Re: eigenvectors and values of K-diagonal matrix and its 2Dim analogues

From: Peter Spellucci (spellucci_at_fb04373.mathematik.tu-darmstadt.de)
Date: 02/14/05 

Date: Mon, 14 Feb 2005 11:57:29 +0000 (UTC)

In article <5y85ojoyi93l@legacy>,
 Alex <alex-math@mail.ru> writes:
>I wrote:
>> Alex <alex-math@mail.ru> writes:
>> >What are eigenvectors and values of 3-diagonal matrix ?
>> >
>> >(a , b, 0 , 0 0 0 0
>> > b , a, b , 0 0 0 0
>> >0 , b, a , b 0 0 0
>> >.................................
> > >???
>> >
>> >
>
>And received kind answer:
>> homeworK?
>It is not a homework - it is related with some work
>in signal processing.
>> gregory karney: a collection of matrices for testing computational >
>algorithms
>> example 7.4
>> lambda(k)= a+2*b*cos(k*pi/(n+1))
>> x(k)(j) = sqrt(2/(n+1))*sin(k*j*pi/(n+1))
>
>Thank You very much !
>But i have a contianuation for this question:
>What are eigenvectors and values of K-diagonal matrix
>of the same kind ?
>
>For example 5-diagonal:
>( a , b_1 , b_2 , 0 0 0 0 0
> b_1 , a , b_1 , b_2 0 , 0 0 0
> b_2 , b_1 , a , b_1 , b_2 , 0 0 0
> 0 , b_2 , b_1 , a , b_1 , b_2 0 0
> 0 , 0 , b_2 , b_1 , a , b_1 , b_2 0 0
>........................................................
>
>Matrix is of finite size. (In practice i need sizes from 10*10 to
>10^4*10^4).
>
>Please pay attention that:
> it is not a ciculant type matrix - it is not of the kind
> a , b_1 , b_2 , 0 0 0 b_2 b1
> .................................................
>
>*******************************
>What can be said about two-dimensional analogue for this
>averaging operator ?
>(I mean the following:
>Let us denote by T the shift operator with the matrix
>
>(0 , 1, 0 , 0 0 0 0
> 0 , 0, 1 , 0 0 0 0
> 0 , 0, 0 , 1 0 0 0
> .......................
>
>Consider the tensor product of the two vector spaces R^n\otimes R^k
>and consider the operator:
> a T_1 \otimes T_2 + b Id_1 \otimes T_2 + c T_1^{t} \otimes T_2
>+ d T_1 \otimes Id_2 + e Id_1 \otimes Id_2 + d T_1^{t} \otimes Id_2
>
>+ c T_1 \otimes T_2^{t} + b Id_1 \otimes T_2^t + a T_1^{t} \otimes
>T_2^t
>
>
>What can be said about its eigenvec and eigenval ?
>And the same about its k-diagonal generalizations ?
>
>
 
the tensor product generalization of the tridiagonal case is quite similar
see again by gregory and karney. but the quindiagonal one is already much harder.
sometimes you can show that the quindiagonal case is the square of a tridiagonal
one, but for your matrix this is not the case (needs modification in the top
left and bottom right elements). the usual way to deal with these cases is to use
the theory of difference equations for the "general row" and then to use the
top and bottom rows to fix the eigenvalues, but this is a hard job for the
quindiagonal case and for the general one impossible by analytic methods.
hth
peter