• :00Days
• :00Hours
• :00Mins
• 00Seconds
A new era for learning is coming soon Suggested languages for you:

Europe

Answers without the blur. Sign up and see all textbooks for free! Q-6-5SE

Expert-verified Found in: Page 331 ### 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 # 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.

See the 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. ### Want to see more solutions like these? 