I think the proof for the closed-form version is accessible to people with a background in linear algebra. The matrix is diagonalizable (not
all matrixes are diagonalizable, but this one is):
M = P Δ P^-1
Here, P is some invertible matrix and Δ is a diagonal matrix. The powers of M reveal how useful this is:
M^N = (P Δ P^-1) * … * (P Δ P^-1)
If you adjust the parentheses, you’ll see N-1 terms of (P^-1 P), which can be removed, giving:
M^N = P Δ^N P^-1
The powers of a diagonal matrix is done by taking powers of the entries on the diagonal.
You see the φ values (1±√5)/2 in the matrix Δ.
Diagonalization of a 2x2 matrix is simple to do on paper. The diagonal of Δ contains the eigenvalues of M, which can be found using the quadratic formula. Proving that this is a correct way to diagonalize any diagonalizable matrix is more of a chore, but for this specific 2x2 matrix, you can just show that that you’ve found the values for P and Δ.
This is very elegant mathematically, but I would not use the closed-form solution if I wanted the exact answer, because you’d need a lot of precision, and that’s inconvenient.
You see the φ values (1±√5)/2 in the matrix Δ.
Diagonalization of a 2x2 matrix is simple to do on paper. The diagonal of Δ contains the eigenvalues of M, which can be found using the quadratic formula. Proving that this is a correct way to diagonalize any diagonalizable matrix is more of a chore, but for this specific 2x2 matrix, you can just show that that you’ve found the values for P and Δ.
This is very elegant mathematically, but I would not use the closed-form solution if I wanted the exact answer, because you’d need a lot of precision, and that’s inconvenient.