Euclidean projection on a set

An Euclidean projection of a point $x_0 in mathbf{R}^n$ on a set $mathbf{S} subseteq mathbf{R}^n$ 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

$min_x : |x-x_0|_2 ~:~ x in mathbf{S}.$

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

$mathbf{S} = left{ x in mathbf{R}^3 ~:~ 2x_1 + x_2 -x_3 = 1 right}.$

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

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 orthogonal to 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 we obtain t = 1/(a^Ta) = 1/6 , so that the projection is x^ast = a/(a^Ta) = (1/3,1/6,-1/6) .

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凸集（Convex Set）：

Definitions

A subset mathbf{C} of $mathbf{R}^n$ is said to be convex if and only if it contains the line segment between any two points in it:

$forall : x_1, x_2 in mathbf{C}, ;; forall : lambda in [0,1] ~:~ lambda x_1 + (1-lambda) x_2 in mathbf{C}.$

Subspaces and affine sets, such as lines, planes and higher-dimensional ‘‘flat’’ sets, are obviously convex, as they contain the entire line passing through any two points, not just the line segment. That is, there is no restriction on the scalar lambda anymore in the above condition.