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Excerpt from Matrices That Generate the Same Krylov Residual Spaces
If the matrix A is normal, then h'(z) is one, and it was recently shown in several different ways that this bound is sharp; i.e., that for each Is, there is an initial vector (depending on k) for which equality holds in In many cases of interest. However. The matrix A is not normal and the factor in (7) may be quite large. (see. For instance. [13] for some interesting physical examples.) In such cases. The bound (7) may be a large overestimate of the actual residual.
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