Power of matrix

If we want to find M^8, the Idea is to rewrite matrix M as (P^-1 )(D)(P) where D is a diagonal matrix (i.e. sth like [ [a1,0],[0,a2]]

Then

M = (P^-1)(D)(P)
M^2 = (P^-1)(D)(P)(P^-1)(D)(P) = (P^-1)(D)^2(P)
M^3 = (P^-1)(D)^2(P)(P^-1)(D)(P) = (P^-1)(D)^3(P)
M^8 = (P^-1)(D)^8(P)

(D)^8 is very easy to find.

then

M^8 can be found.

詳細做法:
http://www.maths.lse.ac.uk/Personal/martin/fme4a.pdf

Power of matrixGiven a matrix A，if I want to find A^8，other than multiplying A 8 times，is there any faster way?
For example，what is〔２　　１〕^8 ?
　　　　　　　　　〔３　－１〕
Sol
｜２－ｘ 　　　１｜
｜３　　　－１－ｘ｜
(2－x)(－1－x)－3=0
x^2－x－2－3=0
x^2=x+5

１　０　０　　０　　０　　０　　　０　　　０　　　　０｜
１ １　　６　１１　４１　　９６　３０１　　　　　｜１
　　　　５　　５　３０　５５　２０５　４８０　１５０５｜５
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１　１　６　１１　４１　９６　３０１｜７８１　１５０５

７８１〔２　　１〕＋１５０５〔１　０〕
　　　〔３　－１〕　　　　　〔０　１〕
＝〔３０６７　７８１〕
　〔２３４３　７２４〕