diagonalizable operator

The expression ”diagonalizable operator” has several meanings in operator theory. The purpose of this entry is to present some commonly used concepts where this terminology appears.

0.1 Definition 1

Let $H$ be a finite dimensional Hilbert space. A linear operator $T:H⟶H$ is said to be diagonalizable if the corresponding matrix (in a given basis) is a diagonalizable matrix (http://planetmath.org/Diagonalizable2).

The above definition is equivalent to: There exists a basis of $H$ consisting of eigenvectors of $T$.

Remark - This is a common definition in linear algebra.

0.2 Definition 2

Let $H$ be a finite dimensional Hilbert space. A linear operator $T:H⟶H$ is said to be diagonalizable if there is an orthonormal basis of $H$ in which $T$ is represented by a diagonal matrix.

The above definition is equivalent to: There exists an orthonormal basis of $H$ consisting of eigenvectors of $T$.

Another equivalent definition is: There exists an orthonormal basis $\{e_1,\mathrm{\dots },e_n\}$ of $H$ and values ${\lambda }_1,\mathrm{\dots },{\lambda }_2\in ℂ$ such that

$$T(x)=\sum _{i=1}^n{\lambda }_i⟨x,e_i⟩e_i$$

Remarks -

• In linear algebra (http://planetmath.org/LinearAlgebra) such operators are also called unitarily diagonalizable.

• Diagonalizable operators (in this sense) are always normal operators. The Spectral theorem for normal operators (http://planetmath.org/SpectralTheoremForHermitianMatrices) assures that the converse is also true.

0.3 Definition 3

Let $H$ be a Hilbert space. A bounded linear operator $T:H⟶H$ is said to be diagonalizable if there exists an orthonormal basis consisting of eigenvectors of $T$.

An equivalent definition is: There exists an orthonormal basis ${\{e_i\}}_{i\in J}$ of $H$ and values ${\{{\lambda }_i\}}_{i\in J}$ such that

$$T(x)=\sum _{i\in J}{\lambda }_i⟨x,e_i⟩e_i$$

Remarks -

• If $H$ is finite dimensional this is the same as definition 2.

• Diagonalizable operators (in this sense) are always normal operators. For compact operators the converse is assured by an appropriate version of the spectral theorem for compact normal operators.

0.4 Definition 4

Let $H$ be a Hilbert space. A linear operator $T:H⟶H$ is said to be diagonalizable if it is to a multiplication operator (http://planetmath.org/MultiplicationOperatorOnMathbbL22) in some $L^2$-space (http://planetmath.org/L2SpacesAreHilbertSpaces), i.e. if there exists

• a measure space $(X,ℬ,\mu )$,

• a unitary operator $U:L^2(X)⟶H$ and

• a function $f\in$ $L^{\mathrm{\infty }}(X)$ (http://planetmath.org/LpSpace) such that

$$T=U{M}_f{U}^{*}$$

where $M_f:L^2(X)⟶L^2(X)$ is the operator of multiplication (http://planetmath.org/MultiplicationOperator) by $f$

$$M_f(\psi )=f.\psi .$$

Remarks -

• If $H={ℂ}^n$ the above definition is equivalent to say that $T$ is unitarily diagonalizable (Definition 2). Indeed, we can think of ${ℂ}^n$ as $L^2(\{1,\mathrm{\dots },n\})$ with the counting measure. In this case, multiplication operators correspond to diagonal matrices.

• Diagonalizable operators (in this sense) are necessarily normal operators (since multiplication operators are so). The discussion about the converse result is the content of general versions of the spectral theorem.