Let $P \in \mathbb{R}^{nxn}$, $|| P ||<1$, where $||\cdot||$ is a consistent norm, $c \in \mathbb{R}^{n}$. Consider the sequence $\{ x^{(k)}\}_{0}^{\infty}$ given by $x^{(k)}=Px^{(k-1)}+c$ and suppose $x^*$ such that $x^*=Px^*+c$. Show that, for all $k=1,2,...$, $||x^{(k)} -x^*|| \leq ||P||^k ||x^0-x^*||$.
As exists a consistent norm such that $||P||<1$ then $\rho(P)<1$ ($\rho(P)$ is the "spectral radius of $P$"), so the sequence $\{x^{(k)}\}$ converges to $x^*$, that is $\lim||x^{(k)}-x^*||=0$. By definition, $\forall \varepsilon>0$, exists $k_0$ such that if $k \geq k_0$, then $||x^{(k)}-x^*||<\varepsilon$. This proves for $k \geq k_0$, what about $k=1,2,...,k_0-1$?
i can only consider the situation of $P\in \mathbb{R},c\in \mathbb{R}$
we can find that $x^{(k)}=P^kx_0+\sum^{k-1}_{j=0}cP^j=P^kx_0+c{1-P^k\over 1-P}=P^k(x_0-{c\over 1-P})+{c\over 1-P}$ then for $x^*=Px^*+c$ we got $x^*={c\over 1-P}$ so that $x^{(k)}=P^k(x_0-x^*)+x^*$
finally $||x^{(k)}-x^*||=||P^k(x_0-x^*)||\leq||P||^k||x_0-x^*||$
by the way,consider $||P||\leq 1$,then $\lim_{k\to\infty}x^{(k)}-x^*=\lim_{k\to\infty}P^k(x_0-x^*)=0$
i hope it would help