Yesterday I was considering the matrix
$$ A = \left( \begin{array}{cc} 1-\alpha & \alpha - \alpha ^2 \\ 1 & 0 \\ \end{array} \right) \approx \left( \begin{array}{cc} -0.47 & -0.68 \\ 1 & 0 \\ \end{array} \right), $$
where $\alpha>0$ is the unique positive root of $x^3 - x^2 - 1 =0$, i.e. $\alpha \approx 1.47$.
Since $A$ has complex conjugate eigenvalues which are inside the unit circle $(\lambda_\pm \approx -0.23\pm0.79\, i,\ |\lambda_\pm| < 0.83)$, I presumed that there was some contraction, i.e.
$$ \|A\,x\| < 0.83 \|x\|,$$
for any $x \in \mathbb R^{2,1}$ and for whatever norm you cared to put on $\mathbb R^{2,1}$ $-$in patrticular, the operator norm of $A$ is less than $1$.
But this is not true for all norms: this is because $A$ has a singular value greater than one $(\approx 1.15)$, so it doesn't contract in the Euclidean norm, for instance. That said, because there is some rotation involved (the eigenvalues are not real), I suspect that there is at least one norm with respect to which, $A$ is a contraction.
My question: is there a norm on $\mathbb R^2$ with respect to which $A$ contracts?
Thanks!
When $A$ has a conjugate pair of (non-real) eigenvalues $re^{\pm i\theta}$ (with $r\ge0$ and $\theta\in(-\pi,\pi)\setminus\{0\}$), let $u-iv$ (with $u,v\in\mathbb R^2$) be an eigenvector corresponding to the eigenvalue $re^{i\theta}$ and let $P$ be the augmented matrix $[u|v]$. Then $A=rPQP^{-1}$, where $Q$ is the rotation matrix for the angle $\theta$. So, if we define a norm on $\mathbb R^2$ by $\|x\|=\|P^{-1}x\|_2$, then $$ \|A\|=\max_{\|x\|=1}\|Ax\| =\max_{\|P^{-1}x\|_2=1}\|(P^{-1}AP)(P^{-1}x)\|_2 =\max_{\|y\|_2=1}\|rQy\|_2=\|rQ\|_2=r $$ and hence $\|Ax\|\color{red}{\le}\|A\|\|x\|=r\|x\|$ for every $x\in\mathbb R^2$.
In your case, $r\approx0.83<1$. Therefore $A$ is a contraction with respect to the norm $\|\cdot\|$.