Effect of column normalization on maximum diagonal entry

Question

Let $\mathbf{A}$ be a $M\times N$ complex matrix, and $\bar{\mathbf{A}}$ be constituted by normalizing each column of $\mathbf{A}$. Therefore, we have $$\mathbf{A}=\bar{\mathbf{A}}\mathbf{\Gamma},$$ where $\mathbf{\Gamma}=\mathrm{diag}(\|\mathbf{a}_n\|)$ is a diagonal matrix and $\mathbf{a}_n$ is the $n$th column of matrix $\mathbf{A}$. Is there any relation between these two following quantities: \begin{align} &\max_m \quad [\mathbf{A}\mathbf{A}^\mathrm{H}]_{m,m},\\ &\max_m \quad [\bar{\mathbf{A}}\bar{\mathbf{A}}^\mathrm{H}]_{m,m}, \end{align} where $[\cdot]_{m,m}$ is the $m$th diagonal entry of $\mathbf{A}$ and $(\cdot)^{\mathrm{H}}$ denotes conjugate transpose operation.

Answer 1

Let $a_{m,n}$ be the elements of matrix $\mathbf{A}$. Then, we have \begin{align} \beta_{\max}&=\max_m \quad [\mathbf{A}\mathbf{A}^\mathrm{H}]_{m,m}=\max_m \sum_{n=1}^{N}|a_{m,n}|^2\\ \bar{\beta}_{\max}&=\max_m \quad [\bar{\mathbf{A}}\bar{\mathbf{A}}^\mathrm{H}]_{m,m}=\max_m \sum_{n=1}^{N}\frac{|a_{m,n}|^2}{\|\mathbf{a}_n\|^2}. \end{align} Therefore, by defining $\alpha_{\max}=\max_n \|\mathbf{a}_n\|$ and $\alpha_{\min}=\min_n \|\mathbf{a}_n\|$, we have \begin{align} \bar{\beta}_{\max}&=\max_m \sum_{n=1}^{N}\frac{|a_{m,n}|^2}{\|\mathbf{a}_n\|^2}\\ &\leq \max_m \sum_{n=1}^{N}\frac{|a_{m,n}|^2}{\alpha_{\min}^2}\\ &=\frac{\beta_{\max}}{\alpha_{\min}^2}. \end{align} Similarly, we obtain \begin{align} \bar{\beta}_{\max}\geq\frac{\beta_{\max}}{\alpha_{\max}^2}. \end{align}