Inverse Eigenvalue Problem for Generalized Arrow-Like Matrices

Applied Mathematics, 2011, 2, 1443-1445 doi:10.4236/am.2011.212204 Published Online December 2011 (http://www.SciRP.org/journal/am)

Inverse Eigenvalue Problem for Generalized Arrow-Like Matrices Zhibin Li, Cong Bu, Hui Wang College of Mathematics, Dalian Jiaotong University, Dalian, China E-mail: [email protected] Received October 14, 2011; revised November 15, 2011; accepted November 23, 2011

Abstract This paper researches the following inverse eigenvalue problem for arrow-like matrices. Give two characteristic pairs, get a generalized arrow-like matrix, let the two characteristic pairs are the characteristic pairs of this generalized arrow-like matrix. The expression and an algorithm of the solution of the problem is given, and a numerical example is provided. Keywords: Generalized Arrow-Like Matrices, Characteristic Value, Inverse Problem, Unique

1. Introduction The Inverse eigenvalue problem for matrices in the problems involved in the field of structural design, pattern recognition, parameter recognition, automatic control and so on, it has a good engineering background, and its research has obvious significance [1]. Many experts and scholars have addressed more extensively and in-depth studied, get a lot of conclusions about inverse eigenvalue problem for Jacobi matrices [2], but there is less research about the inverse eigenvalue problem for arrow- like matrices [3,4]. This paper researches the following inverse eigenvalue problem for generalized arrow-like matrices. Generalized arrow-like matrices refer to the matrix as follows:  a1 b1  bm 1 bm    c a 2  1        c am  (1) J   m 1  cm am 1 bm 1   cm 1 am  2      bn 1      cn 1 an   When m  1 , J becomes generalized Jacobi matrix [1]; when m  n , J is an arrow-like matrice. This article studies the following characteristic value inverse: Question IEPGAM. Given two real numbers  ,  (   ) and two nonzero real vectors Copyright © 2011 SciRes.

x  ( x1 , x2 , , xn )T  R n , y  ( y1 , y2 , , yn )T  R n . Find the n  n real generalized arrow-like matrice J , such that Jx   x, Jy   y . The expression and an algorithm of the solution of the problem is given in Section 2, and a numerical example is provided in Section 3.

2. The Solution of Question IEPGAM Because ( , x) and (  , y ) are two characteristic pairs of the generalized arrow-like matrices J , In it, Let: x0  xn 1  y0  yn 1  b0  bn  c0  cn  0 ,

Di 

xi

yi

xi 1

yi 1

(3)

xi (i  2,3, , m  1) yi

(4)

a1 x1 +b1 x2    bm 1 xm  bm xm 1   x1 ,

(5-1)

c1 x1 +a2 x2   x2 ,

(5-2)

Ei 

x1 y1

(i  0,1, , n) ,

(2)

So

… cm 1 x1 +am xm   xm ,

（5-m） cm x1 +am 1 xm 1  bm 1 xm  2   xm 1 , (5-m+1) cm 1 xm 1 +am  2 xm  2  bm  2 xm  3   xm  2 , (5-m+2)

… AM

Z. B. LI

1444

cn  2 xn  2 +an 1 xn 1  bn 1 xn   xn 1 , cn 1 xn 1 +an xn   xn .

(5-n–1) (5-n)

a1 y1 +b1 y2    bm 1 ym  bm ym 1   y1 , (6-1) c1 y1 +a2 y2   y2 , … cm 1 y1  am ym   ym , cm y1  am 1 ym 1  bm 1 ym  2   ym 1 , cm 1 ym 1  am  2 ym  2  bm  2 ym  3   ym  2 ,

(6-2) (6-m) (6-m+1) (6-m+2)

… cn  2 yn  2  an 1 yn 1  bn 1 yn   yn 1 , cn 1 yn 1  an yn   yn .

(6-n–1) (6-n)

 For inverse bi , ci (i  m  1, m  2, , n  1), ai (i  m  2, m  3, , n) . From (5) and (6), we can get

ET AL.

c j =kb j ,( j  m  1, m  2, , n  1) ,   x j  c j 1 x j 1  b j x j 1 , x j  0;  xj  aj     y j  c j 1 y j 1  b j y j 1 , y  0 . j  yj  ( j  m  2, m  3, , n)

ci 1 xi 1 +ai xi  bi xi 1   xi (i  m  2, m  3, , n) ,

(7) (8)

In order to eliminate ai , multiply by yi on both sides of (7), multiply by xi on both sides of (8), then cut on both sides, we can get bi Di =(   ) xi yi +ci 1 Di 1 (i  m  2, m  3, , n) . (9)

cm Em +1 =(   ) xm 1 ym 1 +bm 1 Dm 1 ,

(15)

am 1 Em+1 = x1 ym 1   xm 1 y1  bm 1 Em+2 ,

(16)

(i  m  2, m  3, , n)

Let i  m  2 ,

ci 1 y1 +ai yi   yi (i  2,3, , m) .

s 0

ci 1 Ei =(   ) xi yi (i  2,3, , m) ,

(20)

ai Ei = x1 yi   xi y1 (i  2,3, , m) ,

(21)

ci (i  2,3, , m  1) . k

(22)

bi 

 For inverse a1 , b1 . From (5) and (6), we can get m

a1 x1 +b1 x2   x1   bs xs 1 ,

(23)

s2

m

a1 y1 +b1 y2   y1   bs ys 1 .

(24)

s2

If D1  0 , from (23) and (24), then we can get a1 

xn  s yn  s ( j  m  1, m  2, , n  1) . k n( s  j ) (11)

(   ) n  ( j 1) xn  s yn  s  k n ( s  j ) ( j  m  1, m  2, , n  1) , Dj s 0

(12) Copyright © 2011 SciRes.

 x1 y2   x2 y1   bs ( xs 1 y2  x2 ys 1 ) s2

D1

, (25)

m

If D j  0 ( j  m  1, m  2, , n  1) , then xi , yi can not be zero at the same time ,so bj 

(19)

From (18) and (19), we can get

Under normal circumstances,



(18)

m

x y x y   x y bm 1 Dm 1  (   )  n n( mn1)  nn1( m n 2)1    m +2 m  2  k k k  n  ( j 1)

(17)

ci 1 x1 +ai xi   xi (i  2,3, , m) ,

(10)

Let i  n , because Dn  0 , so x y bn 1 Dn 1  n n (   ) , k x y  x y Let i  n  1 , bn  2 Dn  2  (   )  n 2 n  n -1 n 1  ; k   k …………

b j D j  (   )

cm . k

 For inverse c1 , ci , bi (i  2,3, , m  1) , ai (i  2,3, , m) . From (5) and 2 to m equation of (6),

To problem A, because ci  kbi , (i  2,3, , n  1) , so (9) become bi Di  (    ) xi yi  kbi 1 Di 1

(14)

 For inverse am 1 , cm , bm . From (5) and the m + 1 equation of (6),

bm 

ci 1 yi 1 +ai yi  bi yi 1   yi (i  m  2, m  3, , n) .

(13)

b1 

     x1 y1   bs  ys 1 x1  y1 xs 1  s2

D1

. (26)

According to the above analysis, to question IEPGAM, we can get the follow theorem. Theorem. If the following conditions are satisfied: 1) D1  0 ; 2) Di  0 (i  m  1, m  2, , n  1) ; 3) Ei  0 (i  2,3, , m  1) Then question IEPGAM has the unique solution, and AM

Z. B. LI

bj 

ET AL.

(   ) n  ( j 1) xn  s yn  s  k n ( s  j ) ( j  m  1, m  2, , n  1) (27) Dj s 0   x j  c j 1 x j 1  b j x j 1 , x j  0;  xj  aj   ,   y j  c j 1 y j 1  b j y j 1 , y  0 j  yj  ( j  m  2, m  3, , n)

bm = am 1 =

bj =

Em +1

(   ) x j 1 y j 1 kE j +1

b1 

3 , 2 1 c3  kb3   , 3 c4  kb4  0 ,

(28)

,

(i  2,3, , m  1) ,

c1 =

(30)

a1   

(31)

a2  a3 

(    ) x1 y1   bs ( ys 1 x1  y1 xs 1 ) s 2

D1

aj =

 x1 y j   x j y1 Ej

( j  2,3,, m) ,

  bs xs 1    s 1 , x1  0; x1  a1   m  bs ys 1   s 1 , yj  0   y1 

(   ) x2 y2  0; E2

(29)

m

b1 

(    ) x1 y1  b2 ( y3 x1  y1 x3 ) 7  ; D1 4 c2  kb2  

(   ) xm 1 ym 1 +bm 1 Dm 1 , kEm +1

 x1 ym 1   xm 1 y1  bm 1 Em+2

1445

( 32)

 x1 y2   x2 y1 E2

a5 

E3

 x4  c3 x3  b4 x5 x4

 x5  c4 x4  b5 x6 x5

m

 7  2   0  3 J   2   0   0 

(34)

c j =kb j ,(i  2,3, , n  1) ,

(35)

(   ) x2 y2 c1 = . E2

(36)

Example 1. Give   1,   2, k  2, m  2, n  5, x  (1,1,1,1,1)T , y  (1, 0, 2, 1, 0)T . It is easy to be calculated D1  1  0, D2  3  0, D4  1  0 ;

From Theorem, the question IEPGAM has the unique solution. And     x5 y5 x4 y4  1 b3  + = ,   2 D3  k k  6

b4 

4 , 3

=1 .

7 4 1 0 0 0

3 4 0 8 3 1 3 0

 0 0  0 0    1 0 6   4 0 3  0 1 

and Jx   x, Jy   y .

4. References [1]

D. J. Wang, “Inverse Eigenvalue Problem in Structural Dynamics,” Journal of Vibration and Shock, No. 2, 1988, pp. 31-43.

[2]

H. Dai, “Inverse Eigenvalue Problem for Jacobi Matrices,” Computation Physics, Vol. 11, No. 4, 1994, pp. 451-456.

[3]

C. H. Wu and L. Z. Lu, “Inverse Eigenvalue Problem for a Kind of Special Matrix,” Journal of Xiamen University (Natural Science), No. 1, 2009, pp. 22-26.

[4]

Q. X. Yin, “Generalized Inverse Eigenvalue Problem for Arrow-Like Matrices,” Journal of Nan Jing University of Aeronautics & Astronautics, Vol. 34, No. 2, 2002, pp. 190-192.

E2  1  0, E3  1  0, E4  2.

3 1 (   ) x3 y3  b3 D3    , kE3 4

=

8 = , 3

So

3. Numerical Examples

b2 

=1 ,

 x1 y3   x3 y1  b3 E4

a4 

(33)

b1 x2  b2 x3 7  , x1 2

    x5 y5 

0, D4  k 

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AM