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 $.

Alternatively, thanks to a comment by @Rodrigo de Azevedo, from Maximize $\langle \mathrm A , \mathrm X \rangle$ subject to $\| \mathrm X \|_2 \leq 1$,
\begin{align}
\|Z\| < 1
&\Leftrightarrow \sigma_{\max}(Z) < 1 \\
&\Leftrightarrow \lambda_{\max}(Z^{\top}Z) < 1 \\
&\Leftrightarrow \lambda_{\max}(ZZ^{\top}) < 1 \quad (\text{SVD decomposition}) \\
&\Leftrightarrow 1 - \lambda_{\max}(ZZ^{\top}) > 0 \\
&\Leftrightarrow \lambda_{\min}(I - ZZ^{\top}) > 0 \\
&\Leftrightarrow (I-ZZ^{\top}) \succ 0_{p\times p}.
\end{align}
