Subnormal operator

In operator theory, subnormal operators are bounded operators on a Hilbert space defined by weakening the requirements for normal operators. Some examples of subnormal operators are isometries and Toeplitz operators with analytic symbols.

Definition

Let H be a Hilbert space. A bounded operator A on H is said to be subnormal if A has a normal extension. In other words, A is subnormal if there exists a Hilbert space K such that H can be embedded in K and there exists a normal operator N of the form

${\displaystyle N={\begin{bmatrix}A&B\\0&C\end{bmatrix}}}$ 

for some bounded operators

${\displaystyle B:H^{\perp }\rightarrow H,\quad {\mbox{and}}\quad C:H^{\perp }\rightarrow H^{\perp }.}$ 

Normality, quasinormality, and subnormality

Normal operators

Every normal operator is subnormal by definition, but the converse is not true in general. An simple class of examples can be obtained by weakening the properties of unitary operators. A unitary operator is an isometry with dense range. Consider now an isometry A whose range is not necessarily dense. A concrete example of such is the unilateral shift, which is not normal. But A is subnormal and this can be shown explicitly. Define an operator U on

${\displaystyle H\oplus H}$ 

by

${\displaystyle U={\begin{bmatrix}A&I-AA^{*}\\0&-A^{*}\end{bmatrix}}.}$ 

Direct calculation shows that U is unitary, therefore a normal extension of A. The operator U is called the unitary dilation of the isometry A.

Quasinormal operators

An operator A is said to be quasinormal if A commutes with A*A. A normal operator is thus quasinormal; the converse is not true. A counter example is given, as above, by the unilateral shift. Therefore the family of normal operators is a proper subset of both quasinormal and subnormal operators. A natural question is how are the quasinormal and subnormal operators related.

We will show that a quasinormal operator is necessarily subnormal but not vice versa. Thus the normal operators is a proper subfamily of quasinormal operators, which in turn are contained by the subnormal operators. To argue the claim that a quasinormal operator is subnormal, recall the following property of quasinormal operators:

Fact: A bounded operator A is quasinormal if and only if in its polar decomposition A = UP, the partial isometry U and positive operator P commute.

Given a quasinormal A, the ideal is to construct dilations for U and P in a sufficiently nice way so everything commutes. Suppose for the moment that U is an isometry. Let V be the unitary dilation of U,

${\displaystyle V={\begin{bmatrix}U&I-UU^{*}\\0&-U^{*}\end{bmatrix}}={\begin{bmatrix}U&D_{U^{*}}\\0&-U^{*}\end{bmatrix}}.}$ 

Define

${\displaystyle Q={\begin{bmatrix}P&0\\0&P\end{bmatrix}}.}$ 

The operator N = VQ is clearly an extension of A. We show it is a normal extension via direct calculation. Unitarity of V means

${\displaystyle N^{*}N=QV^{*}VQ=Q^{2}={\begin{bmatrix}P^{2}&0\\0&P^{2}\end{bmatrix}}.}$ 

On the other hand,

${\displaystyle NN^{*}={\begin{bmatrix}UP^{2}U^{*}+D_{U^{*}}P^{2}D_{U^{*}}&-D_{U^{*}}P^{2}U\\-U^{*}P^{2}D_{U^{*}}&U^{*}P^{2}U\end{bmatrix}}.}$ 

Because UP = PU and P is self adjoint, we have U*P = PU* and D_U*P = D_U*P. Comparing entries then shows N is normal. This proves quasinormality implies normality.

For a counter example that shows the converse is not true, consider again the unilateral shift A. The operator B = A + s for some scalar s remains subnormal. But if B is quasinormal, a straightforward calculation shows that A*A = AA*, which is a contradiction.