Euler-Binet Formula/Proof 2

Theorem: The $n$-th Fibonacci number $F_n$ is given by $F_n = \dfrac{\phi^n -\hat \phi^n}{\sqrt{5}}$ where $\phi$ is the golden mean and $\hat \phi =-1/\phi$.

Proof. Let $A=\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}$. Then $A$ has the eigenvalues $\phi$ and $\hat \phi$.

Let $v_1 \, v_2$ be the eigenvectors of $\phi$ and $\hat \phi$, respectively. Since $\phi$ and $\hat \phi$ are distinct, $v_1$ and $v_2$ form a basis of $\R^2$. Now we have that

So let

Let $B$ be the $2 \times 2$-matrix whose first column is $v_1$, and whose second column is $v_2$. Then \( \displaystyle $B\begin{bmatrix} \phi & 0 \\ 0 & \hat \phi \end{bmatrix} =AB$.\) $b$ is invertible, and we get \( \displaystyle $A=B\begin{bmatrix} \phi & 0\\ 0 & \hat \phi \end{bmatrix}B^{-1}$.\) For each positive integer $n$ we have

(see Cassini's Identity for a proof). Thus we have

Since the first column of $B^{-1}$ is $\begin{bmatrix} \dfrac{1}{\sqrt{5}} \\ \dfrac{-1}{\sqrt{5}} \end{bmatrix}$, we obtain:

Hence the result.