RHS of 15:21
ð
Þ
½
Š ¼
C
2m
m À 1
ð
Þ!
%
e
mÀ1
ffiffiffiffiffi
2π
p
C
2m
m À 1
ð
Þ
mÀ
1
2
¼
C
ffiffiffiffiffiffiffi ffi
2πe
p
eC
2
m À 1
mÀ
1
2 ð15:24Þ
and
lim
m!1
C
ffiffiffiffiffiffiffi ffi
2πe
p
eC
2
m À 1
mÀ
1
2 ¼ 0:
ð15:25Þ
Thus, from (15.20) and (15.21) we get
lim
m!1
R m
j j
j
j ¼ 0:
ð15:26Þ
Remember that jjR m jj ¼ 0 if and only if R m ¼ 0 (zero matrix); see (13.4) and
(13.8) in Sect. 13.1. Finally, taking limit (m ! 1) of both sides of (15.19) we obtain
lim
m!1
X 2m
n¼0
1
n!
A þ B
ð
Þ
n ¼ lim
m!1
X m
n¼0
A
n
n!
X m
n¼0
B
n
n!
þ R m
!
:
Considering (15.26), we get
lim
m!1
X 2m
n¼0
1
n!
A þ B
ð
Þ
n ¼ lim
m!1
X m
n¼0
A
n
n!
X m
n¼0
B
n
n!
:
ð15:27Þ
This implies that (15.17) holds. It is obvious that if A and B commute, then exp A
and exp B commute. That is, exp A exp B ¼ exp B exp A. These complete the
proof.
∎
From Theorem 15.2, we immediately get the following property.
1
ð Þ exp A
ð
Þ
À1 ¼ exp ÀA
ð Þ:
ð15:28Þ
To show this, it suffices to find that A and ÀA commute. That is, A
(ÀA) ¼ À A
2
¼ (ÀA)A. Replacing B with ÀA in (15.17), we have
exp A À A
ð
Þ¼ exp 0 ¼ E ¼ exp A exp ÀA
ð Þ ¼ exp ÀA
ð Þexp A:
ð15:29Þ
This leads to (15.28). Notice here that exp 0 ¼ E from (15.7). We further have
important properties of exponential functions of matrices.
2
ð Þ P
À1 exp A
ð
ÞP ¼ exp P
À1 AP
À
Á :
ð15:30Þ
15.2 Exponential Functions of Matrices and Their Manipulations
587