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Suppose $X$ and $Y$ are nonsingular matrices. Let $\hat{A}= Y^*AX$ and $\hat{B}= Y^*BX$. We say that the pencil $\hat{A}- \lambda \hat{B}$ is equivalent to $A - \lambda B$ and that $X$ and $Y$ are equivalence transformations. $\hat{A}- \lambda \hat{B}$ and $A - \lambda B$ have the same eigenvalues. If $x$ ($y$) is a right (left) eigenvector of $A - \lambda B$, then $\hat{x}=X^{-1}x$ ( $\hat{y} = Y^{-1}y$) is a right (left) eigenvector of $\hat{A}- \lambda \hat{B}$.

Susan Blackford 2000-11-20