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Oct 14, 2012 at 7:59 answer added Federico Poloni timeline score: 1
Oct 14, 2012 at 7:42 answer added Dima Pasechnik timeline score: 1
Oct 14, 2012 at 0:31 comment added R.P. There exists a fixed matrix $S$ such that $SAS^{-1}$ is in Jordan normal form. The same is then true for $B_k = SA_kS^{-1}$. So now consider $x^T A_k^{-1} A x = x^T S^{-1} S A_k^{-1} S^{-1} S A x = x^T S^{-1} (S A_k S^{-1})^{-1} S A x = x^T S^{-1} B_k^{-1} S A x$. And $B_k$ is easy to invert.
Oct 13, 2012 at 23:38 history asked mks CC BY-SA 3.0