10. Linear Systems and the Lanczos Method

To end this book on a lighter note as well as to step back once again to the origin of the method, we review the application of the Lanczos method for solving linear systems. 10.1 Exact solution Let us assume that A is symmetric, positive definite, and of order n and that the system we wish to solve is Ax=b. 10.1 In theory the solution is x= A−1 b; 10.2 however, calculating the inverse operator may be impractical. Again, we exploit the fact that the Lanczos method produces Q n T A Qn = Tn , 10.3 where Qn is an orthogonal matrix. Therefore we can evaluate the inverse of A as A−1 = Qn T n −1 Q n T 10.4 and find the exact solution x= A−1 b= Qn T n −1 Q n T b. 10.5 This certainly seems more practical now, as we need to invert only a tridiagonal matrix. Let us forget for now the minor inconvenience of calculating the tridiagonal form and all the Lanczos vectors.