Timeline for If $A^TA \ge B^TB$ does this imply $AA^T \ge BB^T$?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Jun 23, 2020 at 6:36 | comment | added | Terry Tao | A counterexample has already been given, but one can already guess the answer by noting that the two inequalities have different symmetries. The former inequality is invariant with respect to multiplying A,B on the left by arbitrary orthogonal matrices, whilst the latter is invariant with respect to multiplying A,B on the right by arbitrary orthogonal matrices. | |
| Jun 23, 2020 at 6:18 | review | Close votes | |||
| Jun 30, 2020 at 3:01 | |||||
| Jun 23, 2020 at 6:13 | vote | accept | Andrea Nardi | ||
| Jun 23, 2020 at 5:57 | answer | added | gerw | timeline score: 5 | |
| Jun 23, 2020 at 5:54 | comment | added | vidyarthi | what is the ordering $\ge$ used here? and are the matrices all square? | |
| Jun 23, 2020 at 5:41 | review | First posts | |||
| Jun 23, 2020 at 5:43 | |||||
| Jun 23, 2020 at 5:35 | history | asked | Andrea Nardi | CC BY-SA 4.0 |