- #1

bpet

- 532

- 7

The elements [tex]p_{ij}^{(n)}[/tex] of matrix [tex]P^n[/tex] (size m) are also determined by the Perron formula

[tex]p_{ij}^{(n)} = \sum_{s=1}^r \frac{1}{(v_s-1)!}\left\{\frac{d^{v_s-1}}{d\lambda^{v_s-1}}\left[\frac{\lambda^nA_{ji}(\lambda)(\lambda-\lambda_s)^{v_s}}{|\lambda I_m-P|}\right]\right\}_{\lambda=\lambda_s}[/tex]

where [tex]r[/tex] is the number of distinct eigenvalues, [tex]\lambda_s[/tex] are the distinct eigenvalues with multiplicity [tex]v_s[/tex] (so [tex]v_1+\ldots+v_s=m[/tex]) and [tex]A_{ji}(\lambda)[/tex] is the cofactor of the elements [tex]\lambda\delta_{ji}-p_{ji}[/tex] in the determinant [tex]|\lambda I_m-P|[/tex] and [tex]I_m[/tex] is the identity matrix of the same size as P.

It seems to work for any matrix (not just Markov transition probabilities) and as far as I can tell it's not related to the Perron formula of number theory.

The term [tex]A_{ji}(\lambda)/|\lambda I_m-P|[/tex] would hint that [tex](\lambda I_m-P)^{-1}[/tex] is involved, so I suspect it's done by finding the Laplace transform of [tex]e^{tP}[/tex] and somehow extracting the nth term of the Taylor series. Is this on the right track and if so, how would it be done? In particular, how do they turn the expression into a sum of derivatives at the eigenvalues?