223: Dual of Finite Dimensional Real Vectors Space Constitutes Same Dimensional Vectors Space

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A description/proof of that dual of finite dimensional real vectors space constitutes same dimensional vectors space

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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: Description
• 2: Proof

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Starting Context

• The reader knows a definition of vectors space.
• The reader knows a definition of dual of vectors space.
• The reader admits the proposition that for any 2 vectors spaces over any same field, the set of vectors space homomorphisms constitutes a vectors space.

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Target Context

• The reader will have a description and a proof of the proposition that the dual of any finite dimensional real vectors space constitutes a same dimensional vectors space.

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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: Description

For any finite dimensional real vectors space, $V$, the dual, $V^{\ast }$, constitutes a same dimensional vectors space.

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2: Proof

$V^{\ast }$ is $Hom(V,\mathbb{R})$, and by the proposition that for any 2 vectors spaces over any same field, the set of vectors space homomorphisms constitutes a vectors space, $V^{\ast }$ constitutes a vectors space.

As $V$ is $d$ dimensional, there is a basis, $b_1,b_2,...,b_d$, and for any $v\in V$, $v=v^i{b}_i$. Let us define an element, $b^{\prime i}\in V^{\ast }$, as $b^{\prime i}(v^j{b}_j)=v^i$ where $i=1,2,...,d$, which is indeed linear, because $b^{\prime i}(r_1{v}_1+r_2{v}_2)=b^{\prime i}(r_1{v_1}^j{b}_j+r_2{v_2}^j{b}_j)=r_1{v_1}^i+r_2{v_2}^i=r_1{b}^{\prime i}(v_1)+r_2{b}^{\prime i}(v_2)$. Is $b^{\prime 1},b^{\prime 2},...,b^{\prime d}$ linearly independent? For $c_i{b}^{\prime i}=0$ where $c_i\in \mathbb{R}$, $c_i{b}^{\prime i}(b_j)=c_j=0(b_j)=0$. So, yes, $b^{\prime 1},b^{\prime 2},...,b^{\prime d}$ is linearly independent. Does ${v^{\prime }}_i{b}^{\prime i}$ cover $V^{\ast }$? For any $f\in V^{\ast }$, let us define $f^{\prime }=f(b_i)b^{\prime i}$. For any $v\in V$, $v=v^i{b}_i$, and $f(v)=f(v^i{b}_i)=v^{i}f(b_i)=v^i{f}^{\prime }(b_i)=f^{\prime }(v^i{b}_i)=f^{\prime }(v)$, so, $f$ and $f^{\prime }$ are the same. So, yes, ${v^{\prime }}_i{b}^{\prime i}$ covers $V^{\ast }$. So, $b^{\prime 1},b^{\prime 2},...,b^{\prime d}$ is a basis for $V^{\ast }$, and $V^{\ast }$ is a $d$ dimensional vectors space.

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References

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