Finding dual of matrix algebra

Question

Consider $A=M_n(K)$, the algebra of matrices over the field $K$. Let $\{e_{ij}\}$ be the standard basis for $A$ and $\{X_{ij}\}$ be the basis of $A^*$ dual to $\{e_{ij}\}$.
I need to find the operations on $A^*$, that is, show that comultiplication $\Delta = (m_A)^*$ is given by $$\Delta(X_{ij})=\sum_{k=1}^n X_{ik} \otimes X_{kj}$$ and that counit $\epsilon=(u_A)^*$ is given by $$\epsilon(X_{ij})=\delta_{ij}.$$ By this, it follows that $A^*$ is the same as the matrix coalgebra.
How can I find these formulas for the operations on $A^*$?

Answer 1

By the definition of the comultiplication on the dual, we have that $$\Delta(X_{ij})(e_{kl},e_{mn})=X_{ij}(e_{kl}e_{mn})=\delta_{lm}\delta_{ik}\delta_{nj}.$$ To show your first relation, let us write $$\Delta(X_{ij})=\sum_{klmn}a_{klmn}X_{kl}\otimes X_{mn}.$$ Then the first equation reads at $$a_{klmn}=\delta_{lm}\delta_{ik}\delta_{nj}$$ Inserting this relation yields $$\Delta(X_{ij})=\sum_{l}X_{il}\otimes X_{lj},$$ as desired.

Similarly, the counit is defined so that $$\epsilon(X_{ij})=X_{ij}(u_A)=X_{ij}(I)=\delta_{ij},$$ where here $I$ is the identity matrix.