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:
\(\ds \forall \mathbf x, \mathbf y \in G: \, \) | \(\ds \map {s_\beta} {\mathbf x}\) | \(=\) | \(\ds \map {s_\beta} {\mathbf y}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \beta \, \mathbf x\) | \(=\) | \(\ds \beta \, \mathbf y\) | 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
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Example $28.3$