Character of dual Representation? Let $G$ be a finite group and consider the group ring $\mathbb C[G]$. If $M$ is a $\mathbb C[G]$-module consider the dual representation in $M^*=\operatorname{Hom}(M, \mathbb C)$ given by $$(g\cdot f)(m)=f(g^{-1}\cdot m).$$ How can I show $$\chi_{M^*}(g)=\chi_M(g^{-1}),$$ where $\chi_M$ and $\chi_{M^*}$ are the associated characters of the given representations?
 A: Consider the elements of $M$ as column vectors and let $A(g)$ be the matrix which describes the action of $g$ on $M$ (from the left). So, for a column vector $m \in M$, we have $g \cdot m = A(g)m$. Now consider the elements of $M^*$ also as column vectors. So, for a column vector $f \in M^*$, we have $f(m) = f^Tm$, where the r.h.s. is just the product of the transpose of the column vector $f$ and the column vector $m$. Now, by construction, $(g\cdot f)(m) = f^T(g^{-1}\cdot m) = f^TA(g^{-1})m$, where the r.h.s. is just a product of vectors and matrices. But $f^TA(g^{-1}) = \left(A(g^{-1})^Tf\right)^T$. So $g$ maps the column vector $f \in M^*$ to the column vector $A(g^{-1})^Tf.$ (Note: it's easy to check directly that the map $g \mapsto A(g^{-1})^T$ is a homomorphism, i.e. $A((gh)^{-1})^T = A(g^{-1})^TA(h^{-1})^T$. This is reassuring.) So the matrix of $g$ when acting on $M^*$ from the left is $A(g^{-1})^T$. So, finally, $\chi_{M^*}(g) = tr\ A(g^{-1})^T = tr\ A(g^{-1}) = \chi_M(g^{-1}),$ as desired.
A: Let $V^\vee$ denote the dual of $V$, and for a linear map $f$ denote $f^\vee$ its transpose. Given $\rho: G\to GL(V)$ our representation, the dual is defined as $\rho':G\to GL(V^\vee)$ via, for any $s\in G$, $$ \rho'_s = (\rho_s^{-1})^\vee = \rho_{s^{-1}}^\vee: V^\vee\to V^\vee. $$We are left with proving that the trace of a map equals the trace of its transpose, which can be easily checked e.g. by taking an explicit basis. 
Alternatively for any $s\in G$, notice $\rho_s$ is unitary (since $G$ is finite so $s$ has finite order) and so $V$ has a basis of eigenvectors of $\rho_s$. W.r.t this basis $\rho_s$ is diagonal with entries its eigenvalues, and w.r.t the dual basis on $V^\vee$, $\rho_s^\vee$ is also diagonal with the same entries, thus the traces are equal as desired.
