Quasinormal, subnormal and hyponormal operators

An operator $A$ is called quasinormal if $A$ commutes with $A^{*}A$. Every normal operator is quasinormal, but the converse is false: if $A$ is an isometry, then $A^{*}A = 1$ and therefore $A$ commutes with $A^{*}A$, but if $A$ is not unitary, then $A$ is not normal (a concrete example is the unilateral shift).

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.

If the underlying Hilbert space is finite-dimensional, then it is an easy exercise in linear algebra to prove that $\left\| 1-C\right\| \geq 1$ for all commutators $C$.

The self-commutator of an operator $A$ is the operator $A^{*}A - AA^{*}$.

An operator is called abnormal if no non-zero subspace of the kernel of $A^{*}A - AA^{*}$ reduces $A$ (that is, if it has no normal direct summand).