description/proof of that finite-dimensional 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 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 linear injection between any same-finite-dimensional vectors spaces is a 'vectors spaces - linear morphisms' isomorphism.
Target Context
- The reader will have a description and a proof of the proposition that any finite-dimensional 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:
\(F\): \(\in \{\text{ the fields }\}\)
\(V_1\): \(\in \{\text{ the finite-dimensional } F \text{ vectors spaces }\}\)
\(V_2\): \(\in \{\text{ the finite-dimensional } F \text{ vectors spaces }\}\)
//
Statements:
\(V_1 \cong_{\text{ vectors spaces }} V_2\)
\(\iff\)
\(dim V_1 = dim V_2 = d\)
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2: Proof
Whole Strategy: Step 1: suppose that \(dim V_1 = dim V_2 = d\), and choose any bases for \(V_1\) and \(V_2\) and choose a 'vectors spaces - linear morphisms' isomorphism that maps the basis for \(V_1\) onto the basis for \(V_2\) bijectively; Step 2: suppose that \(V_1 \cong_{\text{ vectors spaces }} V_2\), and choose any 'vectors spaces - linear morphisms' isomorphism and any basis for \(V_1\) and see that \(V_2\) has the image of the basis for \(V_1\) as a basis.
Step 1:
Let us suppose that \(dim V_1 = dim V_2 = d\).
\(V_1\) has a basis, \(B_1 = \{{b_1}_1, ...,, {b_1}_d\}\); \(V_2\) has a basis, \(B_2 = \{{b_2}_1, ...,, {b_2}_d\}\).
Let us define a map, \(f: V_1 \to V_2, v^j {b_1}_j \mapsto v^j {b_2}_j\).
It is well-defined, because \(v^j {b_1}_j\) spans \(V_1\) and the decomposition is unique: the proposition that for any module with any basis, the components set of any element with respect to the basis is unique.
\(f\) is linear, 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 defined by mapping \({b_1}_j\) onto \({b_2}_j\) and linearly expanding the mapping.
\(f\) is injective, because for each \(v^j {b_1}_j \neq v'^j {b_1}_j\), \(f (v^j {b_1}_j) = v^j {b_2}_j \neq v'^j {b_2}_j = f (v'^j {b_1}_j)\).
\(f\) is a 'vectors spaces - linear morphisms' isomorphism, by the proposition that any linear injection between any same-finite-dimensional vectors spaces is a 'vectors spaces - linear morphisms' isomorphism.
So, \(V_1\) and \(V_2\) are 'vectors spaces - linear morphisms' isomorphic.
Step 2:
Let us suppose that \(V_1 \cong_{\text{ vectors spaces }} V_2\).
There is a 'vectors spaces - linear morphisms' isomorphism, \(f: V_1 \to V_2\).
Let us choose any basis for \(V_1\), \(B_1 = \{{b_1}_1, ...,, {b_1}_d\}\).
Let us think of \(B_2 := \{f ({b_1}_1), ...,, f ({b_1}_d)\}\).
Let us see that \(B_2\) is a basis for \(V_2\).
\(v^j f ({b_1}_j)\) spans \(V_2\), because for each \(v_2 \in V_2\), as \(f\) is bijective, there is a \(v \in V_1\) such that \(f (v) = v_2\); \(v = v^j {b_1}_j\); \(f (v) = f (v^j {b_1}_j) = v^j f ({b_1}_j) = v_2\).
\(B_2\) is linearly independent, because for \(c^j f ({b_1}_j) = 0\), \(c^j f ({b_1}_j) = f (c^j {b_1}_j) = 0\), which implies that \(c^j {b_1}_j = 0\), which implies that \(c^j = 0\).
So, \(B_2\) is a basis, and \(V_2\) is \(d\)-dimensional, because \(B_2\) has the \(d\) elements.
