= Solution
\b[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 $x_*$ solve $Ax_*=b$, let $e=x_*-x$, and let $r=Ae$. The quadratic energy error is $E=e^TAe/2$. Moving by $\alpha r$, with $\alpha=(r^Tr)/(r^TAr)$, is <exact line search for a positive-definite quadratic> and gives
$$
E_{\rm new}=E-\frac{(r^Tr)^2}{2r^TAr}.
$$
Writing $\lambda_{\min},\lambda_{\max}>0$ for the extreme <eigenvalues>, one has $r^TAr\le\lambda_{\max}r^Tr$ and $r^Tr=e^TA^2e\ge\lambda_{\min}e^TAe$. Therefore
$$
E_{\rm new}\le\left(1-\frac{\lambda_{\min}}{\lambda_{\max}}\right)E.
$$
The factor lies below one, so the energy tends to zero and positive definiteness forces $x\to x_*$. Here the bound is $E_{\rm new}\le(10/11)E$, since the <eigenvalues> are 1 and 11. This proves convergence for every initial <vector>, rather than merely checking one numerical iteration.
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