Let $C$ be a $k$-coalgebra with basis $\{x_m\}$ where $m \in \{0, 1, ..., n\}$ where $n \geq 0$, with comultiplication defined by
$$\Delta(x_m) = \sum_{t=0}^m x_t \otimes x_{m-t}$$
and counit
$$\epsilon(x_m) = \delta_{0, m}$$
Determine the algebra structure on $C^{*} = \text{Hom}_k(C, k)$
I'm not sure how to get started and I probably need a hint also.
Let $\left\{x_i^*\mid 0\leq i\leq n\right\}$ be the dual basis of the basis $\left\{x_i\mid 0\leq i\leq n\right\}$. Then $x_i^*(x_j)=\delta_{i,j}$. This induces a pairing $\left\langle x,y\right\rangle:=\sum_{i,j}\lambda_{i}\lambda_jx_i^*(x_j)$ where $x=\sum_i\lambda_ix_i^*$ and $y=\sum_j\lambda_jx_j$.
We define an algebra structure on $C^*$ as follows: Let $a,b\in C^*$, then $(a\cdot b)(x):=\sum_{(x)}a(x_{(1)})\otimes b(x_{(2)})$. (so it's applying $(a\otimes b)$ to $\Delta(x)$ componentwise). Also notice that the above is compatible with the pairing in the following sense: $\left\langle a\cdot b,x\right\rangle=\left\langle a\otimes b,\Delta(x)\right\rangle$ and $\left\langle a\otimes b,x\otimes y\right\rangle=\left\langle a,x\right\rangle\left\langle b,y\right\rangle$.
So let's figure out what happens when we caculate $(x_i^*\cdot x_j^*)(x_k)$. We find that \begin{eqnarray} (x_i^*\cdot x_j^*)(x_m) &=& \sum_{t=0}^m x_i^*(x_t)x_j^*(x_{m-t})\\ &=& \begin{cases} x_j^*(x_{m-i}) & \mbox{ if } i\leq m,\\ 0 & \mbox{ else}. \end{cases}\\ &=& \begin{cases} 1 & \mbox{ if } i\leq m \mbox{ and if }i+j=m\\ 0 & \mbox{ else}. \end{cases}\\ &=& \begin{cases} 1 & \mbox{ if }i+j=m\\ 0 & \mbox{ else}. \end{cases} \end{eqnarray} It follows that $x_i^*\cdot x_j^*=x_{i+j}^*$ and we make the convention that $x_{i+j}^*=0$ if $i+j>n$. You can easily see that the unit is given by $x_0^*$ and that $(x_1^*)^i=x_i^*$. It follows that this algebra is generated by $x_1^*$ and that $(x_1^*)^{n+1}=0$. In other words, $C^*\cong k[x]/(x^{n+1})$.
