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$\mathbf{R_1}$ 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.

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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.

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  • $\begingroup$ I see, really thank you. I should focus on the eigenvector instead of the form of the matrix. $\endgroup$
    – lei zhou
    Apr 8 at 10:56

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