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

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Every linear map between finite-dimensional inner product spaces has a unique adjoint

Statement

Every linear map $T : V \to W$ between finite-dimensional real or complex inner product spaces has a unique adjoint $T^{*} : W \to V$ .

Facts & Assumptions

Given: A linear map $T : V \to W$ between finite-dimensional inner product spaces.

[L1]

Every linear functional on a finite-dimensional inner product space has a unique representing vector (Finite-dimensional Riesz representation: every functional is uniquely $v \mapsto ⟨ v, w ⟩$ ).

[L2]

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

[L3]

Inner products separate vectors: equality of all pairings 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

Fix $w \in W$ . The function $v \mapsto ⟨ T v, w ⟩_{W}$ is a linear functional on $V$ , so [L1] supplies a unique vector, call it $T^{*} w$ , satisfying [L2].

2.1step 1.1L3algebra

For scalars $a, b$ and $w_{1}, w_{2} \in W$ , pairing the representatives from step 1.1 shows $T^{*} (a w_{1} + b w_{2})$ and $a T^{*} w_{1} + b T^{*} w_{2}$ have the same pairing with every $v$ . By [L3] they are equal. Thus $T^{*}$ is linear.

3.1step 1.1step 2.1L1∎

Any adjoint must assign to each $w$ the unique representative from step 1.1, so it equals $T^{*}$ . This proves existence and uniqueness, including when either space is zero.

Every linear map between finite-dimensional inner product spaces has a unique adjoint

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources