The following defines the Fibonacci recurrence relation (in matrix form): [F(n+1); F(n)] = [1 1; 1 0]^n * [F(n); F(n-1)] or [F(n+1); F(n)] = [1 1; 1 0]^n * [

The following defines the Fibonacci recurrence relation (in matrix form): [F(n+1); F(n)] = [1 1; 1 0]^n * [F(n); F(n-1)] or [F(n+1); F(n)] = [1 1; 1 0]^n * [1; 0] if we define F(0) = 0 and F(1) = 1. However, we can diagonalize [1 1; 1 0] using its eigenvalues and eigenvectors (see here: http://www.wolframalpha.com/input/?i=eigenvectors+of+%7B%7B1%2C+1%7D%2C+%7B1%2C+0%7D%7D http://www.wolframalpha.com/input/?i=eigenvectors+of+%7B%7B1...). For any diagonalizable matrix A = PDP^-1 where P is the matrix of eigenvectors and D is the matrix of eigenvalues on the diagonal, A^n = PD^nP^-1. Try to expand the A^n if you don't believe it. Using this trick, you have simplified the power of a matrix to a power of two floats, namely the eigenvalues.