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Chapter 6: 8E (page 331)

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

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

Short Answer

Expert verified

The vector\[{\bf{y}}\]is the sum of the vectors in\[W\]as\[{\bf{\hat y}} + {\bf{z}} = \left[ {\begin{aligned}{\frac{3}{2}}\\{\frac{7}{2}}\\1\end{aligned}} \right] + \left[ {\begin{aligned}{ - \frac{5}{2}}\\{\frac{1}{2}}\\2\end{aligned}} \right]\].

A vector orthogonal to \[W\] is \[\left[ {\begin{aligned}{ - \frac{5}{2}}\\{\frac{1}{2}}\\2\end{aligned}} \right]\].

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

Find the product\[{{\bf{u}}_1} \cdot {{\bf{u}}_2}\].

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

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

03

The Orthogonal Decomposition Theorem

If \[\left\{ {{{\bf{u}}_1}, \ldots ,{{\bf{u}}_p}} \right\}\] is any orthogonal basis of \[w\], then \[{\bf{\hat y}} = \frac{{{\bf{y}} \cdot {{\bf{u}}_1}}}{{{{\bf{u}}_1} \cdot {{\bf{u}}_1}}}{{\bf{u}}_1} + \cdots + \frac{{{\bf{y}} \cdot {{\bf{u}}_p}}}{{{{\bf{u}}_p} \cdot {{\bf{u}}_p}}}{{\bf{u}}_p}\], where \[w\] is a subspace of \[{\mathbb{R}^n}\] and \[{\bf{\hat y}} \in w\], for which every \[{\bf{y}}\] can be written as \[{\bf{y}} = {\bf{\hat y}} + {\bf{z}}\] when \[{\bf{z}} \in {w^ \bot }\].

04

Find the Orthogonal projection of \[{\bf{y}}\] onto Span \[\left\{ {{{\bf{u}}_1},{{\bf{u}}_2}} \right\}\]

Find\[{\bf{y}} \cdot {{\bf{u}}_1}\],\[{\bf{y}} \cdot {{\bf{u}}_2}\],\[{{\bf{u}}_1} \cdot {{\bf{u}}_1}\]and\[{{\bf{u}}_2} \cdot {{\bf{u}}_2}\]first, where\[{\bf{y}} = \left[ {\begin{aligned}{ - 1}\\4\\3\end{aligned}} \right]\].

\[\begin{aligned}{\bf{y}} \cdot {{\bf{u}}_1} &= \left[ {\begin{aligned}{ - 1}\\4\\3\end{aligned}} \right] \cdot \left[ {\begin{aligned}1\\1\\1\end{aligned}} \right]\\ &= \left( { - 1} \right)\left( 1 \right) + 4\left( 1 \right) + 3\left( 1 \right)\\ &= 6\end{aligned}\]

\[\begin{aligned}{\bf{y}} \cdot {{\bf{u}}_2} &= \left[ {\begin{aligned}{ - 1}\\4\\3\end{aligned}} \right] \cdot \left[ {\begin{aligned}{ - 1}\\3\\{ - 2}\end{aligned}} \right]\\ &= \left( { - 1} \right)\left( { - 1} \right) + 4\left( 3 \right) + 3\left( { - 2} \right)\\ &= 7\end{aligned}\]

\[\begin{aligned}{{\bf{u}}_1} \cdot {{\bf{u}}_1} &= \left[ {\begin{aligned}1\\1\\1\end{aligned}} \right] \cdot \left[ {\begin{aligned}1\\1\\1\end{aligned}} \right]\\ &= 1\left( 1 \right) + \left( 1 \right)\left( 1 \right) + \left( 1 \right)\left( 1 \right)\\ &= 3\end{aligned}\]

\[\begin{aligned}{{\bf{u}}_2} \cdot {{\bf{u}}_2} &= \left[ {\begin{aligned}{ - 1}\\3\\{ - 2}\end{aligned}} \right] \cdot \left[ {\begin{aligned}{ - 1}\\3\\{ - 2}\end{aligned}} \right]\\ &= - 1\left( { - 1} \right) + \left( 3 \right)\left( 3 \right) + \left( { - 2} \right)\left( { - 2} \right)\\ &= 14\end{aligned}\]

Now, find the projection of\[{\bf{y}}\]onto Span\[\left\{ {{{\bf{u}}_1},{{\bf{u}}_2}} \right\}\]by substituting the obtained values into\[{\bf{\hat y}} = \frac{{{\bf{y}} \cdot {{\bf{u}}_1}}}{{{{\bf{u}}_1} \cdot {{\bf{u}}_1}}}{{\bf{u}}_1} + \cdots + \frac{{{\bf{y}} \cdot {{\bf{u}}_p}}}{{{{\bf{u}}_p} \cdot {{\bf{u}}_p}}}{{\bf{u}}_p}\]

\[\begin{aligned}{\bf{\hat y}} &= \frac{6}{3}{{\bf{u}}_1} + \frac{7}{{14}}{{\bf{u}}_2}\\ &= 2\left[ {\begin{aligned}1\\1\\1\end{aligned}} \right] + \frac{1}{2}\left[ {\begin{aligned}{ - 1}\\3\\{ - 2}\end{aligned}} \right]\\ &= \left[ {\begin{aligned}{\frac{3}{2}}\\{\frac{7}{2}}\\1\end{aligned}} \right]\end{aligned}\]

Hence, \[{\bf{\hat y}} = \left[ {\begin{aligned}{\frac{3}{2}}\\{\frac{7}{2}}\\1\end{aligned}} \right]\].

05

Find a vector orthogonal to \[W\]

Find\[{\bf{z}}\], which is orthogonal to\[W\]by using\[{\bf{z}} = {\bf{y}} - {\bf{\hat y}}\].

\[\begin{aligned}{\bf{z}} = {\bf{y}} - {\bf{\hat y}}\\ &= \left[ {\begin{aligned}{ - 1}\\4\\3\end{aligned}} \right] - \left[ {\begin{aligned}{\frac{3}{2}}\\{\frac{7}{2}}\\1\end{aligned}} \right]\\ &= \left[ {\begin{aligned}{ - \frac{5}{2}}\\{\frac{1}{2}}\\2\end{aligned}} \right]\end{aligned}\]

So,\[{\bf{z}} = \left[ {\begin{aligned}{ - \frac{5}{2}}\\{\frac{1}{2}}\\2\end{aligned}} \right]\].

06

Write \[y\] as a sum of a vector in \[W\]

Write\[{\bf{y}} = {\bf{\hat y}} + {\bf{z}}\], where\[{\bf{\hat y}} \in w\]and\[{\bf{z}} \in {w^ \bot }\].

\[\begin{aligned}{\bf{y}} &= {\bf{\hat y}} + {\bf{z}}\\ &= \left[ {\begin{aligned}{\frac{3}{2}}\\{\frac{7}{2}}\\1\end{aligned}} \right] + \left[ {\begin{aligned}{ - \frac{5}{2}}\\{\frac{1}{2}}\\2\end{aligned}} \right]\end{aligned}\]

Hence, \[{\bf{y}}\] as the sum of a vector in \[W\] is \[{\bf{\hat y}} + {\bf{z}} = \left[ {\begin{aligned}{\frac{3}{2}}\\{\frac{7}{2}}\\1\end{aligned}} \right] + \left[ {\begin{aligned}{ - \frac{5}{2}}\\{\frac{1}{2}}\\2\end{aligned}} \right]\].

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Most popular questions from this chapter

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.

6. \(\left( {\begin{aligned}{{}}3\\{ - 1}\\2\\{ - 1}\end{aligned}} \right),\left( {\begin{aligned}{{}}{ - 5}\\9\\{ - 9}\\3\end{aligned}} \right)\)

A Householder matrix, or an elementary reflector, has the form \(Q = I - 2{\bf{u}}{{\bf{u}}^T}\) where u is a unit vector. (See Exercise 13 in the Supplementary Exercise for Chapter 2.) Show that Q is an orthogonal matrix. (Elementary reflectors are often used in computer programs to produce a QR factorization of a matrix A. If A has linearly independent columns, then left-multiplication by a sequence of elementary reflectors can produce an upper triangular matrix.)

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.

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 1-4, find a least-sqaures solution of \(A{\bf{x}} = {\bf{b}}\) by (a) constructing a normal equations for \({\bf{\hat x}}\) and (b) solving for \({\bf{\hat x}}\).

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