1032: Canonical 'Vectors Spaces - Linear Morphisms' Isomorphism Between Finite-Dimensional Vectors Space and Its Covectors Space w.r.t. Original Space Basis

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definition of canonical 'vectors spaces - linear morphisms' isomorphism between finite-dimensional vectors space and its covectors space w.r.t. original space basis

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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: Note

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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.

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Target Context

• The reader will have a definition of canonical 'vectors spaces - linear morphisms' isomorphism between finite-dimensional vectors space and its covectors space with respect to original space basis.

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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:
$F$: $\in \{\text{ the fields }\}$
$V$: $\in \{\text{ the finite-dimensional }F\text{ vectors spaces }\}$
$V^{\ast }$: $=L(V:F)$
$J$: $\in \{\text{ the finite index sets }\}$
$B$: $\in \{\text{ the bases for }V\}$, $=\{b_j|j\in J\}$
$B^{\ast }$: $=\text{ the dual basis of }B$, $=\{b^j|j\in J\}$
$\ast f$: $:V\to V^{\ast },v^j{b}_j\mapsto \sum _{j\in J}{v}^j{b}^j$, $\in \{\text{ the 'vectors spaces - linear morphisms' isomorphisms }\}$
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Conditions:
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$f$ depends on the choice of $B$ (as will be seen in Note), which is the reason why the concept is called with "with respect to original space basis".

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2: Note

$f$ 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.

Let us see that $f$ depends on the choice of $B$.

Let $B^{\prime }=\{b_j^{\prime }|j\in J\}$ be another 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 $\{{M^{-1}}_l^j{b}^l\}$, 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 $B^{\prime }$, $f^{\prime }:V\to V^{\ast }$, maps $b_j={M^{-1}}_j^l{b}_l^{\prime }$ to $\sum _{l\in J}{M^{-1}}_j^l{b}^{\prime l}=\sum _{l\in J}{M^{-1}}_j^l{M^{-1}}_m^l{b}^m$, which does not equal $b^j$ in general. In fact, when $M$ is orthogonal, $M_l^m={M^{-1}}_m^l$, and $=\sum _{m\in J}{M^{-1}}_j^l{M}_l^m{b}^m=\sum _{m\in J}{\delta }_j^m{b}^m=b^j$.

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References

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