Re: AB=BA

"John D'Errico" <woodchips@xxxxxxxxxxxxxxxx> wrote in
message <fu7hmh$j3s$1@xxxxxxxxxxxxxxxxxx>...
"Arie Driga" <a_driga@xxxxxxxxx> wrote in message
<fu7bn6$qj5$1@xxxxxxxxxxxxxxxxxx>...
"helper " <spamless@xxxxxxxxxx> wrote in message
<fu79jm$mv3$1@xxxxxxxxxxxxxxxxxx>...

I did calculation by hand (for A*B=-B*A), at the end
I get
on my lhs a matrix of [b_21 b_22; b_11 b_12] and on my
rhs I
get [-b_12 -b_11; -b_22 -b_21]

For helper, did you get the same as I did? I am
trying to
understand your explanation as well.

Driga


You are doing correct.

[b_21 b_22; b_11 b_12] = [-b_12 -b_11; -b_22 -b_21]

Now this actually represents 4 different equations. The
first of which would be

b_21 = -b_12

Extract all 4 and you obtain a system of 4 equations.
See
what you can determine from them.

This is what I did to get a system with 4 equations :

0 0 0 1 b_11 -b_11
0 0 1 0 b_12 -b_12
0 1 0 0 * b_21 = -b_21
1 0 0 0 b_22 -b_22

next, I don't know what to do. For John, can you explain it
in a more simplified version, after all I am new in this
thing. Thanks all.

Have fun. Note that I've left out the information
as to when this has a solution for k = -1.

Since its homework, you need to do something.

John


function B = abba(A,k)
% abba: solves the linear system A*B = k*B*A
%
% Note: the solution will not be unique, so a specific,
non-zero
% solution is returned for general A. For k == 1, the
problem will
% always have a solution. For k = -1, the solution will not
% generally exist unless A has a certain property. (I'll
leave this
% up to the student to decide what that is.)
%
% arguments: (input)
% A - nxn (real) array
%
% k - (optional) scalar numeric
% DEFAULT: k = 1

% provide a default for k: k = 1
if (nargin<1) || isempty(k)
k = 1;
elseif numel(k)>1
error('k must be scalar if provided')
end

% verify that A is a (real) square matrix
if (nargin == 0) || isempty(A)
help abba
return
elseif ~isreal(A)
error('A must be a real numeric array')
end
sizeA = size(A);
if (length(sizeA) >2) || (diff(sizeA)~=0)
error('A must be a square matrix')
end
n = sizeA(1);

% create a linear system for A*B - k*B*A
M = kron(eye(n),A) - kron(k*A',eye(n));

% find a non-zero solution to this linear (homogenous) system
Mnull = null(M);
if isempty(Mnull)
% M had full rank
B = [];
return
end
B = reshape(sum(Mnull,2),n,n);





I will read this, and try to understand the function,
because there are a lot that I still do not know. And this
is not my homework, I just want to learn matlab.

Thanks again.

.

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