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TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-28
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Complex spectral theorem: a normal endomorphism of a finite-dimensional complex inner product space has an orthonormal eigenbasis, and conversely

Statement

Let V be a finite-dimensional complex inner product space and let T:VV be linear. Then T is normal if and only if V has an orthonormal basis consisting of eigenvectors of T.

Facts & Assumptions

Given: A finite-dimensional complex inner product space V and a linear endomorphism T:VV.

[L1]

The field C is algebraically closed (The complex numbers are algebraically closed).

[L2]

If the characteristic polynomial splits, then some orthonormal basis makes the matrix upper triangular (If the characteristic polynomial of an endomorphism splits, some orthonormal basis makes its matrix upper triangular).

[L3]

A normal upper-triangular complex matrix is diagonal (A normal upper-triangular matrix is diagonal).

[L4]

In an orthonormal basis, a linear map is normal exactly when its matrix commutes with its conjugate transpose (In an orthonormal basis, self-adjoint means conjugate-transpose symmetry and normal means commuting with the conjugate transpose).

Proof

technique · direct
1.1

Assume T is normal. By [L1], the characteristic polynomial of T splits over C, so [L2] gives an orthonormal basis in which the matrix of T is upper triangular. Because T is normal, [L4] makes that matrix normal, and then [L3] makes it diagonal. Hence the chosen orthonormal basis consists of eigenvectors of T.

L1L2L3L4
2.1

Conversely, assume (e1,,en) is an orthonormal basis of eigenvectors of T, with Tej=λjej. Then the matrix of T in that basis is the diagonal matrix diag(λ1,,λn), and every diagonal matrix commutes with its conjugate transpose. By [L4], T is normal.

L4algebra

Depends on

Used by

Dependency tree · two levels

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Sources