PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-16

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Adjoints satisfy $(S + T)^{} = S^{} + T^{}$ , $(λ T)^{} = \overline{λ} T^{}$ , $(S T)^{} = T^{} S^{}$ , and $T^{* *} = T$

Statement

For compatible linear maps between finite-dimensional real or complex inner product spaces,

$(S + T)^{*} = S^{*} + T^{*}, (λ T)^{*} = \overline{λ} T^{*}, (S T)^{*} = T^{*} S^{*}, T^{* *} = T .$

Also $I^{*} = I$ and $0^{*} = 0$ .

Facts & Assumptions

Given: Compatible finite-dimensional linear maps $S, T$ and a scalar $λ$ .

[L1]

An adjoint is characterised by $⟨ T v, w ⟩ = ⟨ v, T^{*} w ⟩$ (The adjoint $T^{*} : W \to V$ is characterised by $⟨ T v, w ⟩_{W} = ⟨ v, T^{*} w ⟩_{V}$ ).

[L2]

Finite-dimensional adjoints exist and are unique (Every linear map between finite-dimensional inner product spaces has a unique adjoint).

[L3]

Equality of pairings with every vector forces equality of the paired vectors (Inner products separate vectors, and the induced norm is homogeneous: $∥ λ v ∥ = ∣ λ ∣ ∥ v ∥$ ).

Proof

technique · direct

1.1L1L2

For all $v, w$ , linearity in the first argument and [L1] give $⟨ (S + T) v, w ⟩ = ⟨ v, (S^{*} + T^{*}) w ⟩$ . Uniqueness in [L2] proves the sum formula. The same calculation gives $I^{*} = I$ and $0^{*} = 0$ .

1.2L1L2algebra

Likewise $⟨ λ T v, w ⟩ = λ ⟨ v, T^{*} w ⟩ = ⟨ v, \overline{λ} T^{*} w ⟩$ , because the second argument is conjugate-linear. Thus $(λ T)^{*} = \overline{λ} T^{*}$ .

1.3L1L2

For a composite, $⟨ S T v, w ⟩ = ⟨ T v, S^{*} w ⟩ = ⟨ v, T^{*} S^{*} w ⟩$ , so uniqueness gives $(S T)^{*} = T^{*} S^{*}$ .

2.1L1L2L3∎

Conjugate symmetry rewrites [L1] as $⟨ T^{*} w, v ⟩ = ⟨ w, T v ⟩$ , so $T$ is an adjoint of $T^{*}$ . By [L2], $T^{* *} = T$ .

Adjoints satisfy $(S + T)^{} = S^{} + T^{}$ , $(λ T)^{} = \overline{λ} T^{}$ , $(S T)^{} = T^{} S^{}$ , and $T^{* *} = T$

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

Adjoints satisfy (S+T)∗=S∗+T∗(S+T)∗=S∗+T∗, (λT)∗=λ‾T∗(λT)∗=λT∗, (ST)∗=T∗S∗(ST)∗=T∗S∗, and T∗∗=TT∗∗=T

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

Adjoints satisfy $(S + T)^{} = S^{} + T^{}$ , $(λ T)^{} = \overline{λ} T^{}$ , $(S T)^{} = T^{} S^{}$ , and $T^{* *} = T$