# Are all finite dimensional algebras over the real numbers Banach algebra'-able

Suppose that $A$ is a finite dimensional algebra over the real or complex numbers. Then $A$ has a natural topology induced from it being a finite dimensional vector space. Is it always true that there is a norm on $A$ satisfying $\| MN \| \leq \| M \| \| N \|$, or are there some finite dimensional algebras which aren't Banachable'? If we weaken the norm to not being complete, do we obtain stronger results?

• Every finite dimensional normed space is complete, so there is no weakening/strengthening in that direction. Mar 10 '17 at 23:19

Since $A$ is a finite-dimensional vector space, we can define a norm (any $\ell^p$-norm will do). Since $A$ is finite-dimensional, any linear operator is continuous with respect to this norm. To each $a\in A$ we can associate a linear map $x\mapsto ax$. Thus considering $A$ as a subalgebra of $L(A)$ with the operator norm, we obtain a norm on $A$ which is submultiplicative.