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Jun 20, 2012 at 20:42 vote accept user24579
Jun 20, 2012 at 20:41 vote accept user24579
Jun 20, 2012 at 20:42
Jun 20, 2012 at 20:41 vote accept user24579
Jun 20, 2012 at 20:41
Jun 20, 2012 at 20:41 vote accept user24579
Jun 20, 2012 at 20:41
Jun 20, 2012 at 20:38 vote accept user24579
Jun 20, 2012 at 20:41
Jun 20, 2012 at 15:45 answer added Jan Jitse Venselaar timeline score: 1
Jun 20, 2012 at 14:25 history edited François G. Dorais
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Jun 20, 2012 at 12:26 comment added user24579 Non-negative definite is equivalent to semi-positive definite, i.e., I would like to know if for an arbitrary vector $q$, the following relation holds: $q^T(I-J^{\#} J)q \ge 0$
Jun 20, 2012 at 11:56 answer added Denis Serre timeline score: 0
Jun 20, 2012 at 10:14 comment added Denis Serre The matrix $\left(I - J^\# J \right)$ is that of the projection onto $R(M^{-1}J^T)$, parallel to $\ker J$. It is not Hermitian, unless $R(J^T)$ (or equivalently $\ker J$) be stable under $M$. So, what do you mean by being non-negative definite ?
Jun 20, 2012 at 9:45 history edited user24579 CC BY-SA 3.0
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Jun 20, 2012 at 9:24 history asked user24579 CC BY-SA 3.0