For any $x=x_1, \dotsc, x_n$, $y=y_1, \dotsc, y_n$ in $\mathbf E^n$, define $\|x-y\|=\max_{1 \le k \le n}|x_k-y_k|$. Let $f\colon\mathbf E^n \to \mathbf E^n$ be given by $f(x)=y$, where $y_k= \sum_{i=1}^n a_{ki} x_i + b_k$ where $k =1,2, \dotsc,n$. Under what conditions is $f$ a contraction mapping?
Any hint or solution for this question? I am beginner for this course, I can not understand clearly.
I'm not sure if I've understood the question correctly, but for a metric space $(M,d)$ a contraction mapping is usually defined as follows.
Let $f:M\to M$ be a continuous mapping. $f$ is said to be a contraction if there exists a constant $k\in \mathbf{R}$ with $0\le k<1$ and $d(f(x),f(y))\le kd(x,y)$ for all $x,y\in M$.
Intuitively, this means that the application of this mapping "shrinks" or "contracts" the space by bringing every pair of points closer together in a nice manner. Take for example the mapping $$ f:\mathbf{B}^n\to \mathbf{B}^n$$ given by $x\mapsto x/2$.
EDIT: It seems to me that you would like to show this for the given function. Basically, just try to see what relations you get on the coefficients if you want to make it subject to the above definition.
The function $f$ can be written as $$f(x)=Ax+b,$$ where $A$ is the $n\times n$ matrix $$A=\begin{pmatrix} a_{11} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots \\ a_{n1} & \cdots & a_{nn} \end{pmatrix}$$ and $b\in \mathbf E^n$ is the vector $$b=\begin{pmatrix} b_1\\ \vdots \\ b_n \end{pmatrix}.$$ We define the matrix norm $\|A\|=\max_{\|x\|\leq1}\|Ax\|$. With this, we can state a sufficient condition for $f$ being a contraction:
If $\|A\|<1$, then $f$ is a contraction.
To prove this, observe that for any $x\in \mathbf E^n$ we have $\|Ax\|\leq\|A\|\|x\|$. Now, for any $x,y\in \mathbf E^n$ we have $$\|f(x)-f(y)\|=\|Ax-Ay\|=\|A(x-y)\|\leq\|A\|\|x-y\|.$$ Thus if $\|A\|<1$, we have $$\|f(x)-f(y)\|\leq\|A\|\|x-y\|<\|x-y\|,$$ and therefore $f$ is a contraction.
