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Chapter 5: Q5.3-20E (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 \(\left( {{\bf{ - 2,}}\;{\bf{1,}}\;{\bf{2}}} \right)\).

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

Short Answer

Expert verified

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

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 eigenvalues of the matrix

Consider the given matrix \(A = \left( {\begin{array}{*{20}{c}}4&0&0&0\\0&4&0&0\\0&0&2&0\\1&0&0&2\end{array}} \right)\). Since it is given the eigenvalues of the matrix are \(4\), \(4\), \(2\) and \(2\).

03

Find the eigenvector for \(\lambda   = 4\)

Write the matrix form for finding the eigenvector.

\(\begin{array}{c}A - 4I = \left( {\begin{array}{*{20}{c}}0&0&0&0\\0&0&0&0\\0&0&{ - 2}&0\\1&0&0&{ - 2}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}1&0&0&{ - 2}&0\\0&0&{ - 2}&0&0\\0&0&0&0&0\\0&0&0&0&0\end{array}} \right)\;\left\{ \begin{array}{l}{R_1} \leftrightarrow {R_4}\\{R_2} \leftrightarrow {R_3}\end{array} \right\}\\ = \left( {\begin{array}{*{20}{c}}1&0&0&{ - 2}&0\\0&0&1&0&0\\0&0&0&0&0\\0&0&0&0&0\end{array}} \right)\;\left\{ {{R_2} = - \frac{{{R_2}}}{2}} \right\}\end{array}\)

The general solution is,

\(\left\{ {{{\rm{v}}_1},{{\rm{v}}_2}} \right\} = \left\{ {\left( {\begin{array}{*{20}{c}}0\\1\\0\\0\end{array}} \right),\left( {\begin{array}{*{20}{c}}2\\0\\0\\1\end{array}} \right)} \right\}\).

Therefore, the eigenvectors are \(\left\{ {{{\rm{v}}_1},{{\rm{v}}_2}} \right\} = \left\{ {\left( {\begin{array}{*{20}{c}}0\\1\\0\\0\end{array}} \right),\left( {\begin{array}{*{20}{c}}2\\0\\0\\1\end{array}} \right)} \right\}\).

04

Find the Eigenvector for \(\lambda   = {\bf{2}}\)

Write the matrix form for finding the eigenvector.

\(\begin{array}{c}A - 2I = \left( {\begin{array}{*{20}{c}}2&0&0&0\\0&2&0&0\\0&0&0&0\\1&0&0&0\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}1&0&0&0&0\\0&1&0&0&0\\0&0&0&0&0\\1&0&0&0&0\end{array}} \right)\;\left\{ \begin{array}{l}{R_1} \to \frac{{{R_1}}}{2}\\{R_2} \to \frac{{{R_2}}}{2}\end{array} \right\}\\ = \left( {\begin{array}{*{20}{c}}1&0&0&0&0\\0&1&0&0&0\\0&0&0&0&0\\0&0&0&0&0\end{array}} \right)\;\left\{ {{R_4} = {R_4} - {R_1}} \right\}\end{array}\)

The general solution is,

\(\left( {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\\1\\0\end{array}} \right){x_3} + \left( {\begin{array}{*{20}{c}}0\\0\\0\\1\end{array}} \right){x_4}\).

Therefore, the eigenvectors are \(\left\{ {{{\rm{v}}_3},{{\rm{v}}_4}} \right\} = \left\{ {\left( {\begin{array}{*{20}{c}}0\\0\\1\\0\end{array}} \right),\left( {\begin{array}{*{20}{c}}0\\0\\0\\1\end{array}} \right)} \right\}\).

05

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

The matrix\(P\)is formed by eigenvectors

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

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

\(D = \left( {\begin{array}{*{20}{c}}4&0&0&0\\0&4&0&0\\0&0&2&0\\0&0&0&2\end{array}} \right)\)

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

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Most popular questions from this chapter

Let\(B = \left\{ {{{\bf{b}}_{\bf{1}}},{{\bf{b}}_{\bf{2}}},{{\bf{b}}_{\bf{3}}}} \right\}\) and \(D = \left\{ {{{\bf{d}}_{\bf{1}}},{{\bf{d}}_{\bf{2}}}} \right\}\) be bases for vector space \(V\) and \(W\), respectively. Let \(T:V \to W\) be a linear transformation with the property that

\(T\left( {{{\bf{b}}_1}} \right) = 3{{\bf{d}}_1} - 5{{\bf{d}}_2}\), \(T\left( {{{\bf{b}}_2}} \right) = - {{\bf{d}}_1} + 6{{\bf{d}}_2}\), \(T\left( {{{\bf{b}}_3}} \right) = 4{{\bf{d}}_2}\)

Find the matrix for \(T\) relative to \(B\), and\(D\).

Question: Let \(A = \left( {\begin{array}{*{20}{c}}{.6}&{.3}\\{.4}&{.7}\end{array}} \right)\), \({v_1} = \left( {\begin{array}{*{20}{c}}{3/7}\\{4/7}\end{array}} \right)\), \({x_0} = \left( {\begin{array}{*{20}{c}}{.5}\\{.5}\end{array}} \right)\). (Note: \(A\) is the stochastic matrix studied in Example 5 of Section 4.9.)

  1. Find a basic for \({\mathbb{R}^2}\) consisting of \({{\rm{v}}_1}\) and anther eigenvector \({{\rm{v}}_2}\) of \(A\).
  2. Verify that \({{\rm{x}}_0}\) may be written in the form \({{\rm{x}}_0} = {{\rm{v}}_1} + c{{\rm{v}}_2}\).
  3. For \(k = 1,2, \ldots \), define \({x_k} = {A^k}{x_0}\). Compute \({x_1}\) and \({x_2}\), and write a formula for \({x_k}\). Then show that \({{\bf{x}}_k} \to {{\bf{v}}_1}\) as \(k\) increases.

Question: In Exercises \({\bf{3}}\) and \({\bf{4}}\), use the factorization \(A = PD{P^{ - {\bf{1}}}}\) to compute \({A^k}\) where \(k\) represents an arbitrary positive integer.

3. \(\left( {\begin{array}{*{20}{c}}a&0\\{3\left( {a - b} \right)}&b\end{array}} \right) = \left( {\begin{array}{*{20}{c}}1&0\\3&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}a&0\\0&b\end{array}} \right)\left( {\begin{array}{*{20}{c}}1&0\\{ - 3}&1\end{array}} \right)\)

Show that if \({\bf{x}}\) is an eigenvector of the matrix product \(AB\) and \(B{\rm{x}} \ne 0\), then \(B{\rm{x}}\) is an eigenvector of\(BA\).

Let\(G = \left( {\begin{aligned}{*{20}{c}}A&X\\{\bf{0}}&B\end{aligned}} \right)\). Use formula\(\left( {\bf{1}} \right)\)for the determinant in section\({\bf{5}}{\bf{.2}}\)to explain why\(\det G = \left( {\det A} \right)\left( {\det B} \right)\). From this, deduce that the characteristic polynomial of\(G\)is the product of the characteristic polynomials of\(A\)and\(B\).
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