Talk:Frobenius theorem (real division algebras)

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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

If $e_{1},\ldots ,e_{n}$ is orthonormal basis, then $e_{i}e_{j}=0$ by definition of orthonormality, so the claim $e_{i}e_{j}=-e_{j}e_{i}$ is wrong (actually not completely wrong, but doesn't make any sense since $e_{i}e_{j}=0$) and the following arguments about quaternions and case n>2 are also wrong.

Also there's a type in case n=2: $e_{1}e_{2}=-e_{1}e_{2}$, it should be $e_{1}e_{2}=-e_{2}e_{1}$ 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 $B(e_{1},e_{2})$ is zero, not that the algebra product $e_{1}e_{2}$ is zero. The definition of the inner product is $B(e_{1}.e_{2}):=-e_{1}e_{2}-e_{2}e_{1}$. Thus $e_{1}e_{2}=-e_{2}e_{1}$. Mike Stone (talk) 15:32, 11 June 2013 (UTC)

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)

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 $\mathbb {R}$. 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 $\operatorname {tr} :D\to \mathbf {R}$ 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 $\operatorname {tr} (A+B)=\operatorname {tr} (A)+\operatorname {tr} (B)$ and $\operatorname {tr} (\lambda A)=\lambda \operatorname {tr} (A)$.

* In the context of the proof, we can give its domain and codomain as $\operatorname {tr} :D\to \mathbf {R}$.

* 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

“ If k > 2, then D cannot be a division algebra. Assume that k > 2. Let u = e₁e₂e_k. 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: e_k = ∓e₁e₂ and so e₁, ..., e_k−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 = e₁e₂e_k and consider u²=(e₁e₂e_k)*(e₁e₂e_k). 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: e_k = ∓e₁e₂ and so e₁, ..., e_k−1 generate D. This contradicts the minimality of W. ”

Alaexis_¿question? 10:04, 1 May 2024 (UTC)