Why $(A^n)^*\simeq A^n$ as representations of algebra $A$? I'm reading Introduction to representation theory by Etingof et al. and I'm struggling with one statement in the proof of Theorem 2.6. (representations of direct sums of matrix algebras), namely
$$
(A^n)^*\simeq A^n,
$$
where both sides are considered as representations of $A=\bigoplus_i\textrm{Mat}_{d_i}k$, $k$ is any field. As far as I understand, $(A^n)^*$ is a representation of $A^{op}$ (by definition of the dual representation) and $A^{op}\simeq A$ in our case. Thus, we need to construct isomorphism $\varphi$ of vector spaces $(A^n)^*$ and $A^n$ such that for all $a\in A$ and $f\in (A^n)^*$ we have
$$
\varphi(f\circ\rho(a))=\rho(a)(\varphi(f)),
$$
where $\rho(a)$ is corresponding element of $\textrm{End} A^n$.
Isomorphism between $W^*$ and $W$ (as vector spaces) is usually defined as $\varphi(f):=\sum_i f(e_i)\cdot e_i$, where $\{e_i\}$ is an arbitary basis in $W$. However, here we can't apply it here, I think (at least I don't understand how to satisfy the mentioned condition for $\varphi$).
It's not clear for me how such $\varphi$ might be constructed. Any help would be appreciated.
 A: You need to show that there is an isomorphism $A^n\to (A^n)^*$, $v\mapsto v^*$ such that for $a\in A$ you have $(av)^*=v^*a^T$ (having in mind that the isomorphism $A\cong A^{op}$ is the transposition).$\DeclareMathOperator{\tr}{tr}$
This isomorphism can be given explicitly. Let us embed $A^n$ in a large matrix algebra. Notice that trace is linear, so $(v,w)\mapsto \tr(v^Tw)$ is bilinear in $v,w$, and so $v\mapsto v^*=(w\mapsto \tr(v^Tw))$ defines a $k$-linear map $A^n\to (A^n)^*$.
Let us see that this is $A$-linear, i.e. that for all $a\in A$ we have $(av)^*=v^*a^T$, i.e. $(av)^*(w)=v^*(a^Tw)$ for $w\in A^n$. Note that here, $av=\bar a v$, where $\bar a$ is a block diagonal matrix, with $n$ blocks, each in the shape of $a$. It is easy to see that $\bar{a}^T=\overline{a^T}$, so in fact $$(av)^*(w)=(\bar av)^*(w)=\tr(v^T\bar a^Tw)=v^*(\bar a^Tw)=v^*(\overline{a^T}w)=v^*(a^Tw),$$
so indeed $(av)^*=v^*a^T$.
Finally, to show that it is an isomorphism, since $A$ is finite dimensional over $k$, it is enough to show that it is injective, i.e. that for every nonzero $v\in A^n$ there is some $w\in A^n$ such that $\tr(v^Tw)\neq 0$. It is easy to see that it is enough to show that for every nonzero matrix $v$ over $k$, there is a matrix $w$ of the same dimension (also over $k$) such that $\tr(vw)\neq 0$. I feel that this should be well-known, but I can't find a proof, so here is an ad-hoc one.
First, multiply $v$ first by a permutation matrix, to ensure that the product has a nonzero diagonal entry, and then multiply it by a diagonal matrix with all coordinates zero except one to ensure that the product has exactly one nonzero diagonal entry. The result clearly has nonzero trace.
