Linear Transformation from Center of Scalar Ring

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\struct {G, +_G, \circ}_R$ and $\struct {H, +_H, \circ}_R$ be $R$-modules.

Let $\phi: G \to H$ be a linear transformation.

Let $\map Z R$ be the center of the scalar ring $R$.

Let $\lambda \in \map Z R$.


Then $\lambda \circ \phi$ is a linear transformation.


Proof

By definition of linear transformation, we need to show that:

$(1): \quad \forall x, y \in G: \map {\paren {\lambda \circ \phi} } {x +_G y} = \lambda \circ \map \phi x +_H \lambda \circ \map \phi y$
$(2): \quad \forall x \in G: \forall \mu \in R: \map {\paren {\lambda \circ \phi} } {\mu \circ x} = \mu \circ \map {\paren {\lambda \circ \phi} } x$


Let $\lambda \in \map Z R$.

Then:

\(\text {(1)}: \quad\) \(\ds \map {\paren {\lambda \circ \phi} } {x +_G y}\) \(=\) \(\ds \lambda \circ \map \phi {x +_G y}\)
\(\ds \) \(=\) \(\ds \lambda \circ \paren {\map \phi x +_H \map \phi y}\)
\(\ds \) \(=\) \(\ds \lambda \circ \map \phi x +_H \lambda \circ \map \phi y\)
\(\ds \) \(=\) \(\ds \map {\paren {\lambda \circ \phi} } x +_H \map {\paren {\lambda \circ \phi} } y\)


Because $\lambda \in \map Z R$, $\lambda$ commutes with all elements of $R$.

So:

$\forall \mu \in R: \lambda \circ \mu = \mu \circ \lambda$.

Thus:

\(\text {(2)}: \quad\) \(\ds \map {\paren {\lambda \circ \phi} } {\mu \circ x}\) \(=\) \(\ds \lambda \circ \map \phi {\mu \circ x}\)
\(\ds \) \(=\) \(\ds \lambda \circ \mu \circ \map \phi x\)
\(\ds \) \(=\) \(\ds \mu \circ \lambda \circ \map \phi x\) as $\lambda \in \map Z R$
\(\ds \) \(=\) \(\ds \mu \circ \map {\paren {\lambda \circ \phi} } x\)

$\blacksquare$


Sources