I would appreciate if one could help me figure out this problem.

I have a matrix $G$ (for simplicity assume square matrix $n\times n$). I know that if I multiply $G$ with a unitary matrix $U$ as ($A = GU$), the resultant matrix has the same distribution as $G$, i.e., the distribution of the singular values of $G$ and $A$ are equal/identical.

Now let's assume $k<n$ and $V_{n\times k}$ is a semi-unitary matrix ($V^H V = I_{k\times k}$). I am interested in the distribution of the singular values of $B=GV$. Clearly, $B$ has $k$ singular values. Is there anyway to make connection between the singular values of $B$ and $G$.

Any help/hint/reference is appreciated.

  • 2
    $\begingroup$ Isn't it $B=GV$ ? $\endgroup$ – Jean Marie Feb 22 '16 at 21:39
  • $\begingroup$ sorry for the mistake. you are right. I just edited the question. $\endgroup$ – M.X Feb 23 '16 at 13:42

Let $\sigma_1\geq\cdots\geq \sigma_n$ be the singular values of $G$ and let $x\in \mathbb{C}^k$ be a unitary vector. Then $x^*(B^*B)x=(Vx)^*(G^*G)(Vx)=y^*(G^*G)y$ where $y$ is a unitary vector. Then $x^*(B^*B)x\in [\sigma_n^2,\sigma_1^2]$ and the non-zero singular values of $B$ are in $[\sigma_n,\sigma_1]$.

  • $\begingroup$ I see that the assertion "Any help/hint/reference is appreciated" is not true. $\endgroup$ – loup blanc Apr 21 '16 at 9:40

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