Why $(A^n)^*\simeq A^n$ as representations of algebra $A$?

Question

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.

Answer 1

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.