# Comparing iterative methods for linear systems

For a tridiagonal matrix of the from

\begin{bmatrix} a & -b & \newline -b & a & -b \newline & \ddots & \ddots & \ddots \newline & & & & -b \newline & & &-b & a \end{bmatrix}

with $a \geq 2b > 0$ I would like to compare, in funtion of $a$ and $b$, the convergence of the Gauss-Seidel method and the Steepest Descent method. But how to do such a comparison if the information about Gauss-Seidel convergence is given by is spectral radius and the information about the Steepest Descent convergence is given by the ratio between the largest and the smallest eigenvalues?

-
You'll want to see www2.cs.cas.cz/semincm/lectures/2007-05-04-ng.pdf for instance. – J. M. Nov 26 '11 at 18:51
Seems a little bit strange to use Gauß-Seidel or steepest descent, if there is en.wikipedia.org/wiki/Tridiagonal_matrix_algorithm... – Dirk Nov 26 '11 at 20:26
This is a tridiagonal symmetric Toeplitz matrix. There are explicit expressions for its eigenvectors and eigenvalues. For example, they are in Iserles' A First Course in the Numerical Analysis of Differential Equations. Is this for a course you are taking? – Paul Tupper Nov 26 '11 at 21:35
Paul is correct, one can relate the characteristic polynomial of a tridiagonal Toeplitz matrix with the Chebyshev polynomials. – J. M. Nov 27 '11 at 1:54

I think the confusion comes from the statement "Gauss-Seidel convergence is given by its spectral radius", which is incorrect. Gauss-Seidel convergence depends on the spectral radius of a new matrix $L^{-1}R$ where $L-R = A$, for solving $Ax=b$. It is hard to estimate the spectral radius of $L^{-1}R$, but I am pretty sure that it is related to the condition number of $A$ (the ratio between the largest and the smallest eigenvalues).