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In Exercises 19 and 20, all vectors are in \({\mathbb{R}^n}\). Mark each statement True or False. Justify each answer.

  1. \(v \cdot {\rm{v}} = {\left\| {\rm{v}} \right\|^2}\).
  2. For any scalar \(c\),\({\rm{u}} \cdot \left( {c{\rm{v}}} \right) = c\left( {{\rm{u}} \cdot {\rm{v}}} \right)\).
  3. If the distance from \({\rm{u}}\) to \({\rm{v}}\) equals the distance from \({\rm{u}}\) to \( - {\rm{v}}\), then \({\rm{u}}\) and \({\rm{v}}\) are orthogonal.
  4. For a square matrix \(A\), vectors in \({\rm{Col }}A\) are orthogonal to vectors in Nul \(A\).
  5. If vectors \({{\rm{v}}_1}..........,{{\rm{v}}_p}\) span a subspace \(W\) and if \({\rm{x}}\) is orthogonal to each \({{\rm{v}}_j}\) for \(j = 1,..........,p\), then \({\rm{x}}\) is in \({W^ \bot }\)?.

Short Answer

Expert verified
  1. True, by using the definition of the length of the vector.
  2. True, by using Theorem 1(c).
  3. True, by using the definition of orthogonal vectors.
  4. False, as the case fails when the square matrix is not symmetric.
  5. True, by using the definition for the spans of any vector.

Step by step solution

01

Definition of Orthogonal Set

The two vectors \({\rm{u and v}}\) are Orthogonal if:

\(\begin{aligned}{l}{\left\| {{\bf{u}} + {\bf{v}}} \right\|^2} = {\left\| {\bf{u}} \right\|^2} + {\left\| {\bf{v}} \right\|^2}\\{\rm{and}}\\{\bf{u}} \cdot {\bf{v}} = 0\end{aligned}\).

02

 Verification of statement (a)

The definition for the length of any vector states that:

\({\left\| {\bf{v}} \right\|^2} = {\bf{v}} \cdot {\bf{v}}\)

Hence, the given statement is true.

03

 Verification of statement (b)

The Theorem 1(c)resembles that:

\(\left( {c{\bf{u}}} \right) \cdot {\bf{v}} = c\left( {{\bf{u}} \cdot {\bf{v}}} \right) = {\bf{u}} \cdot \left( {c{\bf{v}}} \right)\)

Hence, the given statement is true.

04

 Verification of statement (c)

The definition of the Orthogonal Vectorsstates that:

\(\begin{aligned}{l}{\left\| {{\bf{u}} + {\bf{v}}} \right\|^2} = {\left\| {\bf{u}} \right\|^2} + {\left\| {\bf{v}} \right\|^2}\\{\rm{and}}\\{\bf{u}} \cdot {\bf{v}} = 0\end{aligned}\)

Hence, the given statement is true.

05

 Verification of statement (d)

The given statement is only valid for the particular type of square matrices. Just in case if there is a matrix of type:

\(A = \left( {\begin{aligned}{*{20}{c}}1&1\\0&0\end{aligned}} \right)\)

The given statement would not be true.

Hence, the given statement is False.

06

 Verification of statement (e)

The definition for the spans of any vector states that for the given condition in question, the span:

\({W^ \bot } \subseteq {\mathbb{R}^n}\)

Hence, the given statement is true.

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Most popular questions from this chapter

Let \({{\bf{u}}_1},......,{{\bf{u}}_p}\) be an orthogonal basis for a subspace \(W\) of \({\mathbb{R}^n}\), and let \(T:{\mathbb{R}^n} \to {\mathbb{R}^n}\) be defined by \(T\left( x \right) = {\rm{pro}}{{\rm{j}}_W}x\). Show that \(T\) is a linear transformation.

Question: In Exercises 3-6, verify that\(\left\{ {{{\bf{u}}_{\bf{1}}},{{\bf{u}}_{\bf{2}}}} \right\}\)is an orthogonal set, and then find the orthogonal projection of y onto\({\bf{Span}}\left\{ {{{\bf{u}}_{\bf{1}}},{{\bf{u}}_{\bf{2}}}} \right\}\).

4.\(y = \left[ {\begin{aligned}{\bf{6}}\\{\bf{3}}\\{ - {\bf{2}}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{1}}} = \left[ {\begin{aligned}{\bf{3}}\\{\bf{4}}\\{\bf{0}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{2}}} = \left[ {\begin{aligned}{ - {\bf{4}}}\\{\bf{3}}\\{\bf{0}}\end{aligned}} \right]\)

a. Rewrite the data in Example 1 with new \(x\)-coordinates in mean deviation form. Let \(X\) be the associated design matrix. Why are the columns of \(X\) orthogonal?

b. Write the normal equations for the data in part (a), and solve them to find the least-squares line, \(y = {\beta _0} + {\beta _1}x*\), where \(x* = x - 5.5\).

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

2. \(A = \left( {\begin{aligned}{{}{}}{\bf{2}}&{\bf{1}}\\{ - {\bf{2}}}&{\bf{0}}\\{\bf{2}} {\bf{3}}\end{aligned}} \right)\), \(b = \left( {\begin{aligned}{{}{}}{ - {\bf{5}}}\\{\bf{8}}\\{\bf{1}}\end{aligned}} \right)\)

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