description/proof of that for 'vectors spaces - linear morphisms' isomorphism, its restriction on subspace domain and corresponding range codomain is 'vectors spaces - linear morphisms' isomorphism
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 the range of any linear map between any vectors spaces is a vectors subspace of the codomain.
- 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 for any 'vectors spaces - linear morphisms' isomorphism, its restriction on any subspace domain and the corresponding range codomain 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:
\(F\): \(\in \{\text{ the fields }\}\)
\(V'_1\): \(\in \{\text{ the } F \text{ vectors spaces }\}\)
\(V'_2\): \(\in \{\text{ the } F \text{ vectors spaces }\}\)
\(f'\): \(: V'_1 \to V'_2\), \(\in \{\text{ the 'vectors spaces - linear morphisms' isomorphisms }\}\)
\(V_1\): \(\in \{\text{ the vectors subspaces of } V'_1\}\)
\(V_2\): \(= f' (V_1)\)
\(f\): \(= f \vert_{V_1}: V_1 \to V_2\)
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Statements:
\(f \in \{\text{ the 'vectors spaces - linear morphisms' isomorphisms }\}\)
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2: Proof
Whole Strategy: Step 1: see that \(V_2\) is a vectors subspace of \(V'_2\); Step 2: see that \(f\) is linear; Step 3: see that \(f\) is bijective; Step 4: conclude the proposition.
Step 1:
\(V_2 := f' (V_1)\) is a vectors subspace of \(V'_2\), by the proposition that the range of any linear map between any vectors spaces is a vectors subspace of the codomain, which means that \(V_2\) is an \(F\) vectors space.
So, \(f\) is a map from an \(F\) vectors space into an \(F\) vectors space.
Step 2:
\(f\) is linear, because for each \(v_1, v_2 \in V_1\) and each \(r_1, r_2 \in F\), \(f (r_1 v_1 + r_2 v_2) = f' (r_1 v_1 + r_2 v_2) = r_1 f' (v_1) + r_2 f' (v_2) = r_1 f (v_1) + r_2 f (v_2)\).
Step 3:
\(f\) is bijective, because for each \(v_1, v_2 \in V_1\) such that \(v_1 \neq v_2\), \(f (v_1) = f' (v_1) \neq f' (v_2) = f (v_2)\), because \(f'\) is injective, and \(f\) is obviously surjective.
Step 4:
By the proposition that any bijective linear map between any vectors spaces is a 'vectors spaces - linear morphisms' isomorphism, \(f\) is a 'vectors spaces - linear morphisms' isomorphism.
