# Subnormal operator

In mathematics, especially 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
    $N={\begin{bmatrix}A&B\\0&C\end{bmatrix}}$
for some bounded operators
    $B:H^{\perp }\rightarrow H,\quad {\mbox{and}}\quad C:H^{\perp }\rightarrow H^{\perp }.$

## Normality, quasinormality, and subnormality
### Normal operators

Every normal operator is subnormal by definition, but the converse is not true in general. 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
    $H\oplus H$
by
    $U={\begin{bmatrix}A&I-AA^{*}\\0&-A^{*}\end{bmatrix}}.$
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 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,
    $V={\begin{bmatrix}U&I-UU^{*}\\0&-U^{*}\end{bmatrix}}={\begin{bmatrix}U&D_{U^{*}}\\0&-U^{*}\end{bmatrix}}.$
Define
    $Q={\begin{bmatrix}P&0\\0&P\end{bmatrix}}.$
The operator N = VQ is clearly an extension of A. We show it is a normal extension via direct calculation. Unitarity of V means
    $N^{*}N=QV^{*}VQ=Q^{2}={\begin{bmatrix}P^{2}&0\\0&P^{2}\end{bmatrix}}.$
On the other hand,
    $NN^{*}={\begin{bmatrix}UP^{2}U^{*}+D_{U^{*}}P^{2}D_{U^{*}}&-D_{U^{*}}P^{2}U\\-U^{*}P^{2}D_{U^{*}}&U^{*}P^{2}U\end{bmatrix}}.$
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.

## Minimal normal extension
### Non-uniqueness of normal extensions

Given a subnormal operator A, its normal extension B is not unique. For example, let A be the unilateral shift, on l$^2$(N). One normal extension is the bilateral shift B on l$^2$(Z) defined by
    $B(\ldots ,a_{-1},{{\hat {a}}_{0}},a_{1},\ldots )=(\ldots ,{{\hat {a}}_{-1}},a_{0},a_{1},\ldots ),$
where ˆ denotes the zero-th position. B can be expressed in terms of the operator matrix
    $B={\begin{bmatrix}A&I-AA^{*}\\0&A^{*}\end{bmatrix}}.$
Another normal extension is given by the unitary dilation B' of A defined above:
    $B'={\begin{bmatrix}A&I-AA^{*}\\0&-A^{*}\end{bmatrix}}$
whose action is described by
    $B'(\ldots ,a_{-2},a_{-1},{{\hat {a}}_{0}},a_{1},a_{2},\ldots )=(\ldots ,-a_{-2},{{\hat {a}}_{-1}},a_{0},a_{1},a_{2},\ldots ).$

### Minimality

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*.)
One can show that if two operators B$_1$ and B$_2$ are minimal extensions on K$_1$ and K$_2$, respectively, then there exists a unitary operator
    $U:K_{1}\rightarrow K_{2}.$
Also, the following intertwining relationship holds:
    $UB_{1}=B_{2}U.\,$
This can be shown constructively. Consider the set S consisting of vectors of the following form:
    $\sum _{i=0}^{n}(B_{1}^{*})^{i}h_{i}=h_{0}+B_{1}^{*}h_{1}+(B_{1}^{*})^{2}h_{2}+\cdots +(B_{1}^{*})^{n}h_{n}\quad {\text{where}}\quad h_{i}\in H.$
Let K' ⊂ K$_1$ be the subspace that is the closure of the linear span of S. By definition, K' is invariant under B$_1$* and contains H. The normality of B$_1$ and the assumption that H is invariant under B$_1$ imply K' is invariant under B$_1$. Therefore, K' = K$_1$. The Hilbert space K$_2$ can be identified in exactly the same way. Now we define the operator U as follows:
    $U\sum _{i=0}^{n}(B_{1}^{*})^{i}h_{i}=\sum _{i=0}^{n}(B_{2}^{*})^{i}h_{i}$
Because
    $\left\langle \sum _{i=0}^{n}(B_{1}^{*})^{i}h_{i},\sum _{j=0}^{n}(B_{1}^{*})^{j}h_{j}\right\rangle =\sum _{ij}\langle h_{i},(B_{1})^{i}(B_{1}^{*})^{j}h_{j}\rangle =\sum _{ij}\langle (B_{2})^{j}h_{i},(B_{2})^{i}h_{j}\rangle =\left\langle \sum _{i=0}^{n}(B_{2}^{*})^{i}h_{i},\sum _{j=0}^{n}(B_{2}^{*})^{j}h_{j}\right\rangle ,$
, the operator U is unitary. Direct computation also shows (the assumption that both B$_1$ and B$_2$ are extensions of A are needed here)
    ${\text{if }}g=\sum _{i=0}^{n}(B_{1}^{*})^{i}h_{i},$
    ${\text{then }}UB_{1}g=B_{2}Ug=\sum _{i=0}^{n}(B_{2}^{*})^{i}Ah_{i}.$
When B$_1$ and B$_2$ are not assumed to be minimal, the same calculation shows that above claim holds verbatim with U being a partial isometry.