Supremum of an inner product constrained to an ellipsoid

Question

I am self-studying Boyd's convex optimization textbook, and I am befuddled by Example 9.1 shown here in screenshot. Specifically, I am confused how he gets that the supremum of $q^{\top} z$ over the ellipsoid is equal to that first term on the right-hand side.

I tried to formulate it as a constrained optimization problem on the surface of the ellipsoid, with a Lagrange multiplier, but it doesn't seem to work.

Any insights?

Example 9.1 from Boyd's Convex Optimization textbook

Answer 1

So try centering it first, $\mathcal{E}_0 = \{x| x^TA^{-1}x \leq 1 \}$ then we have that $$\sup_{z \in \mathcal{E}}q^Tz = \sup_{y\in \mathcal{E_0}}(q^Ty) + q^Tx_0.$$

Now consider $\mathcal{E_0}$, since $A$ is positive definite we have that $\mathcal{E} = \{x| \langle A^{-1/2}x, A^{-1/2}x \rangle \leq 1 \} = A^{1/2}B$ where $B$ is the unit ball in whatever Hilbert space you're working in. So for the last part $$\sup_{y \in \mathcal{E}_0}q^Ty = \sup_{\|u\|\leq 1}(q^T A^{1/2}u) = \|A^{1/2}q\|_2$$

The last step again used positive definiteness.