John Stillwell - Naive Lie Theory.pdf - Index of

102 5 The tangent space
Exercises
Another proof of the crucial result det(e A )=e Tr(A) uses less linear algebra but
more calculus. It goes as follows (if you need help with the details, see Tapp
[2005], p. 72 and p. 88).
Suppose B(t) isasmoothpathofn × n complex matrices with B(0)=1, let
b ij (t) denote the entry in row i and column j of B(t), andletB ij (t) denote the
result of omitting row i and column j.
5.3.1 Show that
and hence
d
dt ∣ det(B(t)) =
t=0
n
∑
j=1
det(B(t)) =
n
∑
j=1
(−1) j+1 b 1 j (t)det(B 1 j (t)),
[
(−1) j+1 b ′ 1 j(0)det(B 1 j (0)) + b 1 j (0) d ]
dt ∣ det(B 1 j (t)) .
t=0
5.3.2 Deduce from Exercise 5.3.1, and the assumption B(0)=1,that
d
dt ∣ det(B(t)) = b ′ 11(0)+ d t=0
dt ∣ det(B 11 (t)).
t=0
5.3.3 Deduce from Exercise 5.3.2, and induction, that
d
dt ∣ det(B(t)) = b ′ 11 (0)+b′ 22 (0)+···+ b′ nn (0)=Tr(B′ (0)).
t=0
We now apply Exercise 5.3.3 to the smooth path B(t)=e tA ,forwhichB ′ (0)=A,
and the smooth real function
f (t)=det(e tA ), for which f (0)=1.
By the definition of derivative,
f ′ 1
[
]
(t)=lim det(e (t+h)A ) − det(e tA ) .
h→0 h
5.3.4 Using the property det(MN)=det(M)det(N) and Exercise 5.3.3, show that
f ′ (t)=det(e tA ) d dt ∣ det(e tA )= f (t)Tr(A).
t=0
5.3.5 Solve the equation for f (t) in Exercise 5.3.4 by setting f (t) =g(t)e t·Tr(A)
and showing that g ′ (t)=0, hence g(t)=1. (Why?)
Conclude that det(e A )=e Tr(A) .