# Talk:Frobenius theorem (real division algebras)

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Field: Algebra

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)