According to Frobenius theorem, https://en.wikipedia.org/wiki/Frobenius_theorem_(real_division_algebras) every associative division algebra over $\mathbb{R}$ is either $\mathbb{R}$, or $\mathbb{C}$, or $\mathbb{H}$ where $\mathbb{H}$ is the quaternion algebra. This clearly implies that $\mathbb{R}^3$ does not have an associative division algebra. But is there an easy way to demonstrate this fact without proving Frobenius theorem?
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Yes. If there was such a multiplication on $\mathbb{R}^3$, take same $a\in\mathbb{R}^3$ which is not a multiple of $1$ (the identity element for the multiplication). Consider the map from $\mathbb{R}^3$ into itself defined by $x\mapsto a\times x$. It's an endomorphism of $\mathbb{R}^3$ and therefore it has a real eigenvalue $\lambda$. So, there is some $y\in\mathbb{R}^3\setminus\{0\}$ such that $a\times y=\lambda y$. In other words, $(a-\lambda 1)\times y=0$. But then, since $y\neq0$, $a=\lambda 1$, which is a contradiction.
Of course, the same argument works for $\mathbb{R}^n$ when $n$ is odd and greater than $1$.
José Carlos Santos
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Thanks! I understand everything except why an endomorphism must have a real eigenvalue? – Sid Caroline Feb 27 '18 at 07:29
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2@SidCaroline Because its characteristic polynomial has odd degree and it follows from the intermediate value theorem that every such polynomial has a real root. – José Carlos Santos Feb 27 '18 at 07:30