Maximal scalar product of a vector in sphere of A-Norm.

Question

Given some vector $v\in \mathbb R^n\setminus\{0\}$, a scalar product $\langle \cdot,\cdot \rangle$ and a norm $||\cdot||$ I want to find a unit vector $x$ with $||x||=1$ such that $|\langle v,x \rangle|$ is maximal.

If the norm is induced by the scalar product i.e. $||x||:=\sqrt{\langle x,x \rangle}$ then the solution is simply $\pm \frac{v}{||v||}$, because with Cauchy-Schwarz we attain: $|\langle v,x \rangle| \leq ||v||\cdot ||x||= ||v|| = |\langle v,\pm \frac{v}{||v||} \rangle|$.

But what if this is not the case - in particular, what if we are given the standard scalar product $\langle x,y \rangle := x^Ty$ and A-norm $||x||:=\sqrt{x^TAx}$ for some symmetric, positive definite Matrix $A$?

I read that $x = A^{-1}v$ would be optimal, have not yet seen a proof of this.

Answer 1

$v = |v| e_v$, where $||e_v|| = 1$

Use the Cauchy-Schwartz inequality: $$|\langle v,x \rangle|^2 \le \langle v,v \rangle \cdot \langle x,x \rangle = ||v||^2$$

But $$|\langle v,e_v \rangle|^2 = ||v||^2$$ So it's necessarily a maximum.

Note that I used the induced norm $||.||$ which is not necessarily the same that you provided in your question.

Answer 2

This can be solved using Lagrange multiplier. You want to maximize $f(x) = \langle v, x\rangle$ subject to $g(x) = x^tAx = 1$.

First we use a matrix to represent the inner product $\langle x, y\rangle = x^t B y$. Write $w = Bv$. Then $f(x) = \langle v, x\rangle = w^t x$.

By Lagrange multiplier, $\nabla f = \lambda \nabla g$ as a maximum. Thus $$ w = \lambda A x\Rightarrow x = \frac{1}{\lambda } A^{-1} w = \frac{1}{\lambda} A^{-1} Bv.$$

Thus the maximum/minimum occurs when $$x= \pm \frac{A^{-1}Bv}{\|A^{-1}Bv\|}.$$