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George Tuma

Solution of linear equation

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A real or Hermitian square matrix A is positive semi-definite or non-negative definite if xAx >=0 for all non-zero x.

but that doesnt really say anything about the solution.

all i know about the solution is that you can find it using the conjugate gradients method, but thats about it.

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I am not an expert in linear algebra. It is only a question how to comment some procedure. For example a problem to find the minimum value of the objective function J(x)=x’(kA’A+I) x-y’x-x’y+y’y leads to the formulae x=inv(kA’A+I)*y (x,y are vectores, A is a rectangular matrix). The matrix kA’A+I is square symmetric and positive definite.
More general problem to minimize J(x1,x2)= k1 x1’A1’A1x1+ k2 x2’A2’A2x2+(y’-x1’-x2’) (y-x1-x2) gives a system matrix which is only positive definite. What can I say about the matrix solution? From the theoretical point of view, solution exists or not? I prepared a script for Matlab based on the PCG method for x1,x2,…,xP (arbitrary P) and it works.

Best regards

George

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A mistake:

More general problem to minimize J(x1,x2)= k1 x1’A1’A1x1+ k2 x2’A2’A2x2+(y’-x1’-x2’) (y-x1-x2) gives a system matrix which is only positive semidefinite.

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No, the resulting matrix A'A is only positive semidefinite for any rectangular matrix A (A' is transpose of A). You are right only for the square regular matrix A.

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