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Chapter 6: Q6.2-35E (page 331)

Question: Show that columns of matrix A are orthogonal by making an appropriate matrix calculation. State the calculation you use.

\(A = \left( {\begin{array}{*{20}{c}}{ - {\bf{6}}}&{ - {\bf{3}}}&{\bf{6}}&{\bf{1}}\\{ - {\bf{1}}}&{\bf{2}}&{\bf{1}}&{ - {\bf{6}}}\\{\bf{3}}&{\bf{6}}&{\bf{3}}&{ - {\bf{2}}}\\{\bf{6}}&{ - {\bf{3}}}&{\bf{6}}&{ - {\bf{1}}}\\{\bf{2}}&{ - {\bf{1}}}&{\bf{2}}&{\bf{3}}\\{ - {\bf{3}}}&{\bf{6}}&{\bf{3}}&{\bf{2}}\\{ - {\bf{2}}}&{ - {\bf{1}}}&{\bf{2}}&{ - {\bf{3}}}\\{\bf{1}}&{\bf{2}}&{\bf{1}}&{\bf{6}}\end{array}} \right)\)

Short Answer

Expert verified

It is proved that columns of matrix A are orthogonal.

Step by step solution

01

Find the transpose of matrix A

Use the following MATLAB command to obtain the transpose of matrix A.

\(\begin{array}{c} > > {\rm{A}} = \left( \begin{array}{l}\begin{array}{*{20}{c}}{ - 6}&{ - 3}&6&1\end{array};\,\begin{array}{*{20}{c}}{ - 1}&2&1&{ - 6}\end{array};\,\begin{array}{*{20}{c}}3&6&3&{ - 2;\,\begin{array}{*{20}{c}}6&{ - 3}&6&{ - 1}\end{array}}\end{array};\\\begin{array}{*{20}{c}}2&{ - 1}&2&{3;\,\begin{array}{*{20}{c}}{ - 3}&6&3&{2;\,\,\begin{array}{*{20}{c}}{ - 2}&{ - 1}&2&{ - 3;\,\,\begin{array}{*{20}{c}}1&2&1&6\end{array}}\end{array}}\end{array}}\end{array}\end{array} \right);\\ > > {\rm{A'}} = {\rm{A}};\end{array}\)

The transpose of matrix A is shown below:

\({A^T} = \left( {\begin{array}{*{20}{c}}{ - 6}&{ - 1}&3&6&2&{ - 3}&{ - 2}&1\\{ - 3}&2&6&{ - 3}&{ - 1}&6&{ - 1}&2\\6&1&3&6&2&3&2&1\\1&{ - 6}&{ - 2}&{ - 1}&3&2&{ - 3}&6\end{array}} \right)\)

02

Find the product \({A^T}*A\)

The product \({A^T}*A\) can be calculated by using the following MATLAB code:

\({\rm{P}} = {\rm{A'}}*{\rm{A}}\)

The product \({A^T}*A\) is:

\(\begin{array}{c}{A^T}A = \left( {\begin{array}{*{20}{c}}{100}&0&0&0\\0&{100}&0&0\\0&0&{100}&0\\0&0&0&{100}\end{array}} \right)\\ = 100\left( {\begin{array}{*{20}{c}}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{array}} \right)\\ = 100{I_4}\end{array}\)

As the matrix \({A^T}A\) has only diagonal entries, so the columns of matrix A are orthogonal.

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Most popular questions from this chapter

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

12. \(\left( {\begin{array}{*{20}{c}}{\frac{8}{3}}\\2\end{array}} \right)\)

A certain experiment produce the data \(\left( {1,7.9} \right),\left( {2,5.4} \right)\) and \(\left( {3, - .9} \right)\). Describe the model that produces a least-squares fit of these points by a function of the form

\(y = A\cos x + B\sin x\)

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 \({\bf{\hat x}}\).

1. \(A = \left[ {\begin{aligned}{{}{}}{ - {\bf{1}}}&{\bf{2}}\\{\bf{2}}&{ - {\bf{3}}}\\{ - {\bf{1}}}&{\bf{3}}\end{aligned}} \right]\), \({\bf{b}} = \left[ {\begin{aligned}{{}{}}{\bf{4}}\\{\bf{1}}\\{\bf{2}}\end{aligned}} \right]\)

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

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

16. Use a matrix inverse to solve the system of equations in (7) and thereby obtain formulas for \({\hat \beta _0}\) , and that appear in many statistics texts.

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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