Timeline for How to characterize real square matrices A, such that v'Av >= 0, for all real vectors v with 1'v=0 (1 is the vector of all ones)?
Current License: CC BY-SA 2.5
7 events
| when toggle format | what | by | license | comment | |
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| Nov 5, 2010 at 3:50 | comment | added | Will Jagy | I see your definition, I must agree that you have the condition correct. The matrix you want is in fact not by nature symmetric, so the quickest method is to take the symmetric part of $A$ and apply Suvrit's answer. Note that if the result is semidefinite instead of definite, you will need to check second partial derivative information for the null directions. | |
| Nov 5, 2010 at 3:39 | vote | accept | daizhuo | ||
| Nov 5, 2010 at 3:32 | history | edited | daizhuo | CC BY-SA 2.5 |
added 144 characters in body
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| Nov 4, 2010 at 17:53 | answer | added | Will Jagy | timeline score: 2 | |
| Nov 4, 2010 at 8:28 | answer | added | Suvrit | timeline score: 9 | |
| Nov 4, 2010 at 3:52 | answer | added | Gerry Myerson | timeline score: 0 | |
| Nov 4, 2010 at 3:21 | history | asked | daizhuo | CC BY-SA 2.5 |