Example of an operator that is not subnormal

Question

In some recent questions the term subnormal operator has appeared.

A bounded operator $A$ acting on a Hilbert space $H$ is called subnormal if there exists a Hilbert space $K$ containing $H$ as a subspace and a normal operator $N$ on $K$ (meaning bounded and $N^*N^\mathstrut=N^\mathstrut N^*$) so that:

$$N= \begin{pmatrix}A & B \\0 & C \end{pmatrix}$$

For bounded operators $B: H^\perp \to H$, $C: H^\perp \to H^\perp$.

My question is, what is an example of an operator that is not subnormal?

Answer 1

An aproach is using that every subnormal operator is hyponormal. Then, if we exhibit an non hyponormal, we finish.

Take $H=\ell^2$ and $S$ the right shift. $T=(S^\ast+2S)^2$ is not hyponormal (the prove is straighforward). Thus $T$ is not subnormal.

Answer 2

Fact 1. Every subnormal operator is Hyponormal.

Fact 2. In finite dimension every Hyponormal operator is normal operator.(Since $\operatorname{tr}(A^*A-AA^*)=0$ and only positive definite operator having trace zero is the zero operator.)

So if you consider any non normal operator in finite dimensional Hilbert space, for example any nilpotent matrix, gives you a example of non subnormal operator.

Fact 3. An weighted shift is hyponormal iff weights are nondecreasing. So take any weighted shift whose weights are not nondecreasing, then it won't be hyponormal operator and hence cannot be subnormal.