Powers of a companion matrix and Fibonacci numbers

Eigenvalues. $\det(A-\lambda I)=\lambda^2-\lambda-1$, with roots $\varphi=\frac{1+\sqrt5}{2}$ and $\psi=\frac{1-\sqrt5}{2}$.

Eigenvectors.
- For $\varphi$: $v_1=(1,\varphi)$.
- For $\psi$: $v_2=(1,\psi)$.

Diagonalization. $A=PDP^{-1}$ with $P=\begin{pmatrix}1&1\\\varphi&\psi\end{pmatrix}$ and $D=\textrm{diag}(\varphi,\psi)$.

Fibonacci relation. $\begin{pmatrix}F_{n+1}\\F_n\end{pmatrix}=A^n\begin{pmatrix}F_1\\F_0\end{pmatrix}=A^n\begin{pmatrix}1\\0\end{pmatrix}$.

Closed formula. From $A^n=P D^n P^{-1}$, we get
$$F_n=\frac{\varphi^n-\psi^n}{\sqrt5}.$$

Check. $F_0=0$, $F_1=1$, $F_2=1$, $F_3=2$, consistent with the Binet formula.

Conclusion. $F_n=\frac{1}{\sqrt5}\left[\left(\frac{1+\sqrt5}{2}\right)^n-\left(\frac{1-\sqrt5}{2}\right)^n\right]$.