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I need to bound the eigenvalues of a matrix of the form $D^\top C D$, where:

  • $D \in \mathbb{R}^{n^2 \times n(n-1)/2}$ is a duplication matrix, in particular the matrix for which $D \operatorname{vechh}(X)= vec(X)$, i.e. restore the vectorization of the matrix $X$ from its strictly lower (or upper triangular part). This matrix has exactly one $1$ in each row, and two $1$'s in each column. Its rank is $n(n-1)/2$. Let me call this number $l=n(n-1)/2$.
  • $C$ is a PSD matrix.

Upper Bound. Denote with $\|\cdot\|$ denoting the spectral norm, i.e. the maximum eigenvalue of $\sqrt{(X^\top X)}$, equivalently, the maximum singular value of $X$.

I was starting by $$\|D^\top C D\| \leq \|D^\top\|\| C\|\| D\| \leq \|D^\top\|_F\| C\|\| D\|_F$$

Because $\|D\|_F= \sqrt{\operatorname{trace(D^\top D)}}=\sqrt{\operatorname{trace(2 I_{l})}}=\sqrt{2l}= \sqrt{n(n-1)} $. I can then write

$$\|D^\top C D\| \leq n(n-1)\lambda_{max}(C)$$

But I think this is not really a good bound. Do you have possible improvements in mind?

Lower bound. Assuming the knowledge of the eigenvalues of $C \succ 0$, how does the pre and post multiplication for $D$ shapes the eigenvalues of $C$? I was starting with the definition of max singular value:

$$ \|D^\top C D\| = \sqrt{ \sup_{\|x\|=1} x^\top D^\top C D x} $$.

From here, could an EVD of the matrix $C$ help in moving forward?

yes
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  • The spectral norm $|D|^2=2$ which tightens the bound to $|D^\top C D| \leq 2,|C|;$ – greg Feb 26 '21 at 14:52
  • It comes from the $D^\top D$ property right? – yes Feb 27 '21 at 09:32
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    This link combined with the fact that $\frac D{\sqrt{2}}$ is semi-orthogonal should get you there. – greg Feb 27 '21 at 16:41
  • Thank you for the answer. I would like to accept your answer, if you post it. I have also expanded my question, to try to find a better lower bound than zero. – yes Mar 01 '21 at 10:29
  • @greg Do you have some suggestions on how to move forward in the lower bound? I edited the question for readability. – yes May 31 '21 at 12:41
  • Ostrowski’s theorem may be helpful: $\lambda_i(D^\top CD)=\theta_i\lambda_i(C)$ for some $\theta_i\in[\sigma_n(D)^2,\sigma_1(D)^2]$. – user1551 May 31 '21 at 16:18
  • @user1551 thank you for your suggestion. Do you have some references? – yes May 31 '21 at 17:34
  • See Horn and Johnson's Matrix Analysis. In the 2nd edition, the theorem is stated as theorem 4.5.9 on p.283. – user1551 May 31 '21 at 18:21
  • In that theorem, my matrix D should be non-singular. Here, I use a rectangular matrix, so it is certainly not invertible... – yes May 31 '21 at 18:51

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