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Chapter 6: Q6.2-20E (page 331)

Question: In Exercises 17-22, determine which sets of vectors are orthonormal. If a set is only orthogonal, normalize the vectors to produce an orthonormal set.

20. \(\left( {\begin{array}{*{20}{c}}{ - \frac{2}{3}}\\{\frac{1}{3}}\\{\frac{2}{3}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{\frac{1}{3}}\\{\frac{2}{3}}\\0\end{array}} \right)\)

Short Answer

Expert verified

The set of vectors \(\left\{ {{\bf{u}},{\bf{v}}} \right\}\) is not orthonormal.

The orthormal set of vectors is \(\left\{ {\frac{{\bf{u}}}{{\left\| {\bf{u}} \right\|}},\frac{{\bf{v}}}{{\left\| {\bf{v}} \right\|}}} \right\} = \left\{ {\left( {\begin{array}{*{20}{c}}{ - \frac{2}{3}}\\{\frac{1}{3}}\\{\frac{2}{3}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{\frac{1}{{\sqrt 5 }}}\\{\frac{2}{{\sqrt 5 }}}\\0\end{array}} \right)} \right\}\).

Step by step solution

01

Definition of orthonormal sets

A set \(\left\{ {{{\bf{u}}_1},{{\bf{u}}_2}, \ldots ,{{\bf{u}}_p}} \right\}\) is called anorthonormal set when it is an orthogonal set of unit vectors.

02

Check whether the set of vectors are orthogonal

Consider that \({\bf{u}} = \left( {\begin{array}{*{20}{c}}{ - \frac{2}{3}}\\{\frac{1}{3}}\\{\frac{2}{3}}\end{array}} \right),{\rm{ }}{\bf{v}} = \left( {\begin{array}{*{20}{c}}{\frac{1}{3}}\\{\frac{2}{3}}\\0\end{array}} \right)\).

Check whether the set of the vector is orthogonal, as shown below:

\(\begin{array}{c}{\bf{u}} \cdot {\bf{v}} = \left( {\begin{array}{*{20}{c}}{ - \frac{2}{3}}\\{\frac{1}{3}}\\{\frac{2}{3}}\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}{\frac{1}{3}}\\{\frac{2}{3}}\\0\end{array}} \right)\\ = \left( { - \frac{2}{3}} \right)\left( {\frac{1}{3}} \right) + \frac{1}{3}\left( {\frac{2}{3}} \right) + 0\\ = - \frac{2}{9} + \frac{2}{9}\\ = 0\end{array}\)

It is observed that \(\left\{ {{\bf{u}},{\bf{v}}} \right\}\) is an orthogonal set because \({\bf{u}} \cdot {\bf{v}} = 0\).

03

Check whether the set of vectors is orthonormal

Check whether the set of vectors is orthonormal as shown below:

\(\begin{array}{c}{\left\| {\bf{u}} \right\|^2} = {\bf{u}} \cdot {\bf{u}}\\ = {\left( { - \frac{2}{3}} \right)^2} + {\left( {\frac{1}{3}} \right)^2} + {\left( {\frac{2}{3}} \right)^2}\\ = \left( {\frac{4}{9}} \right) + \left( {\frac{1}{9}} \right) + \left( {\frac{4}{9}} \right)\\ = \frac{9}{9}\\ = 1\\{\left\| {\bf{v}} \right\|^2} = {\bf{v}} \cdot {\bf{v}}\\ = {\left( {\frac{1}{3}} \right)^2} + {\left( {\frac{2}{3}} \right)^2} + 0\\ = \frac{1}{9} + \frac{4}{9}\\ = \frac{5}{9}\end{array}\)

It is observed that \(\left\{ {{\bf{u}},{\bf{v}}} \right\}\) is not an orthonormal set.

04

Normalize the vectors to produce an orthonormal set

Normalize the vectors to produce the orthonormal set as shown below:

\(\begin{array}{c}\left\{ {\frac{{\bf{u}}}{{\left\| {\bf{u}} \right\|}},\frac{{\bf{v}}}{{\left\| {\bf{v}} \right\|}}} \right\} = \left\{ {\frac{1}{{\sqrt 1 }}\left( {\begin{array}{*{20}{c}}{ - \frac{2}{3}}\\{\frac{1}{3}}\\{\frac{2}{3}}\end{array}} \right),\frac{1}{{\sqrt {\frac{5}{9}} }}\left( {\begin{array}{*{20}{c}}{\frac{1}{3}}\\{\frac{2}{3}}\\0\end{array}} \right)} \right\}\\ = \left\{ {\left( {\begin{array}{*{20}{c}}{ - \frac{2}{3}}\\{\frac{1}{3}}\\{\frac{2}{3}}\end{array}} \right),\frac{3}{{\sqrt 5 }}\left( {\begin{array}{*{20}{c}}{\frac{1}{3}}\\{\frac{2}{3}}\\0\end{array}} \right)} \right\}\\ = \left\{ {\left( {\begin{array}{*{20}{c}}{ - \frac{2}{3}}\\{\frac{1}{3}}\\{\frac{2}{3}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{\frac{1}{{\sqrt 5 }}}\\{\frac{2}{{\sqrt 5 }}}\\0\end{array}} \right)} \right\}\end{array}\)

Therefore, the orthonormal set is \(\left\{ {\frac{{\bf{u}}}{{\left\| {\bf{u}} \right\|}},\frac{{\bf{v}}}{{\left\| {\bf{v}} \right\|}}} \right\} = \left\{ {\left( {\begin{array}{*{20}{c}}{ - \frac{2}{3}}\\{\frac{1}{3}}\\{\frac{2}{3}}\end{array}} \right),\left( {\begin{array}{*{20}{c}}{\frac{1}{{\sqrt 5 }}}\\{\frac{2}{{\sqrt 5 }}}\\0\end{array}} \right)} \right\}\).

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Most popular questions from this chapter

In exercises 1-6, determine which sets of vectors are orthogonal.

\(\left[ {\begin{align}{ 2}\\{ - 7}\\{-1}\end{align}} \right]\), \(\left[ {\begin{align}{ - 6}\\{ - 3}\\9\end{align}} \right]\), \(\left[ {\begin{align}{ 3}\\{ 1}\\{-1}\end{align}} \right]\)

In Exercises 1-6, the given set is a basis for a subspace W. Use the Gram-Schmidt process to produce an orthogonal basis for W.

4. \(\left( {\begin{aligned}{{}{}}3\\{ - 4}\\5\end{aligned}} \right),\left( {\begin{aligned}{{}{}}{ - 3}\\{14}\\{ - 7}\end{aligned}} \right)\)

Show that if \(U\) is an orthogonal matrix, then any real eigenvalue of \(U\) must be \( \pm 1\).

In Exercises 17 and 18, all vectors and subspaces are in \({\mathbb{R}^n}\). Mark each statement True or False. Justify each answer.

17. a.If \(\left\{ {{{\bf{v}}_1},{{\bf{v}}_2},{{\bf{v}}_3}} \right\}\) is an orthogonal basis for\(W\), then multiplying

\({v_3}\)by a scalar \(c\) gives a new orthogonal basis \(\left\{ {{{\bf{v}}_1},{{\bf{v}}_2},c{{\bf{v}}_3}} \right\}\).

b. The Gram鈥揝chmidt process produces from a linearly independent

set \(\left\{ {{{\bf{x}}_1}, \ldots ,{{\bf{x}}_p}} \right\}\)an orthogonal set \(\left\{ {{{\bf{v}}_1}, \ldots ,{{\bf{v}}_p}} \right\}\) with the property that for each \(k\), the vectors \({{\bf{v}}_1}, \ldots ,{{\bf{v}}_k}\) span the same subspace as that spanned by \({{\bf{x}}_1}, \ldots ,{{\bf{x}}_k}\).

c. If \(A = QR\), where \(Q\) has orthonormal columns, then \(R = {Q^T}A\).

Suppose the x-coordinates of the data \(\left( {{x_1},{y_1}} \right), \ldots ,\left( {{x_n},{y_n}} \right)\) are in mean deviation form, so that \(\sum {{x_i}} = 0\). Show that if \(X\) is the design matrix for the least-squares line in this case, then \({X^T}X\) is a diagonal matrix.
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