The polar decomposition of the Fibonacci sequence
If we think about the matrix form of the Fibonacci recursion:
tells us about the recursion and the sequence that it generates.
To get the polar decomposition of the matrix A we write it as
A = UP
where U is unitary and P is positive definite.
Writing a matrix in its polar decomposition is much like writing a complex number as
z = r eiθ
where the positive definite part of the polar decomposition corresponds to the magnitude of the complex number and the unitary part of the polar decomposition corresponds to the rotational part of the complex number.
In the case of the Fibonacci recursion we find that
etc., and that U2 = I.
In general, I'd guess that for a nth-order linear recursion, we'll always have that Un = I. That shouldn't be too hard to show.
As we saw before, A has eigenvalues
a = (1 + √5) / 2
b = (1 – √5) / 2
while in this case we find that P has eigenvalues a and –b while U has eigenvalues 1 and -1.
I'm not sure how useful this polar decomposition will be, but I feel fairly sure that I'll find a use for it some day.