Question

Let \(U\) be an \(n \times n\) orthogonal matrix. Show that if \(\left\{ {{{\bf{v}}_1}, \ldots ,{{\bf{v}}_n}} \right\}\) is an orthonormal basis for \({\mathbb{R}^n}\), then so is \(\left\{ {U{{\bf{v}}_1}, \ldots ,U{{\bf{v}}_n}} \right\}\).

Short answer

It is proved that \(\left\{ {U{{\bf{v}}_1}, \ldots ,U{{\bf{v}}_k}} \right\}\) is a basis for \({\mathbb{R}^n}\).

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 if \(\left\{ {{{\bf{v}}_1}, \ldots ,{{\bf{v}}_n}} \right\}\) is an orthonormal basis for \({\mathbb{R}^n}\), then so is \(\left\{ {U{{\bf{v}}_1}, \ldots ,U{{\bf{v}}_n}} \right\}\)

According to Theorem 7, \(\left\{ {U{{\bf{v}}_1}, \ldots ,U{{\bf{v}}_k}} \right\}\) is an orthonormal set in \({\mathbb{R}^n}\). The set \(\left\{ {U{{\bf{v}}_1}, \ldots ,U{{\bf{v}}_k}} \right\}\) forms a basis for\({\mathbb{R}^n}\) because it is a linearly independent set with \(n\) vectors.

Thus, it is proved that \(\left\{ {U{{\bf{v}}_1}, \ldots ,U{{\bf{v}}_k}} \right\}\) is a basis for \({\mathbb{R}^n}\).