Linear Transformation of Generated Module/Proof 2

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 $\phi = \psi$.

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 $

$=$

$\ds \sum_{k \mathop = 1}^n \lambda_k \map \phi {a_k}$

$\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 $

$=$

$\ds \map \psi {\sum_{k \mathop = 1}^n \lambda_k a_k}$

$\ds $

$=$

$\ds \map \psi y$

$\blacksquare$