random generation of a matrix subject to a certain constraint

Hello there,

Is there a way to randomly generate a mx1 matrix A, such that transpose(A)*B*A>=1? where B is an mxm matrix?

Thank you!

[John D'Errico]

What do you know about B? Is it symmetric? Positive definite? Is it real valued or complex?

[John D'Errico]

Anyway, this question is poorly posed, in the same sense that it makes no sense to ask to choose a random positive integer. Without any definition of the distribution of those numbers, it is impossible to generate such a solution.

[Mnr]

B is a real valued positive semi definite matrix.

Answer 1

[John D'Errico]

B is real, semi-positive definite (or non-negative definite) with a complete set of eigenvectors/values.

Ws can always write A as a linear combination of the eigenvectors.

A = sum(c_i*V_i)

Here V_i is the i'th eigenvector, and c_i is the coefficient in the linear combination. I've used sum sloppily here, as a summation over i, not as the MATLAB function sum.

Then what is A'*B*A?

B*A = sum(c_i*lambda_i*V_i)

If we then take the dot product with the vector A, we need to recognize that the vectors V_i are orthogonal, with unit norm. So dot(V_i,V_i) == 1, and dot(V_i*V_j)=0 for i~=j. Therefore we can write

A'*B*A = sum(c_i^2*lambda_i)

Since the matrix is non-negative definite, all of the lambda_i are real, and non-negative. Therefore, if you will choose a vector A such that A'*B*A>=1, all you need do is choose the c_i such that

sum(c_i*lambda_i) >= 1

As I said, this problem is not well-posed, since without definition of a distribution for the c_i, we cannot choose such a random vector. Regardless, this is a sufficient result for you to do as you wish.

[Mnr]

Thank you!