(50g) Nth Fibonacci Number

[Thomas Klemm]

This is the diagonalization of this matrix:



When \(M\) is squared the inner product of \(S^{-1} \cdot S\) cancel:
\(M^2=(S \cdot J \cdot S^{-1})^2=S \cdot J \cdot S^{-1} \cdot S \cdot J \cdot S^{-1}=S \cdot J \cdot J \cdot S^{-1}=S \cdot J^2 \cdot S^{-1}\)
This can be generalized for all powers so that we end up with:
\(M^n=S \cdot J^n \cdot S^{-1}\)
But the power of a diagonalized matrix consists just of the power of its elements on the diagonal (i.e. the eigenvalues).
This brings us back to Eddie's initial solution. Okay, maybe after a little exercise in algebra.

Cheers
Thomas