# Could someone help me to prove or disprove the following inequality?

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 $$(c_{nr})$$.

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

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 $$1/\sigma_{\rm max}$$.

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