Past exam of the mathematics course of the University of Cambridge / 2007 / ib / Paper 4 / 1G / Solution

The dual map reverses the direction of the linear map: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 $W^{\ast }$, so this does define a linear map between the dual spaces.

Let $(v_1,\dots ,v_n)$ and $(w_1,\dots ,w_m)$ be the chosen bases, and let $(v^1,\dots ,v^n)$ and $(w^1,\dots ,w^m)$ be their dual bases. If $A$ represents $\alpha$, then $\alpha (v_i)={\sum }_j{A}_{ji}{w}_j$. ThereforeThe coefficient in row $i$, column $j$ of the matrix of ${\alpha }^{\ast }$ is thus $A_{ji}$, giving 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.