Inverse of Similarity Mapping

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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 Field Axiom $\text M4$: Inverses for Product, 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:

\(\ds \map {\paren {s_\beta}^{-1} } {\map {s_\beta} {\beta^{-1} \, \mathbf x} }\) \(=\) \(\ds \beta^{-1} \mathbf x\) Definition of Inverse Mapping
\(\ds \leadsto \ \ \) \(\ds \map {\paren {s_\beta}^{-1} } {\beta \beta^{-1} \, \mathbf x}\) \(=\) \(\ds \beta^{-1} \, \mathbf x\) Definition of $s_\beta$
\(\ds \leadsto \ \ \) \(\ds \map {\paren {s_\beta}^{-1} } {1_K \, \mathbf x}\) \(=\) \(\ds \beta^{-1} \, \mathbf x\) Field Axiom $\text M4$: Inverses for Product
\(\ds \leadsto \ \ \) \(\ds \map {\paren {s_\beta}^{-1} } {\mathbf x}\) \(=\) \(\ds \beta^{-1} \, \mathbf x\) Field Axiom $\text M3$: Identity for Product
\(\ds \leadsto \ \ \) \(\ds \map {\paren {s_\beta}^{-1} } {\mathbf x}\) \(=\) \(\ds \map {s_{\beta^{-1} } } {\mathbf x}\) Definition of $s_{\beta^{-1} }$

$\blacksquare$


Sources

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