Thread Subject: Matrix Equation solving

Please help me to solve the equation

A'pA-p = -I

A - 3x3 matrix
p - 3x3 unknown symetric matrix
I - identity matrix

"shameer koya" <assigmenteee@yahoo.co.in> wrote in message <fhe92r$m23
$1@fred.mathworks.com>...
> Please help me to solve the equation
>
> A'pA-p = -I
>
> A - 3x3 matrix
> p - 3x3 unknown symetric matrix
> I - identity matrix
----------
This is, after all, a set of nine linear equations in the nine unknown elements of
p, and can therefore be solved by matlab using standard techniques with the
backslash operator, with p temporarily considered a 9 x 1 column matrix. It is
only necessary to determine the 81 elements of the corresponding 9 x 9 matrix.

Assuming there is a unique solution, the above equation is of such a nature that
the matrix p is guaranteed be Hermitian symmetric.

Roger Stafford

"Roger Stafford" <eleanorandrogerxyzzy@mindspring.com.invalid> wrote in
message <fhna35$9u4$1@fred.mathworks.com>...
> "shameer koya" <assigmenteee@yahoo.co.in> wrote in message <fhe92r
$m23
> $1@fred.mathworks.com>...
> > Please help me to solve the equation
> >
> > A'pA-p = -I
> >
> > A - 3x3 matrix
> > p - 3x3 unknown symetric matrix
> > I - identity matrix
> ----------
> This is, after all, a set of nine linear equations in the nine unknown
elements of
> p, and can therefore be solved by matlab using standard techniques with
the
> backslash operator, with p temporarily considered a 9 x 1 column matrix.
It is
> only necessary to determine the 81 elements of the corresponding 9 x 9
matrix.
>
> Assuming there is a unique solution, the above equation is of such a nature
that
> the matrix p is guaranteed be Hermitian symmetric.
--------
  Shameer, I was too lazy Saturday to work out matlab code for automatically
determining those 81 matrix elements I described as necessary in solving
your problem, but now I have done so. Given your 3 x 3 matrix, A, do this:

 D = -eye(3);
 X = repmat(A',3,3);
 Y = reshape(repmat(reshape(repmat(A.',3,1),1,27),3,1),9,9);
 Z = X.*Y-eye(9); % Here is the desired 9 x 9 matrix
 P = reshape(Z\D(:),3,3); % <-- This is the 3 x 3 solution

  Note that the repeated 'reshapes' and 'repmats' used in generating matrix Y
above, perform a variant of the 'repmat' operation. Each individual element
of A is replicated so as to form a local 3 x 3 section of copies, rather than the
entire A being replicated as nine 3 x 3 copies. There may well be a better
way of accomplishing this but I couldn't think of it.

Roger Stafford

Alternative solution:

D=-eye(3);
Z=zeros(3);
BIGA1=[A Z Z; Z A Z; Z Z A]';
BIGA2=reshape(permute(reshape(BIGA1,3,3,3,3),[2 1 4 3]),9,9);
P=reshape((BIGA1*BIGA2-eye(9))\D(:),3,3)

;-)
Bruno

"Roger Stafford" <ellieandrogerxyzzy@mindspring.com.invalid> wrote in
message <fhsnu5$hg7$1@fred.mathworks.com>...

> Shameer, I was too lazy Saturday to work out matlab code for
automatically
> determining those 81 matrix elements I described as necessary in solving
> your problem, but now I have done so. Given your 3 x 3 matrix, A, do this:
>
> D = -eye(3);
> X = repmat(A',3,3);
> Y = reshape(repmat(reshape(repmat(A.',3,1),1,27),3,1),9,9);
> Z = X.*Y-eye(9); % Here is the desired 9 x 9 matrix
> P = reshape(Z\D(:),3,3); % <-- This is the 3 x 3 solution
> .....
------
Here is another alternative that I ought to have thought of before:

 D = -eye(3);
 I = floor((3:11)/3);
 Z = repmat(A',3,3).*A(I,I).'-eye(9);
 P = reshape(Z\D(:),3,3);

Roger Stafford