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Chapter 5: Q5.3-14E (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 \(\left( {{\bf{ - 2,}}\;{\bf{1,}}\;{\bf{2}}} \right)\).

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

Short Answer

Expert verified

The matrix \(A\) is diagonalizable with \(P = \left( {\begin{array}{*{20}{c}}{ - 2}&0&{ - \frac{1}{2}}\\0&1&1\\1&0&0\end{array}} \right)\) and \(D = \left( {\begin{array}{*{20}{c}}5&0&0\\0&5&0\\0&0&4\end{array}} \right)\).

Step by step solution

01

Write the Diagonalization Theorem

The Diagonalization Theorem:An \(n \times n\) matrix \(A\) is diagonalizable if and only if 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 eigenvalues of the matrix

Consider the given matrix \(A = \left( {\begin{array}{*{20}{c}}4&0&{ - 2}\\2&5&4\\0&0&5\end{array}} \right)\). Since from the given matrix eigenvalues of the matrix are \(5,5\) and \(4\). Since the eigenvalues are distinct, the matrix is diagonalizable.\(A\)

03

Find the eigenvalues and eigenvectors

As the sum of all the eigenvalues of is equal to the sum of the diagonal entries of , we can find the third eigenvalue:

\(\begin{array}{c}4 + 5 + x = 4 + 5 + 5\\x = 5\end{array}\)

So, the eigenvalues are\(5,5,4\).

Find eigenvectors.

Write the matrix form for finding the eigenvector for\(\lambda = 5\).

\(\begin{array}{c}\left( {A - 5I} \right){\bf{x}} = 0\\\left( {\begin{array}{*{20}{c}}4&0&{ - 2}\\2&5&4\\0&0&5\end{array}} \right) - 5\left( {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&1&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right)\\\left( {\begin{array}{*{20}{c}}{ - 1}&0&{ - 2}\\2&0&4\\0&0&0\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\\0\end{array}} \right)\end{array}\)

Write the Row-reduced Augmented matrix.

\(\begin{array}{c}M = \left( {\begin{array}{*{20}{c}}{ - 1}&0&{ - 2}&0\\2&0&4&0\\0&0&0&0\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}1&0&2&0\\0&0&0&0\\0&0&0&0\end{array}} \right)\;\left\{ \begin{array}{l}{R_1} = - {R_1}\\{R_2} = {R_2} - 2{R_1}\end{array} \right\}\end{array}\)

Therefore, the parametric form of the solution is shown below:

\(\begin{array}{c}\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{ - 2{x_3}}\\{{x_2}}\\{{x_3}}\end{array}} \right)\\ = {x_2}\left( {\begin{array}{*{20}{c}}0\\1\\0\end{array}} \right) + {x_3}\left( {\begin{array}{*{20}{c}}{ - 2}\\0\\1\end{array}} \right)\end{array}\)

Therefore, the eigenvector for \(\lambda = 5\)are\(\left\{ {{{\rm{v}}_1},{{\rm{v}}_2}} \right\} = \left\{ {\left( {\begin{array}{*{20}{c}}\begin{array}{l} - 2\\0\end{array}\\1\end{array}} \right),\left( {\begin{array}{*{20}{c}}\begin{array}{l}0\\1\end{array}\\0\end{array}} \right)} \right\}\).

Similarly, for \(\lambda = 4\) eigenvector is \({{\rm{v}}_3} = \left( {\begin{array}{*{20}{c}}{ - \frac{1}{2}}\\1\\0\end{array}} \right)\).

04

Find the matrix \(P\) and \(D\)

The matrix\(P\)is formed by eigenvectors

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

The matrix\(D\)is formed by eigenvalues.

\(D = \left( {\begin{array}{*{20}{c}}5&0&0\\0&5&0\\0&0&4\end{array}} \right)\)

05

Find the diagonalizes form of a matrix \(A\)

As the diagonal form of the matrix\(A\)is\(A = PD{P^{ - 1}}\).

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

Thus, the matrix \(A\) is diagonalizable with \(P = \left( {\begin{array}{*{20}{c}}{ - 2}&0&{ - \frac{1}{2}}\\0&1&1\\1&0&0\end{array}} \right)\) and \(D = \left( {\begin{array}{*{20}{c}}5&0&0\\0&5&0\\0&0&4\end{array}} \right)\).

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Most popular questions from this chapter

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

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

Use mathematical induction to show that if \(\lambda \) is an eigenvalue of an \(n \times n\) matrix \(A\), with a corresponding eigenvector, then, for each positive integer \(m\), \({\lambda ^m}\)is an eigenvalue of \({A^m}\), with \({\rm{x}}\) a corresponding eigenvector.

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 \(x\left( 0 \right) = \left( {\begin{aligned}{{20}{c}}{ - 6}\\1\end{aligned}} \right)\).

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.

Use Exercise 12 to find the eigenvalues of the matrices in Exercises 13 and 14.

13. \(A = \left( {\begin{array}{*{20}{c}}3&{ - 2}&8\\0&5&{ - 2}\\0&{ - 4}&3\end{array}} \right)\)
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