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Let $K$ be a field and $V$ a $n$-dimensional $K$-vector space with basis $B=\{b_1, \ldots , b_n\}$. $V^{\star}$ is the dual space of $V$. $B^{\star}$ is the dual basis corresponding to $B$ of $V^{\star}$.

- Let $C=\{c_1, \ldots , c_n\}$ be an other basis of $V$ and $C^{\star}$ its dual basis. Let $S=(s_{ij})_{1\leq i,j\leq n}$ be the transformation matrix of basis from $B$ to $C$, i.e. $\displaystyle{c_j=\sum_{i=1}^ns_{ij}b_i}$ for $j\in \{1, \ldots , n\}$.

Show that $(S^{-1})^T=:A=(a_{ij})_{1\leq i,j\leq n}$ is the transformation matrix of basis from $B^{\star}$ to $C^{\star}$ i.e. that $\displaystyle{c_j^{\star}=\sum_{i=1}^na_{ij}b_i^{\star}}$ for $j\in \{1, \ldots , n\}$.

- Let $\Lambda=\{\lambda_1, \ldots , \lambda_n\}$ a basis of the dual space $V^{\star}$.

Show that there is a basis $C$ of $V$ so that $C^{\star}=\Lambda$.

I have done the following:

- We have that $A$ is the transformation matrix of basis from $B^{\star}$ to $C^{\star}$, i.e., that $\displaystyle{c_j^{\star}=\sum_{i=1}^na_{ij}b_i^{\star}}$ and we want to show that the transformation matrix is equal to $A:=\left (S^{-1}\right )^T$, right? (Wondering)

We have the following:

\begin{equation*}c_j^{\star}(c_k)=c_j^{\star}\left (\sum_{i=1}^{n}s_{ik}b_i\right )\ \overset{ c_j^{\star} \text{ linearform }}{ = } \ \sum_{i=1}^{n}s_{ik}c_j^{\star}\left (b_i\right )=\sum_{i=1}^{n}s_{ik}\sum_{\lambda=1}^na_{\lambda j}b_{\lambda}^{\star}(b_i)=\sum_{i=1}^{n}s_{ik}\sum_{\lambda=1}^na_{\lambda j}\delta_{\lambda i}=\sum_{i=1}^{n}s_{ik}a_{i j}\end{equation*}

From that we get $\displaystyle{\sum_{i=1}^{n}s_{ik}a_{i j}=\delta_{jk}}$, since $c_j^{\star}(c_k)=\delta_{jk}$.

From $\displaystyle{\sum_{i=1}^{n}a_{i j}s_{ik}=\delta_{jk}}$ it follows that $\displaystyle{\sum_{i=1}^{n}a_{ji}^Ts_{ik}=\delta_{jk}}$.

Since $\delta_{jk}=\begin{cases}1 & \text{ if } i=k \\0 & \text{ otherwise } \end{cases}$ it follows that we get the identity matrix.

So we get $A^T\cdot S=I_n$.

Therefore we have that \begin{equation*}A^T=S^{-1}\Rightarrow \left (A^T\right )^T=\left (S^{-1}\right )^T\Rightarrow A=\left (S^{-1}\right )^T\end{equation*}

Is everything correct? (Wondering)

- Could you give me a hint for that? (Wondering)

Do we maybe do the following?

Let $M$ the transformation matrix of basis from $B^{\star}$ to $\Lambda$.

Then do we get from the first question that $M=\left (S^{-1}\right )^T$ ? Or does this only hold for the specific basis $C^{\star}$ ?

(Wondering)