Let $H$ be a Hilbert space and let $S\in B(H)$ be a selfadjoint linear operator. We say that $S$ is positive if for all $x\in H$, $$\langle Sx,x\rangle_H\ge0.$$ If $V\in B(H)$ a not necessarily selfadjoint bounded linear operator, then we say that $V$ is positive-definite if for all $x\in H$, $$\left\langle Vx,x\right\rangle_H\ge c\|x\|_H^2$$ for some $c\in (0,\infty)$.
Now, let $T\in B(H)$ be a bounded linear operator which isn't necessarily selfadjoint. Then it is known that the operator $T^*T\in B(H)$ is selfadjoint and positive in the sense defined above.
My questions:
(1) is it possible to obtain a similar lower bound in terms of a positivity constant $c$ like in the positive-definite definition above when considering $\langle T^*Tx,x\rangle_H$, or does one know only at most that for all $x\in H$ we have $\langle T^*Tx,x\rangle_H\ge0$?
(2) Is there any workaround to obtaining such a bound available by restricting our attention to $x\in H/\{0\}$? If not, that would suggest to me that there are nonzero $x\in H$ for which $\langle Sx,x\rangle_H=0$. Are there examples of positive operators (positive as defined above) for which this happens?
Not without more conditions on $T$! @Daw points out an important example, but for a non-zero counter example, suppose $H$ is separable with orthonormal basis $\{\phi_i\}_{i \in N}$. Then if $A$ is a subset of $\mathbb{N}$, define $T(\cdot) = \sum_{i \in A} c_i \langle \phi_i, \cdot \rangle \phi_i$, so that $T^*T= \sum_{i \in A} c_i^2 \langle \phi_i, \cdot \rangle \phi_i$. Notice that if $x \in span(\phi_i, i\in A^c)$, then $T^*T(x) = 0$, hence $\langle T^*T(x) , x \rangle =0 $ for a host of nonzero $x$'s.
At the minimum you need that $ker(T^*T)={0}$, but even this is not enough-- consider the same example with $A= \mathbb{N}$ and with the $c_i \ne 0$ but $c_i \to 0$. Strict positivity requires stronger assumptions I'm afraid.