The spectrum of a linear operator

Question

Prove that the mapping $$x → Ax + v$$ of the space $R ^ n$ into itself is contractive if and only if the spectrum of the linear operator A lies entirely in the disk {| λ | <1}.

Any hint?

Answer 1

This is not true in general for example $$A = \left(\begin{array}{cc} 0 & \frac{1}{4} \\ 2 & 0\end{array}\right), \, v = (0,0)$$ and $f$ the mapping. You have $$\text{Spec} (A) = \left\{\frac1 {\sqrt 2}, -\frac1 {\sqrt 2}\right\} \subset \{\lambda : |\lambda| < 1\}$$ However $$\left\|f(e_1) - f(0)\right\| = \left\|Ae_1\right\| = \left\|2e_2\right\| = 2 > \left\|e_1\right\|$$ this proves that $f$ is not a contraction.