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 \( \hat{r}_1 \) (see below for definition) 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 \( -\hat{b} \).