806: For Finite-Dimensional Vectors Space and Basis, Vectors Space Is 'Vectors Spaces - Linear Morphisms' Isomorphic to Components Vectors Space

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description/proof of that for finite-dimensional vectors space and basis, vectors space is 'vectors spaces - linear morphisms' isomorphic to components 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: Structured Description
• 2: Natural Language Description
• 3: Note
• 4: Proof

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Starting Context

• The reader knows a definition of %field name% vectors space.
• The reader knows a definition of basis of module.
• The reader knows a definition of %category name% isomorphism.
• The reader admits the proposition that any bijective linear map is a 'vectors spaces - linear morphisms' isomorphism.

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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 and any basis, the vectors space is 'vectors spaces - linear morphisms' isomorphic to the components vectors space with respect to the 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:
$F$: $\in \{\text{ the fields }\}$
$V$: $\in \{\text{ the }d\text{ -dimensional }F\text{ vectors spaces }\}$
$B$: $=\{b_1,...,b_d\}\subseteq V$, $\in \{\text{ the bases of }V\}$
$F^d$: $=\text{ the vectors space }$
$f$: $:V\to F^d,v\to (v^1,...,v^d)$, where $v=v^1{b}_1+...+v^d{b}_d$
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Statements:
$f\in \{\text{ the 'vectors spaces - linear morphisms' isomorphisms }\}$
//

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2: Natural Language Description

For any field, $F$, any $d$-dimensional $F$ vectors space, $V$, any basis, $B=\{b_1,...,b_d\}\subseteq V$, the vectors space, $F^d$, and $f:V\to F^d,v\to (v^1,...,v^d)$, where $v=v^1{b}_1+...+v^d{b}_d$, $f$ is a 'vectors spaces - linear morphisms' isomorphism.

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3: Note

$F$ does not need to be $\mathbb{R}$ or $\mathbb{C}$.

This proposition is usually regarded to be obvious intuitively, but let us be at ease with conscience that we have indeed proved it once and for all.

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4: Proof

Whole Strategy: Step 1: see that $F^d$ is indeed an $F$ vectors space; Step 2: see that $f$ is linear; Step 3: see that $f$ is bijective; Step 4: conclude the proposition.

Step 1:

Let us see that $F^d$ is indeed an $F$ vectors space.

1) for any elements, $v_1,v_2\in F^d$, $v_1+v_2\in F^d$ (closed-ness under addition); 2) for any elements, $v_1,v_2\in F^d$, $v_1+v_2=v_2+v_1$ (commutativity of addition); 3) for any elements, $v_1,v_2,v_3\in F^d$, $(v_1+v_2)+v_3=v_1+(v_2+v_3)$ (associativity of additions); 4) there is a 0 element, $0\in F^d$, such that for any $v\in F^d$, $v+0=v$ (existence of 0 vector); 5) for any element, $v\in F^d$, there is an inverse element, $v^{\prime }\in F^d$, such that $v^{\prime }+v=0$ (existence of inverse vector); 6) for any element, $v\in F^d$, and any scalar, $r\in F$, $r.v\in F^d$ (closed-ness under scalar multiplication); 7) for any element, $v\in F^d$, and any scalars, $r_1,r_2\in F$, $(r_1+r_2).v=r_1.v+r_2.v$ (scalar multiplication distributability for scalars addition); 8) for any elements, $v_1,v_2\in F^d$, and any scalar, $r\in F$, $r.(v_1+v_2)=r.v_1+r.v_2$ (scalar multiplication distributability for vectors addition); 9) for any element, $v\in F^d$, and any scalars, $r_1,r_2\in F$, $(r_1{r}_2).v=r_1.(r_2.v)$ (associativity of scalar multiplications); 10) for any element, $v\in F^d$, $1.v=v$ (identity of 1 multiplication).

Step 2:

$f$ is well-defined, because the decomposition of each vector with respect to any basis is unique: if $v=v^1{b}_1+...+v^d{b}_d=v^{\prime 1}{b}_1+...+v^{\prime d}{b}_d$, $(v^1-v^{\prime 1})b_1+...+(v^d-v^{\prime d})b_d=0$, which implies that $v^j=v^{\prime j}$.

Let us see that $f$ is linear.

Let $v,v^{\prime }\in V$ and $r,r^{\prime }\in F$ be any.

Let $v=v^1{b}_1+...+v^d{b}_d$ and $v^{\prime }=v^{\prime 1}{b}_1+...+v^{\prime d}{b}_d$.

$f(rv+r^{\prime }{v}^{\prime })=f(r(v^1{b}_1+...+v^d{b}_d)+r^{\prime }(v^{\prime 1}{b}_1+...+v^{\prime d}{b}_d))=f((r{v}^1+r^{\prime }{v}^{\prime 1})b_1+...+(r{v}^d+r^{\prime }{v}^{\prime d})b_d)=(r{v}^1+r^{\prime }{v}^{\prime 1},...,r{v}^d+r^{\prime }{v}^{\prime d})=(r{v}^1,...,r{v}^d)+(r^{\prime }{v}^{\prime 1},...,r^{\prime }{v}^{\prime d})=r(v^1,...,v^d)+r^{\prime }(v^{\prime 1},...,v^{\prime d})=rf(v)+r^{\prime }f(v^{\prime })$.

Step 3:

Let us see that $f$ is bijective.

Let $v\ne v^{\prime }\in V$ be any.

$f(v)\ne f(v^{\prime })$, because the decomposition of each vector with respect to any basis is unique: mentioned before.

So, $f$ is injective.

Let $(v^1,...,v^d)\in F^d$ be any.

For $v:=v^1{b}_1+...+v_d{b}_d\in V$, $f(v)=f(v^1{b}_1+...+v_d{b}_d)=(v^1,...,v^d)$.

So, $f$ is surjective.

Step 4:

By the proposition that any bijective linear map is a 'vectors spaces - linear morphisms' isomorphism, $f$ is a 'vectors spaces - linear morphisms' isomorphism.

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References

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