Timeline for Are $n \times n$ special orthogonal matrices, all the entries of which have the same absolute value, possible for $n \neq 4$?
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Aug 29 at 19:13 | comment | added | Oscar Lanzi | Any power of $2$ works, really. | |
| Feb 3, 2020 at 16:51 | answer | added | Gerhard Paseman | timeline score: 0 | |
| Feb 3, 2020 at 14:12 | history | edited | Paul B. Slater | CC BY-SA 4.0 |
"mis-link" hopefully corrected
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| Feb 3, 2020 at 5:15 | review | Close votes | |||
| Feb 14, 2020 at 3:05 | |||||
| Feb 3, 2020 at 0:29 | comment | added | Steven Stadnicki | Note that the construction works both ways; the existence of matrices of the form you're looking for is precisely equivalent to the existence of a Hadamard matrix, by scaling of the entries. The Wikipedia page on Hadamard matrices notes that they can only exist in dimensions $1$, $2$, and $4n$, so you won't find any other examples. | |
| Feb 3, 2020 at 0:21 | history | edited | Paul B. Slater | CC BY-SA 4.0 |
error acknowledged at end
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| Feb 3, 2020 at 0:12 | comment | added | LSpice | If you wish to change your question, then you should edit the question itself. | |
| Feb 2, 2020 at 23:58 | answer | added | Francois Ziegler | timeline score: 1 | |
| Feb 2, 2020 at 23:45 | comment | added | Paul B. Slater | Thanks for the comments! Well, I had taken the Kronecker product of a $2 \times 2$ and a $4 \times 4$ Hadamard matrix to presumably get an $8 \times 8$ one, which proved to be not orthogonal, that is its matrix product with its transpose was not proportional to the identity. But it looks like I should probably recheck my computations/references--in view of the comments---just rechecked. My $H_2$ was miscoded--so mea culpa and the $H_8$ is orthogonal. But what about $n$ not a mulitple of 4. | |
| Feb 2, 2020 at 23:44 | history | edited | Paul B. Slater | CC BY-SA 4.0 |
/edit deleted from linked website
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| Feb 2, 2020 at 22:36 | comment | added | Gerhard Paseman | Also, there are matrices called complex Hadamard matrices of all orders which can supply such orthogonal matrices with entries from the complex numbers. Gerhard "Going Off In Another Dimension" Paseman, 2020.02.02. | |
| Feb 2, 2020 at 22:10 | comment | added | Yemon Choi | Echoing @user44191: Hadamard matrices are precisely the $\pm 1$-valued matrices such that $A^\top A$ is equal to a scalar multiple of the identity, so rescaling one of these always gives an orthogonal matrix whose entries share the same absolute value. The determinant 1 condition should be easy to check in various examples. | |
| Feb 2, 2020 at 21:25 | comment | added | Gerry Myerson | I think you meant to link to math.stackexchange.com/questions/3510189/… | |
| Feb 2, 2020 at 20:56 | comment | added | user44191 | How is an 8 by 8 Hadamard matrix "not orthogonal in character"? | |
| Feb 2, 2020 at 20:51 | history | asked | Paul B. Slater | CC BY-SA 4.0 |