 Solution

Real symmetry and positive definiteness are sufficient, and the preceding parts establish both properties for this matrix. If a residual becomes zero, stop: the solution has been reached and the displayed step quotient should not be evaluated as $0/0$.

For a convergence justification, let $x_{\ast }$ solve $A{x}_{\ast }=b$, let $e=x_{\ast }-x$, and let $r=Ae$. The quadratic energy error is $E=e^{T}Ae/2$. Moving by $\alpha r$, with $\alpha =(r^{T}r)/(r^{T}Ar)$, is exact line search for a positive-definite quadratic and givesWriting ${\lambda }_{\min },{\lambda }_{\max }>0$ for the extreme eigenvalues, one has $r^{T}Ar\le {\lambda }_{\max }{r}^{T}r$ and $r^{T}r=e^T{A}^{2}e\ge {\lambda }_{\min }{e}^{T}Ae$. ThereforeThe factor lies below one, so the energy tends to zero and positive definiteness forces $x\to x_{\ast }$. Here the bound is $E_{\mathrm{new}}\le (10/11)E$, since the eigenvalues are 1 and 11. This proves convergence for every initial vector, rather than merely checking one numerical iteration.