Euler-Binet Formula/Proof 2

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}$

Next is is demonstrated that $A$ has the eigenvalues $\phi$ and $\hat \phi$, where $\hat \phi = 1 - \phi$.

Now we have that:

This shows that:
 * $\begin{pmatrix} \phi \\ 1 \end{pmatrix}$ is an eigenvector of $A$ with eigenvalue $\phi$

and:
 * $\begin{pmatrix} \hat \phi \\ 1 \end{pmatrix}$ is an eigenvector of $A$ with eigenvalue $\hat \phi$.

Thus:
 * $\begin{pmatrix} \frac \phi {\sqrt 5} \\ \frac 1 {\sqrt 5} \end{pmatrix}$ is an eigenvector of $A$ with eigenvalue $\phi$

and:
 * $\begin{pmatrix} \frac {\hat \phi} {\sqrt 5} \\ \frac 1 {\sqrt 5} \end{pmatrix}$ is an eigenvector of $A$ with eigenvalue $\hat \phi$.

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

From $(1)$ we get:
 * $A^n \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} F_{n + 1} \\ F_n \end{pmatrix}$

Substituting:
 * $\begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} \frac \phi {\sqrt 5} \\ \frac 1 {\sqrt 5} \end{pmatrix} - \begin{pmatrix} \frac {\hat \phi} {\sqrt 5} \\ \frac 1 {\sqrt 5} \end{pmatrix}$

we get:

Hence the result.