$ $=$ $\ds \dfrac 1 {\phi - \paren {1 - \phi} } \paren {\phi^n \begin{pmatrix} \phi \\ 1 \end{pmatrix} - \paren {1 - \phi}^n \begin{pmatrix} \paren {1 - \phi} \\ 1 \end{pmatrix} }$ Eigenvalue of Matrix Powers $\ds$ $=$ $\ds \dfrac 1 {\phi - \paren {1 - \phi} } \paren {\begin{pmatrix} \phi^{n + 1} \\ \phi^n \end{pmatrix} - \begin{pmatrix} \paren {1 - \phi}^{n + 1} \\ \paren {1 - \phi}^n \end{pmatrix} }$ Definition of Matrix Product (Conventional)

Hence the result.

$\blacksquare$

## Source of Name

This entry was named for Jacques Philippe Marie Binet and Leonhard Paul Euler.

## Also known as

The Euler-Binet Formula is also known as Binet's formula.