If $f'(0)$ doesn't have eigenvalue $1$, then there is a neighborhood where $f(x)\neq x$ 
Let $f:\mathbb{R}^m\to\mathbb{R}^m$, differentiable, with $f(0) = 0$.
  If the linear transformation $f'(0)$ has no eigenvalue $1$, prove that
  there exists a neighborhood $V$ of $0$ in $\mathbb{R}^m$ such that
  $f(x)\neq x$ for all $x\in V-\{0\}$

My book solves it by the following:

Since the operator $f'(0)$ has no fixed point in the compact
  $S^{n-1}$, there exists $\epsilon>0$ such that $|u|=1\implies
 |f'(0)\cdot u -u|\ge \epsilon$. Being $f$ differentiable, with $f(0) =
 0$, we have $f(x) = f'(0)\cdot x + p(x)\cdot |x| = |x|\left(f'(0)\cdot
 \frac{x}{|x|}+p(x)\right)$ and exists $\delta >0 $ such that
  $0<|x|<\delta \implies |p(x)|<\epsilon$ (why? - 1). So, if
  $0<|x|<\delta$, then $|f(x)-x|\ge |x|\left[\left|f'(x)\cdot
 \frac{x}{|x|}-\frac{x}{|x|}\right|-|p(x)|\right]>0$ and then $f(x)\neq
 x$ (why? - 2)

Also, what the compact $S^{n-1}$ have to do with all of this?
 A: Compactness implies (by the Extreme Value Theorem) that $\inf_{u \in S^{n - 1}} |f'(0)u - u| = |f'(0)u_0 - u_0|$ for some $u_0 \in S^{n - 1}$ (i.e. the minimum is achieved at some point). Since $f'(0)u_0 \ne u_0$ it must be that $|f'(0)u - u| \ge |f'(0)u_0 - u_0| > 0$ for all $u \in S^{n - 1}$.
The existence of $p(x)$ follows from the first order Taylor approximation to $f$ at $0$. That is, $f(x) = f'(0)x + p(x) |x|$ where $|p(x)| \approx |x|$ in the precise sense that is given in the book.
If $f(x) = x$ then $|f(x) - x| = 0$ and the step before this says that $|f(x) - x| > 0$ for $x$ in some neighbourhood of $0$. In case you were wondering about the step before:
$$
\begin{align*}
|f(x) - x| &= \left\lvert |x| \left( f'(0) \cdot \frac{x}{|x|} + p(x) \right) - x \right\rvert \text{(replacing $f(x)$ with its first order approx)} \\
&= \left\lvert |x| \left( f'(0) \cdot \frac{x}{|x|} - \frac{x}{|x|} + p(x) \right) \right\rvert \text{(rearranging)} \\
&\ge |x|\left( \left\lvert f'(0) \cdot \frac{x}{|x|} - \frac{x}{|x|} \right\rvert - |p(x)| \right) \text{(by the reverse triangle inequality)} \\
&\ge |x| (\epsilon - |p(x)|) \quad \text{(since $x/|x| \in S^{n-1}$)} \\
&\ge |x| (\epsilon - \text{\{something $< \epsilon$\}}) \quad \text{(since $|x| < \delta$ implies $|p(x)| < \epsilon$)} \\
&> 0.
\end{align*}
$$
A: Recall that $\mathrm{d}_0f$ is continuous, so that $\mathrm{d}_0f-\textrm{id}_{\mathbb{R}^m}$ also is. By assumption, $\mathrm{d}_0f-\textrm{id}_{\mathbb{R}^n}$ does not vanish. Therefore, since $\mathbb{S}^{m-1}$ is compact, $\mathrm{d}_0f-\textrm{id}_{\mathbb{R}^m}$ reaches a minimum on $\mathbb{S}^{m-1}$ which is positive, this how it is proved that there exists $\varepsilon>0$ such that:
$$\forall u\in\mathbb{S}^{m-1},|\mathrm{d}_0f\cdot u-u|\geqslant\varepsilon.$$
Why 1. The definition of differentiability at $0$ can be written as:
$$|f(x+h)-f(x)-\mathrm{d}_0f\cdot h|=o(h).$$
This is why $\lim\limits_{h\to 0}p(h)=0$ and thus the existence of $\delta$ by definition of the convergence toward $0$.
Why 2. You have proved that $|f(x)-x|>0$, therefore $f(x)$ cannot be equal to $x$, maybe the part you are missing is that $|a-b|\geqslant ||a|-|b||$.
