Lower bound nuclear norm of $A$ by $\mathrm{tr}(|A|)$

Question

Let $A = (a_{ij}) \in \mathbb R^{n \times n}$ be a square matrix. Let $\|A\|_\ast$ be the nuclear norm of $A$. Is the following true?

$$\sum_{i=1}^n |a_{ii}| \le \|A\|_\ast$$

Answer 1

Yes, your statement is true. In general, it can be shown that $$ \operatorname{Tr}(BA) \leq \|A\|_*\|B\|_{\infty} $$ Where $\|B\|_\infty$ is the spectral norm (i.e. the largest singular value) of $B$. Your statement can be attained by taking $B$ to be a suitable diagonal matrix with $\pm 1$ on the diagonal.

See my earlier question here for more on the general inequality.