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Chapter 5: Q20E (page 267)

[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.

20. \[A{\bf{ = }}\left[ {\begin{array}{*{20}{c}}{\bf{1}}&{\bf{2}}&{\bf{3}}&{\bf{2}}\\{\bf{2}}&{{\bf{12}}}&{{\bf{13}}}&{{\bf{11}}}\\{{\bf{ - 2}}}&{\bf{3}}&{\bf{0}}&{\bf{2}}\\{\bf{4}}&{\bf{5}}&{\bf{7}}&{\bf{2}}\end{array}} \right]\]

Short Answer

Expert verified
  1. The largest eigenvalue is\[19.1820368\].
  1. An estimate for the eigenvalue to four decimal places is \[0.01220\] and the corresponding eigenvector is \[\left[ {\begin{array}{*{20}{c}}1\\{0.222577}\\{ - 0.917970}\\{0.660496}\end{array}} \right]\].

Step by step solution

01

Write the function to compute the power matrices

(a)

\[{\rm{function}}\left[ {x,{\rm{lambda}}} \right] = {\rm{powermat}}(A,{x_0},{\rm{nit)}}\]

\[x = {x_0}\];

For\[n = 1:{\rm{nit}}\]

\[{\rm{xnew}} = A*{\rm{x}}\]

\[{\rm{lambda}} = {\rm{norm(xnew,inf)/norm(x,inf);}}\]

\[{\rm{fprintf('n}} = {\rm{\% 4d}}\;{\rm{lambda}} = {\rm{\% gx}} = {\rm{\% g\% g\% g\backslash n',n,lambda,x');}}\]

\[{\rm{x}} = {\rm{xnew;}}\;{\rm{end x}} = {\rm{x/norm(x);\% normalise x fprintf('n}} = {\rm{\% 4d normalised x}} = {\rm{\% g\% g\% g\backslash n',n,x');}}\]

02

Find the middle eigenvalue

(b)

Enter the Matrix\[B\]in MATLAB:

\[ > > \;A = \left[ {1\;2\;3\;2;\;2\;12\;13\;11;\; - 2\;3\;0\;2;\;4\;5\;7\;2} \right]\]

Enter the\[{x_0}\]in MATLAB:

\[ > > {x_0} = \left[ {1\;0\;0\;0} \right]'\]

Now find the eigenvector.

\[ > > {\rm{powermat(}}A,{x_0},7)\]

Construct the data in the table shown below:

\[k\]

\[0\]

\[1\]

\[2\]

\[3\]

\[4\]

\[{{\bf{x}}_k}\]

\[\left[ {\begin{array}{*{20}{c}}1\\0\\0\\0\end{array}} \right]\]

\[\left[ {\begin{array}{*{20}{c}}{.25}\\{.5}\\{ - .5}\\1\end{array}} \right]\]

\[\left[ {\begin{array}{*{20}{c}}{.159091}\\1\\{.272727}\\{.181818}\end{array}} \right]\]

\[\left[ {\begin{array}{*{20}{c}}{.187023}\\1\\{.170483}\\\begin{array}{l}.942201\\.442748\end{array}\end{array}} \right]\]

\[\left[ {\begin{array}{*{20}{c}}{.184166}\\1\\{.180439}\\{.402197}\end{array}} \right]\]

\[A{{\bf{x}}_k}\]

\[\left[ {\begin{array}{*{20}{c}}1\\2\\{ - 2}\\4\end{array}} \right]\]

\[\left[ {\begin{array}{*{20}{c}}{1.75}\\{11}\\3\\2\end{array}} \right]\]

\[\left[ {\begin{array}{*{20}{c}}{3.34091}\\{17.8636}\\{3.04545}\\{7.90909}\end{array}} \right]\]

\[\left[ {\begin{array}{*{20}{c}}{3.58397}\\{19.4606}\\{3.51145}\\{7.82697}\end{array}} \right]\]

\[\left[ {\begin{array}{*{20}{c}}{3.52988}\\{19.1382}\\{3.43606}\\{7.80413}\end{array}} \right]\]

\[{\mu _k}\]

\[4\]

\[11\]

\[17.8636\]

\[19.4606\]

\[19.1382\]

\[k\]

\[5\]

\[6\]

\[7\]

\[{{\bf{x}}_k}\]

\[\left[ {\begin{array}{*{20}{c}}{.184441}\\1\\{.179539}\\{.407778}\end{array}} \right]\]

\[\left[ {\begin{array}{*{20}{c}}{.184414}\\1\\{.179622}\\{.407021}\end{array}} \right]\]

\[\left[ {\begin{array}{*{20}{c}}{.184417}\\1\\{.179615}\\{.407121}\end{array}} \right]\]

\[A{{\bf{x}}_k}\]

\[\left[ {\begin{array}{*{20}{c}}{3.53861}\\{19.1884}\\{3.44667}\\{7.81010}\end{array}} \right]\]

\[\left[ {\begin{array}{*{20}{c}}{3.53732}\\{19.1811}\\{3.44521}\\{7.80905}\end{array}} \right]\]

\[\left[ {\begin{array}{*{20}{c}}{3.53750}\\{19.1822}\\{3.44541}\\{7.80921}\end{array}} \right]\]

\[{\mu _k}\]

\[19.1884\]

\[19.1811\]

\[19.1822\]

The value of \[{\mu _7} = 19.1820 = {\mu _8}\], the largest eigenvalue is\[19.1820368\].

03

Write the function of MATLAB

\[{\rm{function}}\left[ {{\rm{v,lambda}}} \right]{\rm{ = IPM}}\left( {{\rm{B,tol}}} \right)\]

tic;

\[{\rm{A}} = {\rm{inv}}\left( {\rm{B}} \right){\rm{;}}\]

\[{\rm{n}} = {\rm{size}}\left( {{\rm{A}},1} \right);\]

\[{\rm{v}} = {\rm{rand}}\left( {{\rm{n}},1} \right);\]

\[{\rm{v}} = {\rm{v}}/{\rm{norm}}\left( {\rm{v}} \right);\]

\[{\rm{res}} = 1;\]

\[{\rm{while}}\left( {{\rm{rse}} > {\rm{tol}}} \right)\]

\[{\rm{W}} = {\rm{A}}*{\rm{v}};\]

\[{\rm{lambda}} = {\rm{max}}\left( {{\rm{abs}}\left( {\rm{W}} \right)} \right);\]

\[{\rm{V}} = {\rm{W}}/{\rm{lamda}};\]

\[{\rm{res}} = {\rm{norm}}\left( {{\rm{A}}*{\rm{v}} - {\rm{lambda}}*{\rm{v}}} \right);\]

toc

end

04

Find the eigenvector

Enter the Matrix\[B\]in MATLAB:

\[ > > \;A = \left[ {1\;2\;3\;2;\;2\;12\;13\;11;\; - 2\;3\;0\;2;\;4\;5\;7\;2} \right]\]

Now find the Eigenvalue.

\[{\rm{ > > IPM}}\left( {{\rm{B,tol}}} \right)\]

Construct the data in the table shown below:

\[k\]

\[0\]

\[1\]

\[2\]

\[{{\bf{x}}_k}\]

\[\left[ {\begin{array}{*{20}{c}}1\\0\\0\\0\end{array}} \right]\]

\[\left[ {\begin{array}{*{20}{c}}1\\{.226087}\\{ - .921739}\\{.660870}\end{array}} \right]\]

\[\left[ {\begin{array}{*{20}{c}}1\\{.222577}\\{ - .917970}\\{.660496}\end{array}} \right]\]

\[A{{\bf{x}}_k}\]

\[\left[ {\begin{array}{*{20}{c}}{115}\\{26}\\{ - 106}\\{76}\end{array}} \right]\]

\[\left[ {\begin{array}{*{20}{c}}{81.7304}\\{18.1913}\\{ - 75.0261}\\{53.9826}\end{array}} \right]\]

\[\left[ {\begin{array}{*{20}{c}}{81.9314}\\{18.2387}\\{ - 75.2125}\\{54.1143}\end{array}} \right]\]

\[{\mu _k}\]

\[115\]

\[81.7304\]

\[81.9314\]

\[{v_k}\]

\[.00869565\]

\[.0122353\]

\[.0122053\]

Thus, an estimate for the eigenvalue to five decimal places is\[0.01220\]and the corresponding eigenvector is\[\left[ {\begin{array}{*{20}{c}}1\\{0.222577}\\{ - 0.917970}\\{0.660496}\end{array}} \right]\].

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Most popular questions from this chapter

Let \(J\) be the \(n \times n\) matrix of all \({\bf{1}}\)’s and consider \(A = \left( {a - b} \right)I + bJ\) that is,

\(A = \left( {\begin{aligned}{*{20}{c}}a&b&b&{...}&b\\b&a&b&{...}&b\\b&b&a&{...}&b\\:&:&:&:&:\\b&b&b&{...}&a\end{aligned}} \right)\)

Use the results of Exercise \({\bf{16}}\) in the Supplementary Exercises for Chapter \({\bf{3}}\) to show that the eigenvalues of \(A\) are \(a - b\) and \(a + \left( {n - {\bf{1}}} \right)b\). What are the multiplicities of these eigenvalues?

Question: In Exercises 21 and 22, \(A\) and \(B\) are \(n \times n\) matrices. Mark each statement True or False. Justify each answer.

  1. If \(A\) is \(3 \times 3\), with columns \({{\rm{a}}_1}\), \({{\rm{a}}_2}\), and \({{\rm{a}}_3}\), then \(\det A\) equals the volume of the parallelepiped determined by \({{\rm{a}}_1}\), \({{\rm{a}}_2}\), and \({{\rm{a}}_3}\).
  2. \(\det {A^T} = \left( { - 1} \right)\det A\).
  3. The multiplicity of a root \(r\) of the characteristic equation of \(A\) is called the algebraic multiplicity of \(r\) as an eigenvalue of \(A\).
  4. A row replacement operation on \(A\) does not change the eigenvalues.

Question: Find the characteristic polynomial and the eigenvalues of the matrices in Exercises 1-8.

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

Consider an invertible n × n matrix A such that the zero state is a stable equilibrium of the dynamical system x→(t+1)=Ax→(t) What can you say about the stability of the systems

x→(t+1)=A-1x→(t)

Question: Find the characteristic polynomial and the eigenvalues of the matrices in Exercises 1-8.

3. \(\left[ {\begin{array}{*{20}{c}}3&-2\\1&-1\end{array}} \right]\)
See all solutions

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