Image of a sum of positive operators contains the images of each individual operator? In the proof of Proposition 2.52 here: 
https://cs.uwaterloo.ca/~watrous/TQI/TQI.2.pdf,
there is the statement that $\text{im}(\eta(a))\subset\text{im}(\rho)$, where $\rho=\sum_{i=1}^{N}\eta(i)$ is a sum of positive operators and $\rho$ has trace one. 
I don't see this, could someone please help. Thanks!
 A: First consider the simple case where each positive operator $\eta(a)$ is rank one. Each operator may be thus be written as $\eta(a)=x_ax_a^*$ for some choice of vector $x_a$, in which case it clearly holds that $\operatorname{im}(x_ax_a^*)=\operatorname{span}\{x_a\}$. It trivially holds that $\operatorname{span}\{x_a\}\subseteq\operatorname{span}\{x_1,\dots,x_N\}$ for each $a$. The statement in this case therefore follows from the fact that 
$$
\operatorname{im}\Bigl(\sum_{a=1}^N x_ax_a^*\Bigr) = \operatorname{span}\{x_1,\dots,x_N\},
$$
which can be easily verified.
The general case can be seen as follows. As $\eta(a)$ is positive, there exist vectors $\{x_{a,1},\dots,x_{a,m}\}$ such that
$$
\eta(a)= \sum_{b=1}^{m}x_{a,b}x_{a,b}^*
$$
and $\operatorname{im}(\eta(a))=\operatorname{span}\{x_{a,1},\dots,x_{a,m}\}$. Moreover, one has
$$
\rho = \sum_{a=1}^N\sum_{b=1}^m x_{a,b}x_{a,b}^*.
$$
From the argument in the simple case above, it follows that $x_{a,b}\in\operatorname{im}(\rho)$ for each pair of indices $a$ and $b$. In particular, for each $a$ it holds that $x_{a,1},\dots,x_{a,m}\in\operatorname{im}(\rho)$ and thus $\operatorname{span}\{x_{a,1},\dots,x_{a,m}\}\subseteq\operatorname{im}(\rho)$.
