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 \(\Bbb{R}^n\) must be in \(O(n)\). Here are three common ways to define the orthogonal group:

1. \(\{ A \mid A A^\top = I \}\): Orthogonal matrices \(A\) satisfy \(A^{-1} = A^\top\).
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. \(\{ A \mid \langle Av, Aw \rangle = \langle v, w \rangle \}\): Orthogonal transformations preserve the inner product.

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

\[G_0 = \{ m \mid \| m(v) - m(w) \| = \| v - w \| \text{ and } m(0) = 0 \}.\]

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

Proof Outline

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

Let \(A = [m(e_1), m(e_2), \ldots, m(e_n)]\), 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 \(m' = m A^{-1}\) and analyzing its properties.

Step 1: Show that \(m'\) Fixes Each Basis Vector

Since \(m(e_i)\) is the \(i\)th column of \(A\), \(m' (e_i) = e_i\) for each \(i\).

Step 2: Show that \(m'\) Acts as the Identity on Any Vector \(v\)

Consider any vector \(v \in \Bbb{R}^n\). We want to show that \(m'(v) = v\). Notice that

\[ \langle m'(v), e_i \rangle = \langle m'(v), m'(e_i) \rangle = \langle v, e_i \rangle = v_i. \]

Since this holds for each component \(i\), it follows that the \(i\)-th component of \(m'(v)\) matches that of \(v\). Therefore, we conclude that \(m'(v) = v\) for all \(v\), meaning \(m'\) acts as the identity transformation.

Conclusion

Since \(m' = I\), we have \(m = A\), where \(A\) satisfies \(A A^\top = I\), making \(A \in O(n)\). This completes our proof that any transformation \(m\) in \(G_0\) that preserves the inner product is indeed an orthogonal transformation, belonging to \(O(n)\).

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 \(O(n)\) and their role in preserving geometric properties in \(\Bbb{R}^n\).