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An iterative method approximates a solution by repeatedly updating an initial guess. In numerical linear algebra, a stationary iterative method for a linear system uses a fixed iteration map; other iterations can change their update rule or search space at each step.

Minimizing $\Vert y-X\theta {\Vert }^2$ gives the displayed equations by differentiation. Equivalently, the regression residual $r=y-X\hat{\theta }$ satisfies $X^{T}r=0$: it is orthogonal to every column of the design matrix. If $X$ has full column rank, the minimizer is unique and the fitted values are the orthogonal projection of $y$ onto its column space. A rank-deficient design matrix can give multiple coefficient vectors, while the projected fitted vector is still unique.

For an intercept and one predictor, put $S_{xx}={\sum }_i(x_i-\bar{x}{)}^2$ and $S_{xy}={\sum }_i(x_i-\bar{x})(y_i-\bar{y})$. The equations give $\hat{\beta }=S_{xy}/S_{xx}$ and $\hat{\alpha }=\bar{y}-\hat{\beta }\bar{x}$, provided $S_{xx}>0$. In the five-point example in linear regression, $\bar{x}=2$, $\bar{y}=2.8$, $S_{xx}=10$ and $S_{xy}=10$, giving $\hat{\beta }=1$ and $\hat{\alpha }=0.8$.

The normal equations describe the mathematical optimum, but explicitly forming and inverting $X^{T}X$ can amplify numerical problems. A QR decomposition computes the same least-squares solution through orthogonal coordinates; a singular value decomposition also exposes nearly dependent columns. This connects statistical fitting to numerical linear algebra and the choice of a stable algorithm. ^[ref]

Solutions to all three questions of the University of Oxford numerical linear algebra paper, sat on 2 June 2025. Source: released examination PDF, parent PDF pages 1–3. The question statements are omitted. The Gaussian-matrix estimates in parts 2(b)(v) and 3(b)(iv) are used as the paper explicitly permits; their limitations are explained in those solutions.