description/proof of that bijective linear map between vectors spaces 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: Natural Language Description
- 3: Note
- 4: Proof
Starting Context
- The reader knows a definition of %field name% vectors space.
- The reader knows a definition of bijection.
- The reader knows a definition of linear map. The reader knows a definition of %category name% isomorphism.
Target Context
- The reader will have a description and a proof of the proposition that any bijective linear map between any 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:
\(F\): \(\in \{\text{ the fields }\}\)
\(V_1\): \(\in \{\text{ the not necessarily finite-dimensional } F \text{ vectors spaces }\}\)
\(V_2\): \(\in \{\text{ the not necessarily finite-dimensional } F \text{ vectors spaces }\}\)
\(f\): \(: V_1 \to V_2\), \(\in \{\text{ the linear maps }\} \cap \{\text{ the bijections }\}\)
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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 not necessarily finite-dimensional vectors spaces, \(V_1, V_2\), and any bijective linear map, \(f: V_1 \to V_2\), \(f\) is a 'vectors spaces - linear morphisms' isomorphism.
3: Note
In general, a bijective morphism is not necessarily an isomorphism: for example, a bijective continuous map is not necessarily a 'topological spaces - continuous maps' isomorphism, because the inverse is not necessarily continuous, which is the reason why we need to specifically prove this proposition.
4: Proof
Whole Strategy: Step 1: define the inverse, \(f^{-1}: V_2 \to V_1\); Step 2: see that \(f^{-1}\) is linear; Step 3: conclude the proposition.
Step 1:
As \(f\) is bijective, the inverse, \(f^{-1}: V_2 \to V_1\), is well-defined.
Step 2:
Let us see that \(f^{-1}\) is linear.
Let \(v_1, v_2 \in V_2\) and \(r_1, r_2 \in F\) be any. \(f^{-1} (r_1 v_1 + r_2 v_2) = r_1 f^{-1} (v_1) + r_2 f^{-1} (v_2)\)? \(f (r_1 f^{-1} (v_1) + r_2 f^{-1} (v_2)) = r_1 f (f^{-1} (v_1)) + r_2 f (f^{-1} (v_2)) = r_1 v_1 + r_2 v_2\), which means that \(f^{-1} (r_1 v_1 + r_2 v_2) = r_1 f^{-1} (v_1) + r_2 f^{-1} (v_2)\). So, \(f^{-1}\) is linear.
Step 3:
So, \(f\) is a 'vectors spaces - linear morphisms' isomorphism.
