Similarity Mapping is Automorphism

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:

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, $\beta^{-1}$ always exists.

Then:

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

Hence the result.