0
$\begingroup$

Let $(c_{nr})$ be an $N\times R$ complex matrix, then $\forall z_n \in \mathbb{C}$, we have $$ \sum_r \Big|\sum_n c_{nr}z_n\Big|^2 \geq \frac{1}{\sigma_{max}} \sum_n |z_n|^2 $$ where $\sigma_{max}$ is the maximal sigular value of the complex matrix $(c_{nr})$.

$\endgroup$
3
  • 2
    $\begingroup$ This is false: take for $(z_n)$ a nonzero vector in the kernel of the matrix. $\endgroup$ – abx Sep 16 '20 at 12:16
  • 2
    $\begingroup$ the correct inequality should have $\sigma_{\rm min}^2$ instead of $1/\sigma_{\rm max}$ $\endgroup$ – Carlo Beenakker Sep 16 '20 at 12:18
  • 2
    $\begingroup$ This cannot be true on dimensional grounds (imagine scaling the matrix by a factor $\lambda$). $\endgroup$ – gmvh Sep 16 '20 at 12:24
4
$\begingroup$

The $N\times R$ matrix $C$ has elements $c_{nr}$, the $N\times N$ matrix $Z$ has elements $z_n \bar{z}_m$, and $C^\ast$ is the conjugate transpose of $C$. The Hermitian matrix product $CC^\ast$ has eigenvalues $\sigma_n^2$, with $\sigma_n\geq 0$, $n=1,2,\ldots N$ the set of singular values of $C$. Then we have $$\sum_r \left|\sum_n c_{nr}z_n\right|^2 = {\rm tr}\, (CC^\ast Z) = \sum_{n=1}^N \sigma_n^2 |\zeta_n|^2\geq \sigma_{\rm min}^2 \sum_n |\zeta_n|^2=\sigma_{\rm min}^2 \sum_n |z_n|^2,$$ with $\sigma_{\rm min}$ the smallest of the singular values and the vector $\zeta$ obtained from $z$ by a unitary transformation. This is not the inequality in the OP, which should have $\sigma_{\rm min}^2$ instead of $1/\sigma_{\rm max}$.

$\endgroup$
1
  • $\begingroup$ Great! Many thanks to you! $\endgroup$ – Milin Sep 17 '20 at 1:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.