If $B_1 = \{v_1,\ldots,v_n\}$ and $B_2 = \{v_1',\ldots,v_n'\}$ are basis for a vector space V, and $D_1= \{\delta v_1,\ldots, \delta v_n\}$ and $D_2 = \{\delta v_1',\ldots, \delta v_n'\}$ are the corresponding dual bases of $V^*$ prove that, if $P$ is the change-of-basis matrix from $B_1$ to $B_2$, then $(P^{-1})^T$ is the change-of-basis from $D_1$ to $D_2$
(this is a theorem from Schaum's outline of theory and problems of linear algebra)
My knowledge of linear algebra is not great, but so far I have that
\begin{align}
\text{For any }v \in V, [v]_{B_1}&=P[v]_{B_2}\\
\text{so } (\alpha_1,\ldots,\alpha_n)(v_1,\ldots,v_n)^T&=P(\beta_1,\ldots,\beta_n)(v_1',\ldots,v_n')^T
\end{align}
for some $\alpha_1,\ldots,\alpha_n,\beta_1,\ldots,\beta_n \in F$ (where $F$ is a field)
How do I use the fact that $D_1$, $D_2$ are dual basis to further this proof? And also, where does the transpose of $P^{-1}$ come in? Can I use that $D_1 = \delta B_1$ and $D_2 = \delta B_2$ (where I am guessing $\delta$ is the kronecker delta?)
Any hints or suggestions are greatly appreciated, I know that my knowledge is quite limited with regard to this, links to text or sources that could explain a problem like this would be great.
Best Answer
Let $P=(p_{i,j})$ the change of basis matrix from $B_1$ to $B_2$ and $Q=(q_{i,j})$ from $D_1$ to $D_2$ so we have
$$\delta_{j,k}=\delta v'_j(v'_k)=\left(\sum_{i=1}^n q_{i,j}\delta v_i\right)\left(\sum_{l=1}^n p_{l,k}v_l\right)=\sum_{i=1}^n\sum_{l=1}^n q_{i,j}p_{l,k}\delta_{i,l}=\sum_{i=1}^n q_{i,j}p_{i,k}$$ which means that $$Q^TP=I\iff Q=\left(P^{-1}\right)^T$$