vector
fields
are
always
complete.
(vi)
The
one-parameter
group
of
diffeomorphism
X
8
associated
with
the
left
invariant
vector
field
X
is
given
by
x.
= r exp
(8)
•
Theorem
2.6.6
Let~:
H
~
G
be
a
homomorphism,
then
the
following
diagram
commutes:
H - •
~
G
exp
t t
exp
h -
~
~
g
In
the
case
of
groups
of
matrices,
we
shall
see
that
the
exponential
map
exp
:
gl(n,R)
~
GL(n,R)
for
the
general
linear
group
is
given
by
exponentiation
of
matrices.
Let
I
(rather
than
e)
denote
the
identity
matrix
in
GL(n,R),
let
eA
=I+
A+
A
2
/2!
+
•.•
+ Ak/k! +
••..
(2.5-1)
for
A E
gl(n,R).
We
want
to
show
that
the
right
hand
side
of
the
series
converges.
In
fact,
the
series
converges
uniformly
for
A
in
a
bounded
region
of
gl(n,R).
For
a
given
bounded
region
n
of
gl(n,R),
there
is
a~>
o
such
that
for
any
matrix
A
inn,
IAiil 5
~
for
each
element
(or
component)
of
A.
It
follows
by
induction
that
I
(Ak)
ij
I 5
n<k-l>~k.
Then
by
the
Weierstrass
M-
test,
each
of
the
series
of
components
'Ek=O"'
(Ak);/k!,
(1
5
i,
j 5 n)
(2.5-2)
converges
uniformly
for
A
in
n.
Thus
the
series
in
(2.5.1)
converges
uniformly
for
A
in
n.
Let
Sk(A)
be
the
k-th
partial
sum
of
the
series
(2.5.1),
i.e.,
Sk(A)=
I+
A+
A
2
/2!
+
••••
+ Ak/k!
(2.5-3)
and
let
B,C
E
gl(n,R).
Since
c
~
BC
is
a
continuous
map
of
gl(n,R)
into
itself,
it
follows
that
B (limk..., Sk(A)) = limk...,
(BSk(A)).
In
particular,
if
B
is
non-singular,
then
B (limk..., Sk(A))
B-
1
= limk..., BSk(A)B-
1
•
Because
BAkB-
1
=
(BAB-
1
)k,
it
then
follows
that
BeAB-
1
= e
8
A
8
•
From
this,
it
is
easy
to
show
that
if
the
eigenvalues
of
A
are
A.
1
, • • •
A.n,
then
the
eigenvalues
of
eA
are
e
1
•
,
••
, e
1
'!
e
1
•
is
an
eigenvalue
of
eA
follows
easily
if
the
upper
left
81