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], $T^{* *} = T$ .

Depends on

Used by

Dependency tree · two levels

9 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., result 7.5 (standard reference, not scraped)
Sergei Treil, Linear Algebra Done Wrong, §5.5.1, useful formulas for adjoints (standard reference, not scraped)

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