Abstract
AbstractIn this paper we analyze the stability of the problem of performing a rational QZ step with a shift that is an eigenvalue of a given regular pencil $$H-\lambda K$$ H - λ K in unreduced Hessenberg–Hessenberg form. In exact arithmetic, the backward rational QZ step moves the eigenvalue to the top of the pencil, while the rest of the pencil is maintained in Hessenberg–Hessenberg form, which then yields a deflation of the given shift. But in finite-precision the rational QZ step gets “blurred” and precludes the deflation of the given shift at the top of the pencil. In this paper we show that when we first compute the corresponding eigenvector to sufficient accuracy, then the rational QZ step can be constructed using this eigenvector, so that the exact deflation is also obtained in finite-precision.
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