O N S OME J ACOBI M ATRIX I NVERSE E IGENVALUE P ROBLEMS WITH
M IXED G IVEN D ATA
Y ILIN YANG
I NTRODUCTION
M ETHOD OF R ESULT 1
Jacobi Matrices (real symmetric tridiagonal matrices) have a wide range of applications in physics and
engineering, and are closely and non-trivially linked
with many other mathematical objects, such as orthogonal polynomial, one dimensional Schrödinger
operators and the Sturm-Liouville problem. In the
past couple of decades, constructing Jacobi matrices
from different types of data was studied intensively.
In this research, we show that a unique Jacobi matrix can be reconstructed from two types of mixed
data: 1. its two eigenpairs, 2. its eigenvalues and the
eigenvalues of its submatrix obtained by removing
the first two rows and columns and one of its entry.
M AIN R ESULTS
An n × n Jacobi matrix has the form

b 1 a1 0
.
.
.
a1 b2 a2
.
.
.

 0 a2 b3
.
.
.

J =
.
.
.
.
.
.
.
.
. an−2 bn−1 an−1
.
.
.
0
an−1
bn







E XAMPLE
T
T
(1)
Result 1. Let J be an irreducible Jacobi matrix
as in (1). Given two real numbers λ, µ and two
distinct real vectors u = (u1, u2 , . . . , un )T , v =
uk uk+1
T
(v1 , v2 , . . . , vn ) , if
is invertible for all
vk vk+1
k = 1, . . . , n − 1, and
un vn (λ − µ)
=
,
vn un−1 − un vn−1
then a unique Jacobi matrix J can be constructed
such that (λ, u) and (µ, v) are two of its eigenpairs.
Result 2. Let {λi }n
i=1 be the eigenvalues of J and
{νi }n−2
i=1 be the eigenvalues of J(2,2) where λ1 >
. . . > λn and ν1 > . . . > νn−2 . If
Qn
i=1 (b1 − λi )
< 0,
Qn−2
j=1 (b1 − νj )
Plugging λ = 1, u2 = −1 and µ = 2, v2 = 0 respectively in (2.1) and (2.2), we have a1 = 1, b1 = 2.
Thus, we can use a1 and b1 to solve for a2 , b2 in the
next step. Similarly, by use of the iterative method,
we can solve that
where b1 , a1 can be solved by (2.1) and (2.2) since λ 6= µ, u2 6= v2 . Then b2 and a2 can be solved from (2.3) and
(2.4) if and only if the two equations are linearly independent. In general, once we have ak−1 , the sub-system
bk uk + ak uk+1 = λuk − ak−1 uk−1
(3.1)
bk vk + ak vk+1 = µvk − ak−1 vk−1
(3.2)
After plugging a4 to (2.(2n-1)) and (2.2n), the augmented matrix of the equation set has rank 1. So it
satisfies the condition in Result 1.
uk uk+1
only contains the unknown ak . So bk can be solved uniquely if and only if
is invertible for
vk vk+1
k = 1, 2, . . . , n−1. Once we solve an−1 , the last two equations must be linearly dependent, which is equivalent
un vn (λ−µ)
to an−1 = vn un−1 −un vn−1 .
M ETHOD OF R ESULT 2
Let PJ (λ), PJ(1,1) (λ) and PJ(2,2) (λ) be the characteristic polynomial of J, J(1,1) and J(2,2) respectively. Then,
PJ (λ) = (λ − b1 )PJ(1,1) (λ) −
Letting λ = b1 , (4) becomes
2
a1
i
ai
bi
1
1
2
2
1
2
3
1
2
4
1
2
5
1
2
F UTURE W ORK
We are interested in the inverse problems of other
types of band matrices. Specifically, we are working
with the reconstruction of real symmetric pentadiagonal matrix from different types of data. Furthermore, we are exploring the connection between inverse problem of band matrices and A. Horn’s problem, which asks to characterize eigenvalues of Hermitian matrices A, B and A+B.
(4)
[1] Birgler-Konig W. Friedrich, Construction of band
matrices from spectral data. Linear Algebra and
Appl. 40:79-87, 1981.
Qn
i=1 (b1 − λi )
PJ (b1 )
=−
= − Qn−2
.
PJ(2,2) (b1 )
j=1 (b1 − νj )
So a positive a1 can be uniquely determined if
Since we know that
2
a1 PJ(2,2) (λ).
Given λ = 1, µ = 2, u = (1, −1, 0, 1, −1) and v =
(1, 0, −1, 0, 1)T . Find a 5 × 5 Jacobi matrix J such that
(λ, u) and (µ, v) are two eigenpairs of it.
R EFERENCES
Qn
i=1 (b1 −λi )
Qn−2
j=1 (b1 −νj )
< 0.
PJ (λ)
PJ (λ)
PJ (λ)
PJ(1,1) (λ) = c1
+ c2
+ . . . + cn
,
λ − λ1
λ − λ2
λ − λn
(5)
[2] Daniel Boley and Gene H. Golub, A survey of
matrix inverse eigenvalue problems. Inverse Problems, 3(4):595-622, 1987.
(6)
[3] H. Hochstadt. On the construction of a Jacobi matrix from spectural data, Linear Algebra and Appl.,
8:435-446, 1974.
(7)
[4] H. Hochstadt. On the construction of a Jacobi matrix from mixed given data. Linear Algebra and
Appl., 28:113-115, 1979.
we let λ be any eigenvalue λi for i = 1, 2, . . . , n in (4) and (5), thus
0 = PJ (λi ) = (λi − b1 )PJ(1,1) (λi ) −
and
PJ(1,1) (λi ) = ci
n
Y
2
a1 PJ(2,2) (λi )
0
(λi − λj ) = ci PJ (λi ).
j=1,j6=i
0
then a unique irreducible Jacobi matrix J with (1,1)
entry b1 and positive off diagonal entries can be
n
constructed such that σ(J) = {λi }i=1 , σ(J(2,2) ) =
n−2
{νi }i=1
T
We can assume that u = (1, u2 , u3 , . . . , un ) and v = (1, v2 , v3 , . . . , vn ) and u2 6= v2 . Then apply ak−1 uk−1 +
bk uk + ak uk+1 = λuk , where k = 1, 2, . . . , n, and pn+1 = a0 = an = 0, we get a linear system with 2n − 1
unknowns and 2n equations.

b
+
a
u
=
λ
(2.1)
1
1
2




b1 + a1 v2 = µ
(2.2)




a
+
b
u
+
a
u
=
λu
(2.3)
1
2
2
2
3
2


a1 + b2 v2 + a2 v3 = µv2
(2.4)

..
..



.
.




an−1 un−1 + bn un = λun
(2.(2n − 1))


an−1 vn−1 + bn vn = µvn
(2.2n)

where all ai , bi are real. It is called irreducible if ai 6=
0 for i = 1, . . . , n − 1.
an−1
D REXEL U NIVERSITY, D EPARTMENT OF M ATHEMATICS
Combining (6) to (7), we have 0 = (λi − b1 )ci PJ (λi ) −
a21
Qn−2
k=1 (λi −νk )
0
(λi −b1 )PJ (λi )
a21
Qn−2
k=1 (λi
for i = 1, 2, . . . , n. Then according to (5), PJ(1,1) (λ) =
− νk ), and hence we can solve ci =
Pn
i=1
a21
Qn−2
k=1 (λi −νk )
0
(λi −b1 )PJ (λi )
PJ (λ)
λ−λi .
Now we have solved the characteristic polynomial of J(1,1) , so Hochstadt’s method [4] can be applied to finish
the construction of J from σ(J) and σ(J(1,1) ).
A CKNOWLEDGEMENT
I would like to thank my mentor Dr. Lei Cao as well
as Dr. Hugo Woerdeman for giving me exemplary
guidance on this project.