Recipe

Find the closest point in a subspace, and the minimum distance to it

Use when: the problem gives a subspace W and a vector \mathbf v, then asks for the closest point in W, the minimum distance to it, or a proof that no point of W beats a stated distance

Goal: the exact minimum distance from \mathbf v to W, backed by a proof no other point of W can do better

Taught in Orthogonal projection and best approximation

1. 1

   Identify \mathbf v and W, then find \mathbf w, the orthogonal projection of \mathbf v onto W, from the direct-sum decomposition \mathbf v=\mathbf w+\mathbf w^\perp.

   Example

   For \mathbf v=(3,-1,4) and W=\text{span}\{(1,0,0),(0,1,0)\} (the plane z=0), the projection is \mathbf w=(3,-1,0): the first two coordinates carry over, and the third is dropped.

   Why: the rest of this recipe assumes \mathbf w is already known, since the theorem says \mathbf w is unbeatable once you have it, not how to find it in the first place.

2. 2

   Subtract to get \mathbf w^\perp=\mathbf v-\mathbf w, then compute \lVert\mathbf w^\perp\rVert; this is the exact minimum distance from \mathbf v to any point of W.

   Example

   \mathbf w^\perp=(3,-1,4)-(3,-1,0)=(0,0,4), so \lVert\mathbf w^\perp\rVert=4: no point of W can be closer than 4 to \mathbf v.

   • only the minimum distance value is asked for — you're done; report this number.
   • a specific alternate point needs to be shown worse — continue to Step 3.
   • a bound needs proving with no further arithmetic — skip straight to Step 4.

   ⚡ Shortcut

   When W is the orthogonal complement of a single vector \mathbf n (a hyperplane through the origin), skip finding \mathbf w entirely: the minimum distance is \lvert\langle\mathbf v,\mathbf n\rangle\rvert/\lVert\mathbf n\rVert. Here \mathbf n=(0,0,1) and \mathbf v=(3,-1,4) give \lvert4\rvert/1=4, matching the long way exactly.

   Why: this residual is orthogonal to every vector in W by construction, and the theorem's own proof shows its length is exactly the gap to the closest point.

3. 3

   Compute the distance from \mathbf v to the candidate point \mathbf u, then compare it against the distance found in Step 2.

   Example

   For \mathbf u=(3,-2,0)\in W: \mathbf v-\mathbf u=(0,1,4), so \lVert\mathbf v-\mathbf u\rVert=\sqrt{17}\approx4.123, bigger than the 4 found above.

   Why: the theorem's strict inequality guarantees \mathbf w beats every OTHER point of W, so a correctly computed comparison always comes out this way.

4. 4

   State that \lVert\mathbf v-\mathbf u\rVert>\lVert\mathbf v-\mathbf w\rVert for every \mathbf u\in W with \mathbf u\ne\mathbf w; that inequality alone proves the bound.

   Example

   With the minimum distance from Step 2 equal to 4, no point of the plane W can be closer than 4 to \mathbf v, with zero further arithmetic.

   Why: the strict inequality IS the theorem itself, so citing it directly is already a complete proof, not a shortcut around one.