One defines the **finite dual** of a Hopf algebra $H$ as
$$
H^o := \{f \in H^\* ~|~ f(I) = 0, \text{ for some ideal $I$ of $H$ with } \dim\_C(H/I) < \infty \}.
$$
As is well-known, $H^o$ has a well-defined Hopf algebra structure obtained by dualizing the Hopf structure of $H$. 


On the other hand, for any finite-dimensional $H$-module $V$, and element $v \in V$, and a functional $f \in V^\*$,
we can define a functional $c\_{f,v} \in H^\*$ according to
$$
c\_{f,v}(h) := f(hv).
$$
One usually calls any such functional a **matrix coefficient** of $H$. It is not difficult to see that the set of matrix coefficients 
of $H$ forms a Hopf subalgebra of $H^o$, which we will denote by $\operatorname{Mat}(H)$.


What I would like to know is when do we have the equality
$$
H^o = \operatorname{Mat}(H)?
$$



Hopf Duals and Matrix Coefficients
The equality always holds, for any algebra (not just Hopf algebra) $H$. See Kapitel IV, Bemerkung 3.4 **5)** in [Schneider's Hopf algebra class](https://sites.google.com/site/darijgrinberg/hopfalgebren?pli=1) if you can read German (which I assume you can, given your location); it is currently on page 483 (as of version 0.99). Sorry for the bad formatting...


Yes, I just updated the notes because I looked at the proof again and couldn't understand it. If you find similar issues elsewhere in the notes, please let me know!


