description/proof of that vectors spaces over same field are 'vectors spaces - linear morphisms' isomorphic iff they have same dimension
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of %field name% 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 any 'vectors spaces - linear morphisms' isomorphism is a map that maps any basis onto a basis bijectively and expands the mapping linearly.
- The reader admits the proposition that for any vectors space, the bases have the same cardinality.
- The reader admits the proposition that any finite composition of bijections is a bijection, if the codomains of the constituent bijections equal the domains of the succeeding bijections.
Target Context
- The reader will have a description and a proof of the proposition that any vectors spaces over any same field are 'vectors spaces - linear morphisms' isomorphic if and only if they have any same dimension.
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:
\(F\): \(\in \{\text{ the fields }\}\)
\(V_1\): \(\in \{\text{ the } F \text{ vectors spaces }\}\), with any basis, \(B_1 = \{b_{1, j} \vert j \in J_1\}\), where \(J_1\) is a possibly uncountable index set
\(V_2\): \(\in \{\text{ the } F \text{ vectors spaces }\}\), with any basis, \(B_2 = \{b_{2, j} \vert j \in J_2\}\), where \(J_2\) is a possibly uncountable index set
//
Statements:
\(V_1 \cong_{\text{ vectors spaces }} V_2\)
\(\iff\)
\(B_1 \cong_{\text{ sets }} B_2\)
//
When \(V_1\) and \(V_2\) are infinite-dimensional, "have the same dimension" means \(B_1 \cong_{\text{ sets }} B_2\).
2: Proof
Whole Strategy: Step 1: suppose that \(B_1 \cong_{\text{ sets }} B_2\); Step 2: take any bijection, \(f: B_1 \to B_2\), and define \(f': V_1 \to V_2\) as the linear expansion of \(f\) and see that \(f'\) is a 'vectors spaces - linear morphisms' isomorphism; Step 3: suppose that \(V_1 \cong_{\text{ vectors spaces }} V_2\); Step 4: take any 'vectors spaces - linear morphisms' isomorphism, \(f' : V_1 \to V_2\), and see that \(f'\) maps \(B_1\) to the basis \(f' (B_1)\) bijectively; Step 5: see that there is a bijection, \(g: f' (B_1) \to B_2\); Step 6: see that \(g \circ f' \vert_{B_1}: B_1 \to B_2\) is a bijection.
Step 1:
Let us suppose that \(B_1 \cong_{\text{ sets }} B_2\).
Step 2:
Let us take any bijection, \(f: B_1 \to B_2\).
Let us define \(f': V_1 \to V_2\) as the linear expansion of \(f\).
\(f'\) is well-defined and is 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.
So, \(V_1 \cong_{\text{ vectors spaces }} V_2\).
Step 3:
Let us suppose that \(V_1 \cong_{\text{ vectors spaces }} V_2\).
Step 4:
Let us take any 'vectors spaces - linear morphisms' isomorphism, \(f': V_1 \to V_2\).
\(f'\) maps \(B_1\) to the basis \(f' (B_1)\) bijectively, by the proposition that any 'vectors spaces - linear morphisms' isomorphism is a map that maps any basis onto a basis bijectively and expands the mapping linearly.
Step 5:
\(f' (B_1)\) and \(B_2\) have the same cardinality, by the proposition that for any vectors space, the bases have the same cardinality.
That means that there is a bijection, \(g: f' (B_1) \to B_2\).
Step 6:
\(f := g \circ f' \vert_{B_1}: B_1 \to B_2\), is a bijection as a composition of bijections, by the proposition that any finite composition of bijections is a bijection, if the codomains of the constituent bijections equal the domains of the succeeding bijections.
