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Chapter 6: 12E (page 331)

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

12. \[y = \left[ {\begin{aligned}3\\{ - 1}\\1\\{13}\end{aligned}} \right]\], \[{{\bf{v}}_1} = \left[ {\begin{aligned}1\\{ - 2}\\{ - 1}\\2\end{aligned}} \right]\], \[{{\bf{v}}_2} = \left[ {\begin{aligned}{ - 4}\\1\\0\\3\end{aligned}} \right]\]

Short Answer

Expert verified

The closest point is {\bf{\hat y}} = \left[ {\begin{array}{ - 1}\\{ - 5}\\{ - 3}\\9\end{array}} \right].

Step by step solution

01

Write the definition

Orthogonal set: If \[{{\bf{u}}_i} \cdot {{\bf{u}}_j} = 0\] for \[i \ne j\], then the set of vectors \[\left\{ {{{\bf{u}}_1}, \ldots ,{{\bf{u}}_p}} \right\} \in {\mathbb{R}^n}\] is said to be orthogonal.

02

Check the set of given vectors is orthogonal or not

Given vectors are,\[{{\bf{v}}_1} = \left[ {\begin{aligned}1\\{ - 2}\\{ - 1}\\2\end{aligned}} \right]\], and \[{{\bf{v}}_2} = \left[ {\begin{aligned}{ - 4}\\1\\0\\3\end{aligned}} \right]\].

Find \[{{\bf{v}}_1} \cdot {{\bf{v}}_2}\].

\[\begin{aligned}{{\bf{v}}_1} \cdot {{\bf{v}}_2} &= \left[ {\begin{aligned}1\\{ - 2}\\{ - 1}\\2\end{aligned}} \right] \cdot \left[ {\begin{aligned}{ - 4}\\1\\0\\3\end{aligned}} \right]\\ &= 1\left( { - 4} \right) + \left( { - 2} \right)\left( 1 \right) + \left( { - 1} \right)\left( 0 \right) + 2\left( 3 \right)\\ &= 0\end{aligned}\]

As \[{{\bf{v}}_1} \cdot {{\bf{v}}_2} = 0\], so the set of vectors \[\left\{ {{{\bf{v}}_1},{{\bf{v}}_2}} \right\}\] is orthogonal.

03

The Best Approximation Theorem

Here, \[{\bf{\hat y}}\] is said to be the closest pointin \[w\] to \[{\bf{y}}\], if \[{\bf{\hat y}}\] be the orthogonal projection of \[{\bf{y}}\] onto \[w\], where \[w\] is a subspace of \[{\mathbb{R}^n}\], and \[{\bf{y}} \in w\]. \[{\bf{\hat y}}\] can be found as:

\[{\bf{\hat y}} = \frac{{{\bf{y}} \cdot {{\bf{u}}_1}}}{{{{\bf{u}}_1} \cdot {{\bf{u}}_1}}}{{\bf{u}}_1} + \cdots + \frac{{{\bf{y}} \cdot {{\bf{u}}_p}}}{{{{\bf{u}}_p} \cdot {{\bf{u}}_p}}}{{\bf{u}}_p}\]

04

Find the closest point

Find \[{\bf{y}} \cdot {{\bf{v}}_1}\], \[{\bf{y}} \cdot {{\bf{v}}_2}\], \[{{\bf{v}}_1} \cdot {{\bf{v}}_1}\], and \[{{\bf{v}}_2} \cdot {{\bf{v}}_2}\] first, where \[{\bf{y}} = \left[ {\begin{aligned}3\\{ - 1}\\1\\{13}\end{aligned}} \right]\].

\[\begin{aligned}{\bf{y}} \cdot {{\bf{v}}_1} &= \left[ {\begin{aligned}3\\{ - 1}\\1\\{13}\end{aligned}} \right] \cdot \left[ {\begin{aligned}1\\{ - 2}\\{ - 1}\\2\end{aligned}} \right]\\ &= 3\left( 1 \right) + \left( { - 1} \right)\left( { - 2} \right) + 1\left( { - 1} \right) + 13\left( 2 \right)\\ &= 30\end{aligned}\]

\[\begin{aligned}{\bf{y}} \cdot {{\bf{v}}_2} &= \left[ {\begin{aligned}3\\{ - 1}\\1\\{13}\end{aligned}} \right] \cdot \left[ {\begin{aligned}{ - 4}\\1\\0\\3\end{aligned}} \right]\\ &= 3\left( { - 4} \right) + \left( { - 1} \right)\left( 1 \right) + 1\left( 0 \right) + 13\left( 3 \right)\\ &= 26\end{aligned}\]

\[\begin{aligned}{{\bf{v}}_1} \cdot {{\bf{v}}_1} &= \left[ {\begin{aligned}1\\{ - 2}\\{ - 1}\\2\end{aligned}} \right] \cdot \left[ {\begin{aligned}1\\{ - 2}\\{ - 1}\\2\end{aligned}} \right]\\ &= 1\left( 1 \right) + \left( { - 2} \right)\left( { - 2} \right) + \left( { - 1} \right)\left( { - 1} \right) + \left( 2 \right)\left( 2 \right)\\ &= 10\end{aligned}\]

\[\begin{aligned}{{\bf{v}}_2} \cdot {{\bf{v}}_2} &= \left[ {\begin{aligned}{ - 4}\\1\\0\\3\end{aligned}} \right] \cdot \left[ {\begin{aligned}{ - 4}\\1\\0\\3\end{aligned}} \right]\\ &= - 4\left( { - 4} \right) + \left( 1 \right)\left( 1 \right) + 0\left( 0 \right) + \left( 3 \right)\left( 3 \right)\\ &= 26\end{aligned}\]

Now, find the closest point to \[{\bf{y}}\] by substituting the obtained values into \[{\bf{\hat y}} = \frac{{{\bf{y}} \cdot {{\bf{u}}_1}}}{{{{\bf{u}}_1} \cdot {{\bf{u}}_1}}}{{\bf{u}}_1} + \cdots + \frac{{{\bf{y}} \cdot {{\bf{u}}_p}}}{{{{\bf{u}}_p} \cdot {{\bf{u}}_p}}}{{\bf{u}}_p}\].

\[\begin{aligned}{\bf{\hat y}} = \frac{{30}}{{10}}{{\bf{v}}_1} + \frac{{26}}{{24}}{{\bf{v}}_2}\\ = 3\left[ {\begin{aligned}1\\{ - 2}\\{ - 1}\\2\end{aligned}} \right] + \left[ {\begin{aligned}{ - 4}\\1\\0\\3\end{aligned}} \right]\\ = \left[ {\begin{aligned}{ - 1}\\{ - 5}\\{ - 3}\\9\end{aligned}} \right]\end{aligned}\]

Hence,the closest point is \[{\bf{\hat y}} = \left[ {\begin{aligned}{ - 1}\\{ - 5}\\{ - 3}\\9\end{aligned}} \right]\].

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Most popular questions from this chapter

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.

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\[W\].

9.\[y = \left[ {\begin{aligned}4\\3\\3\\{ - 1}\end{aligned}} \right]\],\[{{\bf{u}}_1} = \left[ {\begin{aligned}1\\1\\0\\1\end{aligned}} \right]\],\[{{\bf{u}}_2} = \left[ {\begin{aligned}{ - 1}\\3\\1\\{ - 2}\end{aligned}} \right]\],\[{{\bf{u}}_2} = \left[ {\begin{aligned}{ - 1}\\0\\1\\1\end{aligned}} \right]\]

A healthy child’s 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.

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

In Exercises 1-4, find the equation \(y = {\beta _0} + {\beta _1}x\) of the least-square line that best fits the given data points.

  1. \(\left( {1,0} \right),\left( {2,1} \right),\left( {4,2} \right),\left( {5,3} \right)\)

Find the distance between \({\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{10}\\{ - 3}\end{aligned}} \right)\) and \({\mathop{\rm y}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 1}\\{ - 5}\end{aligned}} \right)\).
See all solutions

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