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Chapter 5: Q5.3-6E (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.

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

Short Answer

Expert verified

The basis for eigenvalue \(\lambda = 5\)is \(\left( {\begin{array}{*{20}{c}}0\\1\\0\end{array}} \right)\)and \(\left( {\begin{array}{*{20}{c}}{ - 2}\\0\\1\end{array}} \right)\)and the basis for eigenvalue \(\lambda = 4\)is \(\left( {\begin{array}{*{20}{c}}{ - 1}\\2\\0\end{array}} \right)\).

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

Consider the given diagonalization matrix of \(A\) as \(A = \left( {\begin{array}{*{20}{c}}4&0&{ - 2}\\2&5&4\\0&0&5\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{ - 2}&0&{ - 1}\\0&1&2\\1&0&0\end{array}} \right)\left( {\begin{array}{*{20}{c}}5&0&0\\0&5&0\\0&0&4\end{array}} \right)\left( {\begin{array}{*{20}{c}}0&0&1\\2&1&4\\{ - 1}&0&{ - 2}\end{array}} \right)\).

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

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

03

Find Eigenvalues and Eigenvectors

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

\({\lambda _1} = 5\), \({\lambda _2} = 4\)and\({\lambda _3} = 5\)

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

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

Thus, the basis for eigenvalue \(\lambda = 5\) is \(\left( {\begin{array}{*{20}{c}}0\\1\\0\end{array}} \right)\) and \(\left( {\begin{array}{*{20}{c}}{ - 2}\\0\\1\end{array}} \right)\) and the basis for eigenvalue \(\lambda = 4\) is \(\left( {\begin{array}{*{20}{c}}{ - 1}\\2\\0\end{array}} \right)\).

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Most popular questions from this chapter

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 \(\left( {{\bf{ - 2,}}\;{\bf{1,}}\;{\bf{2}}} \right)\).

10. \(\left( {\begin{array}{*{20}{c}}{\bf{2}}&{\bf{3}}\\{\bf{4}}&{\bf{1}}\end{array}} \right)\)

For the matrix A, find real closed formulas for the trajectoryx鈫�(t+1)=Ax炉(t)where x鈫�=[01]. Draw a rough sketch

A=[15-27]

Compute the quantities in Exercises 1-8 using the vectors

\({\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 1}\\2\end{aligned}} \right),{\rm{ }}{\mathop{\rm v}\nolimits} = \left( {\begin{aligned}{*{20}{c}}4\\6\end{aligned}} \right),{\rm{ }}{\mathop{\rm w}\nolimits} = \left( {\begin{aligned}{*{20}{c}}3\\{ - 1}\\{ - 5}\end{aligned}} \right),{\rm{ }}{\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}6\\{ - 2}\\3\end{aligned}} \right)\)

2. \({\mathop{\rm w}\nolimits} \cdot {\mathop{\rm w}\nolimits} ,{\mathop{\rm x}\nolimits} \cdot {\mathop{\rm w}\nolimits} ,\,\,{\mathop{\rm and}\nolimits} \,\,\frac{{{\mathop{\rm x}\nolimits} \cdot {\mathop{\rm w}\nolimits} }}{{{\mathop{\rm w}\nolimits} \cdot {\mathop{\rm w}\nolimits} }}\)

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 \(k \to \infty \).

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.
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