TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-16

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An endomorphism is an orthogonal projection exactly when it is idempotent and self-adjoint

Statement

An endomorphism $P$ of a finite-dimensional inner product space is the orthogonal projection onto some subspace if and only if

$P^{2} = P and P^{*} = P .$

In that case it is the orthogonal projection onto $im P$ , along $\ker P = (im P)^{⊥}$ . The cases $P = 0$ and $P = I$ are included.

Facts & Assumptions

Given: An endomorphism $P$ of a finite-dimensional inner product space.

[L1]

Orthogonal projection onto $W$ is linear, idempotent, has image $W$ , and has kernel $W^{⊥}$ (Orthogonal projection is linear, and an orthonormal basis $(e_{i})$ of $W$ gives $P_{W} v = \sum_{i} ⟨ v, e_{i} ⟩ e_{i}$ ).

[L2]

The adjoint is characterised by $⟨ P v, w ⟩ = ⟨ v, P^{*} w ⟩$ and exists uniquely in finite dimension (The adjoint $T^{*} : W \to V$ is characterised by $⟨ T v, w ⟩_{W} = ⟨ v, T^{*} w ⟩_{V}$ , Every linear map between finite-dimensional inner product spaces has a unique adjoint).

[L3]

An orthogonal projection selects the subspace component in the orthogonal direct-sum decomposition (The orthogonal projection $P_{W} v$ is the $W$ -component in $V = W \oplus W^{⊥}$ ).

Proof

technique · direct

1.1L1L2L3

Suppose $P = P_{W}$ . By [L1], $P^{2} = P$ . Decompose both $v$ and $w$ into their $W$ and $W^{⊥}$ components. Orthogonality gives $⟨ P v, w ⟩ = ⟨ P v, P w ⟩ = ⟨ v, P w ⟩$ , so uniqueness in [L2] yields $P^{*} = P$ .

1.2givenalgebra

Conversely, suppose $P^{2} = P$ and $P^{*} = P$ . Every $v$ has the algebraic decomposition $v = P v + (I - P) v$ , where $P v \in im P$ and $P (I - P) v = 0$ , so $(I - P) v \in \ker P$ .

1.3L2given

If $x = P u \in im P$ and $z \in \ker P$ , then [L2] and self-adjointness give $⟨ x, z ⟩ = ⟨ P u, z ⟩ = ⟨ u, P z ⟩ = 0$ . Thus $im P ⊥ \ker P$ .

2.1step 1.2step 1.3L1L3∎

Steps 1.2 and 1.3 show that $P$ selects the $im P$ component in the orthogonal decomposition $V = im P \oplus \ker P$ . By [L3], $P = P_{im P}$ , and [L1] identifies its kernel with $(im P)^{⊥}$ .

Depends on

Used by

Dependency tree · two levels

10 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Sheldon Axler, Linear Algebra Done Right, 4th ed., §§6C and 7A (standard reference, not scraped)