Solving Systems of Linear Equations
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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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采纳的回答
Teja Muppirala
2012-4-26
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.
更多回答(2 个)
bym
2012-4-25
does this answer help? http://www.mathworks.com/matlabcentral/answers/24086-ols-regression-for-multiplr-ys-xs
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Richard Brown
2012-4-25
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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