Timeline for About Sylvester's determinant
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Mar 2 at 18:11 | comment | added | Michael Albanese | @passerby51: You're right. Thanks. | |
| Mar 2 at 18:10 | history | edited | Michael Albanese | CC BY-SA 4.0 |
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| Mar 1 at 23:01 | comment | added | passerby51 | $\text{det}(I_n + v^T A^{-1}v)$ perhaps should just be $1 + v^T A^{-1}v$. At least it should be $I_1$ not $I_n$. | |
| Nov 18, 2014 at 4:05 | history | edited | Michael Albanese | CC BY-SA 3.0 |
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| Nov 10, 2014 at 22:51 | history | edited | Michael Albanese | CC BY-SA 3.0 |
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| Nov 10, 2014 at 18:19 | history | edited | Michael Albanese | CC BY-SA 3.0 |
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| Nov 10, 2014 at 10:08 | comment | added | Branimir Ćaćić | @Anirbit Let $S$ be an $m \times n$ matrix of rank $1$. By the rank-nullity theorem, the nullity of $S$ is $(n-1)$, and hence the orthogonal complement of the nullspace of $S$ is $1$-dimensional; pick a unit vector $v$ in the orthogonal complement of the nullspace of $S$. You can then check that $S = uv^T$ for $u := Sv$. | |
| Nov 10, 2014 at 8:52 | comment | added | Student | @QiaochuYuan :'( That is somehow not very familiar to me. Can you sketch the argument may be? | |
| Nov 10, 2014 at 7:31 | comment | added | Qiaochu Yuan | @Anirbit: that is more or less the definition of being rank one. | |
| Nov 10, 2014 at 5:50 | comment | added | Student | So the crucial thing here seems to be the fact that "A rank one m×n matrix can be written as the outer product of some non-zero u and v." Can you kindly reference this? | |
| Nov 10, 2014 at 5:49 | vote | accept | Student | ||
| Nov 10, 2014 at 3:11 | history | answered | Michael Albanese | CC BY-SA 3.0 |