Talk:Frobenius theorem (real division algebras)
![]() | This article is rated Start-class on Wikipedia's content assessment scale. It is of interest to the following WikiProjects: | ||||||||||
|
Untitled
[edit]What about Octonions? Jim Bowery (talk) 15:59, 24 July 2014 (UTC)
- They're not an associative algebra. Double sharp (talk) 15:31, 14 November 2015 (UTC)
Is it possible someone would be able to include a proof for this?
Serious mistake in proof
[edit]If is orthonormal basis, then by definition of orthonormality, so the claim is wrong (actually not completely wrong, but doesn't make any sense since ) and the following arguments about quaternions and case n>2 are also wrong.
Also there's a type in case n=2: , it should be for the case of quaternions, but it would contradict with orthogonality. — Preceding unsigned comment added by 78.41.194.15 (talk) 11:03, 3 October 2012 (UTC)
There is no mistake: Orthonormality says that the inner product is zero, not that the algebra product is zero. The definition of the inner product is . Thus . Mike Stone (talk) 15:32, 11 June 2013 (UTC)
Proof not encyclopedic
[edit]The proof section, regardless of its accuracy, is not written in an encyclopedic tone. It appears the proof was probably just copied from a paper, but as it stands the only attempt to make it useful to the average reader is the division into arbitrary subsections which are not useful ("The finish" yah I can see that). I lack the knowledge to straighten it out, but I hope someone else can. Integral Python click here to argue with me 17:48, 1 September 2020 (UTC)
- I understood the proof fine. It is in fact an application of the fundamental theorem of algebra and the Cayley-Hamilton theorem, as well as other bits of linear algebra such as the trace of a matrix, the rank-nullity theorem, and basic properties of bilinear forms. The occurrence of the trace of a matrix is due to the Cayley-Hamilton theorem, and the only property of the trace that gets used is that it is linear and maps surjectively to . Anybody with enough of a pure maths education in linear algebra should be able to follow the argument. One well-known textbook which covers all the relevant material is Linear Algebra Done Right by Sheldon Axler. --Svennik (talk) 15:47, 11 October 2021 (UTC)
- I removed some terminology I felt unnecessary like codimension and replaced it with the usual dimension. Also, I made explicit the use of the rank-nullity theorem, and removed the jargon linear form. I've introduced some additional Latex as I wasn't sure how to write using the math template. I hope I haven't introduced any mistakes, or made the proof harder to read. What do people think? I'm still wondering whether the trace map is completely necessary, given that it's only used because:
- * It is the second-leading coefficient of the characteristic polynomial of a linear map. This fact is easily verified.
- * It is a linear map, i.e. it satisfies and .
- * In the context of the proof, we can give its domain and codomain as .
- * It is surjective over its codomain, allowing us to use the rank-nullity theorem to find the dimensionality (dimension?) of its kernel.
- --Svennik (talk) 12:20, 12 October 2021 (UTC)
Proof - finish
[edit]“ | If k > 2, then D cannot be a division algebra. Assume that k > 2. Let u = e1e2ek. It is easy to see that u2 = 1 (this only works if k > 2). If D were a division algebra, 0 = u2 − 1 = (u − 1)(u + 1) implies u = ±1, which in turn means: ek = ∓e1e2 and so e1, ..., ek−1 generate D. This contradicts the minimality of W. | ” |
Possibly I'm being slow, but I don't understand why u2 = 1. Alaexis¿question? 12:31, 30 April 2024 (UTC)
- I think I've realised my error. Maybe we could make this clearer for the reader as follows
“ | If k > 2, then D cannot be a division algebra. Assume that k > 2. Define u = e1e2ek and consider u2=(e1e2ek)*(e1e2ek). By rearranging the elements of this expression and applying the orthonormality relations among the basis elements we find that u2 = 1. If D were a division algebra, 0 = u2 − 1 = (u − 1)(u + 1) implies u = ±1, which in turn means: ek = ∓e1e2 and so e1, ..., ek−1 generate D. This contradicts the minimality of W. | ” |
Alaexis¿question? 10:04, 1 May 2024 (UTC)
- Seems fine to me. NadVolum (talk) 17:18, 6 May 2024 (UTC)
- Thanks! Alaexis¿question? 20:31, 6 May 2024 (UTC)
- Looks fine to me too. This is a good edit. 67.198.37.16 (talk) 03:01, 7 May 2024 (UTC)
- Thank you! Alaexis¿question? 17:02, 7 May 2024 (UTC)
- Looks fine to me too. This is a good edit. 67.198.37.16 (talk) 03:01, 7 May 2024 (UTC)
- Thanks! Alaexis¿question? 20:31, 6 May 2024 (UTC)