Timeline for On nilpotent singular $\mathbb F_2^{n\times n}$ matrices
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Apr 19, 2023 at 0:22 | comment | added | Benjamin Steinberg | Sorry, in my last comment I meant of order n which is a power of 2 | |
| Apr 18, 2023 at 20:12 | comment | added | Benjamin Steinberg | If n is a power of 2 the answer to 3 is k=n. Take I+P where P is the permutation matrix of order 2^n corresponding to the cyclic permutation (1,2,...,n) | |
| Apr 18, 2023 at 19:33 | history | edited | Turbo | CC BY-SA 4.0 |
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| Apr 18, 2023 at 19:33 | history | undeleted | Turbo | ||
| Apr 18, 2023 at 19:28 | history | deleted | Turbo | via Vote | |
| Apr 18, 2023 at 19:25 | comment | added | Benjamin Steinberg | The answer to 4 is no. If n is odd let M be the all ones matrix. | |
| Apr 18, 2023 at 19:13 | comment | added | Benjamin Steinberg | The nilpotent elements do not form a subsemigroup. Take E_12 E_21=E_11 | |
| Apr 18, 2023 at 16:10 | comment | added | Turbo | I think it is called index when the matrix vanishes valnishes. There is no nomenclature for the $k$ introduced. | |
| Apr 18, 2023 at 15:18 | history | edited | Turbo | CC BY-SA 4.0 |
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| Apr 18, 2023 at 7:34 | comment | added | Gerry Myerson | $M=\pmatrix{1&1\cr0&0\cr}$ has determinant zero, but there are no permutations $P,Q$ such that $PMQ$ is nilpotent. | |
| Apr 18, 2023 at 4:16 | history | edited | Turbo | CC BY-SA 4.0 |
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| Apr 18, 2023 at 3:54 | history | asked | Turbo | CC BY-SA 4.0 |