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It is known that not every algebra (over a ground field $\mathbb{R}$ or $\mathbb{C}$) is a Banach algebra. It might be a silly question, but are there examples of finite dimensional ($\mathbb{R}$- or $\mathbb{C}$-)algebras which are not Banach algebras (i.e. for which no submultiplicative norm exists)?

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Every finite dimensional unitary algebra $A$ is isomorphic to a subalgebra of a matrix algebra $M_n(\mathbb R)$. Retricting the norm of the latter gives a norm on $A$.

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    $\begingroup$ This is what I was starting to think. Can you give a reference or short proof as to why such an isomorphism always exists? $\endgroup$
    – Bib-lost
    Apr 3 '16 at 18:47
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    $\begingroup$ @Bib-lost: You can embed $A$ in $End(A)$ since $A$ acts on itself by multiplication. $\endgroup$ Apr 3 '16 at 18:55
  • $\begingroup$ Ok, got it to work, thanks! $\endgroup$
    – Bib-lost
    Apr 3 '16 at 20:25
  • $\begingroup$ Is this result, a formulation of the Wedderburn-Mal'tsev theorem? $\endgroup$
    – user355356
    Jan 25 '19 at 11:29
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I think it's clear that in the finite-dimensional case multiplication is going to be continuous with respect to any norm, just as any linear transformation is bounded. If so then any norm is going to satisfy $||xy||\le c||x||\,||y||$, and then $|||x|||=c||x||$ is submultiplicative.

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  • $\begingroup$ @user43208 I always get that backwards - thanks. $\endgroup$ Dec 3 '18 at 14:54

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