Another estimate can be made for an eigenvalue when an approximate eigenvector is available. Observe that if \(A{\bf{x}} = \lambda {\bf{x}}\), then \({{\bf{x}}^T}A{\bf{x}} = {{\bf{x}}^T}\left( {\lambda {\bf{x}}} \right) = \lambda \left( {{{\bf{x}}^T}{\bf{x}}} \right)\) and the Rayleigh quotient

\(R\left( {\bf{x}} \right){\bf{ = }}\frac{{{{\bf{x}}^T}A{\bf{x}}}}{{{{\bf{x}}^T}{\bf{x}}}}\)

Equals \(\lambda \).If \({\bf{x}}\) is close to an eigenvector for \(\lambda \), then this quotient is close to \(\lambda \). When \(A\) is a symmetric matrix \({A^T} = A\) the Rayleigh quotient \(R\left( {{{\bf{x}}_k}} \right) = \frac{{{\bf{x}}_k^TA{{\bf{x}}_k}}}{{{\bf{x}}_k^T{{\bf{x}}_k}}}\)will have roughly twice as many digits of accuracy as the scaling factor \({\mu _k}\) in the power method. Verify this increased accuracy in Exercises 11 and 12 by computing \({\mu _k}\) and \(R\left( {{{\bf{x}}_k}} \right)\) for \(k{\bf{ = 1,}}....{\bf{,4}}\).

12. \(A = \left( {\begin{aligned}{ {20}{c}}{ - 3}&2\\2&0\end{aligned}} \right)\), \({{\bf{x}}_{\bf{0}}}{\bf{ = }}\left( {\begin{aligned}{ {20}{c}}{\bf{1}}\\{\bf{0}}\end{aligned}} \right)\)

Step by step solution

01

Write the function to compute the power matrices

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

\(x = {x_0}\);

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

Estimate the data

Enter the Matrix\(A\)in MATLAB:

\( > > \;A = \left( { - 3\;2;\;2\;0} \right)\)

Enter the\({{\rm{x}}_0}\)in MATLAB:

\( > > {{\rm{x}}_0} = \left( {1\;\;\;0} \right)';\)

Now compute the power matrices.

\( > > {\rm{powermat(}}A,{x_0},5)\)

Construct the data in the table shown below:

\(k\)	\(0\)	\(0\)	\(0\)	\(0\)	\(0\)
\({{\bf{x}}_k}\)	\(\left( {\begin{aligned}{ {20}{c}}1\\0\end{aligned}} \right)\)	\(\left( {\begin{aligned}{ {20}{c}}{ - 1}\\{.6667}\end{aligned}} \right)\)	\(\left( {\begin{aligned}{ {20}{c}}1\\{ - .4615}\end{aligned}} \right)\)	\(\left( {\begin{aligned}{ {20}{c}}{ - 1}\\{.5098}\end{aligned}} \right)\)	\(\left( {\begin{aligned}{ {20}{c}}1\\{ - .4976}\end{aligned}} \right)\)
\(A{{\bf{x}}_k}\)	\(\left( {\begin{aligned}{ {20}{c}}{ - 3}\\2\end{aligned}} \right)\)	\(\left( {\begin{aligned}{ {20}{c}}{4.3333}\\{ - 2.0000}\end{aligned}} \right)\)	\(\left( {\begin{aligned}{ {20}{c}}{ - 3.9231}\\{2.0000}\end{aligned}} \right)\)	\(\left( {\begin{aligned}{ {20}{c}}{4.0196}\\{ - 2.0000}\end{aligned}} \right)\)	\(\left( {\begin{aligned}{ {20}{c}}{ - 3.9951}\\{2.0000}\end{aligned}} \right)\)
\({\mu _k}\)	\( - 3\)	\( - 4.3333\)	\( - 3.9231\)	\( - 4.0196\)	\( - 3.9951\)
\(R\left( {{{\bf{x}}_k}} \right)\)	\( - 3\)	\( - 3.9231\)	\( - 3.9951\)	\( - 3.9997\)	\( - 3.99998\)

As the actual eigenvalue is \( - 4\) and the values of \({\mu _k}\) and \(R\left( {{{\bf{x}}_k}} \right)\) in the table are estimates the eigenvalue more accurately than \({\mu _k}\).

Hence Proved.

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