Solving under-determined matrix equations

I need to Solve : A1*x+B1*y=c; A2*x+B2*y=d in Matlab where A1, A2, B1 and B2 are m-by-n complex matrices with m<n and rank=m. x, y, c and d are n-by-1 vectors. Hence the system is under-determined. I have seen solution techniques for solving system of equations in the form Ax=b, but how can I apply that to my case? Please let me know..
Thanks in advance S.Paul

2 comentarios

Youssef  Khmou
Youssef Khmou el 18 de Feb. de 2013
hi, are you sure that x,y,c and are nx1? just to make sure
Subhra
Subhra el 19 de Feb. de 2013
You are right. c and d are m x 1. only x and y are n x 1.

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Respuestas (1)

Youssef  Khmou
Youssef Khmou el 19 de Feb. de 2013
Editada: Youssef Khmou el 19 de Feb. de 2013
Hi Subhra,
try to check the lengths you assigned to vectors c,d ; Under-determined system means you have more variables and less equations , in your example c and d must be mx1 vectors to get under-determined system : In this case the matrix A is not square and other techniques must be used : like
1)The Moore-Penrose pseudoinverse matrix or
2)The Least squares algorithm , TLS,..etc
Try this example :
m=4;
n=6; % m<n
A1=randn(m,n)+j*randn(m,n);
B1=randn(m,n)+j*randn(m,n);
A2=randn(m,n)+j*randn(m,n);
B2=randn(m,n)+j*randn(m,n);
rank(A1) % rank(A1)=rank(A2)=rank(B1)=rank(B2)
c=rand(m,1);
d=rand(m,1);
% Matrices concatenation to get AX=B
A=[A1 B1;A2 B2]; % size{A} is 2mx2n and rank{A} is 2m .
B=[c;d]; % size{B}is 2mx1
% Solution
% Moore-Penrose pseudoinverse of matrix since A is not square :
X=pinv(A)*B;
error=X*A-B;
% LS
X2=pinv(A'*A)*A'*B;
% Now get x and y :
%x=X(1:4); y=X(5:end);

4 comentarios

Subhra
Subhra el 19 de Feb. de 2013
Editada: Subhra el 19 de Feb. de 2013
Hi Youssef,
Thanks a lot for your quick reply. That was really a simple idea to convert to AX=B form! Should have got that. One point though. I think, as A is 2m x 2n and rank is 2m, its right pseudoinverse exists, not the left. i.e. A*pinv(A)=I, but pinv(A)*A is not identity matrix. SO, the solution X can be modified as X=A'*pinv(A*A'). However, I have tried this and also your solutions for X. But neither of these is working. Although they are giving very small residual A*X-B, but the final field solution is erratic. My m =31, n=63.
Youssef  Khmou
Youssef Khmou el 19 de Feb. de 2013
Editada: Youssef Khmou el 19 de Feb. de 2013
hi,
ok, did you try the function "linsolve" ?
X=linsolve(A,B); %
Subhra
Subhra el 19 de Feb. de 2013
Nope. Ok I will see.
Subhra
Subhra el 19 de Feb. de 2013
No! Its not working either. My problem is ill-conditioned.

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