1. Ch 7.7: Fundamental Matrices
Suppose that x(1)
(t),…, x(n)
(t) form a fundamental set of solutions
for x' = P(t)x on α < t < β.
The matrix
whose columns are x(1)
(t),…, x(n)
(t), is a fundamental matrix for
the system x' = P(t)x. This matrix is nonsingular since its
columns are linearly independent, and hence detΨ ≠ 0.
Note also that since x(1)
(t),…, x(n)
(t) are solutions of x' = P(t)x,
Ψ satisfies the matrix differential equation Ψ' = P(t)Ψ.
,
)()(
)()(
)(
)()1(
)(
1
)1(
1
=
txtx
txtx
t
n
nn
n
Ψ
2. Example 1:
Consider the homogeneous equation x' = Ax below.
In Chapter 7.5, we found the following fundamental
solutions for this system:
Thus a fundamental matrix for this system is
xx
=′
14
11
tt
etet −
−
=
=
2
1
)(,
2
1
)( )2(3)1(
xx
−
= −
−
tt
tt
ee
ee
t
22
)( 3
3
Ψ
3. Fundamental Matrices and General Solution
The general solution of x' = P(t)x
can be expressed x = Ψ(t)c, where c is a constant vector with
components c1,…, cn:
)()1(
1 )( n
nctc xxx ++=
==
n
n
nn
n
c
c
txtx
txtx
t
1
)()1(
)(
1
)1(
1
)()(
)()(
)( cΨx
4. Fundamental Matrix & Initial Value Problem
Consider an initial value problem
x' = P(t)x, x(t0) = x0
where α < t0 < β and x0
is a given initial vector.
Now the solution has the form x = Ψ(t)c, hence we choose c
so as to satisfy x(t0) = x0
.
Recalling Ψ(t0) is nonsingular, it follows that
Thus our solution x = Ψ(t)c can be expressed as
0
0
10
0 )()( xΨcxcΨ tt −
=⇒=
0
0
1
)()( xΨΨx tt −
=
5. Recall: Theorem 7.4.4
Let
Let x(1)
,…, x(n)
be solutions of x' = P(t)x on I: α < t < β that
satisfy the initial conditions
Then x(1)
,…, x(n)
are fundamental solutions of x' = P(t)x.
=
=
=
1
0
0
0
,,
0
0
1
0
,
0
0
0
1
)()2()1(
n
eee
βα <<== 0
)(
0
)()1(
0
)1(
,)(,,)( ttt nn
exex
6. Fundamental Matrix & Theorem 7.4.4
Suppose x(1)
(t),…, x(n)
(t) form the fundamental solutions given
by Thm 7.4.4. Denote the corresponding fundamental matrix
by Φ(t). Then columns of Φ(t) are x(1)
(t),…, x(n)
(t), and hence
Thus Φ-1
(t0) = I, and the hence general solution to the
corresponding initial value problem is
It follows that for any fundamental matrix Ψ(t),
I=
=Φ
100
010
001
)( 0
t
00
0
1
)()()( xΦxΦΦx ttt == −
)()()()()()( 0
100
0
1
tttttt −−
=⇒== ΨΨΦxΦxΨΨx
7. The Fundamental Matrix Φ
and Varying Initial Conditions
Thus when using the fundamental matrix Φ(t), the general
solution to an IVP is
This representation is useful if same system is to be solved for
many different initial conditions, such as a physical system
that can be started from many different initial states.
Also, once Φ(t) has been determined, the solution to each set
of initial conditions can be found by matrix multiplication, as
indicated by the equation above.
Thus Φ(t) represents a linear transformation of the initial
conditions x0
into the solution x(t) at time t.
00
0
1
)()()( xΦxΦΦx ttt == −
8. Example 2: Find Φ(t) for 2 x 2 System (1 of 5)
Find Φ(t) such that Φ(0) = I for the system below.
Solution: First, we must obtain x(1)
(t) and x(2)
(t) such that
We know from previous results that the general solution is
Every solution can be expressed in terms of the general
solution, and we use this fact to find x(1)
(t) and x(2)
(t).
xx
=′
14
11
tt
ecec −
−
+
=
2
1
2
1
2
3
1x
=
=
1
0
)0(,
0
1
)0( )2()1(
xx
9. Example 2: Use General Solution (2 of 5)
Thus, to find x(1)
(t), express it terms of the general solution
and then find the coefficients c1 and c2.
To do so, use the initial conditions to obtain
or equivalently,
tt
ecect −
−
+
=
2
1
2
1
)( 2
3
1
)1(
x
=
−
+
=
0
1
2
1
2
1
)0( 21
)1(
ccx
=
− 0
1
22
11
2
1
c
c
12. Example 2: Obtain Φ(t) (5 of 5)
The columns of Φ(t) are given by x(1)
(t) and x(2)
(t), and thus
from the previous slide we have
Note Φ(t) is more complicated than Ψ(t) found in Ex 1.
However, now that we have Φ(t), it is much easier to
determine the solution to any set of initial conditions.
+−
−+
=
−−
−−
tttt
tttt
eeee
eeee
t
2
1
2
1
4
1
4
1
2
1
2
1
)(
33
33
Φ
−
= −
−
tt
tt
ee
ee
t
22
)( 3
3
Ψ
13. Matrix Exponential Functions
Consider the following two cases:
The solution to x' = ax, x(0) = x0, is x = x0eat
, where e0
= 1.
The solution to x' = Ax, x(0) = x0
, is x = Φ(t)x0
, where Φ(0) = I.
Comparing the form and solution for both of these cases, we
might expect Φ(t) to have an exponential character.
Indeed, it can be shown that Φ(t) = eAt
, where
is a well defined matrix function that has all the usual
properties of an exponential function. See text for details.
Thus the solution to x' = Ax, x(0) = x0
, is x = eAt
x0
.
∑∑
∞
=
∞
=
+==
10 !! n
nn
n
nn
t
n
t
n
t
e
A
I
AA
14. Example 3: Matrix Exponential Function
Consider the diagonal matrix A below.
Then
In general,
Thus
=
== ∑∑
∞
=
∞
=
t
t
n
n
n
n
nn
t
e
e
t
n
n
n
t
e 2
00 0
0
!/20
0!/1
!
AA
=
20
01
A
,
20
01
20
01
20
01
,
20
01
20
01
20
01
32
3
2
2
=
=
=
= AA
= n
n
20
01
A
15. Coupled Systems of Equations
Recall that our constant coefficient homogeneous system
written as x' = Ax with
is a system of coupled equations that must be solved
simultaneously to find all the unknown variables.
,2211
12121111
nnnnnn
nn
xaxaxax
xaxaxax
+++=′
+++=′
,,
)(
)(
)(
1
1111
=
=
nnn
n
n aa
aa
tx
tx
t
Ax
16. Uncoupled Systems & Diagonal Matrices
In contrast, if each equation had only one variable, solved for
independently of other equations, then task would be easier.
In this case our system would have the form
or x' = Dx, where D is a diagonal matrix:
,00
00
00
21
21112
21111
nnnn
n
n
xdxxx
xxdxx
xxxdx
+++=′
+++=′
+++=′
=
=
nn
n
d
d
d
tx
tx
t
00
00
00
,
)(
)(
)( 22
11
1
Dx
17. Uncoupling: Transform Matrix T
In order to explore transforming our given system x' = Ax of
coupled equations into an uncoupled system x' = Dx, where D
is a diagonal matrix, we will use the eigenvectors of A.
Suppose A is n x n with n linearly independent eigenvectors
ξ(1)
,…, ξ(n)
, and corresponding eigenvalues λ1,…, λn.
Define n x n matrices T and D using the eigenvalues &
eigenvectors of A:
Note that T is nonsingular, and hence T-1
exists.
=
=
n
n
nn
n
λ
λ
λ
ξξ
ξξ
00
00
00
, 2
1
)()1(
)(
1
)1(
1
DT
18. Uncoupling: T-1
AT = D
Recall here the definitions of A, T and D:
Then the columns of AT are Aξ(1)
,…, Aξ(n)
, and hence
It follows that T-1
AT = D.
=
=
=
n
n
nn
n
nnn
n
aa
aa
λ
λ
λ
ξξ
ξξ
00
00
00
,, 2
1
)()1(
)(
1
)1(
1
1
111
DTA
TDAT =
=
)()1(
1
)(
1
)1(
11
n
nnn
n
n
ξλξλ
ξλξλ
19. Similarity Transformations
Thus, if the eigenvalues and eigenvectors of A are known,
then A can be transformed into a diagonal matrix D, with
T-1
AT = D
This process is known as a similarity transformation, and A
is said to be similar to D. Alternatively, we could say that A
is diagonalizable.
=
=
=
n
n
nn
n
nnn
n
aa
aa
λ
λ
λ
ξξ
ξξ
00
00
00
,, 2
1
)()1(
)(
1
)1(
1
1
111
DTA
20. Similarity Transformations: Hermitian Case
Recall: Our similarity transformation of A has the form
T-1
AT = D
where D is diagonal and columns of T are eigenvectors of A.
If A is Hermitian, then A has n linearly independent
orthogonal eigenvectors ξ(1)
,…, ξ(n)
, normalized so that
(ξ(i)
, ξ(i)
) =1 for i = 1,…, n, and (ξ(i)
, ξ(k)
) = 0 for i ≠ k.
With this selection of eigenvectors, it can be shown that
T-1
= T*
. In this case we can write our similarity transform as
T*
AT = D
21. Nondiagonalizable A
Finally, if A is n x n with fewer than n linearly independent
eigenvectors, then there is no matrix T such that T-1
AT = D.
In this case, A is not similar to a diagonal matrix and A is not
diagonlizable.
=
=
=
n
n
nn
n
nnn
n
aa
aa
λ
λ
λ
ξξ
ξξ
00
00
00
,, 2
1
)()1(
)(
1
)1(
1
1
111
DTA
22. Example 4:
Find Transformation Matrix T (1 of 2)
For the matrix A below, find the similarity transformation
matrix T and show that A can be diagonalized.
We already know that the eigenvalues are λ1 = 3, λ2 = -1
with corresponding eigenvectors
Thus
=
14
11
A
−
=
=
2
1
)(,
2
1
)( )2()1(
tt ξξ
−
=
−
=
10
03
,
22
11
DT
23. Example 4: Similarity Transformation (2 of 2)
To find T-1
, augment the identity to T and row reduce:
Then
Thus A is similar to D, and hence A is diagonalizable.
−
=→
−
→
−
→
−−
→
−
−
4/12/1
4/12/1
4/12/110
4/12/101
4/12/110
0111
1240
0111
1022
0111
1
T
D
ATT
=
−
=
−
−
=
−
−
=−
10
03
26
13
4/12/1
4/12/1
22
11
14
11
4/12/1
4/12/11
24. Fundamental Matrices for Similar Systems (1 of 3)
Recall our original system of differential equations x' = Ax.
If A is n x n with n linearly independent eigenvectors, then A
is diagonalizable. The eigenvectors form the columns of the
nonsingular transform matrix T, and the eigenvalues are the
corresponding nonzero entries in the diagonal matrix D.
Suppose x satisfies x' = Ax, let y be the n x 1 vector such that
x = Ty. That is, let y be defined by y = T-1
x.
Since x' = Ax and T is a constant matrix, we have Ty' = ATy,
and hence y' = T-1
ATy = Dy.
Therefore y satisfies y' = Dy, the system similar to x' = Ax.
Both of these systems have fundamental matrices, which we
examine next.
25. Fundamental Matrix for Diagonal System (2 of 3)
A fundamental matrix for y' = Dy is given by Q(t) = eDt
.
Recalling the definition of eDt
, we have
( )
( )
=
=
==
∑
∑
∑∑ ∞
=
∞
=∞
=
∞
=
t
t
n
n
n
n
n
n
n
n
n
n
n
nn
n
e
e
n
t
n
t
n
t
n
t
t
λ
λ
λ
λ
λ
λ
00
00
00
!
00
00
00
!
!
00
00
00
!
)(
1
0
0
1
0
1
0
D
Q
26. Fundamental Matrix for Original System (3 of 3)
To obtain a fundamental matrix Ψ(t) for x' = Ax, recall that
the columns of Ψ(t) consist of fundamental solutions x
satisfying x' = Ax. We also know x = Ty, and hence it follows
that
The columns of Ψ(t) given the expected fundamental solutions
of x' = Ax.
=
==
tn
n
t
n
tnt
t
t
n
nn
n
n
n
n
ee
ee
e
e
λλ
λλ
λ
λ
ξξ
ξξ
ξξ
ξξ
)()1(
)(
1
)1(
1
)()1(
)(
1
)1(
1
1
11
00
00
00
TQΨ
27. Example 5:
Fundamental Matrices for Similar Systems
We now use the analysis and results of the last few slides.
Applying the transformation x = Ty to x' = Ax below, this
system becomes y' = T-1
ATy = Dy:
A fundamental matrix for y' = Dy is given by Q(t) = eDt
:
Thus a fundamental matrix Ψ(t) for x' = Ax is
yyxx
−
=′⇒
=′
10
03
14
11
= −t
t
e
e
t
0
0
)(
3
Q
−
=
−
== −
−
− tt
tt
t
t
ee
ee
e
e
t
220
0
22
11
)( 3
33
TQΨ
