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!

  • $\begingroup$ What aren't you seeing exactly? Why the image of $ \eta (a) $ is included in the image of $ \rho $ ? Or what im refers to? $\endgroup$
    – cnada
    Commented Sep 5, 2018 at 1:20
  • $\begingroup$ yes, why the image is included. $\endgroup$
    – wdc
    Commented Sep 5, 2018 at 1:21

2 Answers 2


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.


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