Question

Show that if an \(n \times n\) matrix satisfies \(\left( {U{\bf{x}}} \right) \cdot \left( {U{\bf{y}}} \right) = {\bf{x}} \cdot {\bf{y}}\) for all x and y in \({\mathbb{R}^n}\), then \(U\) is an orthogonal matrix.

Short answer

It is proved that \(U\) is an orthogonal matrix.

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 \(U\] is an orthogonal matrix

Assume that, \(\left( {U{\bf{x}}} \right) \cdot \left( {U{\bf{y}}} \right) = {\bf{x}} \cdot {\bf{y}}\) for all \({\bf{x}},{\bf{y}}\) in \({\mathbb{R}^n}\) and consider \({{\mathop{\rm e}\nolimits} _1}, \ldots ,{{\mathop{\rm e}\nolimits} _n}\) as the standard basis for \({\mathbb{R}^n}\).

The \(j{\mathop{\rm th}\nolimits} \) column of \(U\) is denoted by \(U{e_j}\), with \(j = 1, \ldots ,n\). The columns of \(U\) are unit vectors because \({\left\| {U{e_j}} \right\|^2} = \left( {U{e_j}} \right) \cdot \left( {U{e_j}} \right) = {e_j} \cdot {e_j} = 1\).

The columns of \(U\) are pairwise orthogonal because \(\left( {U{e_j}} \right) \cdot \left( {U{e_k}} \right) = {e_j} \cdot {e_k} = 0\).

Thus, it is proved that \(U\) is an orthogonal matrix.