Suppose $A \in \mathbb{R}^{nxn}$. Let $\lVert A \rVert_1 = \max \{\lVert Ax\rVert_1: \lVert x \rVert_1 \leq 1\}$ and $\lVert A \rVert_{\infty}= \max \{\lVert Ax\rVert_{\infty}: \lVert x \rVert_{\infty} \leq 1\}$. How do I show that these two norms are dual of one another?
I tried to show this by showing $$\lVert A \rVert_{\infty} = \max\{Tr(Z^TA): \lVert Z\rVert_1 \leq 1\}$$
and vice versa.
I found out that $Tr(Z^TA) = \sum_{i=1}^n z_i^Ta_i$ where $z_i, a_i$ are the columns of $Z, A$ respectively. So we have that
$$\sum_{i=1}^n z_i^Ta_i \leq \sum_{i=1}^n \lVert z_i \rVert_1 \lVert a_i \rVert_{\infty} \leq \sum_{i = 1}^n \lVert a_i \rVert_{\infty}$$
since $$1 \geq \lVert Z \rVert_1 = \max_{1 \leq j \leq n} \sum_{i = 1}^n |z_{ij}|$$
From there on, I am stuck. Also, I have no idea how to show the other direction.
Ideally, I should show this using linear-algebraic techniques. I'm not dealing with dual-spaces or anything of the like.
