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Chapter 6: Q18E (page 331)

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

18. \(y = \left( {\begin{aligned}{ - 3}\\7\\4\\0\end{aligned}} \right),\,\,z = \left( {\begin{aligned}{1}\\{ - 8}\\{15}\\{ - 7}\end{aligned}} \right)\)

Short Answer

Expert verified

The vectors \({\bf{y}}{\rm{ and }}{\bf{z}}\) are not orthogonal to each other.

Step by step solution

01

Definition of orthogonal vectors

The two vectors \({\bf{u}}{\rm{ and }}{\bf{v}}\) are Orthogonal if \({\bf{u}} \cdot {\bf{v}} = 0\).

02

Checking Orthogonality for given vectors.

The given vectors are:

\({\bf{y}} = \left( {\begin{aligned}{*{20}{r}}{ - 3}\\7\\4\\0\end{aligned}} \right),\,\,{\bf{z}} = \left( {\begin{aligned}{*{20}{r}}1\\{ - 8}\\{15}\\{ - 7}\end{aligned}} \right)\)

On having dot products, we get:

\(\begin{aligned}{c}{\bf{y}} \cdot {\bf{z}} = \left( {\begin{aligned}{*{20}{r}}{ - 3}\\7\\4\\0\end{aligned}} \right) \cdot \left( {\begin{aligned}{*{20}{r}}1\\{ - 8}\\{15}\\{ - 7}\end{aligned}} \right)\\ = \left( { - 3} \right)\left( 1 \right) + \left( 7 \right)\left( { - 8} \right) + \left( 4 \right)\left( {15} \right) + \left( 0 \right)\left( { - 7} \right)\\ = - 3 - 56 + 60 + 0\\ = 1\end{aligned}\)

Since \({\bf{y}} \cdot {\bf{z}} \ne 0\).

Hence, the vectors \(y{\rm{ and }}z\) are not orthogonal to each other.

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Most popular questions from this chapter

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

7.\[y = \left[ {\begin{aligned}1\\3\\5\end{aligned}} \right]\],\[{{\bf{u}}_1} = \left[ {\begin{aligned}1\\3\\{ - 2}\end{aligned}} \right]\],\[{{\bf{u}}_2} = \left[ {\begin{aligned}5\\1\\4\end{aligned}} \right]\]

Show that if an \(n \times n\) matrix satisfies \(\left( {U{\bf{x}}} \right) \cdot \left( {U{\bf{y}}} \right) = {\bf{x}} \cdot {\bf{y}}\) for all x and y in \({\mathbb{R}^n}\), then \(U\) is an orthogonal matrix.

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

Let \(\left\{ {{{\bf{v}}_1}, \ldots ,{{\bf{v}}_p}} \right\}\) be an orthonormal set. Verify the following equality by induction, beginning with \(p = 2\). If \({\bf{x}} = {c_1}{{\bf{v}}_1} + \ldots + {c_p}{{\bf{v}}_p}\), then

\({\left\| {\bf{x}} \right\|^2} = {\left| {{c_1}} \right|^2} + {\left| {{c_2}} \right|^2} + \ldots + {\left| {{c_p}} \right|^2}\)

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.
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