Hi, guys. I got a problem in my recent research. It can be sketched as follows:
How to theoretically choose a positive \alpha such that the matrix D=B*B^T/\alpha+B*inv(A)*B^T is nonsingular or well-conditioned? Here B is an m-by-n matrix (n >=m) and A is an n-by-n nonsingular matrix .
I think that you need to have more conditions on B.
For example, if B has all zero entries, then D=0 for any alpha, so there is no alpha that meets your criteria in that case.
I have found that a useful way to find the conditions you need to obtain the results you desire (D non-singular or well-conditioned for some value of alpha) is to try fairly non-restrictive conditions, and then find counter-examples. The counter examples can provide insight on what other conditions are needed.
I think that you need to have more conditions on B.
For example, if B has all zero entries, then D=0 for any alpha, so there is no alpha that meets your criteria in that case.
I have found that a useful way to find the conditions you need to obtain the results you desire (D non-singular or well-conditioned for some value of alpha) is to try fairly non-restrictive conditions, and then find counter-examples. The counter examples can provide insight on what other conditions are needed.
Hello Zhang,
One method is to compute the conditioning number of your D matrix which is norm(D)*norm(inv(D)) if it is closer to 1 it is very well-conditioned else if it is not far from this value it is well-conditioned else it is ill-conditioned. Researchers have used to calculate the two-norm because they have directly the maximum and minimum singular values. You can also use the one-norm or others. It is important to notice that a nonsingular matrix doesn't imply that it is well-conditioned.
If the operator norm of I-D is less than 1 then D would be non-sigular. To compute this norm find the eigenvalues of (I-D)(I-D)*. Then
|| I-D ||^2 = maximum { | \lambda |, \lambda is an eigenvalue of (I-D)(I-D)* }.
Here I denotes the identity matrix. I hope it helps.
Assuming A is regular (INV(A) exists and is unique), your D=B(INV(A) + alpha 1)B^T. The kernel is an eigenvalue problem for alpha:
INV(A)_{ij} + lam 1_{ij} = 0_{ij} or det(INV(A)+lam 1) = 0
If your value for alpha is an eigenvalue, lam, of that eigenproblem, your D will be singular. The further away from any eigenvalue, the better;)
Conditions on B are that it does not reduce the rank of INV(A)+alpha 1.
You may find some useful conditions by expressing INV(A) in terms of its eigenvalues, lam_k, and eigenbases phi_k tensor phi_k, then studying the effect of B on eigenvectors phi_k.
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