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Chapter 6: Q13E (page 331)

Find the distance between \({\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{10}\\{ - 3}\end{aligned}} \right)\) and \({\mathop{\rm y}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 1}\\{ - 5}\end{aligned}} \right)\).

Short Answer

Expert verified

The distance between the vectors x and y is \({\mathop{\rm dist}\nolimits} \left( {{\mathop{\rm x}\nolimits} ,y} \right) = 5\sqrt 5 \).

Step by step solution

01

Distance in \({\mathbb{R}^n}\)

Thelength of the vector is thedistance between u and v in \({\mathbb{R}^n}\), expressed as \({\mathop{\rm dist}\nolimits} \left( {{\mathop{\rm u}\nolimits} ,v} \right)\). Therefore, \({\mathop{\rm dist}\nolimits} \left( {{\mathop{\rm u}\nolimits} ,v} \right) = \left\| {{\mathop{\rm u}\nolimits} - v} \right\|\).

02

Find the distance between the vectors

Compute \({\mathop{\rm x}\nolimits} - {\mathop{\rm y}\nolimits} \) as shown below:

\(\begin{aligned}{c}{\mathop{\rm x}\nolimits} - {\mathop{\rm y}\nolimits} &= \left( {\begin{aligned}{*{20}{c}}{10}\\{ - 3}\end{aligned}} \right) - \left( {\begin{aligned}{*{20}{c}}{ - 1}\\{ - 5}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{*{20}{c}}{10 + 1}\\{ - 3 + 5}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{*{20}{c}}{11}\\{ - 2}\end{aligned}} \right)\end{aligned}\)

Compute the distance between the vectors as shown below:

\(\begin{aligned}{c}{\mathop{\rm dist}\nolimits} \left( {{\mathop{\rm x}\nolimits} ,y} \right) &= \left\| {{\mathop{\rm x}\nolimits} - {\mathop{\rm y}\nolimits} } \right\|\\ &= \sqrt {{{\left( {11} \right)}^2} + {{\left( { - 2} \right)}^2}} \\ &= \sqrt {121 + 4} \\ &= \sqrt {125} \\ &= 5\sqrt 5 \end{aligned}\)

Thus, the distance between the vectors x and y is \({\mathop{\rm dist}\nolimits} \left( {{\mathop{\rm x}\nolimits} ,y} \right) = 5\sqrt 5 \).

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Most popular questions from this chapter

Given \(A = QR\) as in Theorem 12, describe how to find an orthogonal\(m \times m\)(square) matrix \({Q_1}\) and an invertible \(n \times n\) upper triangular matrix \(R\) such that

\(A = {Q_1}\left[ {\begin{aligned}{{}{}}R\\0\end{aligned}} \right]\)

The MATLAB qr command supplies this “full” QR factorization

when rank \(A = n\).

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.)

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

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.

  1. \(\left( {\begin{aligned}{{}{}}3\\0\\{ - 1}\end{aligned}} \right),\left( {\begin{aligned}{{}{}}8\\5\\{ - 6}\end{aligned}} \right)\)

Let u and v be linearly independent vectors in \({\mathbb{R}^n}\) that are not orthogonal. Describe how to find the best approximation to z in \({\mathbb{R}^n}\) by vectors of the form \({{\bf{x}}_1}{\mathop{\rm u}\nolimits} + {{\bf{x}}_2}{\mathop{\rm u}\nolimits} \) without first constructing an orthogonal basis for \({\mathop{\rm Span}\nolimits} \left\{ {{\bf{u}},{\bf{v}}} \right\}\).
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