Past exam of the mathematics course of the University of Cambridge / 2002 / ib / Paper 3 / 16B / a / Solution

Since every eigenvalue has modulus less than one, $1$ is not an eigenvalue and $I-H$ is invertible. Hence the iteration has a unique fixed point $\hat{x}=(I-H{)}^{-1}v$. The stated consistency implication gives $A\hat{x}=b$, and nonsingularity of $A$ identifies $\hat{x}=x^{\ast }$. This justifies using the fixed-point relation even though consistency was stated in only one direction.

Expand the initial error in the eigenvector basis, $x^0-x^{\ast }={\sum }_i{c}_i{w}_i$. Subtract the fixed-point equation from the stationary iterative method for a linear system and iterate:With $q={\max }_i|{\lambda }_i|<1$, any norm obeys $\Vert x^k-x^{\ast }\Vert \le q^k{\sum }_i|c_i|\Vert w_i\Vert \to 0$. ThusThe argument uses the given diagonalizable eigenbasis; the more general convergence criterion is spectral radius less than one.