Question: Let $\mathcal H$ be some finite dimensional Hilbert space and $\mathbb H(\mathcal H)$ the real vector space of Hermitian operators on $\mathcal H$. The authors of this paper: <https://arxiv.org/pdf/quant-ph/0404137.pdf> claim on page 2 that if $\{F\_c \}$ is a set of positive semi-definite operators which span the space of Hermitian operators $\mathbb H(\mathcal H)$, then the operator $$G := \sum\_c F\_c$$ is "trivially non singular". This is not so trivial to me. Can someone please explain why $G$ must be invertible?

Answer: 
$\langle x,Gx \rangle \geq0$. If $\langle x,Gx \rangle = 0$ then $\langle x,F\_c x \rangle = 0$ for each c in which case $\{F\_c\}$ cannot span the space of Hermitian operators; in fact the identity operator is not in the span. Hence $\langle x,Gx \rangle > 0$ and G is invertible.
This is an answer to the original question in which the operators $F\_c$ were stipulated to be positive:

If $A$ is a bounded operator on $\mathcal H$ such that

$\langle x, Ax \rangle > 0 \tag 1$

for all $x \in \mathcal H$, then $A$ must be invertible, since otherwise there exists $y \in \mathcal H$ with

$Ay = 0 \Longrightarrow \langle y, Ay \rangle = 0. \tag 2$

In the light of this fact, we have for every $x \in \mathcal H$,

$\langle x, G x \rangle = \displaystyle \sum\_c \langle x, F\_c x \rangle > 0 \tag 3$

by the positivity of the $F\_c$, and we conclude $G$ is invertible, hence non-singular. 

We note that the $F\_c$ needn't span $\Bbb H(\mathcal H)$ for the above to be valid.

The argument is reasonably trivial.