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Chapter 5: Q5.3-8E (page 267)

Question: Diagonalize the matrices in Exercises \({\bf{7--20}}\), if possible. The eigenvalues for Exercises \({\bf{11--16}}\) are as follows:\(\left( {{\bf{11}}} \right)\lambda {\bf{ = 1,2,3}}\); \(\left( {{\bf{12}}} \right)\lambda {\bf{ = 2,8}}\); \(\left( {{\bf{13}}} \right)\lambda {\bf{ = 5,1}}\); \(\left( {{\bf{14}}} \right)\lambda {\bf{ = 5,4}}\); \(\left( {{\bf{15}}} \right)\lambda {\bf{ = 3,1}}\); \(\left( {{\bf{16}}} \right)\lambda {\bf{ = 2,1}}\). For exercise \({\bf{18}}\), one eigenvalue is \(\lambda {\bf{ = 5}}\) and one eigenvector is \(\left( {{\bf{ - 2,}}\;{\bf{1,}}\;{\bf{2}}} \right)\).

8. \(\left( {\begin{array}{*{20}{c}}{\bf{5}}&{\bf{1}}\\{\bf{0}}&{\bf{5}}\end{array}} \right)\)

Short Answer

Expert verified

The general solution is \({x_1}\left( {\begin{array}{*{20}{c}}1\\0\end{array}} \right)\). Since we cannot generate an eigenvector basis for \({\mathbb{R}^2}\), \(A\) is not diagonalizable.

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

Consider the given matrix \(A = \left( {\begin{array}{*{20}{c}}5&1\\0&5\end{array}} \right)\). We need to diagonalize the given matrix if possible.

As the given matrix is an upper triangular matrix. Therefore, the eigenvalues of\(A\)are its diagonal values.

The eigenvalues are \({\lambda _1} = 5\) and \({\lambda _2} = 5\).

03

Find the matrix form of Equation for finding Eigenvector for \(\lambda {\bf{ = 5}}\)

The matrix equation for finding the eigenvector:

\[\begin{array}{c}\left( {A - \lambda I} \right){\bf{x}} = 0\\\left( {\begin{array}{*{20}{c}}{5 - \lambda }&1\\0&{5 - \lambda }\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\end{array}} \right)\\\left( {\begin{array}{*{20}{c}}{5 - 5}&1\\0&{5 - 5}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\end{array}} \right)\\\left( {\begin{array}{*{20}{c}}0&1\\0&0\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\end{array}} \right)\end{array}\]

Thus, the general solution is \({x_1}\left( {\begin{array}{*{20}{c}}1\\0\end{array}} \right)\). Since we cannot generate an eigenvector basis for \({\mathbb{R}^2}\), \(A\) is not diagonalizable.

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Most popular questions from this chapter

Question: Construct a random integer-valued \(4 \times 4\) matrix \(A\).

  1. Reduce \(A\) to echelon form \(U\) with no row scaling, and use \(U\) in formula (1) (before Example 2) to compute \(\det A\). (If \(A\) happens to be singular, start over with a new random matrix.)
  2. Compute the eigenvalues of \(A\) and the product of these eigenvalues (as accurately as possible).
  3. List the matrix \(A\), and, to four decimal places, list the pivots in \(U\) and the eigenvalues of \(A\). Compute \(\det A\) with your matrix program, and compare it with the products you found in (a) and (b).

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: Diagonalize the matrices in Exercises \({\bf{7--20}}\), if possible. The eigenvalues for Exercises \({\bf{11--16}}\) are as follows:\(\left( {{\bf{11}}} \right)\lambda {\bf{ = 1,2,3}}\); \(\left( {{\bf{12}}} \right)\lambda {\bf{ = 2,8}}\); \(\left( {{\bf{13}}} \right)\lambda {\bf{ = 5,1}}\); \(\left( {{\bf{14}}} \right)\lambda {\bf{ = 5,4}}\); \(\left( {{\bf{15}}} \right)\lambda {\bf{ = 3,1}}\); \(\left( {{\bf{16}}} \right)\lambda {\bf{ = 2,1}}\). For exercise \({\bf{18}}\), one eigenvalue is \(\lambda {\bf{ = 5}}\) and one eigenvector is \(\left( {{\bf{ - 2,}}\;{\bf{1,}}\;{\bf{2}}} \right)\).

12. \(\left( {\begin{array}{*{20}{c}}{\bf{4}}&{\bf{2}}&{\bf{2}}\\{\bf{2}}&{\bf{4}}&{\bf{2}}\\{\bf{2}}&{\bf{2}}&{\bf{4}}\end{array}} \right)\)

Suppose \(A = PD{P^{ - 1}}\), where \(P\) is \(2 \times 2\) and \(D = \left( {\begin{array}{*{20}{l}}2&0\\0&7\end{array}} \right)\)

a. Let \(B = 5I - 3A + {A^2}\). Show that \(B\) is diagonalizable by finding a suitable factorization of \(B\).

b. Given \(p\left( t \right)\) and \(p\left( A \right)\) as in Exercise 5 , show that \(p\left( A \right)\) is diagonalizable.

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