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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: In Exercises \({\bf{5}}\) and \({\bf{6}}\), the matrix \(A\) is factored in the form \(PD{P^{ - {\bf{1}}}}\). Use the Diagonalization Theorem to find the eigenvalues of \(A\) and a basis for each eigenspace.

5. \(\left( {\begin{array}{*{20}{c}}2&2&1\\1&3&1\\1&2&2\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&1&2\\1&0&{ - 1}\\1&{ - 1}&0\end{array}} \right)\left( {\begin{array}{*{20}{c}}5&0&0\\0&1&0\\0&0&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}{\frac{1}{4}}&{\frac{1}{2}}&{\frac{1}{4}}\\{\frac{1}{4}}&{\frac{1}{2}}&{ - \frac{3}{4}}\\{\frac{1}{4}}&{ - \frac{1}{2}}&{\frac{1}{4}}\end{array}} \right)\)

Question: Is \(\left( {\begin{array}{*{20}{c}}{ - 1 + \sqrt 2 }\\1\end{array}} \right)\) an eigenvalue of \(\left( {\begin{array}{*{20}{c}}2&1\\1&4\end{array}} \right)\)? If so, find the eigenvalue.

If \(p\left( t \right) = {c_0} + {c_1}t + {c_2}{t^2} + ...... + {c_n}{t^n}\), define \(p\left( A \right)\) to be the matrix formed by replacing each power of \(t\) in \(p\left( t \right)\)by the corresponding power of \(A\) (with \({A^0} = I\) ). That is,

\(p\left( t \right) = {c_0} + {c_1}I + {c_2}{I^2} + ...... + {c_n}{I^n}\)

Show that if \(\lambda \) is an eigenvalue of A, then one eigenvalue of \(p\left( A \right)\) is\(p\left( \lambda \right)\).

Let \(A{\bf{ = }}\left( {\begin{aligned}{*{20}{c}}{{a_{{\bf{11}}}}}&{{a_{{\bf{12}}}}}\\{{a_{{\bf{21}}}}}&{{a_{{\bf{22}}}}}\end{aligned}} \right)\). Recall from Exercise \({\bf{25}}\) in Section \({\bf{5}}{\bf{.4}}\) that \({\rm{tr}}\;A\) (the trace of \(A\)) is the sum of the diagonal entries in \(A\). Show that the characteristic polynomial of \(A\) is \({\lambda ^2} - \left( {{\rm{tr}}A} \right)\lambda + \det A\). Then show that the eigenvalues of a \({\bf{2 \times 2}}\) matrix \(A\) are both real if and only if \(\det A \le {\left( {\frac{{{\rm{tr}}A}}{2}} \right)^2}\).

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

8. \(\left[ {\begin{array}{*{20}{c}}7&- 2\\2&3\end{array}} \right]\)
See all solutions

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