Direct sum of images of endomorphisms Let $f, g: V\to V$ be $K$-linear functions, $V$ a
$K$-vector space for some field $K$.
Show, that if
$V=\text{im}(f)+\text{im}(g)=\ker(f)+\ker(g)$
and $V$ has finite dimension, then
$V=\text{im}(f) \oplus \text{im}(g) =\ker(f)\oplus \ker(g)$
I tried solving this using the dimensional formula, but sadly didn't manage to incorporate the assumption. Any help would be appreciated!
After pondering for a while I came up with a solution but I'm unsure if it's correct, take a look:
We only need to show $dim(im(f) \cap im(g)) =0$. Assume $dim(im(f) \cap im(g)) \neq 0$. Then
$dim(V) <dim(im(g)) +dim(im(f)) \Rightarrow 
dim(V) - dim(im(g)) <dim(im(f)) \Rightarrow
dim(ker(g)) <dim(im(f)) $
Because $dim(V) \leq dim(ker(g)) +dim(ker(f)) \Rightarrow dim(V) - dim(ker(f)) \leq dim(ker(g)) \Rightarrow
dim(im(f)) \leq dim(ker(g)) $
the transitive property implies
$dim(im(f)) < dim(im(f)) $, a contradiction. Therefore
$dim(V) =dim(im(f)) +dim(im(g)) $
and thus follows the statement.
 A: Let $\dim(V)=n$.
Since $\dim(\operatorname{Im}(f))+\dim(\operatorname{Ker}(f))=n$ and $\dim(U+W)=\dim(U)+\dim(W)-\dim(U\cap W)$ we have
\begin{eqnarray*}
   n&=&\dim(V) \\
   &=&\dim(\operatorname{Im}(f))+\dim(\operatorname{Im}(g))-\dim(\operatorname{Im}(f)\cap\operatorname{Im}(g)) \\
   &=&(n-\dim(\operatorname{Ker}(f)))+(n-\dim(\operatorname{Ker}(g)))-\dim(\operatorname{Im}(f)\cap\operatorname{Im}(g)) \\
   &=& 2n-(\dim(\operatorname{Ker}(f)))+\dim(\operatorname{Ker}(g))-\dim(\operatorname{Ker}(f)\cap\operatorname{Ker}(g)))\\
   &-&\dim(\operatorname{Ker}(f)\cap\operatorname{Ker}(g))-\dim(\operatorname{Im}(f)\cap\operatorname{Im}(g))\\
   &=&2n-\dim(V)-\dim(\operatorname{Ker}(f)\cap\operatorname{Ker}(g))-\dim(\operatorname{Im}(f)\cap\operatorname{Im}(g))\\
   &=&n-\dim(\operatorname{Ker}(f)\cap\operatorname{Ker}(g))-\dim(\operatorname{Im}(f)\cap\operatorname{Im}(g)).\\
\end{eqnarray*}
It follows that
$$
\dim(\operatorname{Ker}(f)\cap\operatorname{Ker}(g))+\dim(\operatorname{Im}(f)\cap\operatorname{Im}(g))=0.
$$
Hence
$$
\operatorname{Ker}(f)\cap\operatorname{Ker}(g)=0\qquad {\rm and }\qquad \operatorname{Im}(f)\cap\operatorname{Im}(g)=0.
$$
So $V=\operatorname{Im}(f)\oplus\operatorname{Im}(g)$ and
$V=\operatorname{Ker}(f)\oplus\operatorname{Ker}(g)$.
