Question

Show that if \(U\) is an orthogonal matrix, then any real eigenvalue of \(U\) must be \( \pm 1\).

Short answer

It is proved that any real eigenvalues of \(U\) is \( \pm 1\).

Step-by-step solution

Statement in Theorem 7

Theorem 7states that consider that \(U\) as an \(m \times n\) matrix with orthonormal columns and assume that x and y are in \({\mathbb{R}^n}\). Then;

1. \(\left\| {U{\bf{x}}} \right\| = \left\| {\bf{x}} \right\|\)
2. \(\left( {U{\bf{x}}} \right) \cdot \left( {U{\bf{y}}} \right) = {\bf{x}} \cdot {\bf{y}}\)
3. \(\left( {U{\bf{x}}} \right) \cdot \left( {U{\bf{y}}} \right) = 0\)such that if \({\bf{x}} \cdot {\bf{y}} = 0\).

Show that real eigenvalues of \(U\) must be \( \pm 1\) 

When \(U{\bf{x}} = \lambda {\bf{x}}\) for some \({\bf{x}} \ne 0\), then according to Theorem 7 and by a property of the norm:

\(\begin{array}{c}\left\| {\bf{x}} \right\| = \left\| {U{\bf{x}}} \right\|\\ = \left\| {\lambda {\bf{x}}} \right\|\\ = \left| \lambda \right|\left\| {\bf{x}} \right\|\end{array}\)

This demonstrates that \(\left| \lambda \right| = 1\), since \({\bf{x}} \ne 0\).

Thus, it is proved that any real eigenvalues of \(U\) is \( \pm 1\).