Page 134 - Matrices theory and applications

7.2. Exponential of a Matrix
(exp A)(exp B) by multiplying term by term the series
∞
1
k
j

(exp A)(exp B)=
j,k=0
In other words,
∞ j!k! A B . 117
1
(exp A)(exp B)= C l ,
l!
l=0
where
l! j k
C l := A B .
j!k!
j+k=l
From the assumption AB = BA, we know that the binomial formula holds.
l
Therefore, C l =(A + B) , which proves the proposition.
Noting that exp 0 n = I n and that A and −A commute, we derive the
following corollary.
Corollary 7.2.1 For every A ∈ M n (CC), exp A is invertible, and its
inverse is exp(−A).
k
k
Given two conjugate matrices B = P −1 AP,wehave B = P −1 A P for
each integer k and thus
exp(P −1 AP)= P −1 (exp A)P. (7.1)
If D = diag(d 1 ,... ,d n ) is diagonal, we have
exp D =diag(exp d 1 ,... , exp d n ).
Of course, this formula, or more generally (7.1), can be combined with
Jordan reduction in order to compute the exponential of a given matrix.
Let us keep in mind, however, that Jordan reduction cannot be carried out
explicitly.
Let us introduce a real parameter t and let us define a function g by
g(t)= exp tA. From Proposition 7.2.1, we see that g satisfies the functional
equation
g(s + t)= g(s)g(t). (7.2)
On the other hand, g(0) = I n ,and we have
∞
t
g(t) − g(0) k−1 k
− A = A .
t k!
k=2
Using any matrix norm, we deduce that
# #
# g(t) − g(0) # e tA − 1 − tA
#
# − A ≤ ,
# t # |t|