Related Eigenproblems

1. If $B$ is nonsingular, then the NHEP $B^{-1}A x = \lambda x$ has the same eigenvalues $\lambda_i$ and corresponding right eigenvectors $x_i$ as $Ax = \lambda Bx$. Similarly, $AB^{-1}z = \lambda z$ has the same eigenvalues $\lambda_i$ as $Ax = \lambda Bx$ and right eigenvectors $z_i = Bx_i$. If $A$ is nonsingular, $A^{-1}B x = \mu x$ has the same right eigenvectors $x_i$ as $Ax = \lambda Bx$, and its eigenvalues are reciprocals $\mu_i = 1/\lambda_i$. Finally, if $A$ is nonsingular, $BA^{-1} z = \mu z$ has reciprocal eigenvalues $\mu_i = 1/\lambda_i$ and right eigenvectors $z_i = Ax_i$. Analogous statements can be made about left eigenvectors.

2. More generally, suppose $Ax = \lambda Bx$ has eigenvalues $\lambda_i$ and corresponding right eigenvectors $x_i$. Let $\alpha_1$, $\alpha_2$, $\beta_1$, and $\beta_2$ be scalars such that $\alpha_1 \beta_2 - \alpha_2 \beta_1 \neq 0$. Then $(\alpha_1 A + \beta_1 B) x = \mu (\alpha_2 A + \beta_2 B)x$ has the same eigenvectors $x_i$ as $Ax = \lambda Bx$ and eigenvalues $\mu_i = (\alpha_1 \lambda_i + \beta_1)/(\alpha_2 \lambda_i + \beta_2)$. If one or both of $\alpha_2 A + \beta_2 B$ and $\alpha_1 A + \beta_1 B$ are nonsingular, then the method in item 1 above can be applied.

3. Let $p( \lambda ) = A_n \lambda_n - \sum_{i=0}^{n-1} A_i \lambda^i$ be an $m$-by-$m$ matrix polynomial, where ${\rm det } p(\lambda)$ is not identically zero. An eigenpair $(\lambda_i, x_i)$ of $p(\lambda)$ satisfies $p(\lambda_i)x_i = 0$. Define the $mn$ by $mn$ block companion pencil of $p(\lambda)$ as
   
   $$A - \lambda B = \bmat{ccccc} 0-\lambda I & I & & & \\ & ... ... A_0 & A_1 & \cdots & A_{n-2} & A_{n-1} -\lambda A_n \emat,$$
   
   
   where all entries are $m$ by $m$ blocks and all entries not explicitly shown are 0. Then $Ax = \lambda Bx$ is a regular generalized eigenvalue problem, and the eigenvalues of $A - \lambda B$ are the eigenvalues of $p(\lambda)$. Note that there are $mn$ eigenvalues. If $(\lambda_i, x_i)$ is an eigenpair of $p(\lambda)$, then $[ x_i^T, \lambda_i x_i^T , \ldots , \lambda_i^{n-1} x_i^T]^T$ is a right eigenvector of $C$ [194].