Affine transformations - the meaning of contractivity An affine transformation $\omega \colon \mathbb{R}^2 \to \mathbb{R}^2$ is a linear mapping followed by a translation, in other words
$$
\omega(x) = Ax+t = 
\begin{pmatrix}
a & b \\ c & d
\end{pmatrix}
\begin{pmatrix}
x_1 \\ x_2
\end{pmatrix}
+
\begin{pmatrix}
e \\ f
\end{pmatrix}.
$$
In general case, it is contractive (in terms of Euclidean metric) when the following is satisfied for some fixed $0 \le s < 1$
$$
\forall x,y \in \mathbb{R}^2 \quad d(\omega(x),\omega(y)) \le s \cdot d(x,y). 
$$

Now according to "Fractals everywhere" by M.F.Barnsley (exercise III.6.4) apparently under the assumption $\det(A-I) \neq 0$ all this holds if the operator norm $|A| \le s <1$. 
Why is this assumption needed? Can't we have a contractive mapping whose fixed point $x_f = (I-A)^{-1}t$ is, like, at infinity?
Another source provides the following conditions
$$
a^2+c^2 < 1; \\ 
b^2+d^2 < 1; \\ 
a^2+b^2+c^2+d^2 - (ad-bc)(ad-bc) < 1;
$$
but what's the meaning of these conditions? And is the inequality $\det(A-I) \neq 0$ automatically enforced in this case?
 A: Let $||.||$ be the induced norm on $M_n(\mathbb{R})$ associated to a
norm $||.||$ (same notation) on $\mathbb{R}^n$. If $\omega(x)=Ax+t$,
then (*): $d(\omega(x),\omega(y))=||\omega(x)-\omega(y)||=||Ax-Ay||\leq||A||||x-y||=||A||d(x,y)$.
We consider the equation $\omega(x)=x$; if $||A||<1$ then, according
to (*) and the Banach theorem, the sequence $x_{n+1}=f(x_n)$ converges
to the unique $z$ s.t. $\omega(z)=z$, that is $(I_n-A)z=t$. Note that
under the above hypothesis, for every $x\not= 0$, $||Ax||<||x||$, that
implies that $Ax\not= x$ and $A-I_n$ is invertible and
$z=(I_n-A)^{-1}t$.
Assume that $n=2,A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ and $||.||$ is the euclidean norm on $\mathbb{R}^2$.
Then $||A||^2$ is the max. of the eigenvalues of $AA^T$. The
characteristic polynomial of $AA^T$ is
$f(X)=X^2-(a^2+b^2+c^2+d^2)X+(ad-bc)^2$. Since $AA^T$ is PSD symmetric, the
solutions of $f(X)=0$ are $\geq 0$; moreover, we want that they are $<1$ and $||Ax||<||x||$.
Then the conditions are $f(1)>0,||Ae_1||<1,||Ae_2||<1$; that is $||A||<1$ iff $1>a^2+b^2+c^2+d^2-(ad-bc)^2,a^2+c^2<1,b^2+d^2<1$.
