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Jul 8, 2015 at 0:26 answer added Igor Rivin timeline score: 2
Jul 7, 2015 at 23:04 comment added Terry Tao If one diagonalises $A$ using a basis of (possibly complex) eigenvectors, then the coefficients of $A^n$ can be computed quite explicitly. (The parabolic case when $A$ has a repeated eigenvalue can be treated separately, and is actually rather easier than the general case as $A^n$ basically depends linearly on $n$ in that case.)
Jul 7, 2015 at 21:53 comment added pre-kidney More precisely, the question is concerned with control of the signs of the matrix entries.
Jul 7, 2015 at 21:51 history edited pre-kidney CC BY-SA 3.0
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Jul 7, 2015 at 19:58 comment added Igor Rivin How is this "non-negativity"?
Jul 7, 2015 at 19:27 history asked pre-kidney CC BY-SA 3.0