Thread View: sci.math.num-analysis
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Started by Nicolas Ratier
Tue, 07 Apr 1992 15:30
Determinant
Author: Nicolas Ratier
Date: Tue, 07 Apr 1992 15:30
Date: Tue, 07 Apr 1992 15:30
46 lines
1976 bytes
1976 bytes
I have to compute the determinant of a block pentadiagonal matrix, exactly
a matrix look like :
/ 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 /
where A is a pentadiagonal matrix
B is a tridiagonal matrix
C is a diagonal matrix
and besides A, B, C are commuting pairwise (i.e. AB = BA, AC = CA, BC = CB)
In fact, it is easy to find a efficient method to compute the determinant
of the pentadiagonal matrix by Gauss elimination way. What i need is a formula
for computing the determinant of a block pentadiagonal matrix from the
determinant of each (commuting) element.
Do you have an idea how to solve this problem ?
Thank you for responding, (please email me)
Nicolas RATIER
-------------------------------------
| L.A.A.S. (France) |
| |
| |
| Nicolas RATIER |
| |
| email : ratier@laas.laas.fr |
| Tel : (33) 61 33 62 46 |
| Fax : (33) 61 33 62 08 |
| Telex : LAASTSE 520930F |
-------------------------------------
Re: Determinant
Author: Peter Montgomery
Date: Wed, 08 Apr 1992 02:10
Date: Wed, 08 Apr 1992 02:10
90 lines
3878 bytes
3878 bytes
In article <7286@laas.laas.fr> ratier@lio.laas.fr (Nicolas Ratier) writes:
>
> I have to compute the determinant of a block pentadiagonal matrix, exactly
> a matrix look like :
>
> / 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 /
>
>where A is a pentadiagonal matrix
> B is a tridiagonal matrix
> C is a diagonal matrix
>
>and besides A, B, C are commuting pairwise (i.e. AB = BA, AC = CA, BC = CB)
>
>
> In fact, it is easy to find a efficient method to compute the determinant
>of the pentadiagonal matrix by Gauss elimination way. What i need is a formula
>for computing the determinant of a block pentadiagonal matrix from the
>determinant of each (commuting) element.
>
# The problem cannot be solved as proposed. In this example,
# A and B are 3 x 3 diagonal matrices (which necessarily
# commute) and C = 0. The determinant expression is not
# a function of det(A) = a1*a2*a3 and det(B) = b1*b2*b3 alone.
with(linalg):
M := array(1..9, 1..9);
for i from 1 to 9 do
for j from 1 to 9 do
dif := abs(i-j);
isub := 1 + ((i-1) mod 3);
if (dif = 0) then
M[i,j] := a.isub;
elif (dif = 3) then
M[i,j] := b.isub;
else
M[i,j] := 0;
fi;
od;
od;
print(M);
print(factor(det(M)));
;quit;
|\^/| MAPLE V
._|\| |/|_. Copyright (c) 1981-1990 by the University of Waterloo.
\ MAPLE / All rights reserved. MAPLE is a registered trademark of
<____ ____> Waterloo Maple Software.
| Type ? for help.
(source listing deleted)
> print(M);
[ a1 0 0 b1 0 0 0 0 0 ]
[ ]
[ 0 a2 0 0 b2 0 0 0 0 ]
[ ]
[ 0 0 a3 0 0 b3 0 0 0 ]
[ ]
[ b1 0 0 a1 0 0 b1 0 0 ]
[ ]
[ 0 b2 0 0 a2 0 0 b2 0 ]
[ ]
[ 0 0 b3 0 0 a3 0 0 b3 ]
[ ]
[ 0 0 0 b1 0 0 a1 0 0 ]
[ ]
[ 0 0 0 0 b2 0 0 a2 0 ]
[ ]
[ 0 0 0 0 0 b3 0 0 a3 ]
> print(factor(det(M)));
2 2 2 2 2 2
a3 a2 a1 (a3 - 2 b3 ) (a2 - 2 b2 ) (a1 - 2 b1 )
> ;quit;
--
Peter L. Montgomery Internet: pmontgom@math.ucla.edu
Department of Mathematics, UCLA, Los Angeles, CA 90024-1555 USA
Go Bruins! This Bruin PhD wants to go solve your problems.
Interests include computational mathematics, compilers, and number theory.
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