Matrix solving via least-squares: Y = A*B where Y, A and B are all matrices

I've been trying to get the lsqcurvefit function to find B such that A*B = Y where B and Y are mxn matrices and A is a mxm matrix of constants. The closest advice I could find is HERE but it doesn't seem to want to work when the "function" is a matrix.

Any advice or direction is greatly appreciated. Is there a better function out there to use? I wrote up a convergence algorithm to solve this but I want to check the results against something that already exists and is proven. Plus, if lsqcurvefit is anything like fit, it'd have the added bonus of returning errors for each value.

[Torsten]

What's wrong with B=A\Y ?

Best wishes

Torsten.

[Scott]

Valid point, I didn't explain it thoroughly enough.

Say we alter B(1,1). This will obviously alter all the values in the first column of Y. The kicker is sum(i,:) for Y and B are locked, thus increasing B(1,1) will increase the values of Y(:,1) and in turn require decreasing of B(1,2:end) and Y(:,2:end).

Thanks for your prompt reply. The lack of information is completely on me. I think, given the conditions, a fitting method perhaps isn't feasible.

[John D'Errico]

You could NEVER have used lsqcurvefit to solve the problem anyway, since it does not allow linear equality constraints. As well, lsqcurvefit does not return statistics on the parameters. As Matt points out, lsqlin will solve the problem, however it too does not return the confidence information about the parameters that you wish.

Answer 1

[Matt J]

Since your model is linear and you have linear constraints on B (its row sums), then LSQLIN is the better tool to use

   b=lsqlin(C,d,[],[],Aeq,beq);
   B=reshape(b,m,n);

where

    C=kron(speye(n),A);
    d=Y(:);
    Aeq=kron(ones(1,n),speye(m));
    beq=rowsums;

[Scott]

That's a huge help, I cannot thank you enough! I was reading through the help file and it appears I can have upper and lower bounds too. Thanks so much for your help!