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Let $\mathbf{F}\in\mathbb{C}^{M\times M}$ and $\mathbf{D} = \operatorname{diag}(\mathbf{d})$ where $\mathbf{d}\in\mathbb{R}^M$. By SVD, $\mathbf{F}\mathbf{D}\mathbf{F}^H=\mathbf{U}\mathbf{S}\mathbf{U}^H$ where $\mathbf{U}$ and $\mathbf{S}=\operatorname{diag}(\mathbf{s})$ contain the singular vectors and singular values. Now, if $\mathbf{F}$ is a constant which is fixed and known and the only variable is $\mathbf{d}$, is there any explicit formula which expresses $\mathbf{s}$ in terms of $\mathbf{F}$ and $\mathbf{d}$?

Thanks a lot.

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  • $\begingroup$ $S=U^HFDF^HU$ --- but presumably that's not what you meant :-) $\endgroup$
    – Suvrit
    Commented Sep 25, 2013 at 12:10
  • $\begingroup$ $FDF^H$ is a congruence, which preserves the signs in $D$, in a sense that the number of the positive/negative/neutral eigenvalues in $FDF^H$ is the same as the number of positive/negative/neutral elements in $D$, respectively. If $USU^H$ is an SVD, then $U$ is unitary, so $USU^H$ is a (unitary) similarity, and the signs in $S$ have to be (up to a permutation) the same as the sings is $D$. Hence, either $D \in [0,\infty\rangle^m$, or $S$ may contain negative elements, which means that $USU^H$ is not an SVD, but an eigenvalue decomposition. You might want to check what do you really have. $\endgroup$
    – Vedran Šego
    Commented Sep 26, 2013 at 10:09

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Your matrix $F D F^H$ is Hermitian, so its singular values are the absolute values of its eigenvalues. The eigenvalues are the same as the eigenvalues of $D F^H F$ or $F^H F D$. Since $F^H F$ can be any positive semidefinite Hermitian matrix, I don't see how you can have a much more "explicit" formula than that.

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