# 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.
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}