The geometry of the normal equations

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In this article, I show that the normal equations define the orthogonal projection of a vector onto a linear subspace.

Setup

Let $\mathcal{S}$ be a vector space and $\mathcal{X} \subseteq \mathcal{S}$ a linear subspace. Let $X$ be a matrix whose column vectors are a basis for $\mathcal{X}$ , i.e. $\mathcal{X} = \mathrm{span}(X)$ . Each vector in $\mathcal{X}$ can be expressed as a linear combination of the column vectors in $X$ :

$\forall \vec{v} \in \mathcal{X}, \quad \exists \vec{w} : \quad \vec{v} = X\vec{w}$

The orthogonal projection

Given a vector $\vec{y} \in \mathcal{S}$ , the orthogonal projection of $\vec{y}$ onto $\mathcal{X}$ is the vector $\vec{y}_{\parallel}$ in $\mathcal{X}$ such that the residual $\vec{y}_{\perp} = \vec{y} - \vec{y}_{//}$ is orthogonal to every vector in $\mathcal{X}$ :

$(\vec{y}_{\parallel}) \in \mathcal{X}$ ,
$\vec{y} = (\vec{y}_{\parallel}) + (\vec{y}_{\perp})$ ,
$\forall w \in \mathcal{S}, \quad (\vec{y}_{\perp}) \perp (X\vec{w})$ .

Solving the equations

A vector $\vec{v}$ is orthogonal to every vector in $\mathcal{X}$ if and only if $X^{\top}\,\vec{v} = 0$ , hence we are looking to solve the following equation:

$\begin{align*} &X^{\top}\,(\vec{y}_{\perp}) = 0 \\ \iff &X^{\top}\,(\vec{y} - \vec{y}_{\parallel}) = 0 \end{align*}$

Since $(\vec{y}_{\parallel}) \in \mathcal{X}$ , we can find a vector $\vec{w} \in \mathcal{S}$ such that:

$\vec{y}_{\parallel} = X\vec{w}$

So the equation to be solved is:

$X^{\top}\,(\vec{y} - X\vec{w}) = 0$

Which is the matrix form of the normal equations.