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Chapter 6: Q23E (page 331)

Let \({\bf{u}} = \left( {\begin{aligned}{2}\\{ - 5}\\{ - 1}\end{aligned}} \right)\) and \({\bf{v}} = \left( {\begin{aligned}{ - 7}\\{ - 4}\\6\end{aligned}} \right)\). Compute and compare \({\bf{u}}.{\bf{v}},\,\,{\left\| {\bf{u}} \right\|^2},\,\,{\left\| {\bf{v}} \right\|^2},{\rm{ and }}{\left\| {{\bf{u}} + {\bf{v}}} \right\|^2}\). Do not use the Pythagorean Theorem.

Short Answer

Expert verified

The required values are:

\(\begin{aligned}{c}{\bf{u}} \cdot {\bf{v}} &= 0\\{\left\| {\bf{u}} \right\|^2} &= 30\\{\left\| {\bf{v}} \right\|^2} &= 101\\{\left\| {{\bf{u}} + {\bf{v}}} \right\|^2} &= 131\end{aligned}\)

On comparing these, it can be observed that the sum of \({\left\| {\bf{u}} \right\|^2}\) and \({\left\| {\bf{v}} \right\|^2}\) is the same as \({\left\| {{\bf{u}} + {\bf{v}}} \right\|^2}\).

Step by step solution

01

 Compute \({\bf{u}} \cdot {\bf{v}},{\left\| {\bf{u}} \right\|^2},\,\,{\left\| {\bf{v}} \right\|^2},{\rm{ and }}{\left\| {{\bf{u}} + {\bf{v}}} \right\|^2}\)

The given vectors are:

\({\bf{u}} = \left( {\begin{aligned}{*{20}{c}}2\\{ - 5}\\{ - 1}\end{aligned}} \right){\rm{ and }}{\bf{v}} = \left( {\begin{aligned}{*{20}{c}}{ - 7}\\{ - 4}\\6\end{aligned}} \right)\)

Find \({\bf{u}} \cdot {\bf{v}}\).

\(\begin{aligned}{c}{\bf{u}} \cdot {\bf{v}} &= 2\left( { - 7} \right) + \left( { - 5} \right)\left( { - 4} \right) + \left( { - 1} \right)6\\ &= 0\end{aligned}\)

Find \({\left\| {\bf{u}} \right\|^2}\) and \({\left\| {\bf{v}} \right\|^2}\).

\(\begin{aligned}{c}{\left\| {\bf{u}} \right\|^2} &= {\bf{u}} \cdot {\bf{u}}\\ &= 2\left( 2 \right) + \left( { - 5} \right)\left( { - 5} \right) + \left( { - 1} \right)\left( { - 1} \right)\\ &= 30\end{aligned}\)

\(\begin{aligned}{c}{\left\| {\bf{v}} \right\|^2} &= {\bf{v}} \cdot {\bf{v}}\\ &= \left( { - 7} \right)\left( { - 7} \right) + \left( { - 4} \right)\left( { - 4} \right) + \left( 6 \right)\left( 6 \right)\\ &= 101\end{aligned}\)

Next, find \({\left\| {{\bf{u}} + {\bf{v}}} \right\|^2}\)

\(\begin{aligned}{c}{\left\| {{\bf{u}} + {\bf{v}}} \right\|^2} &= \left( {{\bf{u}} + {\bf{v}}} \right) \cdot \left( {{\bf{u}} + {\bf{v}}} \right)\\ &= {\left( {2 + \left( { - 7} \right)} \right)^2} + {\left( { - 5 + \left( { - 4} \right)} \right)^2} + {\left( { - 1 + 6} \right)^2}\\ &= 131\end{aligned}\)

02

Make a comparison between obtained values

On comparing the obtained values, it can be observed that the sum of \({\left\| {\bf{u}} \right\|^2}\) and \({\left\| {\bf{v}} \right\|^2}\) is the same as \({\left\| {{\bf{u}} + {\bf{v}}} \right\|^2}\).

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Most popular questions from this chapter

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

10. \(A = \left[ {\begin{aligned}{{}{}}{\bf{1}}&{\bf{2}}\\{ - {\bf{1}}}&{\bf{4}}\\{\bf{1}}&{\bf{2}}\end{aligned}} \right]\), \({\bf{b}} = \left[ {\begin{aligned}{{}{}}{\bf{3}}\\{ - {\bf{1}}}\\{\bf{5}}\end{aligned}} \right]\)

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

In Exercises 17 and 18, all vectors and subspaces are in \({\mathbb{R}^n}\). Mark each statement True or False. Justify each answer.

17. a.If \(\left\{ {{{\bf{v}}_1},{{\bf{v}}_2},{{\bf{v}}_3}} \right\}\) is an orthogonal basis for\(W\), then multiplying

\({v_3}\)by a scalar \(c\) gives a new orthogonal basis \(\left\{ {{{\bf{v}}_1},{{\bf{v}}_2},c{{\bf{v}}_3}} \right\}\).

b. The Gram–Schmidt process produces from a linearly independent

set \(\left\{ {{{\bf{x}}_1}, \ldots ,{{\bf{x}}_p}} \right\}\)an orthogonal set \(\left\{ {{{\bf{v}}_1}, \ldots ,{{\bf{v}}_p}} \right\}\) with the property that for each \(k\), the vectors \({{\bf{v}}_1}, \ldots ,{{\bf{v}}_k}\) span the same subspace as that spanned by \({{\bf{x}}_1}, \ldots ,{{\bf{x}}_k}\).

c. If \(A = QR\), where \(Q\) has orthonormal columns, then \(R = {Q^T}A\).

Exercises 19 and 20 involve a design matrix \(X\) with two or more columns and a least-squares solution \(\hat \beta \) of \({\bf{y}} = X\beta \). Consider the following numbers.

(i) \({\left\| {X\hat \beta } \right\|^2}\)—the sum of the squares of the “regression term.” Denote this number by \(SS\left( R \right)\).

(ii) \({\left\| {{\bf{y}} - X\hat \beta } \right\|^2}\)—the sum of the squares for error term. Denote this number by \(SS\left( E \right)\).

(iii) \({\left\| {\bf{y}} \right\|^2}\)—the “total” sum of the squares of the -values. Denote this number by \(SS\left( T \right)\).

Every statistics text that discusses regression and the linear model \(y = X\beta + \in \) introduces these numbers, though terminology and notation vary somewhat. To simplify matters, assume that the mean of the -values is zero. In this case, \(SS\left( T \right)\) is proportional to what is called the variance of the set of \(y\)-values.

20. Show that \({\left\| {X\hat \beta } \right\|^2} = {\hat \beta ^T}{X^T}{\bf{y}}\). (Hint: Rewrite the left side and use the fact that \(\hat \beta \) satisfies the normal equations.) This formula for is used in statistics. From this and from Exercise 19, obtain the standard formula for \(SS\left( E \right)\):

\(SS\left( E \right) = {y^T}y - \hat \beta {X^T}y\)

Suppose \(A = QR\), where \(Q\) is \(m \times n\) and R is \(n \times n\). Showthat if the columns of \(A\) are linearly independent, then \(R\) mustbe invertible.
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