Timeline for When is the following block matrix invertible?
Current License: CC BY-SA 3.0
10 events
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| Aug 11, 2022 at 14:37 | answer | added | MDR | timeline score: 0 | |
| Sep 11, 2017 at 6:34 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Aug 12, 2017 at 5:51 | answer | added | Rodrigo de Azevedo | timeline score: 2 | |
| Aug 11, 2017 at 6:30 | comment | added | Balaji sb | @AlexM as Robert pointed out i do mean $det(A)$ when A is considered as $dn \times dn$ matrix. | |
| S Aug 10, 2017 at 21:13 | history | suggested | Rodrigo de Azevedo | CC BY-SA 3.0 |
Added tags, minor edits
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| Aug 10, 2017 at 20:46 | review | Suggested edits | |||
| S Aug 10, 2017 at 21:13 | |||||
| Aug 10, 2017 at 20:21 | comment | added | Robert Israel | @AlexM. Presumably what the OP means is $\det(A)$ where $A$ is considered as a $dn \times dn$ matrix over $\mathbb F_q$. | |
| Aug 10, 2017 at 17:58 | comment | added | Alex M. | @Balajisb: "other than saying the $\det(A)$ polynomial in $x_{ij}$ must be non-zero" - careful here, because $A$ is a matrix over $M_n (\Bbb F_q)$ which is not commutative. You need first a notion of determinant for matrices with entries in a non-commutative ring; there are several approaches to define one, one of them being the concept of "quasideterminant". | |
| Aug 10, 2017 at 16:59 | comment | added | Gerhard Paseman | If the Aij are all the identity matrix, (and the xij are also the same), I do not see how A can have determinant different from 0. Indeed, even if the identity matrix is only for I=1,2 and all j (or the transpose) and similarly for the xij, I see A as being singular. Gerhard "And Q Doesn't Matter Here" Paseman, 2017.08.10. | |
| Aug 10, 2017 at 13:34 | history | asked | Balaji sb | CC BY-SA 3.0 |