Talk:Frobenius theorem (real division algebras)

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What about Octonions? Jim Bowery (talk) 15:59, 24 July 2014 (UTC)[reply]

They're not an associative algebra. Double sharp (talk) 15:31, 14 November 2015 (UTC)[reply]

Is it possible someone would be able to include a proof for this?

Serious mistake in proof

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)[reply]

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)[reply]

Proof not encyclopedic

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)[reply]

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)[reply]
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)[reply]

Proof - finish

Possibly I'm being slow, but I don't understand why u2 = 1. Alaexis¿question? 12:31, 30 April 2024 (UTC)[reply]

I think I've realised my error. Maybe we could make this clearer for the reader as follows

Alaexis¿question? 10:04, 1 May 2024 (UTC)[reply]

Seems fine to me. NadVolum (talk) 17:18, 6 May 2024 (UTC)[reply]
Thanks! Alaexis¿question? 20:31, 6 May 2024 (UTC)[reply]
Looks fine to me too. This is a good edit. 67.198.37.16 (talk) 03:01, 7 May 2024 (UTC)[reply]
Thank you! Alaexis¿question? 17:02, 7 May 2024 (UTC)[reply]