One can show that the solution \( \hat{x} \) is also the unique minimizer of the quadratic form

$$ \begin{equation*} f(\hat{x}) = \frac{1}{2}\hat{x}^T\hat{A}\hat{x} - \hat{x}^T \hat{x} , \quad \hat{x}\in\mathbf{R}^n. \end{equation*} $$

This suggests taking the first basis vector \( \hat{p}_1 \) to be the gradient of \( f \) at \( \hat{x}=\hat{x}_0 \), which equals

$$ \begin{equation*} \hat{A}\hat{x}_0-\hat{b}, \end{equation*} $$

and \( \hat{x}_0=0 \) it is equal \( -\hat{b} \). The other vectors in the basis will be conjugate to the gradient, hence the name conjugate gradient method.