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Chapter 6: Q4E (page 331)

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

Short Answer

Expert verified

The value is \(\frac{1}{{{\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} }}{\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{\frac{{ - 1}}{5}}\\{\frac{2}{5}}\end{aligned}} \right)\).

Step by step solution

01

Inner product

Consider \({\mathop{\rm u}\nolimits} ,v,\) and \({\mathop{\rm w}\nolimits} \) as the vectors in \({\mathbb{R}^n}\) and consider \(c\) as the scalar. Then

  1. \({\mathop{\rm u}\nolimits} \cdot {\mathop{\rm v}\nolimits} = {\mathop{\rm v}\nolimits} \cdot {\mathop{\rm u}\nolimits} \)
  2. \(\left( {{\mathop{\rm u}\nolimits} + {\mathop{\rm v}\nolimits} } \right) \cdot {\mathop{\rm w}\nolimits} = {\mathop{\rm u}\nolimits} \cdot {\mathop{\rm v}\nolimits} + {\mathop{\rm v}\nolimits} \cdot {\mathop{\rm w}\nolimits} \)
  3. \(\left( {c{\mathop{\rm u}\nolimits} } \right) \cdot {\mathop{\rm v}\nolimits} = c\left( {{\mathop{\rm u}\nolimits} \cdot {\mathop{\rm v}\nolimits} } \right) = {\mathop{\rm u}\nolimits} \cdot \left( {c{\mathop{\rm v}\nolimits} } \right)\)
  4. \({\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} \ge 0\)and \({\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} = 0\) if and only if \({\mathop{\rm u}\nolimits} = 0\).
02

Compute

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

It is given that \({\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 1}\\2\end{aligned}} \right)\).

Evaluate \({\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} \) as shown below:

\(\begin{aligned}{c}{\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} &= \left( {\begin{aligned}{*{20}{c}}{ - 1}\\2\end{aligned}} \right)\left( {\begin{aligned}{*{20}{c}}{ - 1}\\2\end{aligned}} \right)\\ &= {\left( { - 1} \right)^2} + {\left( 2 \right)^2}\\ &= 1 + 4\\ &= 5\end{aligned}\)

Compute \(\frac{1}{{{\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} }}{\mathop{\rm u}\nolimits} \) as shown below:

\(\begin{aligned}{c}\frac{1}{{{\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} }}{\mathop{\rm u}\nolimits} &= \frac{1}{5}\left( {\begin{aligned}{*{20}{c}}{ - 1}\\2\end{aligned}} \right)\\ &= \left( {\begin{aligned}{*{20}{c}}{\frac{{ - 1}}{5}}\\{\frac{2}{5}}\end{aligned}} \right)\end{aligned}\)

Thus, the value is \(\frac{1}{{{\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} }}{\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{\frac{{ - 1}}{5}}\\{\frac{2}{5}}\end{aligned}} \right)\).

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Most popular questions from this chapter

A healthy child鈥檚 systolic blood pressure (in millimetres of mercury) and weight (in pounds) are approximately related by the equation

\({\beta _0} + {\beta _1}\ln w = p\)

Use the following experimental data to estimate the systolic blood pressure of healthy child weighing 100 pounds.

\(\begin{array} w&\\ & {44}&{61}&{81}&{113}&{131} \\ \hline {\ln w}&\\vline & {3.78}&{4.11}&{4.39}&{4.73}&{4.88} \\ \hline p&\\vline & {91}&{98}&{103}&{110}&{112} \end{array}\)

Find an orthogonal basis for the column space of each matrix in Exercises 9-12.

10. \(\left( {\begin{aligned}{{}{}}{ - 1} & 6 & 6 \\ 3 & { - 8}&3\\1&{ - 2}&6\\1&{ - 4}&{ - 3}\end{aligned}} \right)\)

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.

  1. \(\left( {\begin{aligned}{{}{}}3\\0\\{ - 1}\end{aligned}} \right),\left( {\begin{aligned}{{}{}}8\\5\\{ - 6}\end{aligned}} \right)\)

Let \({\mathbb{R}^{\bf{2}}}\) have the inner product of Example 1. Show that the Cauchy-Schwarz inequality holds for \({\bf{x}} = \left( {{\bf{3}}, - {\bf{2}}} \right)\) and \({\bf{y}} = \left( { - {\bf{2}},{\bf{1}}} \right)\). (Suggestion: Study \({\left| {\left\langle {{\bf{x}},{\bf{y}}} \right\rangle } \right|^{\bf{2}}}\).)

Use the Gram鈥揝chmidt process as in Example 2 to produce an orthogonal basis for the column space of

\(A = \left( {\begin{aligned}{{}{r}}{ - 10}&{13}&7&{ - 11}\\2&1&{ - 5}&3\\{ - 6}&3&{13}&{ - 3}\\{16}&{ - 16}&{ - 2}&5\\2&1&{ - 5}&{ - 7}\end{aligned}} \right)\)
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