Since systems are often represented in terms of matrices and solutions
of system equations often make use of the exponential, it makes sense
to try and understand how these two concepts can be combined. In many
previous applications, we've seen terms like
eat
a
t
come in handy for talking about system behavior. Here, a
was always a scalar quantity. However, what would happen if the scalar aa was replaced by a matrix AA?
The result would be what is known as the matrix exponential.
Recall the definition of the scalar exponential:
eat=1+at1!+a2t22!+a3t33!+…
a
t
1
a
t
1
a
2
t
2
2
a
3
t
3
3
…
(1)
The definition of the matrix exponential is almost identical:
eAt=
I
n
+At1!+A2t22!+A3t33!+…
A
t
I
n
A
t
1
A
2
t
2
2
A
3
t
3
3
…
(2)
Where AA is nn x nn and
I
n
I
n
is the nn x nn
identity matrix. While it is nice to see the resemblance between
these two definitions, applying this infinite series does not turn
out to be very efficient in practice. However, it can be useful
in certain special cases.
Compute
eAt
A
t
where
A=(
01
-10
)
A
0
1
-1
0
.
We can start by taking powers of AA
so that we can use the formal definition.
A=(
01
-10
)
A
0
1
-1
0
(3)
A2=(
01
-10
)(
01
-10
)=(
-10
0-1
)=−I
A
2
0
1
-1
0
0
1
-1
0
-1
0
0
-1
I
(4)
A3=A2A=−A
A
3
A
2
A
A
(5)
A4=A2A2=I
A
4
A
2
A
2
I
(6)
A5=AA2=A
A
5
A
A
2
A
(7)
A6=A2A4=−I
A
6
A
2
A
4
I
(8)
And so the pattern goes, giving:
A4(n−1)+1=A
A
4
n
1
1
A
(9)
A4(n−1)+2=−I
A
4
n
1
2
I
(10)
A4(n−1)+3=−A
A
4
n
1
3
A
(11)
A4(n−1)+4=I
A
4
n
1
4
I
(12)
If we fill in the terms in the definition of
eat
a
t
,
we'll get the following matrix:
eAt=(
1−t22!+t44!−…t−t33!+t55!−…
−t+t33!−t55!+…1−t22!+t44!−…
)
A
t
1
t
2
2
t
4
4
…
t
t
3
3
t
5
5
…
t
t
3
3
t
5
5
…
1
t
2
2
t
4
4
…
(13)
We notice that the sums in this matrix look familiar-in fact,
they are the Taylor Series expansions of the sinusoids.
Therefore, the solution further reduces to:
eAt=(
costsint
−sintcost
)
A
t
t
t
t
t
(14)
The example above illustrates how the use of the true definition to
simplify matrix exponentials might only be easily applied in cases
with inherent repetition. There is a more general method involving
the Laplace Transform. In particular,
ℒeAt=sI−A-1
ℒ
A
t
s
I
A
(15)
We can verify that this is true by inserting the formal definition
of the matrix exponential:
ℒeAt=ℒI+At1!+A2t22!+…=1sI+1s2A+1s3A2+…=sI−A-1
ℒ
A
t
ℒ
I
A
t
1
A
2
t
2
2
…
1
s
I
1
s
2
A
1
s
3
A
2
…
s
I
A
(16)
The jump between the third and fourth equations here may be a
bit hard to believe, but this equality reduces to
I=I
I
I
when both sides are multiplied by
sI−A
s
I
A
.
Taking an inverse Laplace of each side of Laplace Transform of the equation we find an expression for the matrix exponential:
eAt=ℒ-1sI−A-1
A
t
ℒ
s
I
A
(17)
We can do the same example as before, this time using the
Laplace-based method.
A=(
01
-10
)
A
0
1
-1
0
(18)
sI−A-1=(
s-1
1s
)-1=1s2+1(
s1
-11s
)=(
ss2+11s2+1
-1s2+1ss2+1
)
s
I
A
s
-1
1
s
1
s
2
1
s
1
-11
s
s
s
2
1
1
s
2
1
-1
s
2
1
s
s
2
1
(19)
Taking the inverse laplace of this gives us
eAt=(
costsint
−sintcost
)
A
t
t
t
t
t
(20)
In the scalar case, a product of exponentials
eaeb
a
b
reduces to a single exponential whose power is the sum of the
individual exponents' powers,
ea+b
a
b
.
However, in the case of the matrix exponential, this is not
true. If AA and BB
are matrices,
eAeB≠eA+B
A
B
A
B
(21)
unless AA and BB
are commutative (i.e.
AB=BA
A
B
B
A
)
The derivative operates on the matrix exponential the same as
it does on the scalar exponential.
ddteAt=0+A+A2t1!+A3t22!+…=A(I+At1!+A2t22!+…)=AeAt
t
A
t
0
A
A
2
t
1
A
3
t
2
2
…
A
I
A
t
1
A
2
t
2
2
…
A
A
t
(22)