TITLE: To show $\mathrm{ker} f=\{0\}$ for linear mapping $f$.
QUESTION [6 upvotes]: Let $V$ be a vector space over $F$ with basis $\{e_1,e_2,...e_n\}$.
Let $F$ be a linear mapping from $V$ to $V$ such that
$F(e_1) =e_2,...F(e_n)=e_1$.
Show that  $\mathrm{ker} f=\{0\}$. Also find $f^{-1}$.
I just know that $\mathrm{ker} f=\{0\}$ iff $f$ is 1-1. 
So is it enough to show that basic definition for 1-1?
The inverse mapping will be defined as $f^{-1}(e_1) =e_n, f^{-1}(e_2)=e_1,...$
Am I right?

REPLY [4 votes]: Using the rank-nullity theorem it is easy to prove that

if $V$ is a finite dimensional vector space and $f\colon V\to V$ is a linear map, then $f$ is injective (or 1-1) if and only if it is surjective

Prove it and verify your map is surjective.
The inverse function is correct.

REPLY [2 votes]: Recall that for finite dimensional vector spaces you have that 
$$
\dim V = \dim Im F + \dim \ker F,
$$
an by the definition of $F$ you know that 
$$
Im F \in sp\{ e_1, e_2,..., e_n \},
$$
thus $\dim Im F= n$, therefore $\dim \ker F = 0$ so $\ker F = 0_v$.