1012: For Finite-Dimensional Vectors Space, Transition of Dual Bases for Covectors Space w.r.t. Bases for Vectors Space Is This

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description/proof of that for finite-dimensional vectors space, transition of dual bases for covectors space w.r.t. bases for vectors space is this

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Topics

About: vectors space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Proof

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Starting Context

• The reader knows a definition of covectors (dual) space of vectors space.

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Target Context

• The reader will have a description and a proof of the proposition that for any finite-dimensional vectors space, the transition of the dual bases for the covectors space with respect to any bases for the vectors space is this.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$F$: $\in \{\text{ the fields }\}$
$V$: $\in \{\text{ the }d\text{ -dimensional }F\text{ vectors spaces }\}$
$V^{\ast }$: $=\text{ the covectors space of }V$
$B$: $\in \{\text{ the bases for }V\}$, $=\{b_1,...,b_d\}$
$B^{\prime }$: $\in \{\text{ the bases for }V\}$, $=\{b_1^{\prime },...,b_d^{\prime }\}$
$B^{\ast }$: $=\text{ the dual basis of }B$, $=\{b^1,...,b^d\}$
$B^{\prime \ast }$: $=\text{ the dual basis of }{B}^{\prime }$, $=\{b^{\prime 1},...,b^{\prime d}\}$
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Statements:
$b_j^{\prime }=b_k{M}_j^k$
$⟹$
$b^{\prime j}={M^{-1}}_k^j{b}^k$
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2: Proof

Whole Strategy: Step 1: let $b^{\prime j}=N_k^j{b}^k$; Step 2: let the both hand sides operate on $b_l$, and see that $N_k^j={M^{-1}}_k^j$.

Step 1:

Let $b^{\prime j}=N_k^j{b}^k$, which is possible because $B^{\ast }$ is a basis for $V^{\ast }$ and $b^{\prime j}\in V^{\ast }$.

Step 2:

$b^{\prime j}(b_l)=N_k^j{b}^k(b_l)=N_k^j{\delta }_l^k=N_l^j$.

The matrix, $M$, has the inverse, $M^{-1}$, because otherwise, $B^{\prime }$ would not be linearly independent.

From $b_j^{\prime }=b_k{M}_j^k$, $b_j^{\prime }{M^{-1}}_l^j=b_k{M}_j^k{M^{-1}}_l^j=b_k{\delta }_l^k=b_l$.

So, $N_k^j=b^{\prime j}(b_k)=b^{\prime j}(b_l^{\prime }{M^{-1}}_k^l)={M^{-1}}_k^l{b}^{\prime j}(b_l^{\prime })={M^{-1}}_k^l{\delta }_l^j={M^{-1}}_k^j$.

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References

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