λ λ δ δ λ λ

© 2015 al-Farabi Kazakh National University Printed in Kazakhstan UDC 519.6

1*

A.A. Eleuov,

²

N.N. Tungatarov,

³

F.K. Yakhiyayev

1,2 Faculty of Mechanic-Mathematical, Al-Farabi Kazakh National University, Almaty, Kazakhstan;

3Scientific-Research Institute of Mathematics and Mechanics, Al-Farabi Kazakh National University, Almaty, Kazakhstan

*e-mail: Eleuov@mail.ru, Nourmat@mail.ru

A numerucal numerical method for the restoration of the five diagonal symmetric matrices from the spectral data

Abstract. In this work the possibility of restoration of real symmetrical five diagonal final matrixes using four numerically sequences is studied. Three from these four numerically sequences are interpreted as sets of eigenvalues of the considered matrix and else of two matrixes, obtained from considered matrix delet- ing some diagonal elements. The concrete formulas of construction of matrix elements using four sets of eigenvalues are obtained.

Key words: differential equations, band matrix, spectral problems, eigenvalues, five-diagonal symmetric matrix.

Introduction

Strip matrixes appear from the discreteness of boundary-value problems for the linear diffe- rential equations in the section. The algorithm of the restoration of three diagonal matrices from two spectra can be found in the work [1]. Inverse

spectral problems for five diagonal unitary ma- trices are examined in the work [2]. In this work the algorithm of the single-valued restoration of five diagonal symmetrical matrix is indicated. At the beginning of work the properties of the ei- genvalues of the symmetrical five diagonal ma- trixes are investigated

























































n n

n

n n n

n

n n n n n

a b

c

b a b c

c b a b c

c b a b c

c b a b c

c b a b

c b a

A

1 2

1 1 2 3

2 2 2 3 4

3 3 3 2 1

2 2 2 1 0

1 1 1 0

0 0 0











All numbers

a

0

, a

1

,...., a

_n

, b

0

, b

1

,...., b

_n are real, and numbers

c

₀

, c

₁

,...., c

n₂are positive.

Main body

From the work [3] it is known that the eigenva- lues of matrix

A

are real. This is obvious. In our

work we prove the not entirely obvious properties of eigenvalues of matrix

A

. Denote

P

_n1

( ) 



det( A  E )

  _.

Zeros of the polynomial

P

_n1

(  )

determine all eigenvalues of matrix

A

. Consider the equation

) 0 ,..., 0 , 1 ( ,

)

(

_{0 0}

1 





y P

_n ^T

y

A

^



^

 

^



^ (1)

International Journal of Mathematics and Physics 6, №1, 33 (2015) Assertion 1. For any complex value there ex-

ists a vector yy(), which satisfies the equation (1).

Let  be a fixed complex number. Replace the column

j

of matrix (

A



 E

) by the column ^₀^T. By y_j1(), j1,2,...n1denote the determinant

of the modified matrix. Compose the vector ))

( ),..., (

), ( ( )) (

(y  ^T  y₀  y₁  y_n1  . By direct checking we are convinced, that _yj1(



)_{are poly-} nomials from , and vector

_y

^(



)satisfies the equa- tion (1) for any complex .

Note that



















































































n n n

n n

n n

n n

n n n

a b c

b a

b c

c b

a b c

c b a

c

c b a

b

c b a

y

1 2

1 1

2 3

2 2

2 3 4

3 3 3

1

2 2 2

1

1 1 1

0( ) det





















and its zeros coincide with the eigenvalues of the truncated matrix

B

of the dimensionality

n  n

_{, where}

























































n n

n

n n n n

n n n n n

a b

c

b a b c

c b a b c

c b a b c

c b a b

c b a

B

1 2

1 1 2 3

2 2 2 3 4

3 3 3 2 1

2 2 2 1

1 1 1



















 .

The eigenvalues of initial matrix we will number in the order of their growth

n



₀  ₁... (2)

Eigenvalues of matrix

B

analogously number

n



₁  ₂ ... (3)

In the following the zeros of element







































 







n n

n b a

c c b a

b

c b a

c

c b

b

y

1 2 2 3 3

2

2 2 2

0

1 1

0

1( )



















International Journal of Mathematics and Physics 6, №1, 33 (2015) have important role.

)

1(

y is polynomial from the degree

)

1

( n

 with the leading coefficient (b₀(1)^n1). By



n





₂  ₃ 

...

 (4)

denote the zero of polynomial _y1()_.

We will use the spectrum of the matrix C. Ma- trix Cis obtained from the matrix Bby the deletion of the first line and first column.

n



₂  ₃  ...  (5) are eigenvalues of matrix C in the order of their growth.

Formulation of the problem of the restoration All elements

2 1

0 1

0 1

0

, a ,...., a

_n

, b , b ,...., b

_n

, c , c ,...., c

_n

a

of

matrixAto restore on the sequences (2), (3), (4), (5).

The number b0 is assigned, and one of the num- bers of sequence (4) is considered unknown. The algorithm is illustrated below.

Results of the direct problem

Some properties of the sequences (2), (3), (5).

Lemma 1.The following identity is satisfied for any



 











 

 

 

 y( );y( ) P^n1( )y⁰( ) P^n1( )y⁰( )_.

Consequence 1. The equality

) ( ) ( )

( ) ( )

( 

²

P

₁

 y

₀

 P

₁

 y

₀

 y

  _n_  _n  is fulfill for any_.

Proof of the lemma 1.

Accoding to (1) the correspondence

0 1

( ) )

( )

(   ^    ^

  y  P

_n

y

A

(6)

is fulfill for any.

The vector equality

 

₀

)

1

( )

(  

^

  

^

 

y



P

_n

y

A

(7)

is fulfill for any



_.

By scalar multiplication of all members of equality (6) to the vector _y^(



), we obtain:

) ( ) ( )

( ), ( )

( );

( y   y  y  P ₁  y₀  y

A    _n

     (8)

Analogously from (7) we obtain

) ( ) ( )

( ), ( )

( );

( y   y  y P ₁  y₀  y

A    _n

     . (9)

Subtract equality (9) from (8)and using AT

A , we obtain result of lemma 1.

Consequence 1 follow from lemma1 for





 _.

By  denote the set of complex  _for

0 ) (  y _.

Theorem 1. The following assertions are fulfill on the set :

a) All eigenvalues of matrixes

A

, C and

B

, which belong to , simple and real;

b) Eigenvalues of matrixes

A

and

B

,

B

and C, which belong to ,do not coincide;

c) the following inequalities fulfill for eigenva- lues

1 1

2 1 1

0 













...





_n 



_n



Matrixes

A

and

B

have the real simple eigen- values(являются существенно простыми матри- цами), and eigenvalues of matrixes

A

and

B

inter- change.

Proof of theorem 1.

Let



₀is eigenvalues of matrix A, then 0

) ( ₀

1 





Pn . Accoding to consequence 1 we have

0



_y



(  )

² 

_P

_n_1

(  ) _y

₀

(  )

. Then the numbers _Pn₁(



₀) _{and y0}(



₀) are not equal to zero. It means that

_P

_n



_1

( 

₀

)  0

_{, and}



₀ is simple eigenvalue of matrix B.

International Journal of Mathematics and Physics 6, №1, 33 (2015) Assume that



₀is eigenvalue of matrix

B

,

then

_y

₀

( 

₀

)



0

.

Accoding to consequence 1 the following equal- ity is carried out

) ( ) ( )

(

0 _y ₀ ²  _Pn1 _{0 y0} _{0 .}

Consequently, the number

_P

_n1

( 

₀

)

_{is not} equal to zero, thus



₀ cannot be eigenvalue of ma- trix

A

. From other side it follows from the inequali- ty

_y

₀

( 

₀

)



0

_{, that}



₀is simple eigenvalue of matrix

B

. Assertions prove analogously for matrix

C.

Observation. If



₀_then

_y

^

₍ 

₀

₎



₀

_{, that is} all elements of vector

_y

^

( 

₀

)

are equal to zero.

Hence, in particular it follows that

_P

_n1

( 

₀

)



0

_. That is all numbers

P

_n1

( ), 

₀

_y

₀

_{( ),} 

₀

1

( ), ..., ( )

0 _n 0

y  y 

are equal to zero for



₀_. Results on the problem of the restoration In this point we consider that three sequences of the numbers assigned (2), (3), (4), (5). We should find three sequences

1 2

1 2 1

1 0

...., ...., ....,

n n n

c c

c

b b

b

a a

a

Examine the equation

)

0

( )

( A   E x ^    ^

. (10) By

_x

₀

(  ) , _x

₁

(  ) , ..., _x

_n

(  )

denote the elements of vector

x



(  )

_.

For example, for

_x

₀

(  )

the following formula is fulfill

) (

) ) (

(

1

0 0



 





P

n

x y

, if

_P

_n1

(  )  0

_{. (11)}

Analogously, the equality is correct

) (

) ) (

(

1

1 1 

 





 Pn

x y , if P_n1()0. (12)

From the other side, for   A , using a num- ber of Neumann, we obtain the relationship

1 1 1

0 0

2 2

0 0 0 0

2 2 3

( ) ( ) ( 1 )

1 ( 1 1 ....) 1 1 1 ...

x A E E A

E A A A A

    



   

     

  

     

        

 



   

Hence

...

1 ; 1 ;

), 1 ( )

( _{0 2 0 0 3} ² _{0 0}

0           

 

 

 



 x^{ } A ^{ } A ^{ }

x (13)

Examining the equality (11) more detail for



₀

 A  

n , we obtain

International Journal of Mathematics and Physics 6, №1, 33 (2015)











   

 











   

 











   

 

 



 



 

 











 



















1 ...

) (

) .... (

1 ...

) (

) (

1 ...

) (

) ( ) )(

( ) ... (

) )(

( ) (

) )(

( ) ( )

)...(

)(

(

) )....(

)(

( ) (

) ) (

(

3 2 1 2

3 0 12 21 1

0

3 02 20 1

0 1

0 1

1 1

1 0

0 0 1

0 0 1

0

2 1

1 0 0





























































































 

n n n

n

n n

n n

n n

n n

n n

P y P

y

P y P

y P

y

P y P

x y

(14)

Comparing (13) и (14), we obtain the infinite system of the relationships

) (

)

; (

1 0 0 0

0

j n n j

kj k

P A y



 















 ^{ } 

⁽¹⁵⁾

for k=1,2,.. Analogously from (12) we obtain the infinite system of the relationships

) (

)

; (

1 1 1 0

0

j n n j

kj k

P A y



 













 ^{ }



⁽¹⁶⁾

for k=1,2,.. The same system of equations fol- lows for the matrix elements

B

) (

)

; det(

1 0 0

0 j

n j kj k

y

E

B C 

 



 

 





 ^{ }



^{. (17)}

Systems (15), (16), (17) are infinite system for determining the elements of the matrix

A

from the known right sides. Let ^_k is the vector with the zero components, besides

(k



1 )

, which is equal to one. Produce some calculations. Note

2 ,...

3 , 2

, ,

,

2 1

1 1 2 2

3 1 2 1 1 1 0 0 1

2 0 1 0 0 0 0









































n k

c b

a b

c A

c b a b

A

c b a

A

k k k k k k k k k k

k     



















































.

,

1 1 2

2

1 1

1 2

2 3

3 1

n n n

n n

n n

n n n

n n

n n

n n

a b

c A

b a

b c

A



















   

















































By the induction it is easy to prove, that

k k k k

k

k

d d d

A  ^

₀



₀^{( )}

 ^

₀



₁^{( )}

 ^

₁

 .... 

₂^{( )}

 ^

₂ . Following recurrence formulas are true for the coefficients (k1)

d

j :





















































 

 









) ( 2 )

( 1 )

( )

( 1 ) 1

( 2 ) 2

1 (

) 4( ) 2

3( ) 2

2( ) 2

1( ) 1

0( ) 0

1 2(

) 3( ) 1

2( ) 1

1( ) 1

0( ) 0

1 1(

) 2( ) 0

1( ) 0

0( ) 0

1 0(

jk k j

j j jk

k j j j jk

k j j

k k

k k

k k

k k

k k

k

k k

k k

d c d

b d

a d

b d

c d

d c d

b d

a d

b d

c d

d c d

b d

a d

b d

d c d

b d

a d

and

International Journal of Mathematics and Physics 6, №1, 33 (2015)

.

;

;

₁⁽¹⁾ _{0 2}⁽¹⁾ ₀

) 0 1

0(

a d b d c

d   

Here, subsequently we consider that the num- bers

a

_j

, b

_j1

, c

_j2 are equal zero for

n

j 

. Hence it is apparent that coefficients )

(kj

d

depend only on the row elements of matrix

A

with the numbers

(0),(1),(2), (3),...., (2 3),(2 2) k



k

 . Formulate this result in the form of lemma.

Lemma 1. Let

k

is fixed natural number. For

1 2 , , 1 ,

0 

 k

j 

the numbers

d

⁽_j^k1) depends only on the collection

 a

₀

, b

₀

, c

₀

, a

₁

, b

₁

, c

₁

,  , a

₂k₁

, b

₂k₁

, c

₂k₁



. The numbers ₂^{( 1)}

,

₂⁽ ¹₁⁾

,

₂^{( }¹₂⁾



 k

k k k k

k

d d

d

have representa-

tion

k k k k

kk a d f

d₂^{( 1)} _{2 2}^{( )} ₂ ,

1 ) 2

2( ) 2

1 ( 1

2  

  _k ^k_k  _k

kk b d f

d ,

) 2( ) 2

1

( 2

2 k

k k

kk c d

d ^_ 

where

f

₂k

, f

₂k₁ depends only on the collection

 a

₀

, b

₀

, c

₀

, a

₁

, b

₁

, c

₁

,  , a

_2k1

, b

_2k1

, c

_2k1



. Analogously we obtain following representation

1 ) 2 ( 1 2 ) 1

1( ) 0 0(

1



^k



^k

 ... 

^k_{k k}

k

c c c

A  ^  ^  ^  ^

_,

For coefficient c^(k_j ⁾ the following recurrence formulas are true.

) ( 2 )

( 1 )

( 1 )

( 1 ) 1

( 2 ) 2

1 (

) 2( ) 1

2( ) 1

1( ) 1

0( ) 0

1 1(

) 2( ) 0

1( ) 0

0( ) 0

1 0(

, ,

jk k j

j k j

j k j

j k j

j k j

j

k k

k k

k

k k

k k

c c c

b c

a c

b c

c c

c c c

b c

a c

b c

c c c

b c

a c













































and

) 1 1 3( ) 0

1 2( ) 0

1 1( ) 0

1

0(

b ; c a ; c b ; c c

c    

_.

Note that the coefficients c^(k_j ⁾depend only on the row elements of matrix

A

with the num- bers

( 0 ), ( 1 ), ( 2 ),....( 2 k  2 ), ( 2 k  1 )

.

Formulate this result in the form of lemma.

Lemma 2. Let

k

is fixed natural number. For

k

j



0 , 1 ,



, 2

the numbers

c

⁽j^k1) depend only on the collection

 a

0

, b

0

, c

0

, a

1

, b

1

, c

1

,  , a

2_k

, b

2_k

, c

2_k



. The numbers

c

₂^(k_k^_¹₃⁾

, c

₂^(k_k^_¹₁⁾

, c

₂^(k_k^_¹₂⁾ have repre- sentation

1 ) 2

( 1 2 1 ) 2

1 ( 1

2    





_k ^k_k



_k

kk

a c r

c

,

2 ) 2

( 1 2 1 ) 2

1 ( 2

2    





_k ^k_k



_k

kk

b c r

c

,

)

( 1

1 2 ) 2

1

( 3

2 k

k k

kk

c c

c

^_  _{ }

where

r

_2k2

, r

_2k1 depend only on the collection

 a

₀

, b

₀

, c

₀

, a

₁

, b

₁

, c

₁

,  , a

_2k

, b

_2k

, c

_2k



. Lemma 3. For any natural

k

the following re- presentation is true

1 2 ) ( 2 ) ( 2 2 0 0 1

2 ^

;

   _



A

^k



^



^

a

_k

d

^k_k

d

^k_k

g

_k

where

g

₂k₁ depend only on the collection

 a

₀

, b

₀

, c

₀

, a

₁

, b

₁

, c

₁

,  , a

_2k1

, b

_2k1

, c

_2k1



. For the proof of lemma 3 examine the scalar product

International Journal of Mathematics and Physics 6, №1, 33 (2015)

















 











 

















 





















































1 2

0

) 1 ( ) ( )

2( ) ( 1 2 1 ) 2 2( ) ( 2 2 2 ) 2

2( ) 2( 2

1 2

0

) 1 ( ) ( )

2( ) 2

( 1 2 1 ) 2

( 2 2 2 ) 2 2( 1

2

0

) 1 ( ) ( )

1 2( ) 2(

2

0

) 1 ( ) 2 (

2

0 ) 1 2 (

0 ) 0 (

0 1 0

1 0 2

) (

) (

;

;

;

k j

jk jk kk

kk k k

k k k k k

k k k k

k j

jk jk kk

k k k k k

k k k

k k

j

jk jk kk

kk

k j

jk jk k

i k i

i k

j k j

k j k k

d d d

d b d d c d d a

d d d

a d

b d

c d d

d d

d

d d d

d A

A

A ^ ^ ^ ^ ^ ^

(18)

Note that the components which are out of round brackets depend only from the first



2k1



lines of matrix

A

. From (18)and according to lem- ma 1, follows the assertion of the lemma 3.

Lemma 4. For any natural k following represen- tation is true

1 ) 2

2( ) 2( 22 22 22 0 2 0

2  ; (   )  

A ^k ^ ^ a _k b _k c _k d ^k_kd ^k_k h _k where number

h

_2k1 depend only on the collection

 a

₀

, b

₀

, c

₀

, a

₁

, b

₁

, c

₁

,  , a

_2k1

, b

_2k1

, c

_2k1



. For the proof of lemma 4 examine the scalar product

2 2 2 2 2 2

2 2 1 1 ( 1) ( 1) ( 1) ( 1)

0 0 0 0

0 0 0

( 1) ( 1) ( 1) ( 1) ( 1) ( 1) 2 1 ( 1) ( 1)

2 2 2 2 2 1 2 1 2 2

0

; ;

^k

;

^{k k}

k k k k k k k

j j i i j j

j i j

k k k k k k k k k

k k k k k k j j

j

A A A d d d d

d d d d d d d d

   

^



^



^

      

  

        

   



         

   

  



     

According to lemma 1 follows the assertion of the lemma 3.

Lemma 5. For any natural k following represen- tation is true

2 ) 2 1 ( 1 ) 2 ( 1 2 1 ) 2 1 ( 1 ) 2 2( 1 2 1

2 0;  











  





A ^k^ ^ b _k d ^k_k c^k_k a _k d ^k_k c ^k_k h_k

where number

h

₂k₂ depend only on the collection

 a

₀

, b

₀

, c

₀

, a

₁

, b

₁

, c

₁

,  , a

₂k₂

, b

₂k₂

, c

₂k₂



. For the proof of lemma 5 examine the scalar product

2 2 2

2 ( ) ( ) ( ) ( )

0 1 0 1

0 0 0

( ) ( ) ( ) ( ) 2 2 ( ) ( )

2 2 2 1 2 1

0

; ;

^k

;

^{k k}

k k k k k k k

j j i i j j

j i j

k k k k k k k

k k k k j j

j

A A A d c d c

d c d c d c

     

  



 



         

  

  



     

According to lemma 2 we obtain the equality

2 ( ) ( ) ( ) ( ) 2 2 ( ) ( )

0 1 2 2 2 1 2 1

0

( ) ( 1) ( ) ( 1)

2 2 1 2 1 2 1 2 1 2 1 2 2

; ^k

k k k k k k k

k k k k j j

j

k k k k

k k k k k k k

A d c d c d c

d b c d a c h

    ^



 

     

     

  

 



The assertion of lemma 5 follows from the last formula and the lemmas 1 and 2.

Lemma 6. For any natural k following represen- tation is true

2 1 ( ) ( 1)

0

;

1 2 2 2 1 2 1

k k k

k k k k

A

^

  b d c

^_

c

_

     

( ) ( 1)

2 2k 2 1 2 1k 2 2

k k k k k

a d c

^_

b

_

p

_

 

where number

p

₂k₂ depend only on the collection

 a

₀

, b

₀

, c

₀

, a

₁

, b

₁

, c

₁

,  , a

₂k₂

, b

₂k₂

, c

₂k₂



. For the proof of lemma 6 examine the scalar product