# Are there finite dimensional $\mathbb{R}$-algebras which are not Banach algebras?

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)?

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 $A$.

• This is what I was starting to think. Can you give a reference or short proof as to why such an isomorphism always exists? Commented Apr 3, 2016 at 18:47
• @Bib-lost: You can embed $A$ in $End(A)$ since $A$ acts on itself by multiplication. Commented Apr 3, 2016 at 18:55
• Ok, got it to work, thanks! Commented Apr 3, 2016 at 20:25
• Is this result, a formulation of the Wedderburn-Mal'tsev theorem?
– user355356
Commented Jan 25, 2019 at 11:29

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.

• @user43208 I always get that backwards - thanks. Commented Dec 3, 2018 at 14:54