Equivalent Definitions of $O(n)$ and Proof of Membership

In this post, we’ll explore some equivalent definitions for the orthogonal group, $O(n)$, and prove that any transformation preserving distances and the origin in ${\mathbb{R}}^n$ must be in $O(n)$. Here are three common ways to define the orthogonal group:

1. $\{A\mid A{A}^{\mathrm{\top }}=I\}$: Orthogonal matrices $A$ satisfy $A^{-1}=A^{\mathrm{\top }}$.
2. $\{A\mid det(A)=\pm 1\}$: $O(n)$ is a subgroup of the general linear group $GL(n)$, consisting of matrices with determinant $\pm 1$.
3. $\{A\mid ⟨Av,Aw⟩=⟨v,w⟩\}$: Orthogonal transformations preserve the inner product.

Let’s consider $G_0$, the set of transformations on ${\mathbb{R}}^n$ that preserve distances and fix the origin:

$$G_0=\{m\mid \Vert m(v)-m(w)\Vert =\Vert v-w\Vert \text{ and }m(0)=0\}.$$

Goal: We want to show that if $m\in G_0$ preserves the inner product, then $m\in O(n)$.

Proof Outline

Since $m:{\mathbb{R}}^n\to {\mathbb{R}}^n$, we know that $m(v)\in {\mathbb{R}}^n$. Additionally, because $m$ preserves the inner product, the images $m(e_1),\dots ,m(e_n)$ of the standard basis vectors $\{e_i\}$ form another orthonormal basis. This suggests that $m$ can be represented as a linear transformation.

Let $A=[m(e_1),m(e_2),\dots ,m(e_n)]$, which is the matrix formed by taking the vectors $m(e_i)$ as columns. We aim to demonstrate that $m{A}^{-1}=I$, which would imply that $m$ acts as the matrix $A$, and further that $A^{-1}\in O(n)$.

Let’s proceed with the proof by defining $m^{\prime }=m{A}^{-1}$ and analyzing its properties.

Step 1: Show that $$ Fixes Each Basis Vector

Since $$ is the $$th column of $$, $$ for each $$.

Step 2: Show that $$ Acts as the Identity on Any Vector $$

Consider any vector $$. We want to show that $$. Notice that

$$

Since this holds for each component $$, it follows that the $$-th component of $$ matches that of $$. Therefore, we conclude that $$ for all $$, meaning $$ acts as the identity transformation.

Conclusion

Since $$, we have $$, where $$ satisfies $$, making $$. This completes our proof that any transformation $$ in $$ that preserves the inner product is indeed an orthogonal transformation, belonging to $$.

Summary

We demonstrated that the preservation of distance and the origin, along with the inner product, guarantees that a transformation is orthogonal. This property is fundamental in understanding the structure of transformations in $$ and their role in preserving geometric properties in $$.