Asked by GEO GEP
on 25 Apr 2012

Hello:)

I m trying to solve XA=B where both A,B are matrix (instead of B being a vector) Using e.g. LU decomposition ('linsolve' or '/') is possible to obtain such a solution.

However i need to constrain X>0.

Is this an optimization problem (min(||XA-B||),X>0,B), and if it is can someone propose a suitable function ?

Thank you

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Answer by Teja Muppirala
on 26 Apr 2012

Accepted answer

Solving for each row of X is an independent optimization problem that can be solved easily with LSQNONNEG (available from the Optimization Toolbox). Use a loop to solve for each row independently.

Example 1 (test when know the exact answer):

% Set up some data A = rand(5); Xtrue = rand(5); B = Xtrue*A;

% Solve for each row of X using LSQNONNEG X = []; for k = 1:size(B,1) X(k,:) = lsqnonneg(A',B(k,:)'); end

% Verify the result X - Xtrue

Example 2:

A = rand(6,3); B = rand(6,3);

X = []; for k = 1:size(B,1) X(k,:) = lsqnonneg(A',B(k,:)'); end

% Verify that all X are positive X

Note that if your data is very big, this algorithm could easily be sped up by running it in parallel.

Show 1 older comment

Teja Muppirala
on 26 Apr 2012

(I was lazy, but in general you'd probably want to preallocate X here since you do know its size)

Teja Muppirala
on 26 Apr 2012

Good point, Richard

GEO GEP
on 26 Apr 2012

If A=3x3, B=3x3 (and X=3x3), then as Richard said X = BA^{-1}, and either will or will not violate the constraints (there's nothing I can do about it).

However my system can have an arbitrary number of columns where A=3x(3*n), B=3x(3*n), n E R (and X=3x3). If i understand correctly both problems can be tackled with multiple lsqnonneg (or linprog)...

Can this problem be (also) solved by a non negative matrix factorization nnmf (B=W*H, enforcing somehow H=A)

Thank you -so much- for your answers

Answer by bym
on 25 Apr 2012

does this answer help? http://www.mathworks.com/matlabcentral/answers/24086-ols-regression-for-multiplr-ys-xs

Answer by Richard Brown
on 25 Apr 2012

It very much depends on your matrices. What are the dimensions? Rank?

If A square and full rank then `X` is uniquely determined as *X = BA^{-1}*, and either will or will not violate the constraints (there's nothing you can do about it).

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