Let $A, B \in \mathbb{C}^{n\times n}$, and A, B are Hermitian matrix ($A^H = A, B^H = B$). $x \in \mathbb{C}^{n \times 1}$ is the decision variable, then optimization problem is \begin{equation} \text{max}\colon f\left( x \right) = \frac{x^H A x}{x^H B x}, \quad s.t. \lVert x \rVert^2 = 1 \end{equation}
Questions:
- How to solve the maximum value of $f(x)$ and corresponding $x$.
- How to solve the question 1 when the problem is $\text{min}\colon f(x)$.
The restriction $\|x\|^2=1$ is useless, by homogeneity. The max $M$ is the max of $$\frac{y^HB^{-1/2}AB^{-1/2}y}{\|y\|^2}$$ as shown by the change of variable $y=B^{1/2}x.$ The eigenvalues of the positive definite matrix $A_1=B^{-1/2}AB^{-1/2}$ being $0<\lambda_1\leq \ldots\leq \lambda_n$ we have $A_1=U^HDU$ where $U$ is unitary and $D=\mathrm{diag}(\lambda_1,\ldots,\lambda_n).$ Using $z=Uy$ and the fact that $\|z\|^2=\|y\|^2$ we have $$M=\max_{y}\frac{y^HU^HDUy}{\|y\|^2}=\max _{z}\frac{z^HDz}{\|z\|^2}=\max_{\|z\|^2=1}(\lambda_1|z_1|^2+\cdots+\lambda_n|z_n|^2)=\lambda_n.$$ The best $z$, $y$ and $x$ are $z=(0,\ldots,0,1)^T,$ $y=U^Hz$ and $x=B^{-1/2}U^Hz.$
Obviously the minimum $m$ is $\lambda_1.$