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}$

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:

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

We will now determine the eigenvectors of $A$

From the, we have:

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


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

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

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

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

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

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:

Then we notice that:

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

Putting all of the pieces together, we obtain:

Hence the result.