Is there any relation between norm and determinant of matrix? I was going through an IEEE paper on Graph Neural Network Model (page 66) and there is one proof where I read 

$\lVert (\partial\Theta/\partial x)(x,w) \rVert \geq (1-\mu)$ thus determinant
  of $(\partial\Theta/\partial x)(x,w)$ is not null.

Where $\Theta = x -
F_w(x,l)$, $\Theta$ is a vector and $ \mu \in [0,1)$.
Now, I do not understand how they deduce that determinant will be not null on basis of the norm? 
 A: We have $(\partial \Theta/\partial x) = I - (\partial F_w/\partial x)$ in the paper. The authors claim that, since (a) $\|(\partial F_w/\partial x)\|\le\mu$ for some $0\le\mu<1$, we have (b) $\|(\partial \Theta/\partial x)\|\ge1-\mu$ and (in turn?) (c) the determinant of $(\partial \Theta/\partial x)$ is nonzero.
It is true that (a) implies (b). However, as you have already noticed, we cannot infer (c) from (b), because we cannot deduce from a matrix's norm that the determinant of that matrix is nonzero.
That said, we can still infer (c) from (a) from the well-known property of Neumann series: since $\|(\partial F_w/\partial x)\|<1$, the Neumann series
$$
I+(\partial F_w/\partial x)+(\partial F_w/\partial x)^2+\cdots\tag{1}
$$
converges and
$$
\bigg(I-(\partial F_w/\partial x)\bigg)\bigg(I+(\partial F_w/\partial x)+(\partial F_w/\partial x)^2+\cdots\bigg)=I,\tag{2}
$$
meaning that $(\partial \Theta/\partial x) = I - (\partial F_w/\partial x)$ is invertible and its determinant is nonzero.
