# How does banach fixed point theorem related to matrix analysis?

As stated by Banach fixed point theorem, a contraction mapping has only one fixed point. In plain words it means that the contraction mapping T has only one solution that satisfy $Tx = x$.

A stupid question came to my mind : is it just another way to say that eigenvalue is equal to one for some square matrix T ? If so, why would people build the concept of "contraction mapping" rather than just say so.

• Notice that contraction does not need to be a matrix in general. Also, some matrices have a fix point but aren't contraction. – user251257 Aug 16 '15 at 21:04
• If the fixed point is unique, the eigenvalue cannot be $1$, or you will get uncountably many fixed points as solutions to $Ax=x$. And I assume you want a nonzero solution for $Ax=x$. – Gary. Aug 16 '15 at 21:07
• Any linear mapping has a fixed point. It is $x=0$. – A.Γ. Aug 16 '15 at 21:52

Sorry, I am not sure I understood the question, but for a linear map $T$ ,having a unique point is not equivalent to $1$ being an eigenvalue: If $\lambda=1$ , then the associated eigenspace of points {$x: T(x)=x$} is a subspace, and so it will be of dimension one or higher, so it will contain uncountably -many points that will map to themselves, i.e., infinitely -many fixed points, so if you want a unique fixed point then $1$ cannot be an eigenvalue. And, as far as contractions, you need to have the structure of a normed/metric space to talk about contractions. Notice too, as someone wrote in the comments,that contraction maps may not be linear, so they cannot necessarily be represented by a matrix.