Page 135 - Matrices theory and applications

118
from which we obtain
g(t) − g(0)
lim
= A.
t
t→0
We conclude that g has a derivative at t =0, with g (0) = A.Using the

functional equation (7.2), we then obtain that g is differentiable everywhere,
with 7. Exponential of a Matrix, Polar Decomposition, and Classical Groups
g(t)g(s) − g(t)
g (t) = lim = g(t)A.

s→0 s
We observe that we also have
g(s)g(t) − g(t)

g (t) = lim = Ag(t).
s→0 s
From either of these differential equations we see that g is actually infinitely
differentiable. We shall retain the formula
d
exp tA = A exp tA =(exp tA)A. (7.3)
dt
This differential equation is sometimes the most practical way to compute
the exponential of a matrix. This is of particular relevance when A has real
entries but has at least one nonreal eigenvalue if one wishes to avoid the
use of complex numbers.
Proposition 7.2.2 For every A ∈ M n (CC),
det exp A =exp Tr A. (7.4)
Proof
We could deduce (7.4) directly from (7.3). Here is a more elementary
proof. We begin with a reduction of A of the form A = P −1 TP,where T
k
is upper triangular. Since T is still triangular, with diagonal entries equal
k
to t ,exp T is triangular too, with diagonal entries equal to expt jj . Hence
jj

det exp T = exp t jj =exp t jj =exp Tr T.
j j
This is the expected formula, since exp A = P −1 (exp T )P.
∗ k
∗
k ∗
Since (M ) =(M ) , we see easily that (exp M) ∗ =exp(M ). In
particular, the exponential of a skew-Hermitian matrix is unitary, for then
∗
∗
(exp M) exp M =exp(M )exp M =exp(−M)exp M = I n .
Similarly, the exponential of a Hermitian matrix is Hermitian positive
definite, because
∗
1 1
exp M = exp M exp M.
2 2