PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28

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.

In an orthonormal basis, self-adjoint means conjugate-transpose symmetry and normal means commuting with the conjugate transpose

Statement

Let $V$ be a finite-dimensional real or complex inner product space, let $T : V \to V$ be linear, and let $E$ be an orthonormal basis of $V$ . Write

$A = [T]_{E}^{E} .$

Then:

$T$ is self-adjoint if and only if $A = A^{*}$ .
$T$ is normal if and only if $A A^{*} = A^{*} A$ .

Over $R$ , $A^{*}$ is just $A^{T}$ .

Facts & Assumptions

Given: A finite-dimensional real or complex inner product space $V$ , a linear map $T : V \to V$ , an orthonormal basis $E$ , and the matrix $A = [T]_{E}^{E}$ .

[L1]

In orthonormal bases, the matrix of the adjoint is the conjugate transpose (In orthonormal bases, the matrix of the adjoint is the conjugate transpose of the matrix).

[L2]

Matrix representation sends composition to matrix multiplication ( $[S \circ T]_{B}^{D} = [S]_{C}^{D} [T]_{B}^{C}$ ).

Proof

technique · direct

1.1L1

By the definition of self-adjointness in Self-adjoint and normal endomorphisms of a finite-dimensional real or complex inner product space, $T$ is self-adjoint exactly when $T = T^{*}$ . By [L1], this is equivalent to $A = [T]_{E}^{E} = [T^{*}]_{E}^{E} = A^{*}$ .

2.1L1L2∎

By the definition of normality in Self-adjoint and normal endomorphisms of a finite-dimensional real or complex inner product space, $T$ is normal exactly when $T T^{*} = T^{*} T$ ; using [L2] and then [L1], this is equivalent to $[T T^{*}]_{E}^{E} = A [T^{*}]_{E}^{E} = A A^{*} = [T^{*}]_{E}^{E} A = A^{*} A = [T^{*} T]_{E}^{E}$ , hence to $A A^{*} = A^{*} A$ .

Depends on

Dependency tree · 2 levels

Sources

Sheldon Axler, Linear Algebra Done Right, fourth edition