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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 be a finite-dimensional complex inner product space and let be linear. Then is normal if and only if has an orthonormal basis consisting of eigenvectors of .
Facts & Assumptions
Given: A finite-dimensional complex inner product space and a linear endomorphism .
The field is algebraically closed (The complex numbers are algebraically closed).
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).
A normal upper-triangular complex matrix is diagonal (A normal upper-triangular matrix is diagonal).
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
Assume is normal. By [L1], the characteristic polynomial of splits over , so [L2] gives an orthonormal basis in which the matrix of is upper triangular. Because is normal, [L4] makes that matrix normal, and then [L3] makes it diagonal. Hence the chosen orthonormal basis consists of eigenvectors of .
Conversely, assume is an orthonormal basis of eigenvectors of , with . Then the matrix of in that basis is the diagonal matrix , and every diagonal matrix commutes with its conjugate transpose. By [L4], is normal.
Depends on
Self-adjoint and normal endomorphisms of a finite-dimensional real or complex inner product space
In an orthonormal basis, self-adjoint means conjugate-transpose symmetry and normal means commuting with the conjugate transpose
The complex numbers are algebraically closed
If the characteristic polynomial of an endomorphism splits, some orthonormal basis makes its matrix upper triangular
A normal upper-triangular matrix is diagonal