2
MA 36600 LECTURE NOTES: WEDNESDAY, APRIL 22
Now we return to the general case. Consider an
n
×
n
matrix
A
. Say that there exists a nonsingular
n
×
n
matrix
T
such that
T
−
1
A T
=
D
is an
n
×
n
diagonal matrix. Such a matrix
A
is said to be
diagonalizable
.
We will show that
exp (
A
t
) =
T
exp (
D
t
)
T
−
1
.
In practice this is easy to compute because both
T
−
1
and exp (
D
t
) are easy to compute.
Observe that
T
−
1
A
k
T
=
D
k
for any nonnegative integer
k
:
k
= 0:
T
−
1
A
0
T
=
T
−
1
I T
=
I
=
D
0
k
= 1:
T
−
1
A
1
T
=
D
1
k
≥
2:
T
−
1
A
k
T
=
T
−
1
A
k
−
1
T T
−
1
A T
=
T
−
1
A
k
−
1
T
T
−
1
A T
=
D
k
−
1
D
=
D
k
This gives the expression
T
−
1
exp (
A
t
)
T
=
∞
k
=0
T
−
1
A
k
T
t
k
k
!
=
∞
k
=0
D
k
t
k
k
!
= exp (
D
t
)
.
The claim follows.
Example.
Let us review the following homogeneous system considered in the previous lecture:
d
dt
x
=
1
1
4
1
x
.
We found that the general solution is
x
(
t
) =
c
1
1
2
e
3
t
+
c
2
1
−
2
e
−
t
=
Ψ
(
t
)
c
in terms of
Ψ
(
t
) =
e
3
t
e
−
t
2
e
3
t
−
2
e
−
t
,
c
=
c
1
c
2
.
We explain how to find the fundamental matrix
Ψ
(
t
) using exponentials of matrices.
Denote the 2
×
2 matrices
A
=
1
1
4
1
and
Ψ
0
=
1
1
2
−
2
.
The fundamental matrix above is the unique solution to the initial value problem
d
dt
Ψ
=
A
Ψ
,
Ψ
(0) =
Ψ
0
.
We know that this has the unique solution
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 Spring '09
 EdrayGoins
 Exponential Function, Exponentials, Matrix exponential, Exponential map

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