Singular values of a matrix times a diagonal matrix

[jdevita]

Hi,

I have been struggling with this problem for a while, and I have not found the answer in textbooks or google. Any help would be very much appreciated.

Suppose I know the singular value decomposition of matrix B, which is a singular, circulant matrix. That is, I know u_i, v_i, and \sigma_i, such that BB^*v_i = \sigma_i^2v_i and B^*Bu_i = \sigma_i^2u_i. Where B^* is the conjugate transpose.

Now let A = DB, where D is a diagonal matrix. Is there any way to determine the singular values and vectors of A from the singular values and vectors of B?

Thank you,
Jason

 

[marcusl]

You should take advantage of the special and desirable properties of circulant matrices, namely, use an eigenvalue decomposition (EVD) on B instead of a SVD. Every circulant nxn matrix B has the EVD

B=W\Lambda W^H

where \Lambda is a diagonal matrix of eigenvalues and where the columns of W contain the eigenvectors. W contains the complete basis set for the complex discrete Fourier transform of length n, regardless of details of B. Since

W^H = W^{-1}

it is easy to show that

B^{-1}=W\Lambda^{-1} W^H

and B is singular if has one or more zero eigenvalue. To expand on your result, note that

BB^H=W\Lambda \Lambda^HW^H=W|\Lambda|^2W^H

applied to one of the eigenvectors w_i gives

BB^Hw_i=|\lambda_i|^2w_i.

Your singlular values squared {\sigma_i}^2 are known to be the eigenvalues of BB^H, and comparison to the above shows that they are in fact the eigenvalues squared of B.

Multiplying B by a diagonal matrix D removes the circulant symmetry, and I don't see a simple relation between the expansion of A and that of B.