# Maximum of a generalized Rayleigh quotient

Given two symmetric positive definite matrices $A,B\in\mathbb{R}^{n\times n}$ and $x\in\mathbb{R}^n$.

How do I prove that the generalized Rayleigh quotient

$R(A,B,x):=\dfrac{x\cdot A\cdot x}{x\cdot B\cdot x}$

has a maximum $\lambda_{max}$, which corresponds to the largest eigenvalue of the generalized eigenvalue problem

$A\cdot x=\lambda\cdot B\cdot x$?

I have tried to complete the proof by substituting $B=C^2$ and $y=C\cdot x$ to get

$\dfrac{x\cdot A\cdot x}{x\cdot B\cdot x}=\dfrac{y\cdot C^{-T}\cdot A\cdot C^{-1}\cdot y}{y\cdot y}$.

Defining $D:=C^{-T}\cdot A\cdot C^{-1}$ we have $D=D^T$ and then it is simple to show that the maximum eigenvalue of $D$ is the maximum of the quotient by substituting $D=\sum\limits_j\lambda_jv_j\otimes v_j$ and $y=\sum\limits_jy_jv_j$ into the quotient. I can, however, not seem to prove that the eigenvalues of $D$ from

$D\cdot v_j=\lambda_jv_j\quad\Leftrightarrow\quad C^{-T}\cdot A\cdot C^{-1}\cdot v_j=\lambda_jv_j\quad\Leftrightarrow\quad A\cdot C^{-1}\cdot v_j=\lambda_j\cdot C^T\cdot v_j$

correspond to the eigenvalues of

$A\cdot v_j=\lambda_j\cdot C^2\cdot v_j\quad\Leftrightarrow\quad A\cdot v_j=\lambda_j\cdot B\cdot v_j$.

Setting $w_j = C^{-1} v_j$, we have $$A(C^{-1}v_j) = \lambda_j C^T v_j \iff\\ A(w_j) = \lambda_j (C^T C) w_j$$ Making sure that you've chosen $C$ to be normal (verify that such a choice always exists), we note that $C^TC$ and $C$ share eigenvectors.
• I believe this answer suffices to complete the proof. Your last remark is useful, but I believe it is not even necessary to complete the proof, since the eigenvalues $\lambda_j$ do not change under the substitution $w_j:=C^{-1}\cdot v_j$. And thus, $\lambda_{max}$ is still the largest eigenvalue of the generalized eigenproblem $A\cdot x=\lambda\cdot B\cdot x$. – Adriaan Nov 11 '14 at 9:41
• Ah, you're right. I was still thinking in terms of the eigenvalues of $A$. – Omnomnomnom Nov 11 '14 at 12:12