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Singular values of a matrix times a diagonal matrix

  1. Jan 4, 2010 #1

    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 [tex]u_i[/tex], [tex]v_i[/tex], and [tex]\sigma_i[/tex], such that [tex]BB^*v_i = \sigma_i^2v_i[/tex] and [tex]B^*Bu_i = \sigma_i^2u_i[/tex]. Where [tex]B^*[/tex] 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,
  2. jcsd
  3. Jan 4, 2010 #2


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    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

    [tex]B=W\Lambda W^H[/tex]

    where [tex]\Lambda[/tex] 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

    [tex]W^H = W^{-1}[/tex]

    it is easy to show that

    [tex]B^{-1}=W\Lambda^{-1} W^H[/tex]

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

    [tex]BB^H=W\Lambda \Lambda^HW^H=W|\Lambda|^2W^H[/tex]

    applied to one of the eigenvectors [tex]w_i[/tex] gives


    Your singlular values squared [tex]{\sigma_i}^2[/tex] are known to be the eigenvalues of [tex]BB^H[/tex], 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.
    Last edited: Jan 4, 2010
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