I'm trying to compute $$ \|z\|_* := \sup_{\|x\|_A\leq1} |z^T x| $$ for $\|x\|_A^2 := x^T A x$ with $A \succ 0$, symmetric. I see that we're looking at $z^T \in \mathbb{R}^n$ as a linear functional, so this is really the operator norm with $|\cdot|$ over $\mathbb{R}$ and $\|\cdot\|_A$ over $\mathbb{R}^n$.
I make a change of coordinates $y = A^{1/2}x$ and substitute to get $$ \|z\|_* := \sup_{\|y\|_2\leq1} |z^T A^{-1/2}y|. $$ Based on @max_zorn's comment, I use symmetry of $A$ so that $$ \|z\|_* = \sup_{y^T y=1}|A^{-1/2}z^T y| = \sup_{\|y\|_2=1}|A^{-1/2}z^T y| \leq \|A^{-1/2}z\|_2 \|y\|_2 $$ with equality at $y = A^{-1/2}z / \|A^{-1/2}z\|_2$ so $\|z\|_* = \|A^{-1/2}z\|_2$.
What is the functional analysis explanation for why the dual norm should be computable as the euclidean norm of the transformed vector $A^{-1/2}z$? How can I think about duality more intuitively in this case?
When $A$ is symmetric and positive definite, $A^{-1/2}$ is also symmetric and positive definite. Thus $$\|A^{-1/2} z\|_2 = \|z\|_{A^{-1}}.$$ Thus, $\|\cdot\|_A$ and $\|\cdot\|_{A^{-1}}$ are dual to each other. I don't have an intuitive reason for why this is true, but it at least seems quite natural.