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Equivalences (Congruences)

Suppose $X$ is a nonsingular matrix. Let $\hat{A}= X^*AX$ and $\hat{B}= X^*BX$. We say that the pencil $\hat{A}- \lambda \hat{B}$ is congruent to $A - \lambda B$, and that $X$ is a congruence transformation. If $A$ and $B$ are Hermitian, with $B$ positive definite, than $\hat{A}$ and $\hat{B}$ have these same properties. Furthermore, $\hat{A}- \lambda \hat{B}$ and $A - \lambda B$ have the same eigenvalues, and if $x$ is an eigenvector of $A - \lambda B$, so that $Ax = \lambda Bx$, then $\hat{x}=X^{-1}x$ is an eigenvector of $\hat{A}- \lambda \hat{B}$.

Susan Blackford 2000-11-20