International Journal of Emerging Technology and Advanced Engineering
Website: www.ijetae.com (ISSN 2250-2459, ISO 9001:2008 Certified Journal, Volume 6, Issue 7, July 2016)
258
Derivation of an Explicit Function in Non-Singular Hamilton
Symmetric Matrices of Rank 1 Via Linear Quadratics Inverse
Eigenvalue Problem
Oladejo N. K
1, Anang C. R
2University for Development Studies, Navrongo, Ghana
Abstract-- This paper deals with derivation of an explicit function in a non-singular Hamilton symmetric matrices of Rank 1 via linear quadratics inverse eigenvalue problem (LQIEP in the neighborhood of the first type of Hamilton matrices through numerical illustration and examples.
Keywords: Explicit, functions, Hamilton, Symmetric,
Non-singular, Neighborhood,
I. INTRODUCTION
Various solvability and solubility of the inverse eigenvalue problem for Hamilton matrices together with numerical examples are systematically reviewed under certain singular and non-singular Hamilton matrices by Oduro et al (2012) and Oduro (2012a, b),Baah Gyamfi (2013) as well as Oladejo et al (2014) and Oladejo et.al (2015).This paper deals with derivation of an explicit function in a non-singular Hamilton symmetric matrices of Rank 1 via linear quadratics inverse eigenvalue problem (LQIEP in the neighborhood of the first type of Hamilton matrices through numerical illustration and examples.
The linear-quadratic optimal control (LQOC)
Given that
0
:
,
,
nn nm nn TQ
Q
Q
B
A
and0
:
mm TR
R
R
we consider thelinear-quadratic optimal control (LQOC) for the functional;
(
)
(
)
(
)
(
)
1
2
1
)
(
f
i t
t
T T
i
x
u
x
t
Qx
t
u
t
Ru
t
dt
I
Subject to differential equation
,
,
(
)
2
),
(
)
(
)
(
t
Ax
t
Bu
t
t
t
it
fx
t
ix
ix
Then the Hamiltonian function is given by:
3
2
1
,
,
,
x
u
t
x
Qx
u
Ru
p
Ax
Bu
p
H
T
T
T
Given any optimal input
u
and the correspondingstate
x
we solve equation (3) which is LQOCP arising from the Pontryagin minimum principle in equation (3)Thus:
(
(
),
(
),
(
),
)
0
t
x
t
u
t
t
p
u
H
4
0
)
(
)
(
u
t
TR
p
t
TB
Then
u
(
t
)
R
1B
Tp
(
t
)
5
and the adjoint equation is given as:
,
,
(
)
0
6
),
(
),
(
),
(
),
(
f f
i
T
t
p
t
t
t
t
p
t
t
u
t
k
t
p
x
H
,
,
(
)
0
7
),
(
)
(
)
(
f f
i
T T T
t
p
t
t
t
t
p
A
t
p
Q
t
x
We then have;
8
0
)
(
,
,
),
(
)
(
)
(
f
f i T
t
p
t
t
t
t
Qx
t
p
A
t
p
Consequently;
9
0
)
(
,
)
(
,
,
,
)
(
)
(
)
(
)
(
1
f i i
f i T
T
t
p
x
t
x
t
t
t
t
p
t
x
A
Q
B
BR
A
t
p
t
x
dt
d
Equation (9) is a linear, time variant differential equation
International Journal of Emerging Technology and Advanced Engineering
Website: www.ijetae.com (ISSN 2250-2459, ISO 9001:2008 Certified Journal, Volume 6, Issue 7, July 2016)
259
IEP for2
n
2
n
Hamiltonian Matrix of Rank 1 in respect of LQOCPFrom equation (9) and the theory of the Linear quadratic optimal control we have the following (Hamilton’s Equations):
10
0
)
(
,
)
(
,
,
,
)
(
)
(
)
(
)
(
1
f i i f i T Tt
p
x
t
x
t
t
t
t
p
t
x
A
Q
B
BR
A
t
p
t
x
dt
d
We then consider the case where the Hamiltonian matrix
T TA
Q
B
BR
A
H
1Is
2
n
2
n
so that A, Q, BR-1BT are all is
2
2
sub-matrices of H.Inverse Eigenvalue Problem for a Non-Singular
4
4
symmetric matrix Newton’s Method
We construct a Characteristic (Polynomial) function of the diagonal elements from the matrix
2
n
2
n
44 43 42 41 34 33 32 31 24 23 22 21 14 13 12 11 44 43 42 41 34 33 32 31 24 23 22 21 14 13 12 11 a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a A
) 0 ( 44 ) 0 ( 33 ) 0 ( 22 ) 0 ( 11 ) 0 (a
a
a
a
X
In other words, consider the function with independent variables defined on 4 selected elements of matrix
A
, precisely, the diagonal elements:A
trA
a
a
a
a
f
(
11,
22,
33,
44)
2
(
)
det
33 42 21 14 32 43 21 14 42 34 21 13 44 32 21 13 43 34 21 12 44 33 21 12 43 34 22 11 44 33 22 11 42 21 14 32 21 13 44 21 12 33 21 12 43 34 11 44 33 22 43 33 11 44 33 11 44 21 11 33 22 11 2 21 12 43 34 44 33 44 22 33 22 44 11 33 11 22 11 3 44 33 22 11 4 ) ( ) ( ) ( a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a
Thus, given 4 distinct eigenvalues
1,
2,
3,
4 we have the following four (4) functions with 4 independent variables being the diagonal element ofA
.International Journal of Emerging Technology and Advanced Engineering
Website: www.ijetae.com (ISSN 2250-2459, ISO 9001:2008 Certified Journal, Volume 6, Issue 7, July 2016)
260
33 42 21 14 32 43 21 14 42 34 21 13 44 32 21 13 43 34 21 12 44 33 21 12 43 34 22 11 44 33 22 11 4 42 21 14 32 21 13 44 21 12 33 21 12 43 34 11 44 33 22 43 33 11 44 33 11 44 21 11 33 22 11 4 2 21 12 43 34 44 33 44 22 33 22 44 11 33 11 22 11 4 3 44 33 22 11 4 4 44 33 22 11 4)
(
)
(
)
(
)
,
,
,
(
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
a
f
Derivation of an Explicit Formula for the Jacobian in the
4
4
matrices case
22 33 44
1 22 33 33 44 22 44
1 2 1 3 11 1a
a
a
a
a
a
a
a
a
a
f
31 21
11 33 44
1 11 33 33 44 11 44
22 1
a
a
a
a
a
a
a
a
a
a
f
11 22 44
1 11 22 22 44 11 44
1 2 1 3 33 1a
a
a
a
a
a
a
a
a
a
f
31 21
11 22 33
1 11 22 22 33 11 33
44 1
a
a
a
a
a
a
a
a
a
a
f
22 33 44
2 22 33 33 44 22 44
2 2 2 3 11 2a
a
a
a
a
a
a
a
a
a
f
32 22
11 33 44
2 11 33 33 44 11 44
22 2
a
a
a
a
a
a
a
a
a
a
f
11 22 44
2 11 22 22 44 11 44
2 2 2 3 33 2a
a
a
a
a
a
a
a
a
a
f
32 22
11 22 33
2 11 22 22 33 11 33
44 2
a
a
a
a
a
a
a
a
a
a
f
22 33 44
3 22 33 33 44 22 44
3 2 3 3 11 3a
a
a
a
a
a
a
a
a
a
f
33 23
11 33 44
3 11 33 33 44 11 44
22 3a
a
a
a
a
a
a
a
a
a
f
11 22 44
3 11 22 22 44 11 44
3 2 3 3 33 3a
a
a
a
a
a
a
a
a
a
f
33 23
11 22 33
3 11 22 22 33 11 33
44 3
a
a
a
a
a
a
a
a
a
a
f
22 33 44
4 22 33 33 44 22 44
4 2 4 3 11 4a
a
a
a
a
a
a
a
a
a
f
34 24
11 33 44
4 11 33 33 44 11 44
22 4a
a
a
a
a
a
a
a
a
a
f
11 22 33
4 11 22 22 33 11 33
4 2 4 3 44 4a
a
a
a
a
a
a
a
a
a
f
34 24
11 22 44
4 11 22 22 44 11 44
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261
Thus
Computing the Determinant we get
1a
22
2a
11
1a
11
2a
22
trA
1 2
22 11
.
a
a
Det
And the inverse of the Jacobian matrix is given as:
2 11 2 22
1 11 1 22
11 22 2 1
1
1
a
a
a
a
a
a
J
Note that the existence of the inverse of the Jacobian matrix requires that both the target eigenvalues and the diagonal elements of the starting matrix be distinct.
While the (n+1) th iteration of the Newton’s method is given by the following recursive relation
)
(
)
(
( ) ( )1 ) ( ) 1
(n n n n
X
f
X
J
X
X
Under these conditions, the first step of Newton’s method is given
) 0 ( 22
) 0 ( 11
2 ) 0 ( 11 2 ) 0 ( 22
1 ) 0 ( 11 1 ) 0 ( 22 )
0 ( 22
) 0 ( 11
) 1 ( 22
) 1 ( 11
1
a
a
a
a
a
a
Det
a
a
a
a
Numerical Examples
Given that the IEP to be solved is to determine completely the nonsingular symmetric coefficient of
n
n
2
2
matrix of the system4 44 3
2 1 4
4 3 33 2 1 3
4 3 2 22 1 2
4 3 2 1 11 1
16
12
8
4
12
9
6
3
8
6
4
2
4
3
2
x
a
x
x
x
x
x
x
a
x
x
x
x
x
x
a
x
x
x
x
x
x
a
x
A
16
12
8
4
12
9
6
3
8
6
4
2
4
3
2
1
For Non Singular
2
2
symmetric matrices case2 22 1 2
2 1 11 1
4
2
2
x
a
x
x
x
x
a
x
Given the eigenvalues
1
1
,
2
3
. i.e. we let thetarget solution of the form
t t
ve
c
ue
c
x
1
2 3Assuming the initializing matrix
4
2
2
1
;
4
1
) 0 (
X
Moreover,
A
(0) is singularwith
k
2
,
a
11
1
,5
) 0
(
trA
Computing the values of the functions at the initial point:
f
1
a
11,
a
22
12
a
11
a
22
1
a
11a
22
a
122
2
12 22 11 2 22 11 2 2 22 11
2
a
,
a
a
a
a
a
a
f
6
6
)
(
X
(0)f
The inverse of the Jacobian matrix yields:
5
1
2
2
12
1
5
1
2
2
1
1
Det
J
Substituting into the Newton’s equation,
)
(
)
(
( ) ( ) 1) ( ) 1
(n n n n
X
f
X
J
X
X
we get:
6
6
5
1
2
2
12
1
4
1
) 0 (
X
1
1
3
0
4
1
1
1
) 1 (
X
Thus:
1
2
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262
Non Singular2
n
2
n
Symmetric Matrices caseSuppose the IEP to be solved is to determine completely the nonsingular
4
4
symmetric coefficient matrix of the system4 44 3
2 1 4
4 3 33 2 1 3
4 3 2 22 1 2
4 3 2 1 11 1
16
12
8
4
12
9
6
3
8
6
4
2
4
3
2
x
a
x
x
x
x
x
x
a
x
x
x
x
x
x
a
x
x
x
x
x
x
a
x
Given the eigenvalues
1
1
,
2
1
,
3
3
,
4
5
i.e. We let the target solution of the formt t
t
we
c
ve
c
ue
c
x
1
2 2
3 5Given the initializing matrix A=
16
12
8
4
12
9
6
3
8
6
4
2
4
3
2
1
;
16
9
4
1
) 0 (
X
Here,
A
(0) is singular while1
,
4
,
3
,
2
2 3 111
k
k
a
k
and
trA
(0)
30
We compute the values of the functions at the initial point:
Thus;
112
10
28
28
) 0 (
X
f
and
30
20
20
5
6
12
6
3
6
12
8
35
6
12
8
11
J
Estimating determinant and substituting into the Newton’s equation yielded
112
10
28
28
30
6
6
6
20
12
12
12
20
6
8
8
5
3
35
11
23040
1
16
9
4
1
) 1 (
X
1578
.
0
0733
.
0
0803
.
0
0672
.
0
16
9
4
1
) 1 (
X
8422
.
15
0733
.
9
0803
.
4
0672
.
1
Hence,
842
.
15
12
8
4
12
073
.
9
6
3
8
4
080
.
4
2
4
3
2
067
.
1
)
(
X
(1)A
II. CONCLUSION
Various theoretical results have been systematically reviewed and discussed in respect of the inverse eigenvalue problem (IEP). Based on these results we developed the derivation of an explicit function in non-singular Hamilton symmetric matrices of Rank 1 via linear quadratics inverse eigenvalue problem (LQIEP in the neighborhood of the first type of Hamilton matrices through numerical illustration and examples.
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Website: www.ijetae.com (ISSN 2250-2459, ISO 9001:2008 Certified Journal, Volume 6, Issue 7, July 2016)
263
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