Maximise the following expression subject to the constraint

[brendan_foo]

Hi guys,

I wish to maximise the following expression subject to the constraint that $\|\underline{w}\| = 1$, and $\mathbb{R}$ is fixed.

$$P = \underline{w}^H \mathbb{G}^H\mathbb{G}\underline{w}<br /> = \underline{w}^H \mathbb{R} \underline{w}$$

where

$$\mathbb{R} \triangleq \mathbb{G}^H\mathbb{G}$$

I proceed to determine the maximum value, and the value of $\underline{w}$ that achieves it through the general eigenvalue problem and the Cauchy-Schwarz inequality.

Recall that I can only modify w, and its norm is fixed to unity.

$$|\langle \underline{w},\mathbb{R}\underline{w}\rangle| \leq \|\underline{w}\|\|\mathbb{R}\underline{w}\|$$

This achieves equality iff

$$\mathbb{R}\underline{w} = \lambda \underline{w}$$

and will be subsequently maximised if the dominant eigenvector is chosen, such that

$$\mathbb{R}\underline{w} = \lambda_{max} \underline{w}$$

Which then yields a maximum value of $P$ as

$$P = \lambda_{max} \|w\|^2 = \lambda_{max}$$

I just want to doubly check with you guys that this is correct.

Thanks!

 

[fresh_42]

The if clause is ok, I'm not sure about the only if direction, i.e. whether an eigenvector is forced. The above is a quadratic form, so a classical solution with e.g. Lagrange multipliers should do.