# Prove that maximum value is the largest eigenvalue

Given a symmetric positive definite matrix $$A \in \mathbb{R}^{n \times n}$$ and a full rank matrix $$B \in \mathbb{R}^{n \times n}$$. Prove that the maximum value the following optimization problem is the largest eigenvalue of $$B^{-1}B^{-T}A$$.

$$\begin{array}{ll} \text{maximize} & x^T A \, x\\ \text{subject to} & \|Bx\| = 1 \end{array}$$

Can you help me with this one?

• What vector norm are you using in the constraint? – Rodrigo de Azevedo Mar 13 at 14:24
• I think it just says that it is a unit vector. – Lostinspace Mar 13 at 14:26
• Unit in what norm? I assume it's the $2$-norm. – Rodrigo de Azevedo Mar 13 at 14:27
• You could introduce a new variable $y:=Bx$. Or use Lagrange multipliers. What have you tried so far? – Rodrigo de Azevedo Mar 13 at 14:28
• It $B^{-1}B^{-T}A$ should be $B^{-T}AB^{-1}$. – xpaul Mar 13 at 14:29

Let $$y=Bx$$. Then $$x=B^{-1}y$$ and the problem becomes: Prove that the maximum value of $$y^T(B^{-T}AB^{-1})y$$ subject to $$||y||=1$$ is the largest eigenvalue of $$B^{-T}AB^{-1}$$. It is not hard to attain the conclusion.