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Chapter 5: Q5.3-5E (page 267)

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

Short Answer

Expert verified

The basis for eigenvalue \(\lambda = 1\)is \(\left( {\begin{array}{*{20}{c}}1\\0\\{ - 1}\end{array}} \right)\)and \(\left( {\begin{array}{*{20}{c}}2\\{ - 1}\\0\end{array}} \right)\)and the basis for eigenvalue \(\lambda = 5\)is \(\left( {\begin{array}{*{20}{c}}1\\1\\1\end{array}} \right)\).

Step by step solution

01

Write the Diagonalization Theorem

The Diagonalization Theorem:An \(n \times n\) matrix \(A\) is diagonalizable if and only if \(A\) has \(n\) linearly independent eigenvectors. As \(A = PD{P^{ - 1}}\) which has \(D\) a diagonal matrix if and only if the columns of \(P\) are \(n\) linearly independent eigenvectors of \(A\).

02

Find the inverse of the invertible matrix

Write the given diagonalization matrix of \(A\) as:

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

Compare diagonalization matrix of \(A\) with \(A = PD{P^{ - 1}}\).

\[\begin{array}{c}P = \left( {\begin{array}{*{20}{c}}1&1&2\\1&0&{ - 1}\\1&{ - 1}&0\end{array}} \right)\\D = \left( {\begin{array}{*{20}{c}}5&0&0\\0&1&0\\0&0&1\end{array}} \right)\\{P^{ - 1}} = \left( {\begin{array}{*{20}{c}}{\frac{1}{4}}&{\frac{1}{2}}&{\frac{1}{4}}\\{\frac{1}{4}}&{\frac{1}{2}}&{ - \frac{3}{4}}\\{\frac{1}{4}}&{ - \frac{1}{2}}&{\frac{1}{4}}\end{array}} \right)\end{array}\]

03

Find Eigenvalues and Eigenvectors

According to the diagonal entries of the matrix, \(D\) there are three eigenvalues.

\({\lambda _1} = 5\), \({\lambda _2} = 1\)and\({\lambda _3} = 1\)

The three eigenvectors are columns of the matrix \(P\).

\[{{\bf{v}}_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}1\\1\\1\end{array}} \right), {{\bf{v}}_2} = \left( {\begin{array}{*{20}{c}}1\\0\\{ - 1}\end{array}} \right), {{\bf{v}}_3} = \left( {\begin{array}{*{20}{c}}2\\{ - 1}\\0\end{array}} \right)\]

Thus, the basis for eigenvalue \(\lambda = 1\)is \(\left( {\begin{array}{*{20}{c}}1\\0\\{ - 1}\end{array}} \right)\)and \(\left( {\begin{array}{*{20}{c}}2\\{ - 1}\\0\end{array}} \right)\)and the basis for eigenvalue \(\lambda = 5\)is \(\left( {\begin{array}{*{20}{c}}1\\1\\1\end{array}} \right)\).

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Most popular questions from this chapter

Question: Let \(A = \left( {\begin{array}{*{20}{c}}{.5}&{.2}&{.3}\\{.3}&{.8}&{.3}\\{.2}&0&{.4}\end{array}} \right)\), \({{\rm{v}}_1} = \left( {\begin{array}{*{20}{c}}{.3}\\{.6}\\{.1}\end{array}} \right)\), \({{\rm{v}}_2} = \left( {\begin{array}{*{20}{c}}1\\{ - 3}\\2\end{array}} \right)\), \({{\rm{v}}_3} = \left( {\begin{array}{*{20}{c}}{ - 1}\\0\\1\end{array}} \right)\) and \({\rm{w}} = \left( {\begin{array}{*{20}{c}}1\\1\\1\end{array}} \right)\).

  1. Show that \({{\rm{v}}_1}\), \({{\rm{v}}_2}\), and \({{\rm{v}}_3}\) are eigenvectors of \(A\). (Note: \(A\) is the stochastic matrix studied in Example 3 of Section 4.9.)
  2. Let \({{\rm{x}}_0}\) be any vector in \({\mathbb{R}^3}\) with non-negative entries whose sum is 1. (In section 4.9, \({{\rm{x}}_0}\) was called a probability vector.) Explain why there are constants \({c_1}\), \({c_2}\), and \({c_3}\) such that \({{\rm{x}}_0} = {c_1}{{\rm{v}}_1} + {c_2}{{\rm{v}}_2} + {c_3}{{\rm{v}}_3}\). Compute \({{\rm{w}}^T}{{\rm{x}}_0}\), and deduce that \({c_1} = 1\).
  3. For \(k = 1,2, \ldots ,\) define \({{\rm{x}}_k} = {A^k}{{\rm{x}}_0}\), with \({{\rm{x}}_0}\) as in part (b). Show that \({{\rm{x}}_k} \to {{\rm{v}}_1}\) as \(k\) increases.

Let \(J\) be the \(n \times n\) matrix of all \({\bf{1}}\)鈥檚 and consider \(A = \left( {a - b} \right)I + bJ\) that is,

\(A = \left( {\begin{aligned}{*{20}{c}}a&b&b&{...}&b\\b&a&b&{...}&b\\b&b&a&{...}&b\\:&:&:&:&:\\b&b&b&{...}&a\end{aligned}} \right)\)

Use the results of Exercise \({\bf{16}}\) in the Supplementary Exercises for Chapter \({\bf{3}}\) to show that the eigenvalues of \(A\) are \(a - b\) and \(a + \left( {n - {\bf{1}}} \right)b\). What are the multiplicities of these eigenvalues?

Question: Exercises 9-14 require techniques section 3.1. Find the characteristic polynomial of each matrix, using either a cofactor expansion or the special formula for \(3 \times 3\) determinants described prior to Exercise 15-18 in Section 3.1. [Note: Finding the characteristic polynomial of a \(3 \times 3\) matrix is not easy to do with just row operations, because the variable \(\lambda \) is involved.

13. \(\left[ {\begin{array}{*{20}{c}}6&- 2&0\\- 2&9&0\\5&8&3\end{array}} \right]\)

Question: In Exercises \({\bf{1}}\) and \({\bf{2}}\), let \(A = PD{P^{ - {\bf{1}}}}\) and compute \({A^{\bf{4}}}\).

1. \(P{\bf{ = }}\left( {\begin{array}{*{20}{c}}{\bf{5}}&{\bf{7}}\\{\bf{2}}&{\bf{3}}\end{array}} \right)\), \(D{\bf{ = }}\left( {\begin{array}{*{20}{c}}{\bf{2}}&{\bf{0}}\\{\bf{0}}&{\bf{1}}\end{array}} \right)\)

A particle moving in a planar force field has a position vector .\(x\). that satisfies \(x' = Ax\). The \(2 \times 2\) matrix \(A\) has eigenvalues 4 and 2, with corresponding eigenvectors \({v_1} = \left( {\begin{aligned}{{20}{c}}{ - 3}\\1\end{aligned}} \right)\) and \({v_2} = \left( {\begin{aligned}{{20}{c}}{ - 1}\\1\end{aligned}} \right)\). Find the position of the particle at a time \(t\), assuming that \(x\left( 0 \right) = \left( {\begin{aligned}{{20}{c}}{ - 6}\\1\end{aligned}} \right)\).
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