4 ms·
> A less formula driven approach that I expected this to use is just eigenvalue decompose the matrix (with complex eigenvalues that always exists) then raise ea
by JadeNB 3y ago
> A less formula driven approach that I expected this to use is just eigenvalue decompose the matrix (with complex eigenvalues that always exists) then raise each eigenvalue to the power you want and then combine the eigenvectors and eigenvalues together again to get the Nth power.
That's actually what's happening here for the distinct-eigenvalues case: (A - \beta I)/(\alpha - \beta) is the projection on the \alpha-eigenspace, where A^n acts by \alpha^n, with kernel the \beta-eigenspace; and similarly with \alpha and \beta switched. In general, pursuing this same line of attack underlies the proof that the semisimple part of A can be written as a polynomial in A. (Sadly, the polynomial itself depends on A, so this is probably more usefully stated as that the semisimple part of A lies in k[A] if k is perfect.)