Timeline for Solution of a quadratic matrix equation with singular coefficent
Current License: CC BY-SA 3.0
9 events
when toggle format | what | by | license | comment | |
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S Dec 7, 2017 at 20:50 | history | suggested | Rodrigo de Azevedo | CC BY-SA 3.0 |
Minor improvements
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Dec 7, 2017 at 20:17 | comment | added | Robert Israel | In that case your equation can't have positive semidefinite solutions, except in the trivial case $a=0$: $a^T(C S C + C + A) a \ge \|a\|^4 > 0$. | |
Dec 7, 2017 at 20:14 | comment | added | Umer Abdullah | Yes. Apologies for the lack of clarity. $\mathbf{S}$ is symmetric, it is also the product of a column vector with its transpose $\mathbf{S}=s∗s^{T}$ | |
Dec 7, 2017 at 19:50 | comment | added | Rodrigo de Azevedo | is $S$ symmetric? | |
Dec 7, 2017 at 19:44 | review | Suggested edits | |||
S Dec 7, 2017 at 20:50 | |||||
Dec 7, 2017 at 19:17 | history | edited | Umer Abdullah | CC BY-SA 3.0 |
edited body
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Dec 7, 2017 at 18:36 | history | edited | Umer Abdullah | CC BY-SA 3.0 |
added 310 characters in body
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Dec 7, 2017 at 18:21 | review | First posts | |||
Dec 7, 2017 at 18:31 | |||||
Dec 7, 2017 at 18:17 | history | asked | Umer Abdullah | CC BY-SA 3.0 |