description/proof of that for injective linear map between modules, image of linearly independent subset of domain is linearly independent
Topics
About: module
The table of contents of this article
Starting Context
- The reader knows a definition of linear map.
- The reader knows a definition of injection.
- The reader knows a definition of linearly independent subset of module.
Target Context
- The reader will have a description and a proof of the proposition that for any injective linear map between any modules, the image of any linearly independent subset of the domain is linearly independent.
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:
\(R\): \(\in \{\text{ the rings }\}\)
\(M_1\): \(\in \{\text{ the } R \text{ modules }\}\)
\(M_2\): \(\in \{\text{ the } R \text{ modules }\}\)
\(S_1\): \(\subseteq M_1\), \(\in \{\text{ the possibly uncountable sets }\}\)
\(f\): \(: M_1 \to M_2\), \(\in \{\text{ the linear maps }\}\)
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Statements:
\(S_1 \in \{\text{ the linearly independent subsets of } M_1\}\)
\(\implies\)
\(f (S_1) \in \{\text{ the linearly independent subsets of } M_2\}\)
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2: Proof
Whole Strategy: Step 1: take any finite subset of \(f (S_1)\), \(\{f (m_1), ..., f (m_n)\}\), and suppose that \(c^1 f (m_1) + ... + c^n f (m_n) = 0\), and see that \(f (c^1 m_1 + ... + c^n m_n) = 0\) and \(c^1 m_1 + ... + c^n m_n = 0\), and conclude the proposition.
Step 1:
Let us take any finite subset of \(f (S_1)\), \(\{f (m_1), ..., f (m_n)\}\).
Let us suppose that \(c^1 f (m_1) + ... + c^n f (m_n) = 0\).
\(c^1 f (m_1) + ... + c^n f (m_n) = f (c^1 m_1 + ... + c^n m_n)\), because \(f\) is linear.
So, \(f (c^1 m_1 + ... + c^n m_n) = 0\).
As \(f\) is injective, \(c^1 m_1 + ... + c^n m_n = 0\).
As \(S_1\) is linearly independent and \(\{m_1, ..., m_n\}\) is a finite subset of \(S_1\), \(c^1 = ... = c^n = 0\).
That means that \(f (S_1)\) is linearly independent.
