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 Gen eralized
Arrow-Like Matrices
Zhibin Li, Cong Bu, Hui Wang
College of Mathematics, Dalian Jiaotong University, Dalian, China
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 character-
istic 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 pro-
blems 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 in-
verse eigenvalue problem for generalized arrow-like ma-
trices.
Generalized arrow-like matrices refer to the matrix as
follows:
11 1
12
1
11
12
1
1
mm
mm
mmm
mm
n
nn
abb b
ca
ca
Jcab
ca
b
ca
























(1)
When ,
1m
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
T
12
(, ,, ),
n
n
x
xx xRT
12
(, , ,)n
n
y
yy yR.
Find the nn

real generalized arrow-like matrice
J
,
ch that su,
J
xxJyy




(,)
.
The expression and an algorithm of the solution of th e
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:
010nn
xx yy
100 0 nn
bbcc

 , (2)
11
(0,1,,)
ii
iii
xy
Di
xy

n, (3)
1
1
(2,3,, 1
i
ii
xx
Eim
yy )

 (4)
So
11 12111
+b mm mm axxbxb xx



, (5-1)
112 22
+cx axx

, (5-2)
…
11
+
mmm
cxax x
m


111 12mmm mm
axb x
, （5-m）
1
+m cx x





2 22 3m mm m
ax bx
, (5-m+1)
11
+
mm
cx x
2m


 

, (5-m+2)
…

Z. B. LI ET AL. 1444
122 111
+
n nnnnnn
cx axbxx

 

11
+
nn nn
cx ax
, (5-n–1)
n
x

 . (5-n)
11 12111
+b mm mm ayybyb yy


 
112 22
+cy ayy
, (6-1)

, (6-2)
…
11mmm
cyay y
m

, (6-m)
11112mmmmm m
cyaybyy1

 

112 22 3mmm mm mm
cy aybyy
, (6-m+1)
2

  


, (6-m+2)
…
2211 1nnnn nnn
cyay byy
1



11nn nnn
cy ayy
, (6-n–1)


. (6-n)
 For inverse
.
,(1,2,,1),
ii
bci mmn 
,3,,)m n(2ai m
i
From (5) and (6), we can get
11 1
+(2,
iiii iii
cxaxbxxi mmn3,,)

 

11 1
+(2,
iiiiiii
cyay byyimmn
, (7)
3,,)

 
. (8)
In order to eliminate , multiply by i on both
sides of (7), multiply by i
i
a y
x
on both sides of (8), then
cut on both sides, we can get
i11
=()+(2,3,, )
iiiii
bDxyc Dimmn



. (9)
To problem A, because ,(2,3, ,1)
ii
ckbin

, so
(9) become
1
()
(2,3,,)
iiiii i
bDxykb D
im mn


 
 
1
(10)
Let , because , so in0
n
D
n
11 ()
n
nn
xy
bD k




1in
,
Let , -1 1
22 2
()
nnnn
nn
xyx y
bD k
k




 


;
…………
Let , 2im
11+2 2
11 (1) (2)
()
nnn nmm
mm nmnm
xyx yxy
bD k
kk

 
  





Under normal circums ta n ce s,
(1)
()
0
()( 1,2,,
nj ns ns
jj nsj
s
xy
bDj mmn
k

 


 
1).
(11)
If , then 0 (1,2,,1)
j
Djmm n ,
ii
x
y can
not be zero at the same time ,so
(1)
()
0
() (1,2,,
nj ns ns
jnsj
s
j
xy
bjmm
Dk

 




1)
n 
,
(12)
c=,(1,2, ,1)
jj
kbj mmn

 , (13)
11 1
11 1
,0
,0
(2,3,,)
jjjjj j
j
jjjjjj j
j
xcxbx x
x
aycybyy
y
jm mn


 
 








 
;
. (14)
 For inverse ,,.
1mmm
From (5) and the m + 1 equation o f (6),
acb

+11 111
=( )+
mmm mmm
cExyb D


 

, (15)
1m+111111m+2
=
mmmm
aE xyxybE



 , (16)
m
m
c
bk . (17)
 For inverse 1,,(2,3, ,1)
ii
ccbi m


,,)m,
(2,3ai
i

.
From (5) and 2 to m equation of (6),
11
+(2,3,
iiii
cxax xim,)



, (18)
11
+(2,3,
iiii
cyay yim,)



. (19)
From (18) and (19), we can get
1i
=()(2,3,,)
iii
cE xyim




, (20)
i11
=(2,3,
iii
aExyxy im,)



, (21)
(2,3,, 1)
i
i
c
bi m
k

. (22)
 For inverse ,ab.
11
From (5) and (6), we can get
11 1211
2
+m
s
s
s
ax bxxbx 



, (23)
11 1211
2
+m
s
s
s
ay byyby 



. (24)
If 10D

, from (23) and (24), then we can get
1221122 1
2
11
()
m
ss s
s
xyxyb xyxy
aD



 
, (25)


11111 1
2
11
m
sss
s
xyb yxyx
bD



 
. (26)
According to the above analysis, to question IEPGAM,
we can get the follow theorem.
Theorem. If the following conditions are satisfied:
1) 10D

;
2) 0 (Di1,2, ,1)
imm n

 
0 (2, 3,,1)Ei m;
3) i


Then question IEPGAM has the unique solution, and

Z. B. LI ET AL. 1445
(1)
()
0
() (1,2,,
nj nsns
jnsj
s
j
xy
bjmm
Dk

 




1)
n 
(27) 112 3113
11
() () 7
4
xyb yxyx
bD
  

 ;
11 1
111
,0
,0
(2,3,,)
jjjjjj
j
jjjjjjj
j
xcxbx x
x
aycyby y
y
jm mn


 
 








 
22
3
2
ckb

 ,
;
, (28)
33
1
3
ckb

 ,
44
0ckb

,
22
12
()
=0
xy
cE
 ;
111 1
+1
()+
=mmm m
mm
xy bD
bkE

 
, (29)
12 23
11
7
2
bx bx
ax



,
1111 1m+2
1+1
=mm m
mm
xyxyb E
aE



 , (30)
12 21
22
=1
xy xy
aE
 
,
11
+1
()
=(2,3,
jj
jj
xy
bi
kE



,1)m , (31)
13313 4
33
8
=3
xyxy bE
aE
 
,
111111
2
11
() (
m
sss
s
xyb yxyx
bD




)
( 32) 43345
44
4
=3
xcxbx
ax
 
,
11
=(2,3,
jj
jj
xy xy
aj
E
 ,)m
, (33) 54456
55
=1
xcxbx
ax
 
.
1
11
1
1
1
1
1
,0
,0
m
ss
s
m
ss
sj
bx
x
x
a
by
y
y

















So 773
00
24400
010
381
00
236
14
00 0
33
000 01
J







 






;
)
(34)
=,(2,3,,1
jj
ckbin, (35)
22
12
()
=
x
y
cE
 . (36)
and ,
J
xxJyy



.
3. Numerical Examples 4. References
Example 1. Give 1,2, 2,2, 5,kmn

  
TT
(1,0,2,1,0) . [1] D. J. Wang, “Inverse Eigenvalue Problem in Structural
Dynamics,” Journal of Vibration and Shock, No. 2, 1988,
pp. 31-43.
(1,1,1,1,1) ,xy
It is easy to be calculated
124
10,30, 10DDD  ; [2] H. Dai, “Inverse Eigenvalue Problem for Jacobi Matri-
ces,” Computation Physics, Vol. 11, No. 4, 1994, pp.
451-456.
234
10, 10,2.EEE
From Theorem, the question IEPGAM has the unique
solution. And
55 44
32
3
1
+=
6
xy xy
bDk
k
 



,
[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.

23333
3
1
() 4 bxybD
kE
 
3
,
[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.
55
44
0
xy
bDk
 



,