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

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

• "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). – user1551 Mar 23 at 4:36
• Yes, $\lVert \cdot \rVert_2$ is the $2$-norm, but proving the inequality goes through the Frobenius norm. – 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 $$A=0$$.
• Wonderful, so you have proved that $\lVert A_d \rVert_2 \leq \lVert A \rVert_2$, which is clearly the best bound. Thank you. – chaos Mar 23 at 9:19