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)

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)

Might the following be a more "elementary" proof?

Theorem: All finite-dimensional associative division algebras over the real numbers are isomorphic to subalgebras of the quaternions.

Main Proof

Let $A$ be a finite-dimensional associative division algebra over $\mathbb {R}$. We proceed by analyzing possible dimensions:

Case 1: $\dim(A)=1$

In this case, $A\cong \mathbb {R}$.

Case 2: $\dim(A)=2$

By Lemma 1, $A\cong \mathbb {C}$, which is a subalgebra of the quaternions.

Case 3: $\dim(A)=3$

This case is impossible by Lemma 2, as the dimension must be even.

Case 4: $\dim(A)\geq 4$

1. By Lemma 1, there exists $i\in A$ with $i^{2}=-1$.
2. Consider the linear map $f(x)=ixi^{-1}$. This map is diagonalizable with eigenvalues $\pm 1$.
3. The eigenspace of eigenvalue 1 has dimension ≥ 2 (as $i$ commutes with 1 and $i$).
4. This eigenspace cannot have dimension ≥ 3 (would contradict Lemma 3).
5. Therefore the +1 eigenspace of $f$ has dimension 2, and the -1 eigenspace of $f$ has dimension $\dim(A)-2$.
6. Therefore, there exists $J$ in the -1 eigenspace where $iJi^{-1}=-J$, or equivalently, $iJ=-Ji$.
7. By lemma 3, $J^{2}=a+bJ$ for real $a$ and $b$.
8. Since $iJ=-Ji$, we obtain $b=0$.
9. By lemma 1, $a<0$.
10. Let $j=J/{\sqrt {-a}}$, which gives $j^{2}=-1$ and $ij=-ji$.
11. Let $k=ij$. Then ${\text{span}}\{1,i,j,k\}$ is isomorphic to the quaternions, as:
    • $k^{2}=(ij)(ij)=i(ji)j=i(-ij)j=-1$
    • $ijk=ij(ij)=-1$

This leaves us with three subcases:

• $\dim(A)=4$: Then $A$ is isomorphic to the quaternions.
• $\dim(A)=5$: Impossible by Lemma 2.
• $\dim(A)\geq 6$:

1. Let $U=\{z\in A\mid izi^{-1}=-z\}$.
2. Let $V=\{z\in A\mid jzj^{-1}=-z\}$.
3. Step 5 from earlier tell us that $\dim(U)=\dim(V)=\dim(A)-2$.
4. From $\dim(U\cap V)=\dim(U)+\dim(V)-\dim(U+V)$, we obtain $\dim(U\cap V)\geq 2$.
5. It follows that there exists a $z$ that is not a real multiple of $k$ which satisfies $iz=-zi,jz=-zj$.
6. It follows that $z$ commutes with $k$ since $zk=zij=-izj=ijz=kz$, which contradicts lemma 3.

Lemma 1

If $\dim(A)>1$, then for any non-real $z\in A$, ${\text{span}}\{1,z\}$ is an algebra isomorphic to $\mathbb {C}$.

Proof:

1. Let $m$ be the minimal polynomial of $z$.
2. $\deg(m)>1$ (as $z$ is non-real)
3. $\deg(m)<3$ (otherwise $m$ can be factored, creating zero divisors)
4. Therefore $\deg(m)=2$, giving $m(z)=(z-a)^{2}+b=0$
5. If $b\leq 0$, then $(z-a-{\sqrt {-b}})(z-a+{\sqrt {-b}})=0$, creating zero divisors
6. Thus $b>0$, and $(z-a)^{2}$ is real and negative
7. After scaling, we get $i=(z-a)/{\sqrt {b}}$ satisfying $i^{2}=-1$

Lemma 2

$\dim(A)$ is even.

Proof:

1. Assume $\dim(A)$ is odd.
2. For any $z\in A$, consider $f(x)=zx$.
3. The characteristic polynomial of $f$ has odd degree.
4. This implies $f$ has a real eigenvalue $t$.
5. The corresponding eigenvector $v$ satisfies $(z-t)v=0$, making it a zero divisor.
6. This contradicts $A$ being a division algebra.

Lemma 3

If $A$ is commutative, then $\dim(A)\leq 2$.

Proof:

1. Assume $\dim(A)\geq 3$. Choose a linearly independent set $\{1,w,z\}$.
2. By Lemma 1, we can replace $\{1,w,z\}$ with $\{1,i,j\}$ spanning the same subspace of $A$, where $i^{2}=j^{2}=-1$.
3. Consider $ij$. There are three possibilities:
   • Case 1: $ij=1$. Then $i=-j$, contradicting linear independence.
   • Case 2: $ij=-1$. Then $i=j$, contradicting linear independence.
   • Case 3: $ij\neq \pm 1$. Then $(1+ij)(1-ij)=1-(ij)^{2}=0$, creating a non-trivial zero divisor.
4. All cases lead to a contradiction, proving that a commutative division algebra cannot have dimension ≥ 3.

81.96.146.252 (talk) 12:05, 4 February 2025 (UTC)