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Let $p_1,\dots ,p_n$ be conjugate directions for a symmetric positive-definite matrix $A$. Starting from any $x_1$, minimize $x^{T}Ax/2-b^{T}x$ successively along those full lines, stopping early if the linear-system residual $r_j=b-A{x}_j$ vanishes. The quadratic minimization along a line uses ${\alpha }_j=p_j^T{r}_j/(p_j^{T}A{p}_j)$ and gives $r_{j+1}=r_j-{\alpha }_{j}A{p}_j$.

This makes $p_j^T{r}_{j+1}=0$. For $i<j$, conjugacy preserves $p_i^T{r}_{j+1}=p_i^T{r}_j$. By mathematical induction, $r_{j+1}$ is orthogonal to all preceding search vectors. The $n$ conjugate directions are linearly independent and hence form a basis, so $r_{n+1}=0$. Thus the method reaches the unique solution of the linear system in at most $n$ steps in exact arithmetic. The conjugate gradient method generates such directions without storing a complete prescribed basis in advance.

For the quadratic function, write $r_j=b-A{x}_j$ and restrict to the full line $x_j+\alpha p_j$. Symmetry of $A$ givesThe coefficient $p_j^{T}A{p}_j$ is positive by positive definiteness and $p_j\ne 0$. The unique minimizer is thereforeQuadratic minimization along a line also gives the exact changeThus strict decrease holds exactly when $p_j^T{r}_j\ne 0$. The search is over the whole line, so ${\alpha }_j$ can be negative; a positive step size would require the additional descent direction condition $p_j^T{r}_j>0$.