Linear Transformation of Generated Module
Theorem
Let $R$ be a ring.
Let $G$ and $H$ be $R$-modules.
Let $\phi$ and $\psi$ be linear transformations from $G$ into $H$.
Let $S$ be a generator for $G$.
Suppose that:
- $\forall x \in S: \map \phi x = \map \psi x$
Then $\phi = \psi$.
Proof 1
By definitions of generator of module and generated submodule, it follows that:
- $\ds G := \bigcap \set { M' \subseteq G : S \subseteq M', \textrm {$M'$ is a submodule of $G$} }$
Set $S' := \set {x \in G: \map \phi x = \map \psi x}$.
From Elements of Module with Equal Images under Linear Transformations form Submodule, it follows that $S'$ is a submodule of $G$.
By hypothesis, $S \subseteq S'$.
It follows that $S'$ is one of the submodules $M'$.
From Intersection is Subset:Family of Sets, it follows that $G \subseteq S'$.
As $S' \subseteq G$, it follows that $S' = G$.
It follows that for all $x \in G$, we have:
- $\map \phi x = \map \psi x$
which proves that $\phi = \psi$.
$\blacksquare$
Proof 2
This proof assumes that $R$ is a ring with unity, so $G$ and $H$ become unitary modules.
Let $y \in G$ be arbitrary.
Then by definition of generator, $y$ is the linear combination of elements of $S$:
- $\ds y = \sum_{k \mathop = 1}^n \lambda_k a_k$
for $a_1, a_2, \ldots, a_n \in S, \lambda_1, \lambda_2, \ldots, \lambda n \in R$
Then:
\(\ds \map \phi y\) | \(=\) | \(\ds \map \phi {\sum_{k \mathop = 1}^n \lambda_k a_k}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 1}^n \lambda_k \map \phi {a_k}\) | Definition of Linear Transformation | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 1}^n \lambda_k \map \psi {a_k}\) | by hypothesis: $\forall a_k \in S: \map \phi {a_k} = \map \psi {a_k}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \psi {\sum_{k \mathop = 1}^n \lambda_k a_k}\) | Definition of Linear Transformation | |||||||||||
\(\ds \) | \(=\) | \(\ds \map \psi y\) |
$\blacksquare$
Also see
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Theorem $28.3$