description/proof of that kernel of linear map between vectors spaces is vectors subspace of domain
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of kernel of linear map.
- The reader knows a definition of %field name% vectors space.
- The reader admits the proposition that for any vectors space, any nonempty subset of the vectors space is a vectors subspace if and only if the subset is closed under linear combination.
Target Context
- The reader will have a description and a proof of the proposition that the kernel of any linear map between any vectors spaces is a vectors subspace of the domain.
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:
\(F\): \(\in \{\text{ the fields }\}\)
\(V_1\): \(\in \{\text{ the } F \text{ vectors fields }\}\)
\(V_2\): \(\in \{\text{ the } F \text{ vectors fields }\}\)
\(f\): \(: V_1 \to V_2\), \(\in \{\text{ the linear maps }\}\)
\(Ker (f)\): \(= \text{ the kernel of } f\)
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Statements:
\(Ker (f) \in \{\text{ the vectors subspaces of } V_1\}\)
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2: Proof
Whole Strategy: Step 1: see that \(Ker (f)\) is closed under linear combination; Step 2: conclude the proposition.
Step 1:
Let us see that \(Ker (f)\) is closed under linear combination.
Let \(v, v' \in Ker (f)\) and \(r, r' \in F\) be any.
\(f (r v + r' v') = r f (v) + r' f (v') = r 0 + r' 0 = 0\).
So, \(r v + r' v' \in Ker (f)\).
\(0 \in Ker (f)\).
Step 2:
By the proposition that for any vectors space, any nonempty subset of the vectors space is a vectors subspace if and only if the subset is closed under linear combination, \(Ker (f)\) is a vectors subspace of \(V_1\).
