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Chapter 5: Q19E (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.

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

Short Answer

Expert verified
  1. The largest eigenvalue is\[30.288685\].

An estimate for the eigenvalue to four decimal places is \[0.01015\] and the corresponding eigenvector is \[\left[ {\begin{array}{*{20}{c}}{ - 0.603972}\\1\\{ - 0.251135}\\{0.148953}\end{array}} \right]\].

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:

\[ > > \;A = \left[ {10\;7\;8\;7;\;7\;5\;6\;5;\;8\;6\;10\;9;\;7\;5\;9\;10} \right]\]

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:

\[k\]

\[0\]

\[1\]

\[2\]

\[3\]

\[4\]

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

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

\[\left[ {\begin{array}{*{20}{c}}1\\{.7}\\{.8}\\{.7}\end{array}} \right]\]

\[\left[ {\begin{array}{*{20}{c}}{.988679}\\{.709434}\\1\\{.932075}\end{array}} \right]\]

\[\left[ {\begin{array}{*{20}{c}}{.961467}\\{.691491}\\1\\{.942201}\end{array}} \right]\]

\[\left[ {\begin{array}{*{20}{c}}{.958115}\\{.689261}\\1\\{.943578}\end{array}} \right]\]

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

\[\left[ {\begin{array}{*{20}{c}}{10}\\7\\8\\7\end{array}} \right]\]

\[\left[ {\begin{array}{*{20}{c}}{26.2}\\{18.8}\\{26.5}\\{24.7}\end{array}} \right]\]

\[\left[ {\begin{array}{*{20}{c}}{29.0060}\\{20.8675}\\{30.2892}\\{28.5863}\end{array}} \right]\]

\[\left[ {\begin{array}{*{20}{c}}{29.0060}\\{20.8675}\\{30.2892}\\{28.7887}\end{array}} \right]\]

\[\left[ {\begin{array}{*{20}{c}}{29.0505}\\{20.8987}\\{30.3205}\\{28.6097}\end{array}} \right]\]

\[{\mu _k}\]

\[10\]

\[26.5\]

\[30.5547\]

\[30.3205\]

\[30.2927\]

\[k\]

\[5\]

\[6\]

\[7\]

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

\[\left[ {\begin{array}{*{20}{c}}{.957691}\\{.688978}\\1\\{.9433755}\end{array}} \right]\]

\[\left[ {\begin{array}{*{20}{c}}{.957637}\\{.688942}\\1\\{.943778}\end{array}} \right]\]

\[\left[ {\begin{array}{*{20}{c}}{.957630}\\{.688938}\\1\\{.943781}\end{array}} \right]\]

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

\[\left[ {\begin{array}{*{20}{c}}{29.0060}\\{20.8675}\\{30.2892}\\{28.5863}\end{array}} \right]\]

\[\left[ {\begin{array}{*{20}{c}}{29.0054}\\{20.8671}\\{30.2887}\\{28.5859}\end{array}} \right]\]

\[\left[ {\begin{array}{*{20}{c}}{29.0053}\\{20.8670}\\{30.2887}\\{28.5859}\end{array}} \right]\]

\[{\mu _k}\]

\[30.2892\]

\[30.2887\]

\[30.2887\]

The value of \[{\mu _6} = 30.2887 = {\mu _1}\] , the largest eigenvalue is \[30.288685\].

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:

\[ > > \;A = \left[ {10\;7\;8\;7;\;7\;5\;6\;5;\;8\;6\;10\;9;\;7\;5\;9\;10} \right]\]

Now find the Eigenvalue.

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

Construct the data in the table shown below:

\[k\]

\[0\]

\[1\]

\[2\]

\[3\]

\[4\]

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

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

\[\left[ {\begin{array}{*{20}{c}}{ - .609756}\\{.7}\\{ - .243902}\\{.146341}\end{array}} \right]\]

\[\left[ {\begin{array}{*{20}{c}}{ - .604007}\\1\\{ - .251051}\\{.148899}\end{array}} \right]\]

\[\left[ {\begin{array}{*{20}{c}}{ - .603973}\\1\\{ - .251134}\\{.148953}\end{array}} \right]\]

\[\left[ {\begin{array}{*{20}{c}}{ - .603972}\\1\\{ - .251135}\\{.148953}\end{array}} \right]\]

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

\[\left[ {\begin{array}{*{20}{c}}{25}\\{ - 41}\\{10}\\{ - 6}\end{array}} \right]\]

\[\left[ {\begin{array}{*{20}{c}}{ - 59.5610}\\{98.6098}\\{ - 24.7561}\\{14.6829}\end{array}} \right]\]

\[\left[ {\begin{array}{*{20}{c}}{ - 59.5041}\\{98.5211}\\{ - 24.7420}\\{14.6750}\end{array}} \right]\]

\[\left[ {\begin{array}{*{20}{c}}{ - 59.5044}\\{98.5217}\\{ - 24.7423}\\{14.6751}\end{array}} \right]\]

\[\left[ {\begin{array}{*{20}{c}}{ - 59.5044}\\{98.5217}\\{ - 24.7423}\\{14.6751}\end{array}} \right]\]

\[{\mu _k}\]

\[ - 41\]

\[98.6098\]

\[98.5211\]

\[98.5217\]

\[98.5217\]

\[{v_k}\]

\[ - .0243902\]

\[ - .0101410\]

\[ - .0101501\]

\[ - .0101500\]

\[ - .0101500\]

Thus, an estimate for the eigenvalue to five decimal places is\[0.01015\]and the corresponding eigenvector is\[\left[ {\begin{array}{*{20}{c}}{ - 0.603972}\\1\\{ - 0.251135}\\{0.148953}\end{array}} \right]\].

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Most popular questions from this chapter

[M]Repeat Exercise 25 for \[A{\bf{ = }}\left[ {\begin{array}{*{20}{c}}{{\bf{ - 8}}}&{\bf{5}}&{{\bf{ - 2}}}&{\bf{0}}\\{{\bf{ - 5}}}&{\bf{2}}&{\bf{1}}&{{\bf{ - 2}}}\\{{\bf{10}}}&{{\bf{ - 8}}}&{\bf{6}}&{{\bf{ - 3}}}\\{\bf{3}}&{{\bf{ - 2}}}&{\bf{1}}&{\bf{0}}\end{array}} \right]\].

Define\(T:{{\rm P}_3} \to {\mathbb{R}^4}\)by\(T\left( {\bf{p}} \right) = \left( {\begin{aligned}{{\bf{p}}\left( { - 3} \right)}\\{{\bf{p}}\left( { - 1} \right)}\\{{\bf{p}}\left( 1 \right)}\\{{\bf{p}}\left( 3 \right)}\end{aligned}} \right)\).

  1. Show that \(T\) is a linear transformation.
  2. Find the matrix for \(T\) relative to the basis \(\left\{ {1,t,{t^2},{t^3}} \right\}\)for \({{\rm P}_3}\)and the standard basis for \({\mathbb{R}^4}\).

Consider the growth of a lilac bush. The state of this lilac bush for several years (at year鈥檚 end) is shown in the accompanying sketch. Let n(t) be the number of new branches (grown in the year t) and a(t) the number of old branches. In the sketch, the new branches are represented by shorter lines. Each old branch will grow two new branches in the following year. We assume that no branches ever die.

(a) Find the matrix A such that [nt+1at+1]=A[ntat]

(b) Verify that [11]and [2-1] are eigenvectors of A. Find the associated eigenvalues.

(c) Find closed formulas for n(t) and a(t).

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 matrices A in Exercises 1 through 12, find closed formulas for , where t is an arbitrary positive integer. Follow the strategy outlined in Theorem 7.4.2 and illustrated in Example 2. In Exercises 9 though 12, feel free to use technology.

1.A=1203
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