Given a family $A$ of $2\times 2$ matrices, $\rho(A)<1$ (JSR) if and only if there is a norm on $\mathbb{R}^2$ s.t each matrix in $A$ is contracting

Question

Given a family $A$ of $2\times 2$ matrices, $\rho(A)<1$ if and only if there is a norm on $\mathbb{R}^2$ s.t each matrix is contracting.

I encountered this in a paper and it seems obvious but I am struggling a little bit.

For the forward direction. Let $A=\{A_1,A_2,\cdots ,A_n\}$. Then we know that $\rho(A_i)\leq\rho(A)<1$. Thus we know each matrix has eigenvalues of modulus less than 1. How do I conclude there is a norm that makes all of them contracting at the same time?

For the backwards direction: Assume there is such norm $\|\|$. Then by sub-multiplicativity we get that $\rho(A)\leq 1$ but I do not see the strict inequality.

Any help would be appriciated.

Answer 1

What you have to look is the concept of Barabanov norm and this is not restricted to $2\times 2$ matrices. This is not a trivial result.

Consider the family $A:=\{A_1,\ldots,A_N\}$ of $n\times n$ irreducible matrices with joint spectral radius $\rho(A)$. Then, there exists such a norm $V$ on $\mathbb{R}^n$ such that

$$\max_{i=1,...,n} V(A_ix) = \rho(A) V(x),\ \mathrm{for\ all\ }x \in\mathbb{R}^n.$$

This was first proven in N. E. Barabanov, Lyapunov indicator for discrete inclusions, I-III, Autom. Remote Control, 49 (1988), No 2, 152–157.