The matrix exponential is a powerful means for representing the
solution to nn linear, constant
coefficient, differential equations. The initial value problem
for such a system may be written
x′t=Axt
x
t
A
x
t
x0=
x0
x
0
x0
where AA is the n-by-n matrix of
coefficients. By analogy to the 1-by-1 case we might expect
to hold. Our expectations are granted if we properly define
eAt
A
t
. Do you see why simply exponentiating each element of
At will no suffice?
There are at least 4 distinct (but of course equivalent)
approaches to properly defining
eAt
A
t
. The first two are natural analogs of the single
variable case while the latter two make use of heavier matrix
algebra machinery.
Please visit each of these modules to see the definition and a
number of examples.
For a concrete application of these methods to a real dynamical
system, please visit the
Mass-Spring-Damper module.
Regardless of the approach, the matrix
exponential may be shown to obey the 3 lovely properties
-
ddteAt=AeAt=eAtA
t
A
t
A
A
t
A
t
A
-
eA(
t1
+
t2
)=eA
t1
eA
t2
A
t1
t2
A
t1
A
t2
-
eAt
A
t
is nonsingular and
eAt-1=e−(At)
A
t
A
t
Let us confirm each of these on the suite of examples used in
the submodules.
If
A=(
10
02
)
A
1
0
0
2
then
eAt=(
et0
0e2t
)
A
t
t
0
0
2
t
-
ddteAt=(
et0
02e2t
)=(
10
02
)(
et0
0e2t
)
t
A
t
t
0
0
2
2
t
1
0
0
2
t
0
0
2
t
-
(
e
t1
+
t2
0
0e2
t1
+2
t2
)=(
e
t1
e
t2
0
0e2
t1
e2
t2
)=(
e
t1
0
0e2
t1
)(
e
t2
0
0e2
t2
)
t1
t2
0
0
2
t1
2
t2
t1
t2
0
0
2
t1
2
t2
t1
0
0
2
t1
t2
0
0
2
t2
-
eAt-1=(
e−t0
0e−(2t)
)=e−(At)
A
t
t
0
0
2
t
A
t
If
A=(
01
-10
)
A
0
1
-1
0
then
eAt=(
costsint
−sintcost
)
A
t
t
t
t
t
-
ddteAt=(
−sintcost
−cost−sint
)
t
A
t
t
t
t
t
and
AeAt=(
−sintcost
−cost−sint
)
A
A
t
t
t
t
t
-
You will recognize this statement as a basic trig
identity
(
cos
t1
+
t2
sin
t1
+
t2
−sin
t1
+
t2
cos
t1
+
t2
)=(
cos
t1
sin
t1
−sin
t1
cos
t1
)(
cos
t2
sin
t2
−sin
t2
cos
t2
)
t1
t2
t1
t2
t1
t2
t1
t2
t1
t1
t1
t1
t2
t2
t2
t2
-
Using the cofactor expansion we find
eAt-1=(
cost−sint
sintcost
)=(
cos−t−sin−t
sin−tcos−t
)=e−(At)
A
t
t
t
t
t
t
t
t
t
A
t
If
A=(
01
00
)
A
0
1
0
0
then
eAt=(
1t
01
)
A
t
1
t
0
1
-
ddteAt=(
01
00
)=AeAt
t
A
t
0
1
0
0
A
A
t
-
(
1
t1
+
t2
01
)=(
1
t1
01
)(
1
t2
01
)
1
t1
t2
0
1
1
t1
0
1
1
t2
0
1
-
(
1t
01
)-1=(
1−t
01
)=e−(At)
1
t
0
1
1
t
0
1
A
t
