Similarity Mapping is Automorphism

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Theorem

Let $G$ be a vector space over a field $\struct {K, +, \times}$.

Let $\beta \in K$.


Let $s_\beta: G \to G$ be the similarity on $G$ defined as:

$\forall \mathbf x \in G: \map {s_\beta} {\mathbf x} = \beta \mathbf x$


If $\beta \ne 0$ then $s_\beta$ is an automorphism of $G$.


Proof

By definition, a vector space automorphism on $G$ is a vector space isomorphism from $G$ to $G$ itself.

To prove that $s_\beta$ is a automorphism it is sufficient to demonstrate that:

By definition, a vector space isomorphism is a mapping $s_\beta: G \to G$ such that:

$(1): \quad s_\beta$ is a bijection
$(2): \quad \forall \mathbf x, \mathbf y \in G: \map {s_\beta} {\mathbf x + \mathbf y} = \map {s_\beta} {\mathbf x} + \map {s_\beta} {\mathbf y}$
$(3): \quad \forall \mathbf x \in G: \forall \lambda \in K: \map {s_\beta} {\lambda \mathbf x} = \lambda \map {s_\beta} {\mathbf x}$


It has been established in Similarity Mapping is Linear Operator that $s_\beta$ is a linear operator on $G$.

Hence $(2)$ and $(3)$ follow by definition of linear operator.


It remains to prove bijectivity.

That is, that $s_\beta$ is both injective and surjective.


Let $1_K$ denote the multiplicative identity of $K$.

We have:

\(\ds \forall \mathbf x, \mathbf y \in G: \, \) \(\ds \map {s_\beta} {\mathbf x}\) \(=\) \(\ds \map {s_\beta} {\mathbf y}\)
\(\ds \leadsto \ \ \) \(\ds \beta \, \mathbf x\) \(=\) \(\ds \beta \, \mathbf y\) Definition of $s_\beta$
\(\ds \leadsto \ \ \) \(\ds \beta^{-1} \beta \, \mathbf x\) \(=\) \(\ds \beta^{-1} \beta \, \mathbf y\) Field Axiom $\text M4$: Inverses for Product
\(\ds \leadsto \ \ \) \(\ds 1_K \, \mathbf x\) \(=\) \(\ds 1_K \, \mathbf y\) Field Axiom $\text M3$: Identity for Product
\(\ds \leadsto \ \ \) \(\ds \mathbf x\) \(=\) \(\ds \mathbf y\) Vector Space Axiom $\text V 8$: Identity for Scalar Multiplication

Hence it has been demonstrated that $s_\beta$ is injective.


Let $\mathbf y \in G$.

Consider $\beta^{-1} \in K$ defined such that $\beta \beta^{-1} = 1_K$.

By Field Axiom $\text M4$: Inverses for Product, $\beta^{-1}$ always exists.

Then:

\(\ds \forall \mathbf y \in G: \exists \mathbf x \in G: \, \) \(\ds \mathbf x\) \(=\) \(\ds \beta^{-1} \mathbf y\)
\(\ds \leadsto \ \ \) \(\ds \beta \, \mathbf x\) \(=\) \(\ds \beta \beta^{-1} \mathbf y\)
\(\ds \) \(=\) \(\ds 1_K \mathbf y\)
\(\ds \) \(=\) \(\ds \mathbf y\) Field Axiom $\text M3$: Identity for Product
\(\ds \leadsto \ \ \) \(\ds \forall \mathbf y \in G: \exists \mathbf x \in G: \, \) \(\ds \map {s_\beta} {\mathbf x}\) \(=\) \(\ds \mathbf y\) Definition of $s_\beta$

Hence it has been demonstrated that $s_\beta$ is surjective.

Hence the result.

$\blacksquare$


Sources