Rayleigh quotient iteration

The algorithm is very similar to inverse iteration, but replaces the estimated eigenvalue at the end of each iteration with the Rayleigh quotient. Begin by choosing some value ${\displaystyle \mu _{0}}$ as an initial eigenvalue guess for the Hermitian matrix ${\displaystyle A}$. An initial vector ${\displaystyle b_{0}}$ must also be supplied as initial eigenvector guess.

Calculate the next approximation of the eigenvector ${\displaystyle b_{i+1}}$ by

${\displaystyle b_{i+1}={\frac {(A-\mu _{i}I)^{-1}b_{i}}{\|(A-\mu _{i}I)^{-1}b_{i}\|}},}$
where ${\displaystyle I}$ is the identity matrix, and set the next approximation of the eigenvalue to the Rayleigh quotient of the current iteration equal to
${\displaystyle \mu _{i+1}={\frac {b_{i+1}^{*}Ab_{i+1}}{b_{i+1}^{*}b_{i+1}}}.}$

To compute more than one eigenvalue, the algorithm can be combined with a deflation technique.

Note that for very small problems it is beneficial to replace the matrix inverse with the adjugate, which will yield the same iteration because it is equal to the inverse up to an irrelevant scale (the inverse of the determinant, specifically). The adjugate is easier to compute explicitly than the inverse (though the inverse is easier to apply to a vector for problems that aren't small), and is more numerically sound because it remains well defined as the eigenvalue converges.

Consider the matrix

${\displaystyle A=\left[{\begin{matrix}1&2&3\\1&2&1\\3&2&1\\\end{matrix}}\right]}$

for which the exact eigenvalues are ${\displaystyle \lambda _{1}=3+{\sqrt {5}}}$, ${\displaystyle \lambda _{2}=3-{\sqrt {5}}}$ and ${\displaystyle \lambda _{3}=-2}$, with corresponding eigenvectors

${\displaystyle v_{1}=\left[{\begin{matrix}1\\\varphi -1\\1\\\end{matrix}}\right]}$, ${\displaystyle v_{2}=\left[{\begin{matrix}1\\-\varphi \\1\\\end{matrix}}\right]}$ and ${\displaystyle v_{3}=\left[{\begin{matrix}1\\0\\1\\\end{matrix}}\right]}$.

(where ${\displaystyle \textstyle \varphi ={\frac {1+{\sqrt {5}}}{2}}}$ is the golden ratio).

The largest eigenvalue is ${\displaystyle \lambda _{1}\approx 5.2361}$ and corresponds to any eigenvector proportional to ${\displaystyle v_{1}\approx \left[{\begin{matrix}1\\0.6180\\1\\\end{matrix}}\right].}$

We begin with an initial eigenvalue guess of

${\displaystyle b_{0}=\left[{\begin{matrix}1\\1\\1\\\end{matrix}}\right],~\mu _{0}=200}$.

Then, the first iteration yields

${\displaystyle b_{1}\approx \left[{\begin{matrix}-0.57927\\-0.57348\\-0.57927\\\end{matrix}}\right],~\mu _{1}\approx 5.3355}$

the second iteration,

${\displaystyle b_{2}\approx \left[{\begin{matrix}0.64676\\0.40422\\0.64676\\\end{matrix}}\right],~\mu _{2}\approx 5.2418}$

and the third,

${\displaystyle b_{3}\approx \left[{\begin{matrix}-0.64793\\-0.40045\\-0.64793\\\end{matrix}}\right],~\mu _{3}\approx 5.2361}$

from which the cubic convergence is evident.

The following is a simple implementation of the algorithm in Octave.

function x = rayleigh(A, epsilon, mu, x)
  x = x / norm(x);
  % the backslash operator in Octave solves a linear system
  y = (A - mu * eye(rows(A))) \ x; 
  lambda = y' * x;
  mu = mu + 1 / lambda
  err = norm(y - lambda * x) / norm(y)

  while err > epsilon
    x = y / norm(y);
    y = (A - mu * eye(rows(A))) \ x;
    lambda = y' * x;
    mu = mu + 1 / lambda
    err = norm(y - lambda * x) / norm(y)
  end

end

• Power iteration
• Inverse iteration

• Lloyd N. Trefethen and David Bau, III, Numerical Linear Algebra, Society for Industrial and Applied Mathematics, 1997. ISBN 0-89871-361-7.
• Rainer Kress, "Numerical Analysis", Springer, 1991. ISBN 0-387-98408-9