Image of a sum of positive operators contains the images of each individual operator?

In the proof of Proposition 2.52 of John Watrous' QI book, 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 $$\text{im}(\eta(a))\subset\text{im}(\rho)$$, could someone please help explain. Thanks!

• What aren't you seeing exactly? Why the image of $\eta (a)$ is included in the image of $\rho$ ? Or what im refers to? Sep 5 '18 at 1:20
• yes, why the image is included.
– wdc
Sep 5 '18 at 1:21

It suffices to prove that if $P$ and $Q$ are positive semidefinite operators, then $$\operatorname{im}(P) \subseteq \operatorname{im}(P+Q).$$ Once you have this, the statement follows by taking $P = \eta(a)$ and $Q = \rho - \eta(a)$.
Suppose that $u$ is a vector with $u \perp \operatorname{im}(P+Q)$. This implies that $$0 = u^{\ast} (P + Q) u = u^{\ast} P u + u^{\ast} Q u.$$ As $u^{\ast} P u$ and $u^{\ast} Q u$ are both nonnegative and sum to zero, they must both be zero. Because $u^{\ast} P u = 0$, we have that $u \perp \operatorname{im}(P)$. We have just proved that $$\operatorname{im}(P+Q)^{\perp} \subseteq \operatorname{im}(P)^{\perp},$$ which is equivalent to the first containment above that we're aiming to prove, so we're done.
$$\eta(a) \subset U_{a \in \sum} \eta(a)$$ if I can write it this union way. By using the property that if a subset is included in another, the images follow this inclusion too (wikipedia ling for images) :
$$\mathrm{im}(\eta(a)) \subset \mathrm{im}(U_{a \in \sum} \eta(a)) = U_{a \in \sum} \mathrm{im}(\eta(a))$$
By definition, $$\rho$$ is applied over all elements in the alphabet $$\sum$$. So its image will be on the right side of the inclusion.