Deriving the Normal Equations
Suppose we are given a design matrix \(\mathbf{X} \in \mathbb{R}^{N \times K}\) and targets vector \(\mathbf{y} \in \mathbb{R}^N\). We want to find parameter vector \(\mathbf{w} \in \mathbb{R}^K\) which minimizes the residual sum of squares, i.e.
\[\begin{align} \mathop{\arg \min}\limits_\mathbf{w} \ \mathrm{RSS}(\mathbf{w}) &= \sum_i^N (y_i - \mathbf{x}_i^T \mathbf{w})^2 \\ &= (\mathbf{y} - \mathbf{Xw})^T(\mathbf{y} - \mathbf{Xw}) \end{align}\]The solution to the (least squares) problem can be obtained using the normal equations. We can derive the normal equations from first principles as follows. First, let’s expand the \(\mathrm{RSS}\) definition so it’s easier to work with.
\[\begin{align} \mathrm{RSS}(\mathbf{w}) &= (\mathbf{y} - \mathbf{Xw})^T(\mathbf{y} - \mathbf{Xw}) \\ &= (\mathbf{y}^T - \mathbf{w}^T\mathbf{X}^T)(\mathbf{y} - \mathbf{Xw}) && \text{definition of transpose} \\ &= \mathbf{y}^T\mathbf{y} - \mathbf{y}^T\mathbf{Xw} - \mathbf{w}^T\mathbf{X}^T\mathbf{y} + \mathbf{w}^T\mathbf{X}^T\mathbf{Xw} \\ &= \mathbf{y}^T\mathbf{y} - 2\mathbf{y}^T\mathbf{Xw} + \mathbf{w}^T\mathbf{X}^T\mathbf{Xw} \\ \end{align}\]The last step follows because \(\mathbf{w}^T\mathbf{X}^T\mathbf{y}\) is a scalar and so it is equal to its transpose \(\mathbf{y}^T\mathbf{Xw}\).
Now, to minimize \(\mathrm{RSS}(\mathbf{w})\), we take its derivative w.r.t. \(\mathbf{w}\), set it to \(0\), and solve for \(\mathbf{w}\).
\[\begin{align} \nabla_\mathbf{w} \mathrm{RSS}(\mathbf{w}) &= 0 \\ \nabla_\mathbf{w} ( \mathbf{y}^T\mathbf{y} - 2\mathbf{y}^T\mathbf{Xw} + \mathbf{w}^T\mathbf{X}^T\mathbf{Xw} ) &= 0 \\ 0 - 2 \mathbf{X}^T\mathbf{y} + 2 \mathbf{X}^T\mathbf{X}\mathbf{w} &=0 \\ \mathbf{X}^T\mathbf{X}\mathbf{w} &= \mathbf{X}^T\mathbf{y} \\ \mathbf{w} &= (\mathbf{X}^T\mathbf{X})^{-1}\mathbf{X}^T\mathbf{y} \end{align}\]What we have arrived at is exactly the normal equations, where \((\mathbf{X}^T\mathbf{X})^{-1}\mathbf{X}^T\) is known as the (Moore-Penrose) pseudo- inverse.