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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

Question 20: Use a property of determinants to show that \(A\) and \({A^T}\) have the same characteristic polynomial.

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

14. \(A{\bf{ = }}\left( {\begin{array}{*{20}{c}}{\bf{1}}&{\bf{5}}&{{\bf{ - 6}}}&{{\bf{ - 7}}}\\{\bf{2}}&{\bf{4}}&{\bf{5}}&{\bf{2}}\\{\bf{0}}&{\bf{0}}&{{\bf{ - 7}}}&{{\bf{ - 4}}}\\{\bf{0}}&{\bf{0}}&{\bf{3}}&{\bf{1}}\end{array}} \right)\)

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

Mark each statement as True or False. Justify each answer.

a. If \(A\) is invertible and 1 is an eigenvalue for \(A\), then \(1\) is also an eigenvalue of \({A^{ - 1}}\)

b. If \(A\) is row equivalent to the identity matrix \(I\), then \(A\) is diagonalizable.

c. If \(A\) contains a row or column of zeros, then 0 is an eigenvalue of \(A\)

d. Each eigenvalue of \(A\) is also an eigenvalue of \({A^2}\).

e. Each eigenvector of \(A\) is also an eigenvector of \({A^2}\)

f. Each eigenvector of an invertible matrix \(A\) is also an eigenvector of \({A^{ - 1}}\)

g. Eigenvalues must be nonzero scalars.

h. Eigenvectors must be nonzero vectors.

i. Two eigenvectors corresponding to the same eigenvalue are always linearly dependent.

j. Similar matrices always have exactly the same eigenvalues.

k. Similar matrices always have exactly the same eigenvectors.

I. The sum of two eigenvectors of a matrix \(A\) is also an eigenvector of \(A\).

m. The eigenvalues of an upper triangular matrix \(A\) are exactly the nonzero entries on the diagonal of \(A\).

n. The matrices \(A\) and \({A^T}\) have the same eigenvalues, counting multiplicities.

o. If a \(5 \times 5\) matrix \(A\) has fewer than 5 distinct eigenvalues, then \(A\) is not diagonalizable.

p. There exists a \(2 \times 2\) matrix that has no eigenvectors in \({A^2}\)

q. If \(A\) is diagonalizable, then the columns of \(A\) are linearly independent.

r. A nonzero vector cannot correspond to two different eigenvalues of \(A\).

s. A (square) matrix \(A\) is invertible if and only if there is a coordinate system in which the transformation \({\bf{x}} \mapsto A{\bf{x}}\) is represented by a diagonal matrix.

t. If each vector \({{\bf{e}}_j}\) in the standard basis for \({A^n}\) is an eigenvector of \(A\), then \(A\) is a diagonal matrix.

u. If \(A\) is similar to a diagonalizable matrix \(B\), then \(A\) is also diagonalizable.

v. If \(A\) and \(B\) are invertible \(n \times n\) matrices, then \(AB\)is similar to \ (BA\ )

w. An \(n \times n\) matrix with \(n\) linearly independent eigenvectors is invertible.

x. If \(A\) is an \(n \times n\) diagonalizable matrix, then each vector in \({A^n}\) can be written as a linear combination of eigenvectors of \(A\).

Let \(A = \left( {\begin{aligned}{*{20}{c}}{.4}&{ - .3}\\{.4}&{1.2}\end{aligned}} \right)\). Explain why \({A^k}\) approaches \(\left( {\begin{aligned}{*{20}{c}}{ - .5}&{ - .75}\\1&{1.5}\end{aligned}} \right)\) as \(k \to \infty \).
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