# 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.

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.