1E Question: In Exercises 1 and 2, ... [FREE SOLUTION]

91影视

Linear Algebra With Applications

Otto Bretscher

$Math Studyset 91影视 Explanations$ Math

5th Edition 路 442 pages

ISBN: 9780321796974

Chapter 5: 1E (page 202)

Question: In Exercises 1 and 2, you may assume that$\left\{ {{{\bf{u}}_{\bf{1}}},...,{{\bf{u}}_{\bf{4}}}} \right\}$is an orthogonal basis for${\mathbb{R}^{\bf{4}}}$.
1.${{\bf{u}}_{\bf{1}}} = \left[ {\begin{aligned}{{20}{c}}{\bf{0}}\\{\bf{1}}\\{ - {\bf{4}}}\\{ - {\bf{1}}}\end{aligned}} \right]$,${{\bf{u}}_{\bf{2}}} = \left[ {\begin{aligned}{{20}{c}}{\bf{3}}\\{\bf{5}}\\{\bf{1}}\\{\bf{1}}\end{aligned}} \right]$,${{\bf{u}}_{\bf{3}}} = \left[ {\begin{aligned}{{20}{c}}{\bf{1}}\\{\bf{0}}\\{\bf{1}}\\{ - {\bf{4}}}\end{aligned}} \right]$,${{\bf{u}}_{\bf{4}}} = \left[ {\begin{aligned}{{20}{c}}{\bf{5}}\\{ - {\bf{3}}}\\{ - {\bf{1}}}\\{\bf{1}}\end{aligned}} \right]$,${\bf{x}} = \left[ {\begin{aligned}{*{20}{c}}{{\bf{10}}}\\{ - {\bf{8}}}\\{\bf{2}}\\{\bf{0}}\end{aligned}} \right]$
Write x as the sum of two vectors, one in${\bf{Span}}\left\{ {{{\bf{u}}_1},{{\bf{u}}_2},{{\bf{u}}_3}} \right\}$and the other in${\bf{Span}}\left\{ {{{\bf{u}}_{\bf{4}}}} \right\}$.

Short Answer

Expert verified

The vector is ${\bf{x}} = \left[ {\begin{aligned}{*{20}{c}}0\\{ - 2}\\4\\{ - 2}\end{aligned}} \right] + \left[ {\begin{aligned}{*{20}{c}}{10}\\{ - 6}\\{ - 2}\\2\end{aligned}} \right]$.

Step by step solution

Find the orthogonal projection of x

The orthogonal projection of xcan be calculated as follows:

\[\begin{aligned}{c}{\bf{\hat x}} = \frac{{{\bf{x}} \cdot {{\bf{u}}_1}}}{{{{\bf{u}}_1} \cdot {{\bf{u}}_1}}}{{\bf{u}}_1} + \frac{{{\bf{x}} \cdot {{\bf{u}}_2}}}{{{{\bf{u}}_2} \cdot {{\bf{u}}_2}}}{{\bf{u}}_2} + \frac{{{\bf{x}} \cdot {{\bf{u}}_3}}}{{{{\bf{u}}_3} \cdot {{\bf{u}}_3}}}{{\bf{u}}_3}\\ = \frac{{\left[ {\begin{aligned}{*{20}{c}}{10}&{ - 8}&2&0\end{aligned}} \right]\left[ {\begin{aligned}{*{20}{c}}0\\1\\{ - 4}\\{ - 1}\end{aligned}} \right]}}{{\left[ {\begin{aligned}{*{20}{c}}0&1&{ - 4}&{ - 1}\end{aligned}} \right]\left[ {\begin{aligned}{*{20}{c}}0\\1\\{ - 4}\\{ - 1}\end{aligned}} \right]}}{{\bf{u}}_1} + \frac{{\left[ {\begin{aligned}{*{20}{c}}{10}&{ - 8}&2&0\end{aligned}} \right]\left[ {\begin{aligned}{*{20}{c}}3\\5\\1\\1\end{aligned}} \right]}}{{\left[ {\begin{aligned}{*{20}{c}}3&5&1&1\end{aligned}} \right]\left[ {\begin{aligned}{*{20}{c}}3\\5\\1\\1\end{aligned}} \right]}}{{\bf{u}}_2} + \frac{{\left[ {\begin{aligned}{*{20}{c}}{10}&{ - 8}&2&0\end{aligned}} \right]\left[ {\begin{aligned}{*{20}{c}}1\\0\\1\\{ - 4}\end{aligned}} \right]}}{{\left[ {\begin{aligned}{*{20}{c}}1&0&1&{ - 4}\end{aligned}} \right]\left[ {\begin{aligned}{*{20}{c}}1\\0\\1\\{ - 4}\end{aligned}} \right]}}{{\bf{u}}_3}\\ = - \frac{8}{9}\left[ {\begin{aligned}{*{20}{c}}0\\1\\{ - 4}\\{ - 1}\end{aligned}} \right] - \frac{2}{9}\left[ {\begin{aligned}{*{20}{c}}3\\5\\1\\1\end{aligned}} \right] + \frac{6}{9}\left[ {\begin{aligned}{*{20}{c}}1\\0\\1\\{ - 4}\end{aligned}} \right]\\ = \left[ {\begin{aligned}{*{20}{c}}0\\{ - 2}\\4\\{ - 2}\end{aligned}} \right]\end{aligned}\]

Find the orthogonal component

The orthogonal component can be calculated as follows:

$\begin{aligned}{l}z = {\bf{x}} - {\bf{\hat x}}\\ = \left[ {\begin{aligned}{*{20}{c}}{10}\\{ - 8}\\2\\0\end{aligned}} \right] - \left[ {\begin{aligned}{*{20}{c}}0\\{ - 2}\\4\\{ - 2}\end{aligned}} \right]\\ = \left[ {\begin{aligned}{*{20}{c}}{10}\\{ - 6}\\{ - 2}\\2\end{aligned}} \right]\end{aligned}$

The component x as the sum of given vectors is ${\bf{x}} = \left[ {\begin{aligned}{*{20}{c}}0\\{ - 2}\\4\\{ - 2}\end{aligned}} \right] + \left[ {\begin{aligned}{*{20}{c}}{10}\\{ - 6}\\{ - 2}\\2\end{aligned}} \right]$.