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Chapter 5: Q5.3-25E (page 267)

Question: A is a \({\bf{4}} \times {\bf{4}}\) matrix with three eigenvalues. One eigenspace is one-dimensional and one of the other eigenspaces is two-dimensional. Is it possible that A is not diagonalizable? Justify your answer.

Short Answer

Expert verified

According to diagonalize theorem, matrix A of order \(4 \times 4\) is diagonalizable.

Step by step solution

01

Write the given information

Matrix A is of the order \(4 \times 4\) that has 3 eigenvalues and one eigenspace is one-dimensional and one of the other eigenspaces is two-dimensional.

02

Find that A is diagonalizable

Consider \(\left\{ {{{\bf{v}}_1}} \right\}\) is the basis of one-dimensional space, \({{\bf{v}}_2}\) and \({{\bf{v}}_3}\) are the basis of two-dimensional space, and let \({{\bf{v}}_4}\) be an eigenvector in the remaining eigenspace.

As per theorem 7, all the eigenvectors are linearly independent.

Therefore, according to diagonalize theorem, matrix A of order \(4 \times 4\) is diagonalizable.

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Most popular questions from this chapter

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?

In Exercises 9鈥�16, find a basis for the eigenspace corresponding to each listed eigenvalue.

10. \(A = \left( {\begin{array}{*{20}{c}}{10}&{ - 9}\\4&{ - 2}\end{array}} \right)\), \(\lambda = 4\)

Suppose \(A\) is diagonalizable and \(p\left( t \right)\) is the characteristic polynomial of \(A\). Define \(p\left( A \right)\) as in Exercise 5, and show that \(p\left( A \right)\) is the zero matrix. This fact, which is also true for any square matrix, is called the Cayley-Hamilton theorem.

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

Question: Is \(\left( {\begin{array}{*{20}{c}}1\\{ - 2}\\1\end{array}} \right)\) an eigenvector of\(\left){\begin{array}{*{20}{c}}3&6&7\\3&3&7\\5&6&5\end{array}} \right)\)? If so, find the eigenvalue.
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