10 - Normal Operators on Hilbert Spaces

Summary

10.1 Compact Normal Operators

In the final chapter of the book, we will bring to bear all that we have learnt on the study of normal operators. In the process, we will prove a far reaching generalization of the Spectral Theorem for self-adjoint matrices that you would have first encountered in linear algebra. In fact, our first step in the process is to revisit that very result and prove it with more refined machinery.

For now let H denote a finite dimensional complex Hilbert space and let T ∊ B(H). If Λ is an ordered orthonormal basis for H, wewrite [T]_Λ for the matrix of T with respect to Λ. In other words, if Λ = ﹛e₁, e₂, … , e_n﹜, then the (i, j)^th entry of [T]_Λ is 〈T(e_j), e_i〉.

10.1.1 DEFINITION Let H be a finite dimensional Hilbert space. An operator T ∊ B(H) is said to be (unitarily) diagonalizable if H has an orthonormal basis consisting of eigenvectors of T.

Equivalently, T is diagonalizable if there is an orthonormal basis Λ of H such that [T]_Λ is a diagonal matrix.

10.1.2 REMARK A word of warning: Our notion of diagonalizability is stronger than the usual notion from linear algebra. For instance, if T : ℂ² → ℂ² is the operator T(x, y) = (x + y, 2y), then the matrix of T in the standard basis is

This matrix is similar to a diagonal matrix. In other words, there is an invertible matrix P such that P⁻¹AP is diagonal. However, the matrix P cannot be chosen to be a unitary, so T is not unitarily diagonalizable (why this is so is evident from Lemma 10.1.3 below).

Despite this, we will henceforth use the term diagonalizable to mean unitarily diagonalizable. The reason is that this condition is more easily captured in algebraic terms (without reference to the minimal or characteristic polynomial).

10.1.3 LEMMA Let H be a finite dimensional Hilbert space. If T ∊ B(H) is diagonalizable, then T is normal.