If $Sx\neq0$. Why $S^{1/2}Sx \neq0$? where $S\geq0$.

Question

Let $\mathcal{H}$ be a complex Hilbert space. Let $T\in \mathcal{B}(\mathcal{H})$ and let $S\in \mathcal{B}(\mathcal{H})^+$ (i.e. $S^*=S$ and $\langle Sx\;| \;x\rangle \geq0,\;\forall x\in \mathcal{H}$).

Assume that there exists $x\in \mathcal{H}$ such that $Sx\neq0$. Why $S^{1/2}Sx \neq0\,$ ?

Thank you

Answer 1

Assume to the contrary that $ S^{1/2}Sx =0$. Then

$S^2x=S^{1/2}(S^{1/2}Sx)=0$, hence

$||Sx||^2=\langle Sx|Sx\rangle =\langle S^2x|x\rangle=0$ and therefore $Sx=0$, a contradiction.