Transpose matrix dual map How do I see that the representing matrix of the dual map $f^*$ between finite-dimensional dual spaces is given by the transpose of the representing matrix of $f$? Here I want to assume that the matrix $f^*$ is represented with respect to the dual basis.
Apparently this result is very well-known, but I would like to see proof of this.
 A: Let $f: V \to \bar V$ be a linear transformation represented by the matrix $A = [a^i_j]$ relative to the basis $e_1,\dots,e_n$ for $V$ and $\bar e_1, \dots, \bar e_m$ for $\bar V$.  Let $\alpha^1, \dots, \alpha^n$ and $\bar \alpha^1, \dots, \bar \alpha^m$ denote the respective corresponding dual bases.
Suppose that $f^*:\bar V^* \to V^*$ satisfies $f^*(\bar \alpha^i) = \sum_{j} b^i_j \alpha^j$.  That is, suppose that $[b^i_j]^T$ is the matrix of $f^*$ relative to the dual bases.  It follows that for each $i,j$, we have
\begin{align*}
b^i_j & =
\sum_{k} b^i_k \delta^k_j = 
\sum_{k} b^i_k \alpha^k e_j = 
\left(\sum_{k} b^i_k \alpha^k\right) e_j =
f^*(\bar \alpha^i) e_j = 
(f^* \circ \bar \alpha^i) e_j
\\ & =
(\bar \alpha^i \circ f)(e_j) = 
\bar \alpha^i(f(e_j)) = 
\bar \alpha^i\left(\sum_{k}a^k_j \bar e_k\right) = 
\sum_{k}a^k_j \bar \alpha^i \bar e_k = 
\sum_{k}a^k_j \delta^i_k = a^i_j
\end{align*}
as desired.  That is, the matrix of $f^*$ with respect to the dual bases is $[b^i_j]^T = [a^i_j]^T = A^T$.
