Linear Transformation from Center of Scalar Ring
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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
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations