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Forthe gradient is $\nabla F(x)=Ax-b=-r$. The steepest descent method with exact line search for a positive-definite quadratic therefore uses

For $A=\mathrm{diag}(1,\gamma )$ and $b=0$, the minimizer is $x^{\ast }=0$. PutStarting from $x^{(0)}=(\gamma ,1{)}^T$, direct substitution gives ${\alpha }_k=2/(\gamma +1)$ and, inductively,The Euclidean norm is unchanged by the alternating sign, soFor a positive-definite matrix, the spectral condition number of a positive-definite matrix isHere ${\kappa }_2(A)=\gamma$, and hence

The Conjugate gradient method starts with $p_0=r_0$ and updatesBy finite termination of the conjugate gradient method, it reaches the exact solution in at most the number of distinct eigenvalues. This matrix has two, so at mostare required in exact arithmetic.

The exact line search for a positive-definite quadratic givesUsingproduces the stated ratio formula. If $l$ and $L$ are the extreme eigenvalues,Hence the fraction in the ratio is at least $l/L$, and iteration yieldsWhen $n=1$, $l=L$, so the right side is zero after one iteration: exact line search reaches the solution in a single step.