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Chapter 5: Q5.3-7E (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 \(\left( {{\bf{ - 2,}}\;{\bf{1,}}\;{\bf{2}}} \right)\).

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

Short Answer

Expert verified

As the matrix product \(AP\) and \(PD\) are same, that is, the matrix \(A\) is 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 \(A = \left( {\begin{array}{*{20}{c}}1&0\\6&{ - 1}\end{array}} \right)\). 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\({\lambda _1} = 1\)and\({\lambda _2} = - 1\).

Thus, the diagonal matrix is \(D = \left( {\begin{array}{*{20}{c}}1&0\\0&{ - 1}\end{array}} \right)\).

03

Find Eigenvectors

Eigenvectors for \({\lambda _1} = 1\),

\[\begin{array}{c}A{\bf{x}} = \lambda {\bf{x}}\\A{\bf{x}} = {\bf{x}}\\\left( {A - I} \right){\bf{x}} = 0\\\left( {\left( {\begin{array}{*{20}{c}}1&0\\6&{ - 1}\end{array}} \right) - \left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\end{array}} \right)\\\left( {\begin{array}{*{20}{c}}0&0\\6&{ - 2}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\end{array}} \right)\end{array}\]

Assume \({x_2} = t\)then we get:

\[\begin{array}{l}\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}t\\{3t}\end{array}} \right)\\\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1\\3\end{array}} \right)t\end{array}\]

Eigenvectors for \({\lambda _1} = - 1\),

\[\begin{array}{c}A{\bf{x}} = \lambda {\bf{x}}\\A{\bf{x}} = {\bf{x}}\\\left( {A - I} \right){\bf{x}} = 0\\\left( {\left( {\begin{array}{*{20}{c}}1&0\\6&{ - 1}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\end{array}} \right)\\\left( {\begin{array}{*{20}{c}}2&0\\6&0\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\end{array}} \right)\end{array}\]

Assume \({x_2} = t\)then we get:

\[\begin{array}{l}\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\t\end{array}} \right)\\\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\1\end{array}} \right)t\end{array}\]

04

Find the matrix \(P\)

\[\begin{array}{c}P = \left\{ {{v_1},{v_2}} \right\}\\ = \left( {\begin{array}{*{20}{c}}1&0\\3&1\end{array}} \right)\end{array}\]

05

Find the matrix \({P^{ - {\bf{1}}}}\)

\[\begin{array}{l}{P^{ - 1}} = \frac{1}{{1\left( 1 \right) - 0}}\left( {\begin{array}{*{20}{c}}1&0\\{ - 3}&3\end{array}} \right)\\{P^{ - 1}} = \left( {\begin{array}{*{20}{c}}1&0\\{ - 3}&3\end{array}} \right)\end{array}\]

06

Find the matrix product \(AP\)

\[\begin{array}{c}AP = \left( {\begin{array}{*{20}{c}}1&0\\6&{ - 1}\end{array}} \right)\left( {\begin{array}{*{20}{c}}1&0\\3&1\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}1&0\\3&{ - 1}\end{array}} \right)\end{array}\]

07

Find the matrix product \(PD\)

\[\begin{array}{c}PD = \left( {\begin{array}{*{20}{c}}1&0\\3&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}1&0\\0&{ - 1}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}1&0\\3&{ - 1}\end{array}} \right)\end{array}\]

Thus, as matrix product \(AP\) and \(PD\) are same, that is, the matrix \(A\) is diagonalizable.

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Most popular questions from this chapter

Question: Let \(A = \left( {\begin{array}{*{20}{c}}{.5}&{.2}&{.3}\\{.3}&{.8}&{.3}\\{.2}&0&{.4}\end{array}} \right)\), \({{\rm{v}}_1} = \left( {\begin{array}{*{20}{c}}{.3}\\{.6}\\{.1}\end{array}} \right)\), \({{\rm{v}}_2} = \left( {\begin{array}{*{20}{c}}1\\{ - 3}\\2\end{array}} \right)\), \({{\rm{v}}_3} = \left( {\begin{array}{*{20}{c}}{ - 1}\\0\\1\end{array}} \right)\) and \({\rm{w}} = \left( {\begin{array}{*{20}{c}}1\\1\\1\end{array}} \right)\).

  1. Show that \({{\rm{v}}_1}\), \({{\rm{v}}_2}\), and \({{\rm{v}}_3}\) are eigenvectors of \(A\). (Note: \(A\) is the stochastic matrix studied in Example 3 of Section 4.9.)
  2. Let \({{\rm{x}}_0}\) be any vector in \({\mathbb{R}^3}\) with non-negative entries whose sum is 1. (In section 4.9, \({{\rm{x}}_0}\) was called a probability vector.) Explain why there are constants \({c_1}\), \({c_2}\), and \({c_3}\) such that \({{\rm{x}}_0} = {c_1}{{\rm{v}}_1} + {c_2}{{\rm{v}}_2} + {c_3}{{\rm{v}}_3}\). Compute \({{\rm{w}}^T}{{\rm{x}}_0}\), and deduce that \({c_1} = 1\).
  3. For \(k = 1,2, \ldots ,\) define \({{\rm{x}}_k} = {A^k}{{\rm{x}}_0}\), with \({{\rm{x}}_0}\) as in part (b). Show that \({{\rm{x}}_k} \to {{\rm{v}}_1}\) as \(k\) increases.

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


For the matrix A,find real closed formulas for the trajectory x鈫�(t+1)=Ax炉(t)wherex鈫�=[01]. Draw a rough sketchA=[-0.51.5-0.61.3]

Show that if \(A\) is diagonalizable, with all eigenvalues less than 1 in magnitude, then \({A^k}\) tends to the zero matrix as \(k \to \infty \). (Hint: Consider \({A^k}x\) where \(x\) represents any one of the columns of \(I\).)

M] In Exercises 19 and 20, find (a) the largest eigenvalue and (b) the eigenvalue closest to zero. In each case, set \[{{\bf{x}}_{\bf{0}}}{\bf{ = }}\left( {{\bf{1,0,0,0}}} \right)\] and carry out approximations until the approximating sequence seems accurate to four decimal places. Include the approximate eigenvector.

19.\[A{\bf{=}}\left[{\begin{array}{*{20}{c}}{{\bf{10}}}&{\bf{7}}&{\bf{8}}&{\bf{7}}\\{\bf{7}}&{\bf{5}}&{\bf{6}}&{\bf{5}}\\{\bf{8}}&{\bf{6}}&{{\bf{10}}}&{\bf{9}}\\{\bf{7}}&{\bf{5}}&{\bf{9}}&{{\bf{10}}}\end{array}} \right]\]
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