Bound on the $2$-norm of a diagonal sub-matrix

Question

Let $A \in \mathbb{R}^{n \times n}$ be an invertible real matrix and write $A_d$ for the sub-matrix consisting of its diagonal part only, namely $(A_d)_{ij} = A_{ij}$ if $i = j$ and $0$ otherwise. I can prove that $$\lVert A_d \rVert_2 \leq \lVert A_d \rVert_F \leq \lVert A \rVert_F \leq \sqrt{n}\lVert A \rVert_2 $$ but can this inequality be improved? In other words, can we find $A$ invertible such that $\lVert A_d \rVert_2 = \sqrt{n}\lVert A \rVert_2$?

Answer 1

Let $|a_{ii}|$ be the diagonal entry of $A$ with the largest magnitude and let $e_i$ be the $i$-th vector in the standard basis of $\mathbb R^n$. Then $$ \|A_d\|_2=|a_{ii}|\le\|Ae_i\|_2\le\max_{\|u\|_2=1}\|Au\|_2=\|A\|_2. $$ Clearly this inequality is tight when $A$ is a diagonal matrix. It also implies that $\|A_d\|_2=\sqrt{n}\|A\|_2$ only if $\sqrt{n}\|A\|_2\le\|A\|_2$. Therefore, provided that $n\ge1$, $\|A_d\|_2$ can possibly be equal to $\sqrt{n}\|A\|_2$ only when $n=1$ or $A=0$.