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Q-6-5SE

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Linear Algebra and its Applications
Found in: Page 331
Linear Algebra and its Applications

Linear Algebra and its Applications

Book edition 5th
Author(s) David C. Lay, Steven R. Lay and Judi J. McDonald
Pages 483 pages
ISBN 978-03219822384

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Short Answer

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.

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

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Step by Step Solution

Statement in Theorem 7

Theorem 7 states 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.

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