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 $(c\_{nr})$.



Could someone help me to prove or disprove the following inequality?
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 $1/\sigma\_{\rm max}$.


