Sign in to follow this  
George Tuma

Solution of linear equation

Recommended Posts

Quote:

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.

Share this post


Link to post
Share on other sites
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

Share this post


Link to post
Share on other sites
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.

Share this post


Link to post
Share on other sites
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.

Share this post


Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now

Sign in to follow this