comparision of two rank one operators Suppose $S,T$ are two rank one operators on some Hilbert space $H$,If $S\leq T$,can we conclude that $T=kS$ for some $k\in \Bbb C$?
 A: Contrary to what the comments say, the answer is no when $S$ is not required to be positive (even if selfadjoint). For instance can take any two rank-one positive operators $R,T$, and then $S=-R$ satisfies $S\leq T$ while not being a multiple of $T$. Say,
$$
\begin{bmatrix}0&\phantom{-}0\\0&-1\end{bmatrix} \leq \begin{bmatrix} 1&0\\0&0\end{bmatrix}. 
$$
When the requirement is $0\leq S\leq T$ then yes, if $T$ is rank-one then $S$ is rank-one and $S=kT$ for some $k\in\mathbb C$. Indeed, if $\operatorname{ran}T=\mathbb Cx$, then $\ker T=(\operatorname{ran}T^*)^\perp=(\operatorname{ran}T)^\perp=\{x\}^\perp$. So if $T\ne0$ (if $T=0$ then $S=0$) we have $Tx\ne0$. So $Tx=\lambda x$ with $\lambda\ne0$. It follows that $T=\lambda P$, where $P$ is the orthogonal projection onto $\mathbb Cx$. Now if $y\perp x$, then $$\langle Sy,y\rangle\leq \langle Ty,y\rangle=0,$$ so $Sy=0$ (here we use that $S\geq0$). So $\ker S\subset \{x\}^\perp$, and since $S$ is selfadjoint it follows that $\operatorname{ran}S=\mathbb Cx$. So $S=\mu P$, and now we can compare $\mu$ and $\lambda$. 
The result can also obtained from the fact that $0\leq S\leq T$ implies that $S^{1/2}=RT^{1/2}$ for some contraction $R$. Here, as $T^{1/2}$ is rank-one, so is $RT^{1/2}=S$. 
