Timeline for Non-zero diagonal through permutation of rows
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| May 6, 2011 at 18:56 | history | edited | Ralph | CC BY-SA 3.0 |
added 198 characters in body
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| May 6, 2011 at 7:29 | answer | added | Denis Serre | timeline score: 3 | |
| May 6, 2011 at 5:49 | comment | added | Ralph | Thanks for the hint. I'll do it this way (by referring to the Leibniz formula as proposed by Gerhard above). | |
| May 6, 2011 at 4:33 | comment | added | Dima Pasechnik | if I needed to use such a statement, I would just say that "it is easy to show by induction on the size of A that..." | |
| May 5, 2011 at 22:48 | comment | added | Gerhard Paseman | I think I stated it poorly. More specifically, for the class of 0-1 nxn matrices A such that A is invertible, and such that A has a zero in every row and a zero in every column, there are examples of such A where PAQ has at most (n-1) zeros on the diagonal, where P and Q are allowed to range over all nxn permutation matrices. So your example is a counterexample to the poor statement, and says nothing about the (hopefully correctly formulated) statement above. If I fixed n to be 2 (or 3) however, then the statement above is false. Gerhard "Hope It's Right This Time" Paseman, 2011.05.05 | |
| May 5, 2011 at 22:35 | comment | added | Ralph | @Gerhard: Isn't the matrix $$(0,1) $$ $$(1,0)$$ a counterexample for your 2. comment ? | |
| May 5, 2011 at 22:13 | comment | added | Gerhard Paseman | Oops. in the above, A should be a square matrix of order n. Gerhard "Ask Me About System Design" Paseman, 2011.05.05 | |
| May 5, 2011 at 22:11 | comment | added | Gerhard Paseman | On a related note, if A is invertible (the determinant is nonzero) and there is a zero in every row and in every column, one can get at most (n-1) zeros on the diagonal of PAQ in general even allowing for permutation matrices P and Q. Gerhard "Ask Me About System Design" Paseman, 2011.05.05 | |
| May 5, 2011 at 22:08 | comment | added | Gerhard Paseman | I don't know of a reference, but you can also appeal to the determinant formula as a sum of (permuted) products for a shorter proof. Then you may not need a reference. Gerhard "Ask Me About System Design" Paseman, 2011.05.05 | |
| May 5, 2011 at 21:56 | comment | added | Ralph | @Ricky: I assume that the identity isn't zero and still hope for a reference. | |
| May 5, 2011 at 21:21 | comment | added | darij grinberg | No, Tony. Otherwise algebra would be swamped in statements like "if $I$ is an ideal of $R$, then $R/I$ is a ring or zero". | |
| May 5, 2011 at 21:19 | comment | added | Tony Huynh | Doesn't a ring with identity have at least two elements? | |
| May 5, 2011 at 21:13 | comment | added | user5810 | I'm hoping you won't find a reference of that statement, since it is false. (Consider the 1x1 matrix over the zero ring.) | |
| May 5, 2011 at 21:00 | history | asked | Ralph | CC BY-SA 3.0 |