A particular Generalized Eigenvalue Problem Data
I have three $N \times N$ complex hermitian matrices $A=xx^{H}$,$R=rr^{H}$ and a positive-definite matrix $B$. Here $x$ and $r$ are two $N \times 1$ complex vectors. Let $\lambda_{i}, 1\leq i\leq N$ denotes the N eigenvalues of B which are also positive. Clearly $A$ and $R$ are two rank one positive semi-definite matrices. $B$ is invertible. 
What I need to find


*

*What is the largest eigenvalue of the GEVP?


\begin{align}
(A\otimes R)v=\gamma (B\otimes R)v
\end{align}


*

*Will the maximum eigenvalue be (seemingly nice) $||r||^{2}x^{H}B^{-1}x$?


What I know


*

*Consider the generalized Eigenvalue problem (GEVP)
\begin{align}
Av=\gamma Bv
\end{align}
Since $B$ is invertible, this is equivalent to find the eigenvalues of $B^{-1}A$, in fact , since $A$ is rank one matrix, there is only one eigenvalue which will be positive, and it will be given by $x^{H}B^{-1}x$ ($A=xx^{H}$).

*Now I am interested in the matrices, $A \otimes R$ and $B \otimes R$ which are $N^{2} \times N^{2}$ in dimension. Now $A \otimes R$ is a rank one matrix, and its only non-zero eigenvalue is $||x||^{2}||r||^{2}$. $B \otimes R$ is a positive semi-definite matrix with $N$ of its eigenvalues being $\lambda_{i}||r||^{2}, 1\leq i \leq N$ and rest of the $N^{2}-N$ eigenvalues being zero. 

 A: Use the mixed multiplication property of the Kronecker product with $v=x\otimes z$:
\begin{align}
(A\otimes R)(x\otimes z)= & \gamma (B\otimes R)(x\otimes z)  \\
 (Ax)\otimes(Rz) = & \gamma(Bx)\otimes(Rz)\\
\end{align}
 and notice that any $z$ such that $Rz=0$ solves it.
EDIT The following is some of the origianl derivations, left here since I feel the question of the values of $\gamma$ is still unclear in my mind. I hope something here helps, give me your thoughts later and maybe I will see something more.
\begin{align}
(A\otimes R)v= & \gamma (B\otimes R)v & \text{the original equation}\\
(A\otimes R)v- \gamma (B\otimes R)v = & 0 & \text{subtraction of right term }\\
\left[A\otimes R- \gamma (B\otimes R)\right]v = & 0 & \text{(reverse) distribution of $v$}\\
\left[A\otimes R+ (-\gamma B)\otimes R\right]v = & 0 & \text{scalar distributes to either term of Kronecker}\\
\left[(A-\gamma B)\otimes R\right]v = & 0 & \text{Kronecker (reverse) distribution of $R$}\\
\end{align}
The use if $B^{-1}$ is still possible here. Use $B^{-1} \otimes I$ and the mixed multiplaction property of the Kronecker:
\begin{align}
(B^{-1} \otimes I)\left[(A-\gamma B)\otimes R\right]v = & 0 \\
\left[B^{-1}(A-\gamma B)\otimes IR\right]v = & 0 \\
\left[(B^{-1}A-\gamma I)\otimes R\right]v = & 0
\end{align}
Eigenvectors for both may be used here. Let $(B^{-1}A-\gamma I)x=\lambda_x x$ so that $x$ is the right eigenvector, and $Ry=\lambda_y y$ similarly for $R$. Then we have using the mixed multiplication property
\begin{align}
& \left[(B^{-1}A-\gamma I)  \otimes R\right]\left(x \otimes y\right) \\
=& \left[(B^{-1}A-\gamma I)x\right]  \otimes \left(Ry\right) \\
=& \left(\lambda_x x\right) \otimes \left(\lambda_y y\right) \\
=& \left(\lambda_x I x\right) \otimes \left(\lambda_y Iy\right) \\
=& \left(\lambda_x I  \otimes \lambda_y I \right) \left(x \otimes y\right) \\
=& \lambda_x \lambda_y(I \otimes I)\left(x \otimes y\right) \\
=& \lambda_x \lambda_y\left(x \otimes y\right)
\end{align}
The eigenvalues of $R$ multiply with those of the original system
