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Transcript
NOTES
567
" 1 2" " l 9" " 1 4 "
.0 1. . 0 1 . . 0 1 .
" 1 7" " l 5" " 1 3 "
. 0 1. . 0 1 . . 0 1 .
" 1 6" " l l " " 1 8"
0 1
. 0 1. . 0 1 .
which has constant sum
For this square, A =
and AJ =
9 135
0 9
3 15
1 15
and constant product
0 3
0 0
3 30
0 3
1 1 1
1 1 1
1 1 1
1 1 1
1 1 1
1 1 1
In fact, it is easily shown by
induction that
A" =
3"- 1 5n3fl-»
0
3"-1
1 1 1
1 1 1
1 1 1
forn > 2.
The fact that A has distinct entries, but A3 does not, shows that the above
theorem does not hold for 3 x 3 magic squares of matrices (with magic
squares defined to satisfy (i) to (iv)).
References
1. J. M. H. Peters, Inverses and cubes, Math. Gaz. 65 (1982), pp. 253-4.
2. T. J. Fletcher, Linear algebra through its applications, van Nostrand
Reinhold (1972).
RAY HILL
Department of Mathematics and Computer Science, University ofSalford M5 4WT
S. M. ELZAIDI
Department of Mathematics, University of Nottingham NG7 2RD
80.47
An iterative algorithm for matrix inversion which is
always convergent
The problem of numerically solving a set of linear equations, or
equivalently of inverting a matrix, can be tackled either by a direct method
such as Gaussian elimination or by an iterative technique [1]. For a general
matrix the former is usually more efficient, but for particular types, such as
sparse matrices, an iterative approach can be more efficient. This is because
the latter can be advantageous, both in terms of the total amount of
computation involved and also in the amount of computer storage required,
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THE MATHEMATICAL GAZETTE
568
as discussed for example in [1]. Furthermore, iterative techniques can be
more accurate as they are not affected by the cumulative effect of roundingoff errors and the corresponding pivoting problems inherent in direct
methods.
Now, a well-known iterative procedure is the Gauss-Seidel method, but
this is restricted in application to those situations for which it is convergent;
this corresponds to the relevant matrix being in some suitable sense
diagonally dominant. The purpose of the present note is to point out a
simple way by which any matrix inversion problem may be transformed into
one for which the Gauss-Seidel method converges.
We consider a set of n simultaneous linear equations of the form
n
^cipfy
= bp
(1 < p < ri)
q=\
for the n unknowns xp (1 < p < n). If xpr) is the rth iterate, the GaussSeidel method gives
X
P
(r+l)
=
1
n
7> - 1
2^apqxq
+
U=i
2 , <WV - bP >
i=p+i
and it is known that this yields a convergent algorithm if the n x n matrix
A = [apq] is both symmetric and positive-definite (see, for example, [1]).
Let us now consider the set of linear equations
Cx = d
where C is non-singular, in a case where the Gauss-Seidel algorithm applied
to the matrix C does not converge. Operating on both sides of this equation
with the transposed matrix CT immediately transforms it into
Ax = b
where A = C C and b = C d, and we can now easily show that A is both
symmetric and positive definite. To prove the former, we note that
r
T
AT = (CTC)T = C^C™ = CTC = A
while the latter property follows from yTAy = yTCTCy = (Cy)TCy > 0,
for y non-zero, since Cy = 0 => y = 0, because C is non-singular. It
follows immediately that A = CTC is a positive definite matrix and hence
that the Gauss-Seidel method applied to the above equation Ax = b will
always converge and will yield the solution of our original equation
Cx = d.
This technique may easily be extended to calculating C~l, the inverse of
C, which satisfies CC~l = I. Since
AC~X = CTCC~l = CTI = CT
we consider in turn each column of C_1 and the corresponding column of
CT, and use the Gauss-Seidel method to solve for the former.
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NOTES
569
Finally we give an example of the above technique applied to a 3 x 3
matrix for which the Gauss-Seidel method cannot even be formulated, let
alone converge, since the diagonal elements of the matrix are all zero. The
equations we take are
l5
/ 0 6 l \ lx\
4 0 3 y = 13
\5 2 o / \z,
l\
H
which, in fact, have the exact solution (x, y, z) = (1, 2, 3).
Following our technique we multiply both sides of the above equation by
the transpose of the matrix above and hence obtain
' 41 10 12 \ M
10 40 6
12 6 10/
/
97\
108
54
We begin the iteration by taking x ( ' = 0 , followed by repeated
\0/
application of the Gauss-Seidel iteration formula as given above. The first
/ 2.366 \
iterate is x(1) = 2.109 which clearly differs significantly from the exact
\ 1.300/
solution. Further iterations, however, converge quite rapidly leading to
/1.002\
x<7) = 2.001 with the error decreasing by a factor of about 3 at each step.
\2.998/
We conclude that our modification of the Gauss-Seidel method is successful
in yielding an iterative algorithm which always converges.
0 6 1 T1
The calculation of | 4 0 3 by solving
5 2 0
' 41 10 12 \ X[ X2 X-i
10 40 6 y\ yi yi
,12 6 10 , z l z 2 z 3
0 4 5
6 0 2
1 3 0
is left as an exercise to the reader.
Reference
1. A. Ralston and P. Rabinowitz, A first course in numerical analysis,
McGraw-Hill (1978).
S. SIMONS
School of Mathematical Sciences, Queen Mary and Westfield College,
Mile End Road, London El 4NS
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