Show that there is a basis C of V so that C* = Λ

[mathmari]

Hey!

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 $V^{\star}$.

1. 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 $j\in \{1, \ldots , n\}$.
   
2. 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:

1. 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)
   
2. 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)
   

 

[I like Serena]

I have done the following:

1. 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*}

Hey mathmari!

It seems correct to me. (Nod)

2. 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 $C^{\star}$ ?

Isn't it that we can rather construct $S = (M^{-1})^T$ by the reverse argument of the first question? (Wondering)

$M$ must exist since $V^*$ is a vector space.
And with the constructed $S$ we can find a basis $C$ in $V$ that is the dual of $\Lambda$. (Thinking)

 

[mathmari]

Isn't it that we can rather construct $S = (M^{-1})^T$ by the reverse argument of the first question? (Wondering)

$M$ must exist since $V^*$ is a vector space.
And with the constructed $S$ we can find a basis $C$ in $V$ that is the dual of $\Lambda$. (Thinking)

Why must $M$ exist? Could you explain it further to me? (Wondering)

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

Let $S:=(M^{-1})^T$.

Let $C$ be the basis that we get by the transformation matrix $S$ from the basis $B$.

From the first question we have then that $\left (S^{-1}\right )^T$ is the transformation matrix of basis from the dual basis $B^{\star}$ to the dual basis $C^{\star}$. Since $\left (S^{-1}\right )^T=M$ and $M$ is the transformation matrix from $B^{\star}$ to $\Lambda$, it follows that $\Lambda=C^{\star}$.


Is everything correct? Could we improve something? (Wondering)

 

[I like Serena]

Why must $M$ exist? Could you explain it further to me?

Lemma: The matrix for the transformation of a basis B to a basis C in a vector space must exist.
Proof: Observe that every basis vector in C must have a unique linear representation of basis vectors in B.
When we combine those linear representations into a matrix we have the matrix for the basis transformation. (Nerd)

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

Let $S:=(M^{-1})^T$.

Let $C$ be the basis that we get by the transformation matrix $S$ from the basis $B$.

From the first question we have then that $\left (S^{-1}\right )^T$ is the transformation matrix of basis from the dual basis $B^{\star}$ to the dual basis $C^{\star}$. Since $\left (S^{-1}\right )^T=M$ and $M$ is the transformation matrix from $B^{\star}$ to $\Lambda$, it follows that $\Lambda=C^{\star}$.

Is everything correct? Could we improve something?

It looks fine to me. (Nod)