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An Euclidean projection of a point \(x_0\) in \(\mathbb{R}^n\) on a set \(\mathbf{S} \subseteq \mathbb{R}^n\) is a point that achieves the smallest Euclidean distance from \(x_0\) to the set. That is, it is any solution to the optimization problem

\[\min_x \, \left\| x – x_0 \right\|_2 \, : \, x \in \mathbf{S}.\]

When the set \(\mathbf{S}\) is convex, there is a unique solution to the above problem. In particular, the projection on an affine subspace is unique.

Example: assume that \(\mathbf{S}\) is the hyperplane

\[\mathbf{S} = \left\{ x \in \mathbb{R}^3 \, : \, 2x_1 + x_2 -x_3 = 1 \right\}.\]

The projection problem reads as a linearly constrained least-squares problem, of particularly simple form:

\[\min_x \, \left\| x \right\|_2 \, : \, 2x_1 + x_2 -x_3 = 1.\]

The projection of \(x_0 = 0\) on \(\mathbf{S}\) turns out to be aligned with the coefficient vector \(a = (2,1,-1)\). Indeed, components of \(x\) orthogonal to \(a\) don’t appear in the constraint, and only increase the objective value. Setting \(x = t a\) in the equation defining the hyperplane and solving for the scalar \(t\) we obtain

\[t = \frac{1}{a^T a} = \frac{1}{6},\]

so that the projection is

\[x^* = \frac{a}{a^T a} = \left(\frac{1}{3},\frac{1}{6},-\frac{1}{6}\right).\]