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Chapter 6: 6E (page 331)

In exercises 1-6, determine which sets of vectors are orthogonal.

\(\left[ {\begin{array}{*{20}{c}}5\\{ - 4}\\0\\3\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}{ - 4}\\1\\{ - 3}\\8\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}3\\3\\5\\{ - 1}\end{array}} \right]\)

Short Answer

Expert verified

The given set is orthogonal.

Step by step solution

01

Definition of an 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 for orthogonality of vectors

Let the given vectors be, \({u_1} = \left[ {\begin{align}{ 5}\\{-4}\\{ 0}\\3\end{align}} \right]\), \({u_2} = \left[ {\begin{align}{ - 4}\\1\\{ - 3}\\8\end{align}} \right]\) and \({u_3} = \left[ {\begin{align}3\\3\\5\\{ - 1}\end{align}} \right]\).

First find \({u_1} \cdot {u_2}\):

\(\begin{align}{c}{u_1} \cdot {u_2} = \left( 5 \right)\left( { - 4} \right) + \left( { - 4} \right)\left( 1 \right) + \left( 0 \right)\left( { - 3} \right) + \left( 3 \right)\left( 8 \right)\\ = - 20 - 4 + 0 + 24\\ = 0\end{align}\)

Now, find \({u_2} \cdot {u_3}\):

\(\begin{align}{c}{u_2} \cdot {u_3} = \left( { - 4} \right)\left( 3 \right) + \left( 1 \right)\left( 3 \right) + \left( { - 3} \right)\left( 5 \right) + \left( 8 \right)\left( { - 1} \right)\\ = - 12 + 3 - 15 - 8\\ = - 32\end{align}\)

Since \({u_2} \cdot {u_3} \ne 0\), hence, the given set is not orthogonal.

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Most popular questions from this chapter

Suppose \(A = QR\) is a \(QR\) factorization of an \(m \times n\) matrix

A (with linearly independent columns). Partition \(A\) as \(\left[ {\begin{aligned}{{}{}}{{A_1}}&{{A_2}}\end{aligned}} \right]\), where \({A_1}\) has \(p\) columns. Show how to obtain a \(QR\) factorization of \({A_1}\), and explain why your factorization has the appropriate properties.

In Exercises 5 and 6, describe all least squares solutions of the equation \(A{\bf{x}} = {\bf{b}}\).

5. \(A = \left( {\begin{aligned}{{}{}}{\bf{1}}&{\bf{1}}&{\bf{0}}\\{\bf{1}}&{\bf{1}}&{\bf{0}}\\{\bf{1}}&{\bf{0}}&{\bf{1}}\\{\bf{1}}&{\bf{0}}&{\bf{1}}\end{aligned}} \right)\), \({\bf{b}} = \left( {\begin{aligned}{{}{}}{\bf{1}}\\{\bf{3}}\\{\bf{8}}\\{\bf{2}}\end{aligned}} \right)\)

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

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.

3. \(\left( {\begin{aligned}{{}{}}2\\{ - 5}\\1\end{aligned}} \right),\left( {\begin{aligned}{{}{}}4\\{ - 1}\\2\end{aligned}} \right)\)

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.
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