Inverse of Similarity Mapping

Theorem
Let $G$ be a vector space over a field $K$.

Let $\beta \in K$ such that $\beta \ne 0$.

Let $s_\beta: G \to G$ be the similarity on $G$ defined as:
 * $\forall \mathbf x \in G: \map {s_\beta} {\mathbf x} = \beta \mathbf x$

Let $\paren {s_\beta}^{-1}$ denote the inverse of $s_\beta$.

Then:
 * $\paren {s_\beta}^{-1} = s_{\beta^{-1} }$

where $\beta^{-1}$ is the multiplicative inverse in $K$ of $\beta$.

Proof
From Similarity Mapping is Automorphism, $s_\beta$ is an automorphism of $G$.

Hence $s_\beta$ is an vector space isomorphism from $G$ to $G$ itself.

So by definition $s_\beta$ is a bijection.

Hence the existence of this inverse $\paren {s_\beta}^{-1}$ follows from Bijection iff Left and Right Inverse.

By, we have that there exists a multiplicative inverse $\beta^{-1}$ for $\beta$ such that:
 * $\beta \beta^{-1} = 1_K = \beta^{-1} \beta$

where $1_K$ is the multiplicative identity of $K$.

Then: