Two notes from Jiřì Matoušek‘s book Thirty-three Miniatures: Mathematical and Algorithmic Applications of Linear Algebra [1,2].

Fibonacci numbers in O(lgn) steps

1.1 Matrix formulation for recursive calculation

$\left(\begin{array}{c}F_{n+2}\\ F_{n+1}\end{array}\right) = M\left(\begin{array}{c}F_{n+1}\\ F_{n}\end{array}\right)\\for\ M\ =\left(\begin{array}{cc}1 & 1 \\ 1 & 0\end{array}\right)\\ \therefore \left(\begin{array}{c}F_{n+1}\\ F_{n}\end{array}\right) = M^{n}\left(\begin{array}{c}1\\ 0\end{array}\right)$

1.2 At most log2n multiplications needed size z <= log2n

$n = 2^a + 2^b +\ ...+ 2^z\ for\ a < b <\ ...,\ M^n=M^{2^a}M^{2^b}...M^{2^z}$

This can be extended to any recursive formula.

Fibonacci formula

2.1 Given the formula

$F_{n+2} = F_{n+1} + F_{n}$

2.2 If we start with the ansatz for either of the two sequences, un and vn, that compose Fn

$u_n = \tau^{n} \\ \therefore \tau^{n+2} = \tau^{n+1} + \tau^{n} \\ \Rightarrow \tau^{2} = \tau + 1$

2.3 This yields two distinct roots

$\tau = (1 \pm \sqrt{5}) / 2$

2.4 The two roots individually form two sequences, u and v, that are linearly independent. Thus Fibonacci numbers can be written in terms of these basis vectors.

$\mathbf{F}=\alpha \mathbf{u}+\beta \mathbf{v}$

2.5 The values of α and β can be evaluation by solving the linear systems, and eventually

$F_n = \frac{1}{\sqrt{5}}\left\{(\frac{1+\sqrt{5}}{2})^{n}-(\frac{1-\sqrt{5}}{2})^n\right\}$