How to find non-zero solutions to underdetermined systems of equations

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fa wu
fa wu 2024년 10월 25일
댓글: fa wu 2024년 10월 25일
I am solving an underdetermined system of equations and Matlab gives me a zero vector. Zero vector is indeed a solution to the underdetermined system of equations, but it is not helpful. I would like to get a non-zero solution. It would be best if I could specify the positive or negative values ​​of the components of the solution vector. For example, x1>0
A=[0.719600000000000,0.069400000000000,0.526400000000000,0.110700000000000;1.833200000000000,0.377900000000000,1.469000000000000,0.454700000000000];
b=[0;0];
x = A\b

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Torsten
Torsten 2024년 10월 25일
편집: Torsten 2024년 10월 25일
The null() function gives you a basis for ker(A):
A=[0.719600000000000,0.069400000000000,0.526400000000000,0.110700000000000;1.833200000000000,0.377900000000000,1.469000000000000,0.454700000000000];
N = null(A)
N = 4×2
-0.5051 0.0193 -0.3585 -0.6051 0.7700 -0.1124 -0.1531 0.7879
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A*N
ans = 2×2
1.0e-15 * -0.0173 0.0139 0.0833 0.1110
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So every linear combination of these two vectors will give a vector n with A*n = 0.
"lsqlin" could be used instead if you want to set lower and upper bounds in the coordinates of n. But note that it might happen that no solution exists.
The following code solves for a vector with positive first component n1:
C=[0.719600000000000,0.069400000000000,0.526400000000000,0.110700000000000;1.833200000000000,0.377900000000000,1.469000000000000,0.454700000000000];
d=[0;0];
Aeq = [1 0 0 0];
beq = 1;
n = lsqlin(C,d,[],[],Aeq,beq)
Minimum found that satisfies the constraints. Optimization completed because the objective function is non-decreasing in feasible directions, to within the value of the optimality tolerance, and constraints are satisfied to within the value of the constraint tolerance.
n = 4×1
1.0000 0.6629 -1.5306 0.3624
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C*n
ans = 2×1
1.0e-09 * 0.8290 -0.2444
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