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.