Linear map with spectrum radius $<1$ is contraction in some norm if the basis is well-chosen Given a norm $x\mapsto \lVert x \rVert$ on $\mathbb{R}^n$, a linear map $f$ is contracting if $\exists \varepsilon >0$ such that $\forall x \in \mathbb{R}^n,\lVert f(x) \rVert \leq (1-\varepsilon)\lVert x \rVert$. Assume that the norm is either the sup norm
$\lVert x \rVert_{\infty}=\max\limits_{1\leq i \leq n}|x_i|$ 
or the $\ell^1$norm 
$\lVert x \rVert_{1}=|x_1|+...+|x_n|$
or the Euclidean norm 
$\lVert x \rVert_{2}=\sqrt{x_{1}^2+...+x_{n}^2}$ and assume that $Spec_{\mathbb{c}}(f)\subset B(0,1)$.
Prove that after a suitable linear change of variable, $f$ can be assumed to be contracting for the chosen norm. Find an example where the change of variable is necessary. 
In my opinion, the problem is to find a suitable basis, under which the matrix $A$ of $f$ can meet the requirement that the norm of $A$ $<1$. However, no mater in which case above, I have difficulty finding the suitable linear change of variable. I only figure out the case in which $A$ can be diagonalized. In this case I can use the Jordan normal form of $f$ to work out the Euclidean norm. But in other cases, I have difficulties. 
 A: The Jordan normal form of $f$ (extended to $\Bbb C^n$) has eigenvalues with $|\lambda|<1$ on the diagonal and some $1$'s next to the diagonal. By scaling basis vectors, we can turn those $1$'s into positive $\varepsilon$'s of our choice. We choose $\epsilon<1-\max|\lambda|$ and let
$q:=\max\{\max|\lambda|+\epsilon, \sqrt{\max|\lambda|^2+\epsilon^2},\max\{ \max|\lambda|,\epsilon\}\}<1$. So now we have a complex basis $v_1,\ldots,v_n$ where $f(v_k)=\lambda v_k$ or $f(v_k)=\lambda v_k+\epsilon v_{k-1}$. With respect to this basis, 
$$\|f(v_k)\|_1\le |\lambda|+\epsilon<1,\quad \|f(v_k)\|_2\le \sqrt{|\lambda|^2+\epsilon^2}<1,\quad\|f(v_k)\|_\infty\le \max\{|\lambda|,\epsilon\}<1,  $$
i.e., 
$$ \|f(v_k)\|\le q\|v_k\|.$$
Now transfer this to general $v=\sum c_kv_k$, and finally use that we may assume wlog that if $v_k$ is in the base, then so is $\overline{v_k}$. Take the $v_k+\overline{v_k}$ and (for $\lambda\notin \Bbb R$) the $i(v_k-\overline{v_k})$ as a real basis and show that our inequaloty still holds.
