Hyponormal operator

Remarks

Example: for the unilateral shift $S(x_0, x_1, \ldots) = (0, x_0, x_1, \ldots)$ on $\ell^2(\N)$, one has $S^*S = I$ and $SS^*$ is the projection onto the vectors with $x_0 = 0$. So $S^*S - SS^*$ is the projection onto the first coordinate, which is positive; $S$ is hyponormal.

Non-example: the backward shift $S^*$ gives $(S^*)^* S^* - S^*(S^*)^* = SS^* - S^*S$, the negative of that projection, and $\ip{(SS^* - S^*S)e_0}{e_0} = -1 < 0$ for $e_0 = (1, 0, 0, \ldots)$.

Equivalent form: $T$ is hyponormal if and only if $\norm{T^*x} \le \norm{Tx}$ for all $x \in H$, since $\ip{(T^*T - TT^*)x}{x} = \norm{Tx}^2 - \norm{T^*x}^2$. The operator $T$ is normal exactly when both $T$ and $T^*$ are hyponormal.