Question

[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

1. The largest eigenvalue is\[19.1820368\].

2. 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

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

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

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

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

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

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\[19.1820368\].

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

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\[\left[ {\begin{array}{*{20}{c}}1\\{0.222577}\\{ - 0.917970}\\{0.660496}\end{array}} \right]\].