description/proof of that linear map is injective iff kernel is \(0\)-subspace
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.
Target Context
- The reader will have a description and a proof of the proposition that any linear map is injective if and only if its kernel is the \(0\)-subspace.
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:
\(R\): \(\in \{\text{ the rings }\}\)
\(M_1\): \(\in \{\text{ the } R \text{ modules }\}\)
\(M_2\): \(\in \{\text{ the } R \text{ modules }\}\)
\(f\): \(: M_1 \to M_2\), \(\in \{\text{ the linear maps }\}\)
//
Statements:
\(f \in \{\text{ the injections }\}\)
\(\iff\)
\(ker (f) = \{0\}\)
//
2: Proof
Whole Strategy: Step 1: suppose that \(ker (f) = \{0\}\); Step 2: suppose that \(f (v_1) = f (v_2)\) for any \(v_1 \neq v_2\), and find a contradiction; Step 3: suppose that \(f\) is an injection; Step 4: see that \(ker (f) = \{0\}\).
Step 1:
Let us suppose that \(ker (f) = \{0\}\).
Step 2:
Let \(v_1, v_2 \in M_1\) be any such that \(v_1 \neq v_2\).
Let us suppose that \(f (v_1) = f (v_2)\).
\(f (v_1) - f (v_2) = 0\).
But \(f (v_1) - f (v_2) = f (v_1 - v_2)\), because \(f\) was linear.
So, \(f (v_1 - v_2) = 0\), which would mean that \(v_1 - v_2 \in ker (f) = \{0\}\), which would mean that \(v_1 - v_2 = 0\), which would mean that \(v_1 = v_2\), a contradiction.
So, \(f (v_1) \neq f (v_2)\).
Step 3:
Let us suppose that \(f\) is injective.
\(f (0) = 0\), because \(f\) is linear.
There is no other \(v \in M_1\) such that \(f (v) = 0\), because \(f\) is injective.
So, \(ker (f) = \{0\}\).
