Conditions for affine transform to be contractive

Question

What are conditions for affine transform $ w \begin{pmatrix} x_{1}\\ x_{2} \end{pmatrix} = A x + t= \begin{pmatrix} a & b \\ c & d\end{pmatrix} \begin{pmatrix} x_{1} \\ x_{2} \end{pmatrix} + \begin{pmatrix} e \\ f \end{pmatrix} $ to be a contraction on $\mathbb{R}^2$ ?

First, I found in Barnsely, "Fractals Everywhere", that if $S\subset R^2$, then $area(A(S)) = |det A| ( area(S))$. Is it then sufficient to hold: $|det A|< 1$?

Second, how is this connected with a condition that $w$ is contractive when $ ||A||_{2} < 1$? Are these conditions equivalent?

Answer 1

The first condition is not equivalent, as it uses the area, whereas contraction is about length. To see why, consider the matrix $$\begin{pmatrix} a&0\\0&\frac1{a^2}\end{pmatrix}.$$

We can make $a$ arbitrarily large and the determinant will get arbitrarily small, but it still won't be a contraction - things still get stretched in one direction.

You can see this in another light by considering the eigenvalues of the matrix. To be a contraction you need all of the eigenvalues to be less than $1$, while the determinant only tells you about their product.

Answer 2

Let $f:(E_1,d_1)\rightarrow (E_2,d_2)$. $f$ is a contractive mapping iff there is $k\in [0,1)$ s.t. for every $x,y$, $d_2(f(x),f(y))\le kd_1(x,y)$; $f$ is a contraction mapping when $E_1=E_2,d_1=d_2$.

Here, $E_1=E_2$ and assume that $w$ is a contraction for $d(x,y)=||x-y||_2$. Then for every $x,y$, $||A(x-y)||_2\leq k||x-y||_2$; that is equivalent to $\sup_{||u||=1}||Au||_2\leq k$; that is, there is $k<1$ s.t. $||A||_2\leq k$.

Conclusion. $w$ is a contraction iff $\rho(AA^T)<1$.

Remark. The condition $\rho(A)<1$, given by @Carmeister, is necessary but it does not suffice; indeed (for example) the matrix $A=\begin{pmatrix}0&a\\0&0\end{pmatrix}$ is a contraction only for $|a|<1$.