1031: Between Vectors Spaces, Map That Maps Basis onto Basis Bijectively and Expands Mapping Linearly Is 'Vectors Spaces - Linear Morphisms' Isomorphism

<The previous article in this series | The table of contents of this series | The next article in this series>

description/proof of that between vectors spaces, map that maps basis onto basis bijectively and expands mapping linearly is 'vectors spaces - linear morphisms' isomorphism

---

Topics

About: vectors space

---

The table of contents of this article

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

---

Starting Context

• The reader knows a definition of %field name% vectors space.
• The reader knows a definition of %category name% isomorphism.
• The reader knows a definition of basis of module.
• The reader admits the proposition that for any module with any basis, the components set of any element with respect to the basis is unique.
• The reader admits the proposition that from any module with any basis into any module, a linear map can be defined by mapping the basis and linearly expanding the mapping.
• The reader admits the proposition that any bijective linear map between any vectors spaces is a 'vectors spaces - linear morphisms' isomorphism.

---

Target Context

• The reader will have a description and a proof of 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.

---

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.

---

Main Body

---

1: Structured Description

Here is the rules of Structured Description.

Entities:
$F$: $\in \{\text{ the fields }\}$
$V_1$: $\in \{\text{ the }F\text{ vectors spaces }\}$
$V_2$: $\in \{\text{ the }F\text{ vectors spaces }\}$
$J$: $\in \{\text{ the possibly uncountable index sets }\}$
$B_1$: $=\{{b_1}_j|j\in J\}$, $\in \{\text{ the bases for }{V}_1\}$
$B_2$: $\in \{\text{ the bases for }{V}_2\}$
$f$: $:B_1\to B_2$, $\in \{\text{ the bijections }\}$
$g$: $:V_1\to V_2,\sum _{j\in S}{v}^j{b_1}_j\mapsto \sum _{j\in S}{v}^{j}f({b_1}_j)$, where $S\in \{\text{ the finite subsets of }J\}$
//

Statements:
$g\in \{\text{ the 'vectors spaces - linear morphisms' isomorphisms }\}$
//

---

2: Note

$V_1$ or $V_2$ does not need to be finite-dimensional.

---

3: Proof

Whole Strategy: Step 1: see that $g$ is well-defined; Step 2: see that $g$ is a linear map; Step 3: see that $g$ is injective; Step 4: see that $g$ is surjective; Step 5: conclude the proposition.

Step 1:

Let us see that $g$ is well-defined.

For each element, $v\in V_1$, $v=\sum _{j\in S}{v}^j{b_1}_j$ where $S$ is a finite subset of $J$, by the definition of basis of module.

The decomposition is unique, by the proposition that for any module with any basis, the components set of any element with respect to the basis is unique.

So, $g$ is determined for each element of $V_1$ unambiguously.

Step 2:

$g$ is a linear map, by the proposition that from any module with any basis into any module, a linear map can be defined by mapping the basis and linearly expanding the mapping: $f$ is mapping the basis, $B_1$, and $g$ is linearly expanding $f$.

Step 3:

Let us see that $g$ is injective.

Let $v,v^{\prime }\in V_1$ be any such that $v\ne v^{\prime }$.

$v=\sum _{j\in S}{v}^j{b_1}_j$ and $v^{\prime }=\sum _{j\in S^{\prime }}{v}^{\prime j}{b_1}_j$.

$g(v)=\sum _{j\in S}{v}^{j}f({b_1}_j)$ and $g(v^{\prime })=\sum _{j\in S^{\prime }}{v}^{\prime j}f({b_1}_j)$.

Let us suppose that $g(v)=g(v^{\prime })$.

$\{f({b_1}_j)|j\in S\cup S^{\prime }\}\subseteq B_2$ was distinct, because $f$ was bijective.

By the proposition that for any module with any basis, the components set of any element with respect to the basis is unique, $S=S^{\prime }$ and $v^j=v^{\prime j}$, a contradiction against $v\ne v^{\prime }$.

So, $g(v)\ne g(v^{\prime })$.

Step 4:

Let us see that $g$ is surjective.

For each $v_2\in V_2$, $v_2=\sum _{j\in \{1,...,n\}}{v}^j{b_2}_j$, where ${b_2}_j\in B_2$.

As $f$ is bijective, $\{f^{-1}({b_2}_1),...,f^{-1}({b_2}_n)\}=\{{b_1}_{j_1},...,{b_1}_{j_n}\}\subseteq B_1$ is valid and distinct.

$v^1{b_1}_{j_1}+...+v^n{b_1}_{j_n}\in V_1$ and $g(v^1{b_1}_{j_1}+...+v^n{b_1}_{j_n})=v^{1}f({b_1}_{j_1})+...+v^{n}f({b_1}_{j_n})=v^1{b_2}_1+...+v^n{b_2}_n=v_2$.

Step 5:

So, $g$ is a 'vectors spaces - linear morphisms' isomorphism, by the proposition that any bijective linear map between any vectors spaces is a 'vectors spaces - linear morphisms' isomorphism.

---

References

<The previous article in this series | The table of contents of this series | The next article in this series>