You may recall from Calculus that for any numbers
aa and
tt one may achieve
eat
a
t
via
eat=limit k→∞1+atkk
a
t
k
1
a
t
k
k
(1)
The natural matrix definition is therefore
eAt=limit k→∞I+Atkk
A
t
k
I
A
t
k
k
(2)
where
II is the n-by-n identity
matrix.
The easiest case is the diagonal case, e.g.,
A=(
10
02
)
A
1
0
0
2
for then
I+Atkk=(
1+tkk0
01+2tkk
)
I
A
t
k
k
1
t
k
k
0
0
1
2
t
k
k
and so (recalling Equation 1 above)
eAt=(
et0
0e2t
)
A
t
t
0
0
2
t
Note that this is NOT the exponential of each
element of AA.
As a concrete example let us suppose
A=(
01
-10
)
A
0
1
-1
0
From
I+At=(
1t
−t1
)
I
A
t
1
t
t
1
I+At22=(
1t2
−t21
)(
1t2
−t21
)=(
1−t24t
−t1−t24
)
I
A
t
2
2
1
t
2
t
2
1
1
t
2
t
2
1
1
t
2
4
t
t
1
t
2
4
I+At33=(
1−t23t−t327
−t+t3271−t23
)
I
A
t
3
3
1
t
2
3
t
t
3
27
t
t
3
27
1
t
2
3
I+At44=(
-3t28+t4256+1t−t316
−t+t316-3t28+t4256+1
)
I
A
t
4
4
-3
t
2
8
t
4
256
1
t
t
3
16
t
t
3
16
-3
t
2
8
t
4
256
1
I+At55=(
-2t25+t4125+1t−2t325+t53125
−t+2t325−t53125-2t25+t4125+1
)
I
A
t
5
5
-2
t
2
5
t
4
125
1
t
2
t
3
25
t
5
3125
t
2
t
3
25
t
5
3125
-2
t
2
5
t
4
125
1
We discern a pattern: the diagonal elements are equal even
polynomials while the off diagonal elements are equal but
opposite odd polynomials. The degree of the polynomial will
grow with kk and in the limit we
'recognize'
eAt=(
costsint
−sintcost
)
A
t
t
t
t
t
If
A=(
01
00
)
A
0
1
0
0
then
I+Atkk=(
1t
01
)
I
A
t
k
k
1
t
0
1
for each value of kk and so
eAt=(
1t
01
)
A
t
1
t
0
1
