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Feb 13, 2018 at 23:46 comment added SIM2 Oh now I can understand what's going on. Thank you so much Irving!
Feb 13, 2018 at 23:29 comment added Geoffrey Irving Apologies: fixed ${n \choose 2}$ to $O(n^2)$.
Feb 13, 2018 at 23:28 history edited Geoffrey Irving CC BY-SA 3.0
deleted 7 characters in body
Feb 13, 2018 at 23:21 comment added SIM2 So do I have to consider ${ n \choose 2} + n$ minors? And am I undetstanding minor of intervals properly : A[i:j][i:j] for some i and j?
Feb 13, 2018 at 23:10 comment added Geoffrey Irving If you have a diagonal matrix with a -1, the interval minor that is just -1 will have negative determinant -1. In general, the determinant of any minor of a tridiagonal matrix will be a product of determinants of minors of intervals. If all interval minors are nonnegative, all their products will be.
Feb 13, 2018 at 23:07 comment added SIM2 I'm not sure considering minors given by intervals is enough. Diagonal matrix of (0,-1,0,...) looks like a counterexample. Maybe additionally checking diagonal elements gives wanted results, but it will be really thankful if you give me a reference.
Feb 13, 2018 at 21:44 comment added Geoffrey Irving I edited the post to handle semidefiniteness more efficiently.
Feb 13, 2018 at 21:43 history edited Geoffrey Irving CC BY-SA 3.0
Described how to do positive semidefiniteness.
Feb 13, 2018 at 21:34 comment added SIM2 But I'm not sure that I can avoid floating-point error if I use $\epsilon $. The reason I prefer Sylvester's criterion is that I can ensure determinant of integer matrix is integer, so use $\ round(det(M))$ instead of $\ det(M)$. I can still use reflections, thanks for your answer again.
Feb 13, 2018 at 21:14 comment added SIM2 I have to make sure that I need positive 'semi'-definiteness, not just definiteness.Householder reflection preserves positive semi-definiteness, not only positive definiteness? Thanks for your answer!
Feb 13, 2018 at 21:04 history answered Geoffrey Irving CC BY-SA 3.0