Eigenvectors and injectivity

Injectivity and kernel. A linear map is injective iff $\ker f=\{0\}$: indeed $f(u)=f(v)\iff f(u-v)=0\iff u-v\in\ker f$.

$0$ is an eigenvalue $\iff\ker f\neq\{0\}$. By definition, $0$ is an eigenvalue iff there exists $v\neq0$ with $f(v)=0\cdot v=0$, i.e. iff $\ker f$ contains a nonzero vector.

Conclusion. $f$ is injective $\iff\ker f=\{0\}\iff0$ is not an eigenvalue.

Remark. If $E$ is finite-dimensional, injective is equivalent to invertible (rank-nullity), so $f$ is invertible iff $0$ is not an eigenvalue.