222: Between Finite-Dimensional Vectors Space and Its Double Dual, There Is Canonical 'Vectors Spaces - Linear Morphisms' Isomorphism

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description/proof of that between finite-dimensional vectors space and its double dual, there is canonical 'vectors spaces - linear morphisms' isomorphism

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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 dual basis for covectors (dual) space of basis for finite-dimensional vectors space.
• The reader knows a definition of %category name% isomorphism.
• The reader admits the proposition that between any vectors spaces, any map that maps any basis onto any basis bijectively and expands the mapping linearly is a 'vectors spaces - linear morphisms' isomorphism.
• The reader admits 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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Target Context

• The reader will have a description and a proof of the proposition that between any finite-dimensional vectors space and its double dual, there is the canonical 'vectors spaces - linear morphisms' isomorphism.

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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:
$F$: $\in \{\text{ the fields }\}$
$V$: $\in \{\text{ the finite-dimensional }F\text{ vectors spaces }\}$
$V^{\ast }$: $=L(V:F)$
${V^{\ast }}^{\ast }$: $=L(L(V:F):F)$
$J$: $\in \{\text{ the finite index sets }\}$
$B$: $\in \{\text{ the bases for }V\}$, $=\{b_j|j\in J\}$
$B^{\ast }$: $=\text{ the dual basis of }B$, $=\{b^j|j\in J\}$
${B^{\ast }}^{\ast }$: $=\text{ the dual basis of }{B}^{\ast }$, $=\{{\tilde{b}}_j|j\in J\}$
$f$: $:V\to {V^{\ast }}^{\ast },v^j{b}_j\mapsto \sum _{j\in J}{v}^j{\tilde{b}}_j$


Statements:
$f\in \{\text{ the 'vectors spaces - linear morphisms' isomorphisms }\}$
$\wedge$
$f$ does not depend on the choice of $B$
$\wedge$
$\mathrm{\forall }v\in V,\mathrm{\forall }w\in V^{\ast }(w(v)=f(v)(w))$
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2: Proof

Whole Strategy: Step 1: see that $f$ is a 'vectors spaces - linear morphisms' isomorphism; Step 2: see that $f$ does not depend on the choice of $B$ by taking another basis $B^{\prime }$ and seeing that the map constructed by $B^{\prime }$ is $f$: Step 3: see that $\mathrm{\forall }v\in V,\mathrm{\forall }w\in V^{\ast }(w(v)=f(v)(w))$ by extending $v$, $w$, and $f(v)$ with the bases.

Step 1:

$B^{\ast }$ is indeed a basis for $V^{\ast }$, by Note for the definition of dual basis for covectors (dual) space of basis for finite-dimensional vectors space.

${B^{\ast }}^{\ast }$ is indeed a basis for ${V^{\ast }}^{\ast }$, by Note for the definition of dual basis for covectors (dual) space of basis for finite-dimensional vectors space.

$f$ is indeed a 'vectors spaces - linear morphisms' isomorphism, by the proposition that between any vectors spaces, any map that maps any basis onto any basis bijectively and expands the mapping linearly is a 'vectors spaces - linear morphisms' isomorphism.

Step 2:

Let us see that $f$ does not depend on the choice of $B$.

Let $B^{\prime }=\{b_j^{\prime }|j\in J\}$ be any other basis for $V$.

$b_j^{\prime }=b_l{M}_j^l$ for an invertible matrix, $M$. So, $b_j=b_l^{\prime }{M^{-1}}_j^l$.

The dual basis of $B^{\prime }$, $B^{\prime \ast }=\{b^{\prime j}|j\in J\}$, is $\{{M^{-1}}_l^j{b}^l\}$, by 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.

The dual basis of $B^{\prime \ast }$, ${B^{\prime \ast }}^{\ast }=\{{\widetilde{b^{\prime }}}_j|j\in J\}$, is $\{{\tilde{b}}_l{M}_j^l\}$, by 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.

The canonical 'vectors spaces - linear morphisms' isomorphism with respect to $B^{\prime }$, $f^{\prime }:V\to {V^{\ast }}^{\ast }$, maps $b_j=b_l^{\prime }{M^{-1}}_j^l$ to ${\widetilde{b^{\prime }}}_l{M^{-1}}_j^l={\tilde{b}}_m{M}_l^m{M^{-1}}_j^l={\tilde{b}}_m{\delta }_j^m={\tilde{b}}_j$.

So, $f^{\prime }=f$.

Step 3:

Let us see that $\mathrm{\forall }v\in V,\mathrm{\forall }w\in V^{\ast }(w(v)=f(v)(w))$.

$v=v^j{b}_j$ and $w=w_l{b}^l$, and $w(v)=w_l{b}^l(v^j{b}_j)=w_l{v}^j{b}^l(b_j)=w_l{v}^j{\delta }_j^l=w_j{v}^j$.

$f(v)(w)=f(v^j{b}_j)(w_l{b}^l)=v^j{\tilde{b}}_j(w_l{b}^l)=v^j{w}_l{\tilde{b}}_j(b^l)=v^j{w}_l{\delta }_j^l=v^j{w}_j$.

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References

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