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 $\lVert A_d \rVert_2 = \sqrt{n}\lVert A \rVert_2$?

  • $\begingroup$ "I can prove that $\lVert A_d \rVert_2 \leq \sqrt{n}\lVert A \rVert_2$ by using inequalities with the Frobenius norm". What is the norm $\|\cdot\|_2$ in your question? The standard notation for Frobenius norm is $\|\cdot\|_F$, while $\|\cdot\|_2$ conventionally denotes the induced $2$-norm (i.e. the largest singular value). $\endgroup$ – user1551 Mar 23 at 4:36
  • $\begingroup$ Yes, $\lVert \cdot \rVert_2$ is the $2$-norm, but proving the inequality goes through the Frobenius norm. $\endgroup$ – chaos Mar 23 at 8:20

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 $A=0$.

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  • $\begingroup$ Wonderful, so you have proved that $\lVert A_d \rVert_2 \leq \lVert A \rVert_2$, which is clearly the best bound. Thank you. $\endgroup$ – chaos Mar 23 at 9:19

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