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Chapter 6: 17E (page 331)

Let \({\rm{y}} = \left( {\begin{aligned}{{}}4\\8\\1\end{aligned}} \right)\), \({{\bf{u}}_1} = \left( {\begin{aligned}{{}}{{2 \mathord{\left/ {\vphantom {2 3}} \right.} 3}}\\{{1 \mathord{\left/

{\vphantom {1 3}} \right.} 3}}\\{{2 \mathord{\left/{\vphantom {2 3}} \right.} 3}}\end{aligned}} \right)\),\({{\bf{u}}_1} = \left( {\begin{aligned}{{}}{{{ - 2} \mathord{\left/{\vphantom {{ - 2} 3}} \right.} 3}}\\{{2 \mathord{\left/{\vphantom {2 3}} \right.} 3}}\\{{1 \mathord{\left/{\vphantom {1 3}} \right.} 3}}\end{aligned}} \right)\)and \(W = {\bf{Span}}\left\{ {{{\bf{u}}_1},{{\bf{u}}_2}} \right\}\).

  1. Let\(U = \left( {\begin{aligned}{{}}{{{\bf{u}}_1}}&{{{\bf{u}}_2}}\end{aligned}} \right)\). Compute \({U^T}U\) and \(U{U^T}\).
  2. Compute \({\bf{Pro}}{{\bf{j}}_W}{\bf{y}}\) and \(\left( {U{U^T}} \right){\bf{y}}\).

Short Answer

Expert verified
  1. The matrices are \[{U^T}U = \left[ {\begin{aligned}1&0\\0&1\end{aligned}} \right]\], and \[U{U^T} = \left[ {\begin{aligned}{\frac{8}{9}}&{\frac{{ - 2}}{9}}&{\frac{2}{9}}\\{\frac{{ - 2}}{9}}&{\frac{5}{9}}&{\frac{4}{9}}\\{\frac{2}{9}}&{\frac{4}{9}}&{\frac{5}{9}}\end{aligned}} \right]\].
  2. \[{\rm{Pro}}{{\rm{j}}_W}{\bf{y}} = U{U^T}{\bf{y}} = \left[ {\begin{aligned}2\\4\\5\end{aligned}} \right]\].

Step by step solution

01

Form \[U = \left[ {\begin{aligned}{{{\bf{u}}_1}}&{{{\bf{u}}_2}}\end{aligned}} \right]\]

(a)

Given vectors are,\[{{\bf{u}}_1} = \left[ {\begin{aligned}{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}\\{{1 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}}\\{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}\end{aligned}} \right]\], and \[{{\bf{u}}_2} = \left[ {\begin{aligned}{{{ - 2} \mathord{\left/

{\vphantom {{ - 2} 3}} \right.

\kern-\nulldelimiterspace} 3}}\\{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}\\{{1 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}}\end{aligned}} \right]\].

Find \[U = \left[ {\begin{aligned}{{{\bf{u}}_1}}&{{{\bf{u}}_2}}\end{aligned}} \right]\].

\[U = \left[ {\begin{aligned}{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}&{{{ - 2} \mathord{\left/

{\vphantom {{ - 2} 3}} \right.

\kern-\nulldelimiterspace} 3}}\\{{1 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}}&{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}\\{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}&{{1 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}}\end{aligned}} \right]\]

02

Determine \[{U^T}U\] and \[U{U^T}\]

First, find the transpose of \[U\], that is \[{U^T}\].

\[{U^T} = \left[ {\begin{aligned}{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}&{{1 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}}&{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}\\{{{ - 2} \mathord{\left/

{\vphantom {{ - 2} 3}} \right.

\kern-\nulldelimiterspace} 3}}&{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}&{{1 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}}\end{aligned}} \right]\]

Now, compute \[{U^T}U\] and \[U{U^T}\].

\[\begin{aligned}{U^T}U = \left[ {\begin{aligned}{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}&{{1 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}}&{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}\\{{{ - 2} \mathord{\left/

{\vphantom {{ - 2} 3}} \right.

\kern-\nulldelimiterspace} 3}}&{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}&{{1 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}}\end{aligned}} \right]\left[ {\begin{aligned}{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}&{{{ - 2} \mathord{\left/

{\vphantom {{ - 2} 3}} \right.

\kern-\nulldelimiterspace} 3}}\\{{1 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}}&{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}\\{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}&{{1 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}}\end{aligned}} \right]\\ = \left[ {\begin{aligned}{\frac{4}{9} + \frac{1}{9} + \frac{4}{9}}&{\frac{{ - 4}}{9} + \frac{2}{9} + \frac{2}{9}}\\{\frac{{ - 4}}{9} + \frac{2}{9} + \frac{2}{9}}&{\frac{4}{9} + \frac{4}{9} + \frac{1}{9}}\end{aligned}} \right]\\ = \left[ {\begin{aligned}1&0\\0&1\end{aligned}} \right]\end{aligned}\]

\[\begin{aligned}U{U^T} = \left[ {\begin{aligned}{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}&{{{ - 2} \mathord{\left/

{\vphantom {{ - 2} 3}} \right.

\kern-\nulldelimiterspace} 3}}\\{{1 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}}&{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}\\{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}&{{1 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}}\end{aligned}} \right]\left[ {\begin{aligned}{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}&{{1 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}}&{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}\\{{{ - 2} \mathord{\left/

{\vphantom {{ - 2} 3}} \right.

\kern-\nulldelimiterspace} 3}}&{{2 \mathord{\left/

{\vphantom {2 3}} \right.

\kern-\nulldelimiterspace} 3}}&{{1 \mathord{\left/

{\vphantom {1 3}} \right.

\kern-\nulldelimiterspace} 3}}\end{aligned}} \right]\\ = \left[ {\begin{aligned}{\frac{4}{9} + \frac{4}{9}}&{\frac{2}{9} - \frac{4}{9}}&{\frac{4}{9} - \frac{2}{9}}\\{\frac{2}{9} - \frac{4}{9}}&{\frac{1}{9} + \frac{4}{9}}&{\frac{2}{9} + \frac{2}{9}}\\{\frac{4}{9} - \frac{2}{9}}&{\frac{2}{9} + \frac{2}{9}}&{\frac{4}{9} + \frac{1}{9}}\end{aligned}} \right]\\ = \left[ {\begin{aligned}{\frac{8}{9}}&{\frac{{ - 2}}{9}}&{\frac{2}{9}}\\{\frac{{ - 2}}{9}}&{\frac{5}{9}}&{\frac{4}{9}}\\{\frac{2}{9}}&{\frac{4}{9}}&{\frac{5}{9}}\end{aligned}} \right]\end{aligned}\]

Hence, \[{U^T}U = \left[ {\begin{aligned}1&0\\0&1\end{aligned}} \right]\] and \[U{U^T} = \left[ {\begin{aligned}{\frac{8}{9}}&{\frac{{ - 2}}{9}}&{\frac{2}{9}}\\{\frac{{ - 2}}{9}}&{\frac{5}{9}}&{\frac{4}{9}}\\{\frac{2}{9}}&{\frac{4}{9}}&{\frac{5}{9}}\end{aligned}} \right]\].

03

Write the Theorem

Let \[{\bf{y}} \in {\mathbb{R}^n}\], then \[{\rm{Pro}}{{\rm{j}}_W}{\bf{y}} = U{U^T}{\bf{y}}\], if \[U = \left[ {\begin{aligned}{{{\bf{u}}_1}}&{{{\bf{u}}_2}}& \cdots &{{{\bf{u}}_p}}\end{aligned}} \right]\].

04

Find \[{\bf{Pro}}{{\bf{j}}_W}{\rm{y}}\] and \[\left( {{U^T}U} \right){\rm{y}}\] 

(b).

Since \[U = \left[ {\begin{aligned}{{{\bf{u}}_1}}&{{{\bf{u}}_2}}\end{aligned}} \right]\] from part (a). then \[{\rm{Pro}}{{\rm{j}}_W}{\bf{y}} = U{U^T}{\bf{y}}\].

From part (a) \[U{U^T} = \left[ {\begin{aligned}{\frac{8}{9}}&{\frac{{ - 2}}{9}}&{\frac{2}{9}}\\{\frac{{ - 2}}{9}}&{\frac{5}{9}}&{\frac{4}{9}}\\{\frac{2}{9}}&{\frac{4}{9}}&{\frac{5}{9}}\end{aligned}} \right]\], then find \[U{U^T}{\bf{y}}\], where \[{\bf{y}} = \left[ {\begin{aligned}4\\8\\1\end{aligned}} \right]\].

\[\begin{aligned}U{U^T}{\bf{y}} = \left[ {\begin{aligned}{\frac{8}{9}}&{\frac{{ - 2}}{9}}&{\frac{2}{9}}\\{\frac{{ - 2}}{9}}&{\frac{5}{9}}&{\frac{4}{9}}\\{\frac{2}{9}}&{\frac{4}{9}}&{\frac{5}{9}}\end{aligned}} \right]\left[ {\begin{aligned}4\\8\\1\end{aligned}} \right]\\ = \left[ {\begin{aligned}{\frac{{32}}{9} - \frac{{16}}{9} + \frac{2}{9}}\\{\frac{{ - 8}}{9} + \frac{{40}}{9} + \frac{4}{9}}\\{\frac{8}{9} + \frac{{32}}{9} + \frac{5}{9}}\end{aligned}} \right]\\ = \left[ {\begin{aligned}2\\4\\5\end{aligned}} \right]\end{aligned}\]

So, \[{\rm{Pro}}{{\rm{j}}_W}{\bf{y}} = \left[ {\begin{aligned}2\\4\\5\end{aligned}} \right]\], as \[{\rm{Pro}}{{\rm{j}}_W}{\bf{y}} = U{U^T}{\bf{y}}\].

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Most popular questions from this chapter

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

Determine which pairs of vectors in Exercises 15-18 are orthogonal.

15. \({\mathop{\rm a}\nolimits} = \left( {\begin{aligned}{*{20}{c}}8\\{ - 5}\end{aligned}} \right),{\rm{ }}{\mathop{\rm b}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 2}\\{ - 3}\end{aligned}} \right)\)

In Exercises 9-12, find a unit vector in the direction of the given vector.

9. \(\left( {\begin{aligned}{*{20}{c}}{ - 30}\\{40}\end{aligned}} \right)\)

A certain experiment produces the data \(\left( {1,1.8} \right),\left( {2,2.7} \right),\left( {3,3.4} \right),\left( {4,3.8} \right),\left( {5,3.9} \right)\). Describe the model that produces a least-squares fit of these points by a function of the form

\(y = {\beta _1}x + {\beta _2}{x^2}\)

Such a function might arise, for example, as the revenue from the sale of \(x\) units of a product, when the amount offered for sale affects the price to be set for the product.

a. Give the design matrix, the observation vector, and the unknown parameter vector.

b. Find the associated least-squares curve for the data.

Let \(X\) be the design matrix used to find the least square line of fit data \(\left( {{x_1},{y_1}} \right), \ldots ,\left( {{x_n},{y_n}} \right)\). Use a theorem in Section 6.5 to show that the normal equations have a unique solution if and only if the data include at least two data points with different \(x\)-coordinates.
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