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Bounds on the error inevitably rely on bounds for , since . There is a large number of problem dependent ways to estimate ; we mention a few here.
When a splitting is used to get an iteration
then the matrix whose inverse norm we need is . Often, we know how to estimate if the splitting is a standard one such as Jacobi or SOR, and the matrix has special characteristics such as Property A. Then we may estimate .
When is symmetric positive definite, and Chebyshev acceleration with adaptation of parameters is being used, then at each step the algorithm estimates the largest and smallest eigenvalues and of anyway. Since is symmetric positive definite, .
This adaptive estimation is often done using the Lanczos algorithm, which can usually provide good estimates of the largest (rightmost) and smallest (leftmost) eigenvalues of a symmetric matrix at the cost of a few matrix-vector multiplies. For general nonsymmetric , we may apply the Lanczos method to or , and use the fact that .
It is also possible to estimate provided one is willing to solve a few systems of linear equations with and as coefficient matrices. This is often done with dense linear system solvers, because the extra cost of these systems is , which is small compared to the cost of the LU decomposition (see Hager , Higham  and Anderson, et al. ). This is not the case for iterative solvers, where the cost of these solves may well be several times as much as the original linear system. Still, if many linear systems with the same coefficient matrix and differing right-hand-sides are to be solved, it is a viable method.
The approach in the last paragraph also lets us estimate the alternate error bound . This may be much smaller than the simpler in the case where the rows of are badly scaled; consider the case of a diagonal matrix with widely varying diagonal entries. To compute , let denote the diagonal matrix with diagonal entries equal to the entries of ; then (see Arioli, Demmel and Duff ). can be estimated using the technique in the last paragraph since multiplying by or is no harder than multiplying by and and also by , a diagonal matrix.