This question is related to my previous question The algorithm to find the largest eigenvalue and one of its eigenvector of a symmetric tridiagonal matrix? The matrix in question is a symmetric tridiagonal matrix in the form of

$$\left( {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}&{}&{} \\ {{y_1}}&{{x_2}}& \ddots &{} \\ {}& \ddots & \ddots &{{y_{n - 1}}} \\ {}&{}&{{y_{n - 1}}}&{{x_n}} \end{array}} \right)$$

where $y_1,...,y_{n-1}$ are all positive numbers. The question is

Suppose now we already know an eigenvalue $\lambda$ of it, what is the right way to compute an eigenvector associated with $\lambda$?

There seems to be an naive method, but I am very not sure if it is correct. Suppose the eigenvector is $a=(a_1,...,a_n)$, Set $a_1$ to arbitrary non-zero number if $y_1 \neq \lambda$, and set $a_1 = 0$ otherwise, then solve $a_2$ by $x_1a_1+y_1a_2=\lambda (a_1+a_2)$, then solve $a_3$ by $y_1a_1+x_2a_2+y_2a_3=\lambda (a_1+a_2+a_3)$, and so on so forth. Will it really be so simple?

  • $\begingroup$ Yes, but it’s more tedious than it needs to be. $\endgroup$ – k.stm Jul 31 '17 at 15:48
  • $\begingroup$ Er, are you planning on implementing this? That is, is a computer supposed to do the algorithm? $\endgroup$ – k.stm Jul 31 '17 at 15:55
  • $\begingroup$ @k.stm Thanks for your comment and answer. Yes, however, it is desired to be done with a computer. $\endgroup$ – Ralph B. Jul 31 '17 at 17:54
  • $\begingroup$ I do not understand your method. To me, it seems relevant to compare $x_1$, rather than $y_1$ to $\lambda$. I believe the right hand side of the first equation and the second equations should be $\lambda a_1$ and $\lambda a_2$. $\endgroup$ – Carl Christian Aug 1 '17 at 20:30

If you already know an approximate eigenvalue $\hat{\lambda}$ of a (tridiagonal) symmetric matrix, then the "standard" way to refine it and compute the corresponding eigenvector is to apply the inverse power method to the matrix $A - \hat{\lambda}I$., i.e., execute the power method using the action of the matrix $(A - \hat{\lambda} I)^{-1}$ in place of $A$

You need not worry about the fact that your coefficient matrix $A - \hat{\lambda} I$ is nearly singular. It is known that the while the error of the solve can be quite large, it tends to point in the correct direction. Beresford Parlett has a discussion of this phenomenon in his textbook: "The symmetric eigenvalue problem".

If the spectrum of $A$ contains a cluster around $\lambda$, then the rate of convergence of the shifted-inverse power method may be insufficient. In this case you should try to apply the inverse subspace iteration with Rayleigh-Ritz acceleration to the matrix $A - \hat{\lambda} I$.


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