Showing that g exists, so that ker(g)=Im(f) and ker(f)=Im(g) Let $F$ be a field and $X$ a $F$-linear space with dim$_FX=n<\infty$. Let $f:X\rightarrow X$ be a $F$-linear function.
Show that there exists a $F$-linear function $g:X\rightarrow X$, so that ker(g)=Im(f) and ker(f)=Im(g).
Since it's $X\rightarrow X$ for both:
$ker(g)=\{x\in X|g(x)=0\in X\}$
$ker(f)=\{x\in X|f(x)=0\in X\}$
$Im(g)=\{g(x)|x\in X\}$
$Im(f)=\{f(x)|x\in X\}$
But i have trouble to find such a function $g$, since $f$ is not explicit stated.
ker(g)=Im(f) $\rightarrow$ $\{x\in X|g(x)=0\in X\}=\{f(x)|x\in X\}$
and now? i'm lost here.
Also if someone could exchange the brackets with {} that would be nice.
 A: Let $\{y_1,y_2,\dots,y_k\}$ be a basis for $\operatorname{Im}(f)$. Choose $\{x_1,x_2,\dots,x_k\}$ so that $f(x_i)=y_i$, for $i=1,2,\dots,k$.
Consider a basis $\{u_1,\dots,u_r\}$ of $\ker(f)$.

Prove that $\{x_1,x_2,\dots,x_k,u_1,\dots,u_r\}$ is a basis for $X$. In particular $n=k+r$ (this is the rank-nullity theorem).

Consider $\{v_1,\dots,v_r\}$ such that $\{y_1,\dots,y_k,v_1,\dots,v_r\}$ is a basis for $X$.
There exists a unique linear map $g\colon X\to X$ such that
$$
g(y_1)=0,\quad
g(y_2)=0,\quad
\dots,\quad
g(y_k)=0,\quad
g(v_1)=u_1,\quad
\dots\quad,
g(v_r)=u_r
$$

Prove this $g$ has the required properties.

A: Let $dim Ker(g)=dimImg(f)=k\le n,\\ dimKer(f)=dimImg(g)=n-k;\\$
Let $B_1=\{x_i\}_{i\le n}$ a base of $X$ where $\{x_i\}_{i\ge (k+1)}$  is a base of $Ker(f)$. And let $B_2=\{h_i\}_{i\le n} $ a base of X where $h_i \in Ker(g) \space \forall i\le k.$
Now put : $\space f(x_i)=h_i\space \forall i \le k \\g(h_i)=x_i \space \forall i\ge k+1$
These linear functions do what we want. They exist because are defined on a base. Hope this help you.
I hope you know that if $V,W$ are two $F$-vector space and $B=\{x_i\}_{i\le n}$ is a base of $V$ and $w_1,...w_n$ are vectors of $W$ then there exists an unique linear function $f:V\to W$ such that $f(x_i)=w_i \space \forall i.$.
