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Feb 7, 2016 at 17:48 history edited user9072
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Oct 16, 2015 at 20:02 comment added Joe Silverman Write $J=UBU^{-1}$ in Jordan normal form. Then your limit is $U^{-1}\left(\lim_{x\to\infty}(UAU^{-1}+xJ)^{-1}\right)U$. So if $B$ is diagonalizable, you're reduced to thediagonal case that you already analyzed. If $J$ is not diagonal, it's a bit more complicated,but should be computable. Presumably the eigenvalue zero blocks of $J$ are the ones that are important.
Nov 28, 2013 at 16:21 history asked MrX CC BY-SA 3.0