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.

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

  • $\begingroup$ thanks a lot!, let me go through it thoroughly again, I also came through the initial steps and was thinking about posing a question on null spaces of kronecker product of two matrices. Your last steps are bit involved. Let me check it:) $\endgroup$ – dineshdileep Nov 4 '12 at 15:12
  • $\begingroup$ Okay, I'll pause the edits for now, as I am attempting my own better understanding as well. $\endgroup$ – adam W Nov 4 '12 at 15:14
  • $\begingroup$ I just saw the edit. $R$ is a rank-one positive definite matrix with $R=rr^{H}$. So its only non-zero eigen value is $||r||^{2}$. The highest eigen-value of the original system is $x^{H}B^{-1}x$. So do you mean to say answer is $||r||^{2}x^{H}B^{-1}x$ $\endgroup$ – dineshdileep Nov 4 '12 at 15:17
  • $\begingroup$ Yes, though $\otimes R$ makes the GEVP singular, and that may give some other issues. Such as the $x\otimes z$ where $x$ is free and $Rz=0$. A trivial solution? $\gamma$ is free there too. $\endgroup$ – adam W Nov 4 '12 at 15:37
  • $\begingroup$ if it is of interest to you , please go to the link given below, it deals with GEVP containing singular positive semi-definite matrices. I believe we can apply those results here. google.co.in/… $\endgroup$ – dineshdileep Nov 4 '12 at 16:08

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