 Solution

The algebra must be finite-dimensional, and there is no universal bound on its dimension. Let $e=\epsilon m$ and write its coevaluation morphism as a finite sumFor any $v\in A$, one snake identity givesThus the finitely many $b_i$ span $A$, making it a finite-dimensional vector space. The other snake identity shows that $e$ is nondegenerate in the other variable as well, so the maps to the dual space determined by this pairing are isomorphisms. In the standard algebraic terminology this is a Frobenius algebra.

Every positive finite dimension of a vector space occurs: take $A=k^d$ with coordinatewise multiplication and $\epsilon (a_1,\dots ,a_d)={\sum }_i{a}_i$. The coordinate idempotents $e_i$ give $e(e_i,e_j)={\delta }_{ij}$ and $n(1)={\sum }_i{e}_i\otimes e_i$. If zero unital algebras are allowed, the zero object gives dimension zero too.

The pairing need not make $A$ a semisimple algebra. For example, the dual numbers $k[t]/(t^2)$ with $\epsilon (a+bt)=b$ have pairing matrixand coevaluation morphism $1\mapsto 1\otimes t+t\otimes 1$. This pairing is nondegenerate over every field, despite the nilpotent ideal $(t)$. Finite-dimensionality is the dimension conclusion; separability is not part of the given definition.