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

*No products are associated with this question.*

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)

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).

MATLAB and Simulink resources for Arduino, LEGO, and Raspberry Pi test

Learn moreOpportunities for recent engineering grads.

Apply Today
## 0 Comments