If AAA is diagonalizable with A=PDP−1A = PDP^{-1}A=PDP−1 where D=diag(2,3)D = \text{diag}(2, 3)D=diag(2,3), then A10=A^{10} =A10=
PD10P−1=Pdiag(210,310)P−1PD^{10}P^{-1} = P\text{diag}(2^{10}, 3^{10})P^{-1}PD10P−1=Pdiag(210,310)P−1
10PDP−110PDP^{-1}10PDP−1
Pdiag(2+10,3+10)P−1=Pdiag(12,13)P−1P\text{diag}(2 + 10, 3 + 10)P^{-1} = P\text{diag}(12, 13)P^{-1}Pdiag(2+10,3+10)P−1=Pdiag(12,13)P−1
P(DT)10(PT)−1P(D^T)^{10}(P^T)^{-1}P(DT)10(PT)−1