fixed points of an affine transformation is unique iff $1 \notin SP(\vec{f} )$

Question

Let $f$ be Affine transformation from $E$ to $E$ (always we assume it finite dimensional ) and $\overrightarrow{f}$ is the linear mapping associated to $f$. Then the map $f$ has a unique fixed point if and only if $1$ is not an eigenvalue of $\overrightarrow{f}$

Proof : If $f$ has a unique fixed point, according to $7.2$ , all of its fixed points is a affine subspace direction $\{\vec{0}\}=\operatorname{Ker}(\overrightarrow{f}-Id_{\vec{E}})$, so $1$ is not unique value of $\overrightarrow{f}$.

the origine of text

i don't understand why $\operatorname{Ker}(\overrightarrow{f}-Id_{\vec{E}})$ equal $\{\vec{0}\}$ and after that why we get $1\notin \operatorname{Sp}(f)$

any help would be appreciated

Answer 1

A single point is an affine space with tangent space (i.e., direction) of dimension$~0$. The result 7.2 apparently says that if there are any fixed points, they form an affine space with direction $\ker(\vec f-Id)$. So if there is a unique fixed point, that subspace must be of dimension$~0$, which means precisely that $1$ is not an eigenvalue of$~\vec f$ (for if it were, its eigenspace would be $\ker(\vec f-Id)$ and of nonzero dimension).

Answer 2

The set of fixed points of $f$ is the kernel of $f - Id$, and the latter is indeed an affine subspace with direction $\text{Ker}(\vec{f} - Id)$. Since by hypothesis there is only one fixed point, the affine subspace is reduced to a point and so must be $\text{Ker}(\vec f - Id)$. Saying that $1$ is not a proper value of $f$ is exactly the same as saying that $f-Id$ only vanishes in $0$.