Infimum of an Matrix-induced inner product Hello this is my first post here!
Does anyone have a proof for this standard fact?

If $H$ is an $n \times n$ positive definite matrix and $\mathbf n \in \Bbb R \setminus \{0\}$ then
$$
\inf_{\mathbf y \in \Bbb R^n, \langle \mathbf n, \mathbf y \rangle=1} 
\langle \mathbf y,H \mathbf y \rangle = 
\frac{1}{\langle \mathbf n, H^{-1} \mathbf n\rangle}
\quad \text{and} \quad
\underset{\mathbf y \in \Bbb R^n, \langle \mathbf n, \mathbf y \rangle=1}{\operatorname{argmin}} = \frac{H^{-1} \mathbf n}{\langle \mathbf n, H^{-1}\mathbf n \rangle}.
$$

Couldn't find it anywhere. Thank you!
 A: Note: I assume in this context that "positive definite" implies symmetric, so that $H$ is symmetric.

One approach is to use the method of Lagrange multipliers. We want to minimize $f(\mathbf y )= \langle \mathbf y,H \mathbf y\rangle$ subject to the constraint that $g(\mathbf  y) = \langle \mathbf n,\mathbf y \rangle = 1$. We compute
$$
\nabla f = 2H \mathbf y, \quad \nabla g = \mathbf n.
$$
Thus, if $\mathbf y$ for which the minimum is achieved, then there must be a $\lambda$ such that
$$
\nabla f = \lambda \nabla g \implies 2H \mathbf y = \lambda \mathbf n \implies \mathbf y = \frac{\lambda}2 H^{-1}\mathbf n.
$$
On the other hand, because $g(\mathbf y) = 0$, we have
$$
\left \langle \mathbf n,
\frac{\lambda}{2}H^{-1} \mathbf n
\right \rangle = 1 \implies \lambda = \frac{2}{\langle \mathbf n, H^{-1}\mathbf n\rangle}.
$$
Thus, if a minimum is achieved, it must be achieved at
$$
\mathbf y = \frac {\lambda}{2}H^{-1}\mathbf n = \frac{H^{-1} \mathbf n}{\langle \mathbf n,H^{-1} \mathbf n\rangle}
$$
and the minimum must be
$$
f \left( \frac{H^{-1} \mathbf n}{\langle \mathbf n,H^{-1} \mathbf n\rangle} \right) =
\frac {1}{\langle \mathbf n, H^{-1}\mathbf n \rangle}.
$$

An alternative proof: let $M$ be such that $H = M^TM$ (such an $M$ could be found via Cholesky decomposition for instance). Make the substitution $\mathbf x = M\mathbf y$. We can rewrite the minimization as
$$
\inf_{\langle \mathbf n,  M^{-1}\mathbf x \rangle = 1} \langle M^{-1}\mathbf x, H M^{-1}\mathbf x\rangle = \\
\inf_{\langle M^{-T}\mathbf n, \mathbf x \rangle = 1} \langle M^{-1}\mathbf x, M^T(M M^{-1})\mathbf x\rangle = \\
\inf_{\langle M^{-T}\mathbf n, \mathbf x \rangle = 1} \langle \mathbf x, \mathbf x\rangle.
$$
