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Chapter 6: Q7E (page 331)

Compute the quantities in Exercises 1-8 using the vectors

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

7. \(\left\| {\mathop{\rm w}\nolimits} \right\|\)

Short Answer

Expert verified

The length of w is \(\left\| {\mathop{\rm w}\nolimits} \right\| = \sqrt {35} \).

Step by step solution

01

The length of a vector

The nonnegative scalar \(\left\| {\mathop{\rm v}\nolimits} \right\|\) is defined aslength (ornorm) as shown below:

\(\left\| {\mathop{\rm v}\nolimits} \right\| = \sqrt {{\mathop{\rm v}\nolimits} \cdot {\mathop{\rm v}\nolimits} } = \sqrt {v_1^2 + v_2^2 + \cdots + v_n^2} ,\)and \({\left\| {\mathop{\rm v}\nolimits} \right\|^2} = {\mathop{\rm v}\nolimits} \cdot {\mathop{\rm v}\nolimits} \)

02

Compute

\(\left\| {\mathop{\rm w}\nolimits} \right\|\)

It is given that \({\mathop{\rm w}\nolimits} = \left( {\begin{aligned}{*{20}{c}}3\\{ - 1}\\{ - 5}\end{aligned}} \right)\).

Compute the length of \({\mathop{\rm w}\nolimits} \) as shown below:

\(\begin{aligned}{c}\left\| {\mathop{\rm w}\nolimits} \right\| &= \sqrt {{\mathop{\rm w}\nolimits} \cdot {\mathop{\rm w}\nolimits} } \\ &= \sqrt {{3^2} + {{\left( { - 1} \right)}^2} + {{\left( { - 5} \right)}^2}} \\ &= \sqrt {9 + 1 + 25} \\ &= \sqrt {35} \end{aligned}\)

Thus, the length of w is \(\left\| {\mathop{\rm w}\nolimits} \right\| = \sqrt {35} \).

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Most popular questions from this chapter

To measure the take-off performance of an airplane, the horizontal position of the plane was measured every second, from \(t = 0\) to \(t = 12\). The positions (in feet) were: 0, 8.8, 29.9, 62.0, 104.7, 159.1, 222.0, 294.5, 380.4, 471.1, 571.7, 686.8, 809.2.

a. Find the least-squares cubic curve \(y = {\beta _0} + {\beta _1}t + {\beta _2}{t^2} + {\beta _3}{t^3}\) for these data.

b. Use the result of part (a) to estimate the velocity of the plane when \(t = 4.5\) seconds.

In Exercises 11 and 12, find the closest point to\[{\bf{y}}\]in the subspace\[W\]spanned by\[{{\bf{v}}_1}\], and\[{{\bf{v}}_2}\].

11.\[y = \left[ {\begin{aligned}3\\1\\5\\1\end{aligned}} \right]\],\[{{\bf{v}}_1} = \left[ {\begin{aligned}3\\1\\{ - 1}\\1\end{aligned}} \right]\],\[{{\bf{v}}_2} = \left[ {\begin{aligned}1\\{ - 1}\\1\\{ - 1}\end{aligned}} \right]\]

In Exercises 7–10, let\[W\]be the subspace spanned by the\[{\bf{u}}\]’s, and write y as the sum of a vector in\[W\]and a vector orthogonal to\[W\].

7.\[y = \left[ {\begin{aligned}1\\3\\5\end{aligned}} \right]\],\[{{\bf{u}}_1} = \left[ {\begin{aligned}1\\3\\{ - 2}\end{aligned}} \right]\],\[{{\bf{u}}_2} = \left[ {\begin{aligned}5\\1\\4\end{aligned}} \right]\]

Given data for a least-squares problem, \(\left( {{x_1},{y_1}} \right), \ldots ,\left( {{x_n},{y_n}} \right)\), the following abbreviations are helpful:

\(\begin{aligned}{l}\sum x = \sum\nolimits_{i = 1}^n {{x_i}} ,{\rm{ }}\sum {{x^2}} = \sum\nolimits_{i = 1}^n {x_i^2} ,\\\sum y = \sum\nolimits_{i = 1}^n {{y_i}} ,{\rm{ }}\sum {xy} = \sum\nolimits_{i = 1}^n {{x_i}{y_i}} \end{aligned}\)

The normal equations for a least-squares line \(y = {\hat \beta _0} + {\hat \beta _1}x\) may be written in the form

\(\begin{aligned}{c}{{\hat \beta }_0} + {{\hat \beta }_1}\sum x = \sum y \\{{\hat \beta }_0}\sum x + {{\hat \beta }_1}\sum {{x^2}} = \sum {xy} {\rm{ (7)}}\end{aligned}\)

Derive the normal equations (7) from the matrix form given in this section.

Let \(\left\{ {{{\bf{v}}_1}, \ldots ,{{\bf{v}}_p}} \right\}\) be an orthonormal set. Verify the following equality by induction, beginning with \(p = 2\). If \({\bf{x}} = {c_1}{{\bf{v}}_1} + \ldots + {c_p}{{\bf{v}}_p}\), then

\({\left\| {\bf{x}} \right\|^2} = {\left| {{c_1}} \right|^2} + {\left| {{c_2}} \right|^2} + \ldots + {\left| {{c_p}} \right|^2}\)
See all solutions

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