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Chapter 6: Q6.2-32E (page 331)

Question: Let \(\left\{ {{{\bf{v}}_{\bf{1}}},{{\bf{v}}_{\bf{2}}}} \right\}\) be an orthogonal set of nonzero vectors, and let \({c_{\bf{1}}}\), \({c_{\bf{2}}}\) be any nonzero scalars. Show that \(\left\{ {{c_{\bf{1}}}{{\bf{v}}_{\bf{1}}},{c_{\bf{2}}}{{\bf{v}}_{\bf{2}}}} \right\}\) is also an orthogonal set. Since orthgonality of a set is defined in terms of pairs of vectors, this shows that if the vectors in an orthogonal set are normalized, the new set will still be orthogonal.

Short Answer

Expert verified

It is proved that the set \(\left\{ {{c_1}{{\bf{v}}_1},{c_2}{{\bf{v}}_2}} \right\}\) is an orthogonal set.

Step by step solution

01

Write the relation for the vectors \(\left\{ {{{\bf{v}}_{\bf{1}}},{{\bf{v}}_{\bf{2}}}} \right\}\)

Since the set \(\left\{ {{{\bf{v}}_1},{{\bf{v}}_2}} \right\}\) is an orthogonal set of nonzero vectors, so \({{\bf{v}}_1} \cdot {{\bf{v}}_2} = 0\).

02

Check the orthogonality for \(\left\{ {{c_{\bf{1}}}{{\bf{v}}_{\bf{1}}},{c_{\bf{2}}}{{\bf{v}}_{\bf{2}}}} \right\}\)

As \({c_1}\) and \({c_2}\) are non-zero scalars, then the dot product for the set of vectors \(\left\{ {{c_1}{{\bf{v}}_1},{c_2}{{\bf{v}}_2}} \right\}\) is shown below:

\(\begin{array}{c}\left( {{c_1}{{\bf{v}}_1}} \right) \cdot \left( {{c_2}{{\bf{v}}_2}} \right) = {c_1}\left( {{{\bf{v}}_1} \cdot \left( {{c_2}{{\bf{v}}_2}} \right)} \right)\\ = {c_1}{c_2}\left( {{{\bf{v}}_1} \cdot {{\bf{v}}_2}} \right)\\ = {c_1}{c_2}\left( 0 \right)\\ = 0\end{array}\)

Therefore, the set \(\left\{ {{c_1}{{\bf{v}}_1},{c_2}{{\bf{v}}_2}} \right\}\) is an orthogonal set.

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Most popular questions from this chapter

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\[W\].

10.\[y = \left[ {\begin{aligned}3\\4\\5\\6\end{aligned}} \right]\],\[{{\bf{u}}_1} = \left[ {\begin{aligned}1\\1\\0\\{ - 1}\end{aligned}} \right]\],\[{{\bf{u}}_2} = \left[ {\begin{aligned}1\\0\\1\\1\end{aligned}} \right]\],\[{{\bf{u}}_3} = \left[ {\begin{aligned}0\\{ - 1}\\1\\{ - 1}\end{aligned}} \right]\]

Find the distance between \({\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}0\\{ - 5}\\2\end{aligned}} \right)\) and \({\mathop{\rm z}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 4}\\{ - 1}\\8\end{aligned}} \right)\).

A simple curve that often makes a good model for the variable costs of a company, a function of the sales level \(x\), has the form \(y = {\beta _1}x + {\beta _2}{x^2} + {\beta _3}{x^3}\). There is no constant term because fixed costs are not included.

a. Give the design matrix and the parameter vector for the linear model that leads to a least-squares fit of the equation above, with data \(\left( {{x_1},{y_1}} \right), \ldots ,\left( {{x_n},{y_n}} \right)\).

b. Find the least-squares curve of the form above to fit the data \(\left( {4,1.58} \right),\left( {6,2.08} \right),\left( {8,2.5} \right),\left( {10,2.8} \right),\left( {12,3.1} \right),\left( {14,3.4} \right),\left( {16,3.8} \right)\) and \(\left( {18,4.32} \right)\), with values in thousands. If possible, produce a graph that shows the data points and the graph of the cubic approximation.

Find an orthonormal basis of the subspace spanned by the vectors in Exercise 3.

Question: In Exercises 9-12, find (a) the orthogonal projection of b onto \({\bf{Col}}A\) and (b) a least-squares solution of \(A{\bf{x}} = {\bf{b}}\).

11. \(A = \left( {\begin{aligned}{{}{}}{\bf{4}}&{\bf{0}}&{\bf{1}}\\{\bf{1}}&{ - {\bf{5}}}&{\bf{1}}\\{\bf{6}}&{\bf{1}}&{\bf{0}}\\{\bf{1}}&{ - {\bf{1}}}&{ - {\bf{5}}}\end{aligned}} \right)\), \({\bf{b}} = \left( {\begin{aligned}{{}{}}{\bf{9}}\\{\bf{0}}\\{\bf{0}}\\{\bf{0}}\end{aligned}} \right)\)
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