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?

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    $\begingroup$ Every finite dimensional normed space is complete, so there is no weakening/strengthening in that direction. $\endgroup$ Mar 10, 2017 at 23:19

1 Answer 1


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.

  • $\begingroup$ I'm currently away from the PC, I'll expand on this when I return $\endgroup$
    – Aweygan
    Mar 10, 2017 at 23:20
  • $\begingroup$ Of course! I forgot that any Banach space with a continuous multiplication structure has an equivalent norm upon which it is a Banach algebra. I should really reread Rudin at some point. $\endgroup$ Mar 10, 2017 at 23:42
  • $\begingroup$ @JacobDenson yeah I definitely got the idea from the first few pages of chapter 10 $\endgroup$
    – Aweygan
    Mar 10, 2017 at 23:44

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