solve $\underset{d \in \mathbb{R}^n}{min}$ $g^Td$ subject to $d^THd = 1$ where H $\in \mathbb{R}^{n \times n}$ is positive definite and symmetric

Question

I would like to solve the optimization problem $$\underset{d \in \mathbb{R}^n}{min} \ g^Td$$ subject to $$d^THd = 1$$ where H $\in \mathbb{R}^{n \times n}$ is positive definite and symmetric $\textbf{without}$ using Lagrange multipliers.

Does someone know of a way to obtain the solution without relying on Lagrange Multipliers? It would be sufficient to show that $d^* := -\frac{H^{-1}g}{||H^{-1}g||_H}$ where $|| \cdot ||_H = \sqrt{d^THd}$ is the optimal solution (i.e. $g^Td^* \leq g^Td \ \ \ \forall \ d \in \mathbb{R}^n$)

Answer 1

Let us introduce a new inner product $[\cdot,\cdot]$ on $\mathbb{R}^n$ by setting $[u,v]=(u,Hv)\equiv u^THv$. This is possible, because $H$ is symmetric and positive definite. We have $||u||^2_H=[u,u]$. Now the constraints say that $d$ lies on the unit sphere (with respect to the new inner product), and we have $g^Td=[d,H^{-1}g]$. Where does $[d,H^{-1}g]$ have minimum on the unit sphere? Obviously, the minimum is attained at the unit vector that has the direction opposite to $H^{-1}g$, i.e., at the vector $$d^*=-H^{-1}g/||H^{-1}g||_H.$$ Indeed, we have $[d^*,H^{-1}g]=-||H^{-1}g||_H$, and for any $d$ on the unit sphere one has $|[d,H^{-1}g]|\le ||d||_H||H^{-1}g||_H=||H^{-1}g||_H$ and hence $[d,H^{-1}g]\ge-||H^{-1}g||_H$.