Question

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 \(k{\bf{ = 1,}}....{\bf{,4}}\).

11. \(A{\bf{ = }}\left( {\begin{aligned}{ {20}{c}}{\bf{5}}&{\bf{2}}\\{\bf{2}}&{\bf{2}}\end{aligned}} \right)\), \({{\bf{x}}_{\bf{0}}}{\bf{ = }}\left( {\begin{aligned}{ {20}{c}}{\bf{1}}\\{\bf{0}}\end{aligned}} \right)\)

Short answer

As the actual eigenvalue is \(6\) and the values of \({\mu _k}\) and \(R\left( {{{\bf{x}}_k}} \right)\) in the table are estimates the eigenvalue more accurately than \({\mu _k}\).

Step-by-step solution

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\(n = 1:{\rm{nit}}\)

\({\rm{xnew}} = A {\rm{x}}\)

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

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

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

Estimate the data

Enter the Matrix\(A\)in MATLAB:

\( > > \;A = \left( {5\;2;\;2\;2} \right)\)

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

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

Now compute the power matrices.

\( > > {\rm{powermat(}}A,{{\rm{x}}_0},5)\)

Construct the data in the table shown below:

\(k\)	\(0\)	\(0\)	\(0\)	\(0\)	\(0\)
\({{\bf{x}}_k}\)	\(\left( {\begin{aligned}{ {20}{c}}1\\0\end{aligned}} \right)\)	\(\left( {\begin{aligned}{ {20}{c}}1\\{.4}\end{aligned}} \right)\)	\(\left( {\begin{aligned}{ {20}{c}}1\\{.4828}\end{aligned}} \right)\)	\(\left( {\begin{aligned}{ {20}{c}}1\\{.4971}\end{aligned}} \right)\)	\(\left( {\begin{aligned}{ {20}{c}}1\\{.4995}\end{aligned}} \right)\)
\(A{{\bf{x}}_k}\)	\(\left( {\begin{aligned}{ {20}{c}}5\\2\end{aligned}} \right)\)	\(\left( {\begin{aligned}{ {20}{c}}{5.8}\\{2.8}\end{aligned}} \right)\)	\(\left( {\begin{aligned}{ {20}{c}}{5.9655}\\{2.9990}\end{aligned}} \right)\)	\(\left( {\begin{aligned}{ {20}{c}}{5.9942}\\{2.9942}\end{aligned}} \right)\)	\(\left( {\begin{aligned}{ {20}{c}}{5.9990}\\{2.9990}\end{aligned}} \right)\)
\({\mu _k}\)	\(5\)	\(5.8\)	\(5.9655\)	\(5.9942\)	\(5.9990\)
\(R\left( {{{\bf{x}}_k}} \right)\)	\(5\)	\(5.9655\)	\(5.9990\)	\(5.99997\)	\(5.9999993\)

As the actual eigenvalue is \(6\) 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.