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Chapter 6: Q8E (page 331)

Find an orthonormal basis of the subspace spanned by the vectors in Exercise 4.

Short Answer

Expert verified

An orthonormal basis is \(\left\{ {\left( {\begin{aligned}{{}{}}{\frac{3}{{\sqrt {50} }}}\\{\frac{{ - 4}}{{\sqrt {50} }}}\\{\frac{5}{{\sqrt {50} }}}\end{aligned}} \right),\left( {\begin{aligned}{{}{}}{\frac{1}{{\sqrt 6 }}}\\{\frac{2}{{\sqrt 6 }}}\\{\frac{1}{{\sqrt 6 }}}\end{aligned}} \right)} \right\}\).

Step by step solution

01

Compute \(\left\| {{{\bf{v}}_1}} \right\|\) and \(\left\| {{{\bf{v}}_2}} \right\|\)

Let the vectors \({{\bf{v}}_1} = \left( {\begin{aligned}{{}{}}3\\{ - 4}\\5\end{aligned}} \right),{{\bf{v}}_2} = \left( {\begin{aligned}{{}{}}3\\6\\3\end{aligned}} \right)\) from Exercise 4.

Compute \(\left\| {{{\bf{v}}_1}} \right\|\) and \(\left\| {{{\bf{v}}_2}} \right\|\) as shown below:

\(\begin{aligned}{}\left\| {{{\bf{v}}_1}} \right\| &= \sqrt {{3^2} + {{\left( { - 4} \right)}^2} + {5^2}} \\ &= \sqrt {9 + 16 + 25} \\ &= \sqrt {50} \\\left\| {{{\bf{v}}_2}} \right\| &= \sqrt {{3^2} + {6^2} + {3^2}} \\ &= \sqrt {9 + 36 + 9} \\ &= \sqrt {54} \\ &= 3\sqrt 6 \end{aligned}\)

02

Determine an orthonormal basis

Obtain an orthonormal basis as shown below:

\(\begin{aligned}{}\left\{ {\frac{{{{\bf{v}}_1}}}{{\left\| {{{\bf{v}}_1}} \right\|}},\frac{{{{\bf{v}}_2}}}{{\left\| {{{\bf{v}}_2}} \right\|}}} \right\} &= \left\{ {\frac{1}{{\sqrt {50} }}\left( {\begin{aligned}{{}{}}3\\{ - 4}\\5\end{aligned}} \right),\frac{1}{{3\sqrt 6 }}\left( {\begin{aligned}{{}{}}3\\6\\3\end{aligned}} \right)} \right\}\\ & = \left\{ {\left( {\begin{aligned}{{}{}}{\frac{3}{{\sqrt {50} }}}\\{\frac{{ - 4}}{{\sqrt {50} }}}\\{\frac{5}{{\sqrt {50} }}}\end{aligned}} \right),\left( {\begin{aligned}{{}{}}{\frac{1}{{\sqrt 6 }}}\\{\frac{2}{{\sqrt 6 }}}\\{\frac{1}{{\sqrt 6 }}}\end{aligned}} \right)} \right\}\end{aligned}\)

Hence, an orthonormal basis is \(\left\{ {\left( {\begin{aligned}{{}{}}{\frac{3}{{\sqrt {50} }}}\\{\frac{{ - 4}}{{\sqrt {50} }}}\\{\frac{5}{{\sqrt {50} }}}\end{aligned}} \right),\left( {\begin{aligned}{{}{}}{\frac{1}{{\sqrt 6 }}}\\{\frac{2}{{\sqrt 6 }}}\\{\frac{1}{{\sqrt 6 }}}\end{aligned}} \right)} \right\}\).

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Most popular questions from this chapter

In Exercises 5 and 6, describe all least squares solutions of the equation \(A{\bf{x}} = {\bf{b}}\).

6.\(A = \left( {\begin{aligned}{{}{}}{\bf{1}}&{\bf{1}}&{\bf{0}}\\{\bf{1}}&{\bf{1}}&{\bf{0}}\\{\bf{1}}&{\bf{1}}&{\bf{0}}\\{\bf{1}}&{\bf{0}}&{\bf{1}}\\{\bf{1}}&{\bf{0}}&{\bf{1}}\\{\bf{1}}&{\bf{0}}&{\bf{1}}\end{aligned}} \right)\),\({\bf{b}} = \left( {\begin{aligned}{{}{}}{\bf{7}}\\{\bf{2}}\\{\bf{3}}\\{\bf{6}}\\{\bf{5}}\\{\bf{4}}\end{aligned}} \right)\)

Find an orthonormal basis of the subspace spanned by the vectors in Exercise 3.

Compute the quantities in Exercises 1-8 using the vectors

\({\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 1}\\2\end{aligned}} \right),{\rm{ }}{\mathop{\rm v}\nolimits} = \left( {\begin{aligned}{*{20}{c}}4\\6\end{aligned}} \right),{\rm{ }}{\mathop{\rm w}\nolimits} = \left( {\begin{aligned}{*{20}{c}}3\\{ - 1}\\{ - 5}\end{aligned}} \right),{\rm{ }}{\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}6\\{ - 2}\\3\end{aligned}} \right)\)

4. \(\frac{1}{{{\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} }}{\mathop{\rm u}\nolimits} \)

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\({\mathbb{R}^{\bf{4}}}\).

2.\({{\bf{u}}_{\bf{1}}} = \left[ {\begin{aligned}{\bf{1}}\\{\bf{2}}\\{\bf{1}}\\{\bf{1}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{2}}} = \left[ {\begin{aligned}{ - {\bf{2}}}\\{\bf{1}}\\{ - {\bf{1}}}\\{\bf{1}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{3}}} = \left[ {\begin{aligned}{\bf{1}}\\{\bf{1}}\\{ - {\bf{2}}}\\{ - {\bf{1}}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{4}}} = \left[ {\begin{aligned}{ - {\bf{1}}}\\{\bf{1}}\\{\bf{1}}\\{ - {\bf{2}}}\end{aligned}} \right]\),\({\bf{x}} = \left[ {\begin{aligned}{\bf{4}}\\{\bf{5}}\\{ - {\bf{3}}}\\{\bf{3}}\end{aligned}} \right]\)

Write v as the sum of two vectors, one in\({\bf{Span}}\left\{ {{{\bf{u}}_1}} \right\}\)and the other in\({\bf{Span}}\left\{ {{{\bf{u}}_2},{{\bf{u}}_3},{{\bf{u}}_{\bf{4}}}} \right\}\).

Let \(\left\{ {{{\bf{v}}_1}, \ldots ,{{\bf{v}}_p}} \right\}\) be an orthonormal set in \({\mathbb{R}^n}\). Verify the following inequality, called Bessel’s inequality, which is true for each x in \({\mathbb{R}^n}\):

\({\left\| {\bf{x}} \right\|^2} \ge {\left| {{\bf{x}} \cdot {{\bf{v}}_1}} \right|^2} + {\left| {{\bf{x}} \cdot {{\bf{v}}_2}} \right|^2} + \ldots + {\left| {{\bf{x}} \cdot {{\bf{v}}_p}} \right|^2}\)
See all solutions

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