Subnormal Operators

It is trivial that if an operator belongs to a commutative algebra that is closed under the formation of adjoints, then that operator is normal. If, conversely, an algebra is closed under the formation of adjoints and consists of normal operators only, then that algebra is commutative. (Proof: if $A+iB$ and $C+iD$ are in the algebra, with $A$, $B$, $C$, and $D$ Hermitian, then $A$, $B$, $C$, and $D$ are in the algebra, because adjoints are formable, and therefore so are $A+iC$, $A+iD$, $B+iC$, and $B+iD$; the assumed normality implies that everything commutes.) This problem asks what happens if the condition that the algebra be closed under adjoints is dropped.

Here, as usual, $\chi$ denotes the characteristic function of the set indicated by its subscript.

An operator $A$ on a Hilbert space $\mathbf{H}$ is subnormal if there exists a normal operator $B$ on a Hilbert space $\mathbf{K}$ such that $\mathbf{H}$ is a subspace of $\mathbf{K}$, the subspace $\mathbf{H}$ is invariant under the operator $B$, and the restriction of $B$ to $\mathbf{H}$ coincides with $A$; such a $B$ is called a normal extension of $A$. An operator $A$ is called quasinormal if $A$ commutes with $A^{*}A$.

Normality implies quasinormality, but not conversely (witness the unilateral shift). The present assertion is that quasinormality implies subnormality, but, again, the converse is false. To get a counterexample, add a non-zero scalar to the unilateral shift. The result is just as subnormal as the unilateral shift, but a straightforward computation shows that if it were also quasinormal, then the unilateral shift would be normal.

An operator $A$ is called quasinormal if $A$ commutes with $A^{*}A$. A subspace is invariant under an operator if the operator maps it into itself; "non-trivial" means different from both $0$ and the whole space.

An operator $A$ on $\mathbf{H}$ is subnormal if there exists a normal operator $B$ on a Hilbert space $\mathbf{K}$ such that $\mathbf{H}$ is a subspace of $\mathbf{K}$, the subspace $\mathbf{H}$ is invariant under $B$, and the restriction of $B$ to $\mathbf{H}$ coincides with $A$. A normal extension $B$ (on $\mathbf{K}$) of a subnormal operator $A$ (on $\mathbf{H}$) is minimal if there is no reducing subspace of $B$ between $\mathbf{H}$ and $\mathbf{K}$; that is, whenever $\mathbf{M}$ reduces $B$ and $\mathbf{H} \subset \mathbf{M}$, it follows that $\mathbf{M} = \mathbf{K}$.

This result shows that it is permissible to speak of "the" minimal normal extension of a subnormal operator. Typical example: the minimal normal extension of the unilateral shift is the bilateral shift.

An operator $A$ on $\mathbf{H}$ is subnormal if there exists a normal operator $B$ on a Hilbert space $\mathbf{K}$ such that $\mathbf{H}$ is a subspace of $\mathbf{K}$, the subspace $\mathbf{H}$ is invariant under $B$, and the restriction of $B$ to $\mathbf{H}$ coincides with $A$; such a $B$ is the minimal normal extension of $A$ if there is no reducing subspace of $B$ strictly between $\mathbf{H}$ and $\mathbf{K}$.

The first non-trivial example of a subnormal operator shows that the spectrum need not simply grow along with the operator: if $A$ is the unilateral shift and $B$ is the bilateral shift, then $\operatorname {spec} A$ is the unit disc, whereas $\operatorname {spec} B$ is only the perimeter of the unit disc.

An operator $A$ on $\mathbf{H}$ is subnormal if there exists a normal operator $B$ on a Hilbert space $\mathbf{K}$ such that $\mathbf{H}$ is a subspace of $\mathbf{K}$, the subspace $\mathbf{H}$ is invariant under $B$, and the restriction of $B$ to $\mathbf{H}$ coincides with $A$; such a $B$ is the minimal normal extension of $A$ if there is no reducing subspace of $B$ strictly between $\mathbf{H}$ and $\mathbf{K}$. A hole in a compact subset of the complex plane is a bounded component of its complement.

The spectral inclusion theorem (Problem 200) for subnormal operators can be sharpened by this result: the spectrum of a subnormal operator is always obtained from the spectrum of its minimal normal extension by "filling in some of the holes."

An operator $A$ on $\mathbf{H}$ is subnormal if there exists a normal operator $B$ on a Hilbert space $\mathbf{K}$ such that $\mathbf{H}$ is a subspace of $\mathbf{K}$, the subspace $\mathbf{H}$ is invariant under the operator $B$, and the restriction of $B$ to $\mathbf{H}$ coincides with $A$; such a $B$ is called a normal extension of $A$.

If $A$ (on $\mathbf{H}$) is subnormal, with normal extension $B$ (on $\mathbf{K}$), where an operator $A$ on $\mathbf{H}$ is subnormal if there exists a normal operator $B$ on a Hilbert space $\mathbf{K}$ such that $\mathbf{H}$ is a subspace of $\mathbf{K}$, the subspace $\mathbf{H}$ is invariant under $B$, and the restriction of $B$ to $\mathbf{H}$ coincides with $A$, then the relation between $A^{*}$ and $B^{*}$ is best expressed in terms of the projection $P$ from $\mathbf{K}$ onto $\mathbf{H}$. If $f$ and $g$ are in $\mathbf{H}$, then

Since the operator $PB^{*}$ on $\mathbf{K}$ leaves $\mathbf{H}$ invariant, its restriction to $\mathbf{H}$ is an operator on $\mathbf{H}$, and, according to the preceding chain of equations, that restriction is equal to $A^{*}$. That is the answer:

for every $f$ in $\mathbf{H}$.

This relation between $A^{*}$ and $B^{*}$ has a curious consequence. If $f \in \mathbf{H}$, then

The result ($\left\| A^{*}f\right\| \leq \left\| Af\right\|$) can be reformulated in another useful way; it is equivalent to the operator inequality

(Indeed: $\left\| A^{*}f\right\|^2 = \left\langle AA^{*}f, f \right\rangle$ and $\left\| Af\right\|^2 = \left\langle A^{*}Af, f \right\rangle$.)

An operator $A$ such that $A^{*}A \geq AA^{*}$ is called hyponormal. The result of the discussion above is that every subnormal operator is hyponormal. On a finite-dimensional space every hyponormal operator is normal: since $\operatorname {tr}(AB)$ is always equal to $\operatorname {tr}(BA)$, it follows that $\operatorname {tr}(A^{*}A - AA^{*})$ is always $0$; if $A^{*}A \geq AA^{*}$, then $A^{*}A - AA^{*}$ is a positive operator with trace $0$, and therefore $A^{*}A - AA^{*} = 0$. This is a generalization of the statement that on a finite-dimensional space every subnormal operator is normal (cf. Problem 195).

In both cases, one or the other of the direct summands may be absent.

An operator $A$ such that $A^{*}A \geq AA^{*}$ is called hyponormal.

An operator $A$ such that $A^{*}A \geq AA^{*}$ is called hyponormal. It follows from the discussion of hyponormal operators on finite-dimensional spaces (Problem 203) that a hyponormal operator of finite rank (on a possibly infinite-dimensional space) is always normal; this problem asks about limits of operators of finite rank.

An operator $A$ is compact if and only if both $\re A$ and $\im A$ are compact. An operator $A$ such that $A^{*}A \geq AA^{*}$ is called hyponormal; Problem 206 can be stated in this form: if $A$ is hyponormal and both $\re A$ and $\im A$ are compact, then $A$ is normal. It should be kept in mind that even if $A$ is hyponormal, $A^{*}$ need not be. What is trivially true, however, is that $iA$ is hyponormal at the same time as $A$ is and $\im (iA) = \re A$.

An operator $A$ is called quasinormal if $A$ commutes with $A^{*}A$; an operator $A$ on $\mathbf{H}$ is subnormal if there exists a normal operator $B$ on a Hilbert space $\mathbf{K}$ such that $\mathbf{H}$ is a subspace of $\mathbf{K}$, the subspace $\mathbf{H}$ is invariant under $B$, and the restriction of $B$ to $\mathbf{H}$ coincides with $A$; an operator $A$ such that $A^{*}A \geq AA^{*}$ is called hyponormal. Normality implies quasinormality, quasinormality implies subnormality, and subnormality implies hyponormality; a normal idempotent must be Hermitian (and hence a projection).

An operator $A$ such that $A^{*}A \geq AA^{*}$ is called hyponormal. Every power of a normal operator is normal, and, as an almost equally trivial consequence, every power of a subnormal operator is subnormal; this problem shows that the corresponding fact for hyponormal operators is false. Any solution is automatically also an example of a hyponormal operator that is not subnormal (Problem 203), since a subnormal operator's square would have to be subnormal, hence hyponormal; the converse is not true, since the hyponormal operator used in Solution 203 has the property that all its powers are hyponormal also.