# Spectral Theorem for normal operators on finite dimensional vector spaces

In this post, I will follow up the discussion of my last post by proving the spectral theorem for normal operators and as corollaries the spectral theorem for hermitian, unitary and symmetric operators.

Theorem (Spectral Theorem for Normal Operators):For any normal operator on a (finite dimensional) hermitian space there is an orthonormal basis of consisting of eigenvectors for Equivalently, for any normal matrix there is a unitary matrix such that is diagonal.

**Proof:** Let be an eigenvector for of length By the last theorem of last post, is also an eigenvector for so (the 1-dimensional space spanned by ) is -invariant. Then by the proposition before that theorem, is -invariant. Since the restriction of to any -invariant subspace of is also a normal operator on that subspace, we may use induction on dimension to prove the theorem:

Assume has an orthonormal basis of eigenvectors Then since the form is non-degenerate (), we have thus the set is an orthonormal basis of of eigenvectors for

To prove the theorem in matrix form, assume first that the form is the standard hermitian form (which we can do without loss of generality since every positive definite form is conjugate to this one) so that the canonical basis is orthonormal. Let be the normal matrix of the normal operator in the canonical basis. Then there exists an orthonormal basis consisting of eigenvectors for The the change of basis matrix from the canonical basis to is unitary and the matrix of in this new basis is Since the elements of are eigenvectors for this matrix is diagonal.

QED

To see why a change of basis matrix between two orthonormal basis is unitary, let be two vectors in your space. Then

and

But we have seen in last post that a matrix is unitary exactly when for all

Corollary (Spectral Theorem for Hermitian Operators):For any Hermitian operator on a Hermitian space there is an orthonormal basis of consisting of eigenvectors for moreover, the eigenvalues of are real. Equivalently, for any hermitian matrix there is a unitary matrix such that is a real diagonal matrix.

**Proof:** Since a hermitian operator is a normal operator, we only need to show that its eigenvalues are all real. Let be an eigenvector for with eigenvalue and the matrix of in some basis. Then

Note that and that Then also

So we get and since we must have i.e.

QED

Corollary (Spectral Theorem for Unitary Matrix):For any unitary matrix there is a unitary matrix such that is diagonal. Equivalently, every conjugacy class in the unitary group contains a diagonal matrix.

**Proof:** Since unitary matrices are normal, there is nothing to prove.

QED

Corollary (Spectral Theorem for Symmetric Operators):For any symmetric operator on a Euclidean space there is an orthonormal basis of consisting of eigenvectors for moreover, the eigenvalues of are all real. Equivalently, for any real symmetric matrix there is an orthogonal matrix such that is a real diagonal matrix and the eigenvalues of are real.

**Proof:** A Symmetric operator on an Euclidean space can be seen as an hermitian operator on a hermitian space by extending to a complex vector space and by extending the inner product to a hermitian product.

QED

**Reference:** M. Artin, *Algebra*, 2nd edition.