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PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16
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Orthogonal projection is linear, and an orthonormal basis (ei) of W gives PWv=∑i⟨v,ei⟩ei

Statement

Let (e0,…,er−1) be an orthonormal basis of a subspace W of a finite-dimensional inner product space V. Then

PWv=∑i<r⟨v,ei⟩ei.

The map PW is linear, with image W, kernel W⊥, and PW2=PW. Moreover, regarding both projections as endomorphisms of V,

I−PW=PW⊥.

Facts & Assumptions

Given: A subspace W, an orthonormal basis (ei)i<r of W, and v∈V.

[L1]

The orthogonal projection PWv is the unique w∈W for which v−w∈W⊥ (The orthogonal projection PWv is the W-component in V=W⊕W⊥).

[L2]

In an orthonormal basis, the coefficient of a vector is its inner product with the corresponding basis vector (Bessel's inequality for a finite orthonormal list and Parseval's identity for an orthonormal basis).

Proof

technique · direct
1.1L2L3algebra

Put p=∑i<r⟨v,ei⟩ei. Then p∈W, and for every basis vector ej, [L3] and orthonormality give ⟨v−p,ej⟩=0. By [L2], this makes v−p orthogonal to all of W.

1.2L1

The defining decomposition shows PWw=w for w∈W and PWz=0 for z∈W⊥. Hence im⁡PW=W, ker⁡PW=W⊥, and PW2=PW.

2.1step 1.1L1L3

The uniqueness clause in [L1] gives PWv=p, proving the formula. Linearity follows immediately from [L3] and the formula.

3.1L1step 1.2∎

For the decomposition v=PWv+(v−PWv), the second summand lies in W⊥. Its projection onto W⊥ is itself, so PW⊥v=v−PWv.

Depends on

Used by

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Sources