Quasinormal operator

In operator theory, quasinormal operators is a class of bounded operators defined by weakening the requirements of a normal operator.
Every quasinormal operator is a subnormal operator. Every quasinormal operator on a finite-dimensional Hilbert space is normal.
Definition and some properties.
Definition.
Let "A" be a bounded operator on a Hilbert space "H", then "A" is said to be quasinormal if "A" commutes with "A*A", i.e.
Properties.
A normal operator is necessarily quasinormal.
Let "A" = "UP" be the polar decomposition of "A". If "A" is quasinormal, then "UP = PU". To see this, notice that
the positive factor "P" in the polar decomposition is of the form ("A*A"), the unique positive square root of "A*A". Quasinormality means "A" commutes with "A*A". As a consequence of the continuous functional calculus for self adjoint operators, "A" commutes with "P" = ("A*A") also, i.e.
So "UP = PU" on the range of "P". On the other hand, if "h" ∈ "H" lies in kernel of "P", clearly "UP h" = 0. But "PU h" = 0 as well. because "U" is a partial isometry whose initial space is closure of range "P". Finally, the self-adjointness of "P" implies that "H" is the direct sum of its range and kernel. Thus the argument given proves "UP" = "PU" on all of "H".
On the other hand, one can readily verify that if "UP" = "PU", then "A" must be quasinormal. Thus the operator "A" is quasinormal if and only if "UP" = "PU".
When "H" is finite dimensional, every quasinormal operator "A" is normal. This is because that in the finite dimensional case, the partial isometry "U" in the polar decomposition "A" = "UP" can be taken to be unitary. This then gives
In general, a partial isometry may not be extendable to a unitary operator and therefore a quasinormal operator need not be normal. For example, consider the unilateral shift "T". "T" is quasinormal because "T*T" is the identity operator. But "T" is clearly not normal.
Quasinormal invariant subspaces.
It is not known that, in general, whether a bounded operator "A" on a Hilbert space "H" has a nontrivial invariant subspace. However, when "A" is normal, an affirmative answer is given by the spectral theorem. Every normal operator "A" is obtained by integrating the identity function with respect to a spectral measure "E" = {"EB"} on the spectrum of "A", "σ"("A"):
For any Borel set "B" ⊂ "σ"("A"), the projection "EB" commutes with "A" and therefore the range of "EB" is an invariant subspace of "A".
The above can be extended directly to quasinormal operators. To say "A" commutes with "A*A" is to say that "A" commutes with ("A*A"). But this implies that "A" commutes with any projection "EB" in the spectral measure of ("A*A"), which proves the invariant subspace claim. In fact, one can conclude something stronger. The range of "EB" is actually a "reducing subspace" of "A", i.e. its orthogonal complement is also invariant under "A".