Why is the condition $\|Z\| < 1$ equivalent to $I - ZZ^{\top} > 0$? As the title says, for a matrix $Z \in \mathbb{R}^{p \times q}$, the condition $\begin{Vmatrix}Z\end{Vmatrix} < 1$ equivalent to $I - ZZ^{\top} > 0$. How can I show the equivalence?
Attempt:
$\begin{Vmatrix}Z\end{Vmatrix} = \sup_{|x| = 1} \begin{Vmatrix}Zx\end{Vmatrix}$
$\implies$ $\begin{Vmatrix}Zx\end{Vmatrix} < 1$ $\implies$ $(Zx)^{\top}(Zx) < 1$ $\implies$ $x^{\top}Z^{\top}Zx < 1$.
Multiplying by the identity matrix on both sides,
$\implies$ $(x^{\top}Z^{\top}Zx)I < I \implies I - (x^{\top}Z^{\top}Zx)I > 0 $.

You are almost there.
\begin{align}
I-ZZ^T\succ0
&\Leftrightarrow x^T(I-ZZ^T)x>0 \text{ for every unit vector } x\\
&\Leftrightarrow 1-\|Zx\|^2>0 \text{ for every unit vector } x\\
&\Leftrightarrow \|Zx\|<1 \text{ for every unit vector } x\tag{1}\\
&\Leftrightarrow \|Z\|<1\tag{2}.
\end{align}
In $(1)\Rightarrow(2)$, we have used the fact that the value of $f:x\mapsto\|Zx\|$ attains maximum on the unit sphere because the unit sphere is compact and $f$ is continuous.
