ABSTRACT
The answer lies in the diagonalization. Suppose we can find a non-singular matrix P such that P−1AP is diagonal, say
P−1AP = D =
λ1 0. . . 0 λn
. (6.2)
Then λ m 1 0
. . .
0 λmn
= Dm = (P−1AP )m = (P−1AP )(P−1AP ) . . . (P−1AP ) = P−1AmP,
which gives Am = PDmP−1. Therefore, if we write Ao = I,
eA =
m! Am =
m! PDmP−1 = P
( ∞∑ m=0
m! Dm
) P−1 =
P
e λ1 0
. . .
