# what is wrong with my equivalent transformation on optimization below

$$\mathbf{R_1}$$ and $$\mathbf{R_2}$$ are positive definite symmetric matrices, $$\mathbf{q}\in\mathbb{C}^{N\times1}$$, and I want to maximize the objective function below $$J(\mathbf{q})=\frac{\mathbf{q}^{H}\mathbf{R}_{1}\mathbf{q}}{\mathbf{q}^{H}\mathbf{R}_1\mathbf{q}+\mathbf{q}^{H}\mathbf{R}_2\mathbf{q}}=\frac{\mathbf{q}^{H}\mathbf{R}_{1}\mathbf{q}}{\mathbf{q}^{H}(\mathbf{R}_1+\mathbf{R}_2)\mathbf{q}}. \tag{1}\label{eq1}$$ Since $$\mathbf{q}^{H}\mathbf{R}_{1}\mathbf{q}$$ is a scalar and a positive number, I can obtain, $$J(\mathbf{q})=\frac{1}{1+\frac{\mathbf{q}^{H}\mathbf{R}_2\mathbf{q}}{\mathbf{q}^{H}\mathbf{R}_{1}\mathbf{q}}} \tag{2}\label{eq2},$$ then maximizing $$J(\mathbf{q})$$ equals to maximizing $$\frac{\mathbf{q}^{H}\mathbf{R}_1\mathbf{q}}{\mathbf{q}^{H}\mathbf{R}_{2}\mathbf{q}}$$, which is the generalized Rayleigh quotient, so $$\mathbf{q}_{\text{opt}}=$$ eigenvector corresponding to the largest eigenvalue of $$\mathbf{R}_2^{-1}\mathbf{R}_1$$.

But when reviewing \eqref{eq1}, I realize I can directly use the generalized Rayleigh quotient and obtain $$\mathbf{q}_{\text{opt}}=$$ eigenvector corresponding to the largest eigenvalue of $$({\mathbf{R}_1+\mathbf{R}_2})^{-1}\mathbf{R}_1$$. So I get confused, which is the optimal solution for \eqref{eq1} and why? I feel something is wrong with the transformation from \eqref{eq1} to \eqref{eq2}, but I cannot tell.

Note that if $$A$$ and $$B$$ are positive definite, then $$A+B$$ is positive definite.

Suppose $$v$$ is an eigenvector corresponding to eigenvalue $$\lambda$$ of $$B^{-1}A$$, $$B^{-1}Av = \lambda v$$, then we have $$Av=\lambda Bv$$

then we have $$(A+B)v=Av + Bv =Av+\frac1{\lambda} Av = (1+\frac1\lambda ) Av$$ or

$$(A+B)^{-1}Av=(1+\frac1{\lambda})^{-1}v$$

Hence if $$v$$ is an eigenvector to $$B^{-1}A$$ with eigenvalue $$\lambda$$, then $$v$$ is an eigenvector to $$(A+B)^{-1}A$$ with eigenvalue $$\left(1+\frac1{\lambda}\right)^{-1}$$ which is an increasing function of $$\lambda$$.

Hence the optimization is equivalent.

• I see, really thank you. I should focus on the eigenvector instead of the form of the matrix. Apr 8 at 10:56