= Solution
The <dual map> reverses the direction of the <linear map>:
$$
\boxed{\alpha^*:W^*\longrightarrow V^*,\qquad \alpha^*(\lambda)=\lambda\circ\alpha.}
$$
Here the <dual spaces> consist of complex-linear <functionals>. Composition with $\alpha$ is linear both as a <functional> on $V$ and as an operation on $W^*$, so this does define a <linear map> between the <dual spaces>.
Let $(v_1,\ldots,v_n)$ and $(w_1,\ldots,w_m)$ be the chosen <bases>, and let $(v^1,\ldots,v^n)$ and $(w^1,\ldots,w^m)$ be their <dual bases>. If $A$ represents $\alpha$, then $\alpha(v_i)=\sum_jA_{ji}w_j$. Therefore
$$
(\alpha^*w^j)(v_i)=w^j(\alpha(v_i))=A_{ji},\qquad\alpha^*w^j=\sum_iA_{ji}v^i.
$$
The coefficient in row $i$, column $j$ of the <matrix> of $\alpha^*$ is thus $A_{ji}$, giving \b[the <matrix transpose> $A^{\mathsf T}$]. This calculation introduces no <complex conjugation>: the algebraic <dual map> uses complex-linear <functionals>, whereas a <Hermitian transpose> belongs to the inner-product adjoint construction.
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