Elements of Module with Equal Images under Linear Transformations form Submodule

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Theorem

Let $G$ and $H$ be $R$-modules.

Let $\phi$ and $\psi$ be linear transformations from $G$ into $H$.


Then the set $S = \set {x \in G: \map \phi x = \map \psi x}$ is a submodule of $G$.


Proof

Let $x, y \in S$.

Let $\lambda \in R$.


Then:

\(\ds \map \phi {x + y}\) \(=\) \(\ds \map \phi x + \map \phi y\) Definition of Linear Transformation
\(\ds \) \(=\) \(\ds \map \psi x + \map \psi y\) $x, y \in S$
\(\ds \) \(=\) \(\ds \map \psi {x + y}\) Definition of Linear Transformation
\(\ds \map \phi {\lambda \circ x}\) \(=\) \(\ds \lambda \circ \map \phi x\) Definition of Linear Transformation
\(\ds \) \(=\) \(\ds \lambda \circ \map \psi x\) $x \in S$
\(\ds \) \(=\) \(\ds \map \psi {\lambda \circ x}\) Definition of Linear Transformation

Hence $x + y, \lambda \circ x \in S$.

By Submodule Test, $S$ is a submodule of $G$.

$\blacksquare$


Also see


Sources

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