dual basis and eigenvectors How can one show that given a diagonalizable matrix $M$, with basis v_1, v_2, ..., v_n as eigenvectors, the dual basis elements are also eigenvectors of $M^*$, the dual space of M? I think I need to use the biorthogonal property on basis, but don't know how. Thanks for suggestions.
 A: Let $\eta_1, \dotsc,\eta_n$ be the dual basis; that is,
$$ V^* \ni \eta_i:V \to F, \\
\eta_i(v_j) = \delta_{ij} $$
(definition of dual basis).
Now, $M^*$ is defined by
$$ (M^*\eta)(v) = \eta(Mv), $$
so if $Mv_j=\lambda_j v_j$, we have
$$ (M^*\eta_i)(v_j) = \eta_i(Mv_j) = \lambda_j \eta_i(v_j) = \lambda_j \delta_{ij} = \lambda_i \delta_{ij} = \lambda_i\eta_i (v_j). $$
Therefore $M^*\eta_i$ and $\lambda_i \eta_i$ act in the same way on the basis, and by linearity, on any $v \in V$. Therefore they are the same element of $V^*$, so
$$ M^* \eta_i = \lambda_i \eta_i, $$
as required.
A: See $M$ as a linear map $V \to V$. Then, by definition, $M^*:V^* \to V^*$ is the map sending $f \in V^*$ to $f\circ M$. Let $\delta_i$ be the dual basis element dual to $v_i$ and let $\lambda_i$ be the eigenvalue associated to $v_i$. Then $M^*(\delta_i)=\delta_i \circ M=\lambda_i\cdot \delta_i$, as you can easily check by evaluating on the basis $\{v_1,...,v_n\}$. Therefore the $\delta_i$'s are eigenvectors for $M^*$.
