Talk:Frobenius theorem (real division algebras)

Untitled
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 $$e_ie_j = 0$$ by definition of orthonormality, so the claim $$e_ie_j = -e_je_i$$ is wrong (actually not completely wrong, but doesn't make any sense since $$e_ie_j = 0$$) and the following arguments about quaternions and case n>2 are also wrong.

Also there's a type in case n=2: $$e_1e_2 = - e_1e_2$$, it should be $$e_1e_2 = - e_2e_1$$ 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 $$e_1e_2$$ is zero. The definition of the inner product is $$ B(e_1.e_2):=  -e_1e_2-e_2e_1$$. Thus $$e_1e_2=-e_2e_1$$. 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 $u = e_{1}e_{2}e_{k}$. It is easy to see that $u^{2} = 1$ (this only works if $k > 2$). If $D$ were a division algebra, $0 = u^{2} − 1 = (u − 1)(u + 1)$ implies $u = ±1$, which in turn means: $e_{k} = ∓e_{1}e_{2}$ and so $e_{1}, ..., e_{k−1}$ generate $D$. This contradicts the minimality of $W$.

Possibly I'm being slow, but I don't understand why $u^{2} = 1$. 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 $u = e_{1}e_{2}e_{k}$ and consider $u^{2}=(e_{1}e_{2}e_{k})*(e_{1}e_{2}e_{k})$. By rearranging the elements of this expression and applying the orthonormality relations among the basis elements we find that $u^{2} = 1$. If $D$ were a division algebra, $0 = u^{2} − 1 = (u − 1)(u + 1)$ implies $u = ±1$, which in turn means: $e_{k} = ∓e_{1}e_{2}$ and so $e_{1}, ..., e_{k−1}$ 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)