One can show that the solution \( \boldsymbol{x} \) is also the unique minimizer of the quadratic form $$ \begin{equation*} f(\boldsymbol{x}) = \frac{1}{2}\boldsymbol{x}^T\boldsymbol{A}\boldsymbol{x} - \boldsymbol{x}^T \boldsymbol{x} , \quad \boldsymbol{x}\in\mathbf{R}^n. \end{equation*} $$ This suggests taking the first basis vector \( \boldsymbol{p}_1 \) to be the gradient of \( f \) at \( \boldsymbol{x}=\boldsymbol{x}_0 \), which equals $$ \begin{equation*} \boldsymbol{A}\boldsymbol{x}_0-\boldsymbol{b}, \end{equation*} $$ and \( \boldsymbol{x}_0=0 \) it is equal \( -\boldsymbol{b} \). The other vectors in the basis will be conjugate to the gradient, hence the name conjugate gradient method.