To show $\mathrm{ker} f=\{0\}$ for linear mapping $f$.

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?

If $x = x_1 e_1 + \cdots x_1 e_n \in \mathsf{ker}(f)$ then

$0 = f(x) = x_1f(e_1) + \cdots +x_{n-1}f(x_{n-1})+ x_nf(e_n)$

$= x_1e_2 + \cdots + x_{n-1}e_{n} + x_ne_1$

$= x_ne_1 + x_1 e_2 + \cdots + x_{n-1}e_n$

$\Longrightarrow \quad x_n =x_1 = \cdots =x_{n-1} = 0 \quad \Longrightarrow \quad x=0$.

Because the domain and codomain of the injective transformation $f$ have equal dimension, $f$ is an isomorphism.

A function $f:A \to B$ between two sets is injective if and only if it has a left inverse with respect to composition; that is, if and only if there exists a function $g:B \to A$ such that $g \circ f = \mathrm{Id}_A$. Similarly, a function $f:A \to B$ is surjective if and only if has a right inverse, or rather there exists a function $g:B \to A$ such that $f \circ g = \mathrm{Id}_{B}$.

In your example, $A,B = V$, and the left/right inverse $g$ you should look at is $g = f \circ \stackrel{n-1}{\cdots}\circ f$, since

$g \circ f = f \circ g = f \circ \stackrel{n}{\cdots} \circ f = \mathrm{Id}_{V}$

Of course, you should show why the composite of $f$ with itself $n$ times gives you the identity map (think permutation of the basis elements).

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.

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 $\ker F = 0_v$.

Hints :

• if $f,g:X\to X$ are applications such that $f\circ g$ is injective (which means "one to one"), then $g$ is injective.

• id $E$ is a finite dimensional space and if $f\in\mathcal{L}(E)$ is injective or surjective, then $f$ is a bijection.

• Try to understand what can be said about $F^n$ (which means of course $F\circ\cdots\circ F)$