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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?

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

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  • $\begingroup$ What should I need to show to prove it? $\endgroup$ – Lostinspace Mar 13 at 14:38
  • $\begingroup$ @Lostinspace, you can use the normal form theory of quadratic forms to prove it. $\endgroup$ – xpaul Mar 13 at 14:46

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