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Nov 20 at 12:41 history edited ADNNNNNNNNNNN CC BY-SA 4.0
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Nov 20 at 12:40 comment added ADNNNNNNNNNNN Yup exactly ! Q is (strictly) positive definite
Nov 20 at 5:34 answer added msantama timeline score: 0
Nov 20 at 5:12 comment added msantama That would be true without the linear equality constraints. In the original post, though, don’t we need $V^\top V - W^\top W + D^2 \succ 0$ instead of the stated condition to ensure (strict) convexity?
Nov 20 at 2:33 comment added copper.hat Surely the solution is $x=-{1 \over 2} Q^{-1} c$?
Nov 20 at 0:09 history edited RobPratt CC BY-SA 4.0
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Nov 19 at 21:56 history edited ADNNNNNNNNNNN
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Nov 19 at 21:42 history edited ADNNNNNNNNNNN CC BY-SA 4.0
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Nov 19 at 21:42 history edited ADNNNNNNNNNNN CC BY-SA 4.0
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S Nov 19 at 21:40 review First questions
Nov 19 at 21:40
S Nov 19 at 21:40 history asked ADNNNNNNNNNNN CC BY-SA 4.0