TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-16

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The orthogonal projection is the unique nearest point in the subspace

Statement

Let $W$ be a subspace of a finite-dimensional inner product space $V$ . For every $v \in V$ , the vector $P_{W} v$ is the unique point of $W$ nearest to $v$ : for every $w \in W$ ,

$∥ v - P_{W} v ∥ \leq ∥ v - w ∥,$

with equality if and only if $w = P_{W} v$ .

Facts & Assumptions

Given: A subspace $W$ , a vector $v \in V$ , and $w \in W$ .

[L1]

The residual $v - P_{W} v$ lies in $W^{⊥}$ , while $P_{W} v$ lies in $W$ (The orthogonal projection $P_{W} v$ is the $W$ -component in $V = W \oplus W^{⊥}$ ).

[L2]

Orthogonal vectors $x, y$ satisfy $∥ x + y ∥^{2} = ∥ x ∥^{2} + ∥ y ∥^{2}$ (Pythagoras, the parallelogram identity, and the real and complex polarisation identities).

Proof

technique · direct

1.1

Decompose $v - w = (v - P_{W} v) + (P_{W} v - w)$ . The first term lies in $W^{⊥}$ and the second in $W$ , so they are orthogonal by [L1].

L1algebra

2.1

By [L2], $∥ v - w ∥^{2} = ∥ v - P_{W} v ∥^{2} + ∥ P_{W} v - w ∥^{2} \geq ∥ v - P_{W} v ∥^{2}$ . Nonnegativity gives the asserted inequality.

step 1.1L2

3.1

Equality holds exactly when $∥ P_{W} v - w ∥^{2} = 0$ , which by positive definiteness is exactly $w = P_{W} v$ .

step 2.1∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 12 results over 6 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Sheldon Axler, Linear Algebra Done Right, 4th ed., result 6.61 (standard reference, not scraped)
Sergei Treil, Linear Algebra Done Wrong, Theorem 5.3.2 (standard reference, not scraped)