description/proof of that linear injection between same-finite-dimensional vectors spaces is 'vectors spaces - linear morphisms' isomorphism
Topics
About: vectors space
About: category
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description
- 2: Natural Language Description
- 3: Proof
Starting Context
- The reader knows a definition of linear map.
- The reader knows a definition of dimension of vectors space.
- The reader knows a definition of injection.
- The reader knows a definition of %category name% isomorphism.
- The reader admits the proposition that for any finite-dimensional vectors space, there is no linearly independent subset that has more than the dimension number of elements.
- The reader admits the proposition that any linear surjection from any finite-dimensional vectors space to any same-dimensional vectors space is a 'vectors spaces - linear morphisms' isomorphism.
Target Context
- The reader will have a description and a proof of the proposition that any linear injection between any same-finite-dimensional vectors spaces 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:
\(F\): \(\in \{\text{ the fields }\}\)
\(V_1\): \(\in \{\text{ the } d \text{ -dimensional } F \text{ vectors spaces }\}\)
\(V_2\): \(\in \{\text{ the } d \text{ -dimensional } F \text{ vectors spaces }\}\)
\(f\): \(: V_1 \to V_2\), \(\in \{\text{ the linear injections }\}\)
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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 spaces, \(V_1, V_2\), and any linear injection, \(f: V_1 \to V_2\), \(f\) is a 'vectors spaces - linear morphisms' isomorphism.
3: Proof
Whole Strategy: see that \(f\) is surjective; Step 1: choose any basis of \(V_1\), \(\{e_1, ..., e_d\}\); Step 2: see that \(\{f (e_1), ..., f (e_d)\}\) is linearly independent on \(V_2\); Step 3: see that \(\{f (e_1), ..., f (e_d)\}\) spans \(V_2\); Step 4: conclude the proposition.
Step 1:
Let us choose any basis of \(V_1\), \(\{e_1, ..., e_d\}\).
Step 2:
Let us see that \(\{f (e_1), ..., f (e_d)\}\) is linearly independent on \(V_2\).
Any element, \(v \in V_1\), is \(v = c^j e_j\) (with the Einstein convention).
If \(v \neq 0\), \(f (v) = c^j f (e_j) \neq 0\), because otherwise, \(f (v' + v) = f (v') + f (v) = f (v')\), a contradiction against \(f\)'s being injective. That means that \(c^j f (e_j) = 0\) only if \(v = 0\), which implies that \(c^j\) s are all \(0\). So, \(\{f (e_1), ..., f (e_d)\}\) is linearly independent on \(V_2\).
Step 3:
Let us see that \(\{f (e_1), ..., f (e_d)\}\) spans \(V_2\).
\(\{f (e_1), ..., f (e_d)\}\) spans \(V_2\), because otherwise, there would be an element, \(v' \in V_2\), that would be linearly independent from \(\{f (e_1), ..., f (e_d)\}\), a contradiction against the proposition that for any finite-dimensional vectors space, there is no linearly independent subset that has more than the dimension number of elements.
So, \(f\) is surjective.
Step 4:
So, \(f\) is a 'vectors spaces - linear morphisms' isomorphism, by the proposition that any linear surjection from any finite-dimensional vectors space to any same-dimensional vectors space is a 'vectors spaces - linear morphisms' isomorphism.
