Timeline for Null space vs. semi-positive definite matrix
Current License: CC BY-SA 3.0
12 events
when toggle format | what | by | license | comment | |
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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: $q^T(I-J^{\#} J)q \ge 0$
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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 |
added 14 characters in body
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Jun 20, 2012 at 9:24 | history | asked | user24579 | CC BY-SA 3.0 |