# The square of an hyponormal operator

If $$A$$ is an hyponormal operator on a complex Hilbert space $$F$$ (i.e. $$AA^* \leq A^*A$$).

Is $$A^2$$ also hyponormal? i.e. is $$A^2(A^2)^* \leq (A^2)^*A^2?$$

Recall that $$AA^* \leq A^*A$$ iff $$\langle (AA^*-A^*A)x,x\rangle\leq 0$$.

Note also that $$A$$ is hyponormal iff $$\|Ax\|\geq \|A^*x\|,$$ for all $$x\in F$$

In general no; if $$S$$ is the unilateral shift then e.g. the operator $$A=S^*+2S$$ is hyponormal but $$A^2$$ is not. This example appears as an exercise in Chapter II.4 of Conway's "Subnormal Operators" book, and can be checked by hand via a mildly tedious calculation. It may also follow from more general results of Ito and Wong ("Subnormality and Quasinormality of Toeplitz operators, Proc. Amer. Math. Soc., 34 (1972), pp157-164, MR0303334) --Conway gives this as a reference but I haven't read through it.