Subnormal operator

Every normal operator is subnormal by definition, but the converse is not true in general. A 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.

An operator A is said to be quasinormal if A commutes with A*A.^[2] 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.^[3]

Given a quasinormal A, the idea 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 subnormality.

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.

Given a subnormal operator A, its normal extension B is not unique. For example, let A be the unilateral shift, on l²(N). One normal extension is the bilateral shift B on l²(Z) defined by

${\displaystyle B(\ldots ,a_{-1},{{\hat {a}}_{0}},a_{1},\ldots )=(\ldots ,{{\hat {a}}_{-1}},a_{0},a_{1},\ldots ),}$

where ˆ denotes the zero-th position. B can be expressed in terms of the operator matrix

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

Another normal extension is given by the unitary dilation B' of A defined above:

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

whose action is described by

${\displaystyle B'(\ldots ,a_{-2},a_{-1},{{\hat {a}}_{0}},a_{1},a_{2},\ldots )=(\ldots ,-a_{-2},{{\hat {a}}_{-1}},a_{0},a_{1},a_{2},\ldots ).}$

Thus one is interested in the normal extension that is, in some sense, smallest. More precisely, a normal operator B acting on a Hilbert space K is said to be a minimal extension of a subnormal A if K' ⊂ K is a reducing subspace of B and H ⊂ K' , then K' = K. (A subspace is a reducing subspace of B if it is invariant under both B and B*.)^[4]

One can show that if two operators B₁ and B₂ are minimal extensions on K₁ and K₂, respectively, then there exists a unitary operator

${\displaystyle U:K_{1}\rightarrow K_{2}.}$

Also, the following intertwining relationship holds:

${\displaystyle UB_{1}=B_{2}U.\,}$

This can be shown constructively. Consider the set S consisting of vectors of the following form:

${\displaystyle \sum _{i=0}^{n}(B_{1}^{*})^{i}h_{i}=h_{0}+B_{1}^{*}h_{1}+(B_{1}^{*})^{2}h_{2}+\cdots +(B_{1}^{*})^{n}h_{n}\quad {\text{where}}\quad h_{i}\in H.}$

Let K' ⊂ K₁ be the subspace that is the closure of the linear span of S. By definition, K' is invariant under B₁* and contains H. The normality of B₁ and the assumption that H is invariant under B₁ imply K' is invariant under B₁. Therefore, K' = K₁. The Hilbert space K₂ can be identified in exactly the same way. Now we define the operator U as follows:

${\displaystyle U\sum _{i=0}^{n}(B_{1}^{*})^{i}h_{i}=\sum _{i=0}^{n}(B_{2}^{*})^{i}h_{i}}$

Because

${\displaystyle \left\langle \sum _{i=0}^{n}(B_{1}^{*})^{i}h_{i},\sum _{j=0}^{n}(B_{1}^{*})^{j}h_{j}\right\rangle =\sum _{ij}\langle h_{i},(B_{1})^{i}(B_{1}^{*})^{j}h_{j}\rangle =\sum _{ij}\langle (B_{2})^{j}h_{i},(B_{2})^{i}h_{j}\rangle =\left\langle \sum _{i=0}^{n}(B_{2}^{*})^{i}h_{i},\sum _{j=0}^{n}(B_{2}^{*})^{j}h_{j}\right\rangle ,}$

, the operator U is unitary. Direct computation also shows (the assumption that both B₁ and B₂ are extensions of A are needed here)

${\displaystyle {\text{if }}g=\sum _{i=0}^{n}(B_{1}^{*})^{i}h_{i},}$

${\displaystyle {\text{then }}UB_{1}g=B_{2}Ug=\sum _{i=0}^{n}(B_{2}^{*})^{i}Ah_{i}.}$

When B₁ and B₂ are not assumed to be minimal, the same calculation shows that above claim holds verbatim with U being a partial isometry.