description/proof of that for vectors space endomorphism, if there is map into domain surjective w.r.t. basis s.t. map is endomorphism after map, endomorphism is identity
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of %structure kind name% endomorphism.
- The reader knows a definition of %field name% vectors space.
- The reader knows a definition of linear map.
Target Context
- The reader will have a description and a proof of the proposition that for any vectors space endomorphism, if there is any map into the domain of the endomorphism surjective with respect to any basis such that the map is the endomorphism after the map, the endomorphism is the identity.
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:
\(F\): \(\in \{\text{ the fields }\}\)
\(V\): \(\in \{\text{ the } F \text{ vectors spaces }\}\)
\(f\): \(: V \to V\), \(\in \{\text{ the vectors space endomorphisms }\}\)
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Statements:
\(\exists S' \in \{\text{ the sets }\}, \exists B \in \{\text{ the bases of } V\}, \exists f': S' \to V, \text{ such that } B \subseteq f' (S') (f' = f \circ f')\)
\(\implies\)
\(f = id: V \to V\)
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2: Proof
Whole Strategy: Step 1: take any \(v \in V\) and see that \(v = v^1 b_1 + ... + v^k b_k\) where \(\{b_1, ..., b_k\} \subseteq B\); Step 2: see that for each \(j\), \(b_j = f' (s'_j)\) for an \(s'_j \in S'\); Step 3: see that \(f (v) = v\).
Step 1:
Let \(v \in V\) be any.
\(v = v^1 b_1 + ... + v^k b_k\), where \(\{b_1, ..., b_k\} \subseteq B\) and \(\{v^1, ..., v^k\} \subseteq F\).
Step 2:
As \(B \subseteq f' (S')\), for each \(j \in \{1, ..., k\}\), there is an \(s'_j \in S'\) such that \(b_j = f' (s'_j)\).
So, \(v = v^1 f' (s'_1) + ... + v^k f' (s'_k)\).
Step 3:
As \(f\) is linear, \(f (v) = f (v^1 f' (s'_1) + ... + v^k f' (s'_k)) = v^1 f (f' (s'_1)) + ... + v^k f (f' (s'_k))\), but as \(f' = f \circ f'\), \( = v^1 f' (s'_1) + ... + v^k f' (s'_k) = v\).
As \(v\) is arbitrary, \(f\) is the identity map.
3: Note
For the case that \(f\) is just a set endomorphism, there is a proposition that requires more for \(f'\).
