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Chapter 5: Q3E (page 267)

Compute the quantities in Exercises 1-8 using the vectors

\({\mathop{\rm u}\nolimits} = \left( {\begin{array}{*{20}{c}}{ - 1}\\2\end{array}} \right),{\rm{ }}{\mathop{\rm v}\nolimits} = \left( {\begin{array}{*{20}{c}}4\\6\end{array}} \right),{\rm{ }}{\mathop{\rm w}\nolimits} = \left( {\begin{array}{*{20}{c}}3\\{ - 1}\\{ - 5}\end{array}} \right),{\rm{ }}{\mathop{\rm x}\nolimits} = \left( {\begin{array}{*{20}{c}}6\\{ - 2}\\3\end{array}} \right)\)

3. \(\frac{1}{{{\mathop{\rm w}\nolimits} \cdot {\mathop{\rm w}\nolimits} }}{\mathop{\rm w}\nolimits} \)

Short Answer

Expert verified

The value is \(\frac{1}{{{\mathop{\rm w}\nolimits} \cdot {\mathop{\rm w}\nolimits} }}{\mathop{\rm w}\nolimits} = \left( {\begin{array}{*{20}{c}}{\frac{3}{{35}}}\\{ - \frac{1}{{35}}}\\{ - \frac{1}{7}}\end{array}} \right)\).

Step by step solution

01

Inner product

Consider \({\mathop{\rm u}\nolimits} ,v,\) and \({\mathop{\rm w}\nolimits} \) as the vectors in \({\mathbb{R}^n}\) and consider \(c\) as the scalar. Then

  1. \({\mathop{\rm u}\nolimits} \cdot {\mathop{\rm v}\nolimits} = {\mathop{\rm v}\nolimits} \cdot {\mathop{\rm u}\nolimits} \)
  2. \(\left( {{\mathop{\rm u}\nolimits} + {\mathop{\rm v}\nolimits} } \right) \cdot {\mathop{\rm w}\nolimits} = {\mathop{\rm u}\nolimits} \cdot {\mathop{\rm v}\nolimits} + {\mathop{\rm v}\nolimits} \cdot {\mathop{\rm w}\nolimits} \)
  3. \(\left( {c{\mathop{\rm u}\nolimits} } \right) \cdot {\mathop{\rm v}\nolimits} = c\left( {{\mathop{\rm u}\nolimits} \cdot {\mathop{\rm v}\nolimits} } \right) = {\mathop{\rm u}\nolimits} \cdot \left( {c{\mathop{\rm v}\nolimits} } \right)\)
  4. \({\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} \ge 0\)and \({\mathop{\rm u}\nolimits} \cdot {\mathop{\rm u}\nolimits} = 0\) if and only if \({\mathop{\rm u}\nolimits} = 0\).
02

Compute

\(\frac{1}{{{\mathop{\rm w}\nolimits} \cdot {\mathop{\rm w}\nolimits} }}{\mathop{\rm w}\nolimits} \)

It is given that \({\mathop{\rm w}\nolimits} = \left( {\begin{array}{*{20}{c}}3\\{ - 1}\\{ - 5}\end{array}} \right)\).

Evaluate \({\mathop{\rm w}\nolimits} \cdot {\mathop{\rm w}\nolimits} \) as shown below:

\(\begin{array}{c}{\mathop{\rm w}\nolimits} \cdot {\mathop{\rm w}\nolimits} &= \left( {\begin{array}{*{20}{c}}3\\{ - 1}\\{ - 5}\end{array}} \right)\left( {\begin{array}{*{20}{c}}3\\{ - 1}\\{ - 5}\end{array}} \right)\\ &= {3^2} + {\left( { - 1} \right)^2} + {\left( { - 5} \right)^2}\\ &= 9 + 1 + 25\\ = 35\end{array}\)

Compute \(\frac{1}{{{\mathop{\rm w}\nolimits} \cdot {\mathop{\rm w}\nolimits} }}{\mathop{\rm w}\nolimits} \) as shown below:

\(\begin{array}{c}\frac{1}{{{\mathop{\rm w}\nolimits} \cdot {\mathop{\rm w}\nolimits} }}{\mathop{\rm w}\nolimits} &= \frac{1}{{35}}\left( {\begin{array}{*{20}{c}}3\\{ - 1}\\{ - 5}\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}{\frac{3}{{35}}}\\{\frac{{ - 1}}{{35}}}\\{\frac{{ - 5}}{{35}}}\end{array}} \right)\\ &= \left( {\begin{array}{*{20}{c}}{\frac{3}{{35}}}\\{\frac{{ - 1}}{{35}}}\\{\frac{{ - 1}}{7}}\end{array}} \right)\end{array}\)

Thus, the value is \(\frac{1}{{{\mathop{\rm w}\nolimits} \cdot {\mathop{\rm w}\nolimits} }}{\mathop{\rm w}\nolimits} = \left( {\begin{array}{*{20}{c}}{\frac{3}{{35}}}\\{ - \frac{1}{{35}}}\\{ - \frac{1}{7}}\end{array}} \right)\).

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Most popular questions from this chapter

Question: Find the characteristic polynomial and the eigenvalues of the matrices in Exercises 1-8.

3. \(\left[ {\begin{array}{*{20}{c}}3&-2\\1&-1\end{array}} \right]\)

Question: Find the characteristic polynomial and the eigenvalues of the matrices in Exercises 1-8.

2. \(\left[ {\begin{array}{*{20}{c}}5&3\\3&5\end{array}} \right]\)

Show that if \(A\) is diagonalizable, with all eigenvalues less than 1 in magnitude, then \({A^k}\) tends to the zero matrix as \(k \to \infty \). (Hint: Consider \({A^k}x\) where \(x\) represents any one of the columns of \(I\).)

Question: In Exercises 21 and 22, \(A\) and \(B\) are \(n \times n\) matrices. Mark each statement True or False. Justify each answer.

  1. If \(A\) is \(3 \times 3\), with columns \({{\rm{a}}_1}\), \({{\rm{a}}_2}\), and \({{\rm{a}}_3}\), then \(\det A\) equals the volume of the parallelepiped determined by \({{\rm{a}}_1}\), \({{\rm{a}}_2}\), and \({{\rm{a}}_3}\).
  2. \(\det {A^T} = \left( { - 1} \right)\det A\).
  3. The multiplicity of a root \(r\) of the characteristic equation of \(A\) is called the algebraic multiplicity of \(r\) as an eigenvalue of \(A\).
  4. A row replacement operation on \(A\) does not change the eigenvalues.

Let\(B = \left\{ {{{\bf{b}}_{\bf{1}}},{{\bf{b}}_{\bf{2}}},{{\bf{b}}_{\bf{3}}}} \right\}\) and \(D = \left\{ {{{\bf{d}}_{\bf{1}}},{{\bf{d}}_{\bf{2}}}} \right\}\) be bases for vector space \(V\) and \(W\), respectively. Let \(T:V \to W\) be a linear transformation with the property that

\(T\left( {{{\bf{b}}_1}} \right) = 3{{\bf{d}}_1} - 5{{\bf{d}}_2}\), \(T\left( {{{\bf{b}}_2}} \right) = - {{\bf{d}}_1} + 6{{\bf{d}}_2}\), \(T\left( {{{\bf{b}}_3}} \right) = 4{{\bf{d}}_2}\)

Find the matrix for \(T\) relative to \(B\), and\(D\).
See all solutions

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