Conditions for the dimension of rank of sum of linear transforms be equal the sum of ranks Let $E,F$ vector spaces of finite dimension over $K$ and $f,g$ linear tranforms from $E$ to $F$. Consider the linear tranform $f+g:E \to F$, $(f+g)(u) = f(u)+g(u) \;\;\forall u \in E$.
Prove that are equivalent:
(i)$\dim Im(f+g) = \dim Im(f) + \dim Im(g)$
(ii)$Im(f)\cap Im(g)=\{0\},f(Nuc(g))=Im(f), g(Nuc(f)) = Im(g)$
I tried using rank nulity theorem for $f,g$ and $f+g$, but did not go anywhere.. 
 A: Let $F=Im(f)$, $G=Im(g)$, $S=Im(f+g)$.
Let also $\Gamma$ be a supplement of $Nuc(g)$ in $E$, so that $\dim(\Gamma)=rk(g)$.
Proof of $(i) \Rightarrow (ii)$ : suppose (i) is true.
On one hand, we have $S \subseteq F+G$ ; but on the other hand we have
$dim(S)=rk(f+g)=rk(f)+rk(g)=dim(F)+dim(G) \geq dim(F+G)$ by (i). So $S=F+G$
and $dim(S)=dim(F+G)$. Using the formula $dim(F+G)=dim(F)+dim(G)-dim(F\cap G)$, we deduce $dim(F\cap G)=0$, i.e. $F\cap G=\lbrace 0 \rbrace$ which is the first part of (ii).
Now, $f+g$ must be injective on $\Gamma$ (otherwise we would have a nonzero $\gamma\in \Gamma$ such that $(f+g)(\gamma)=0$, and then $f(\gamma)=-g(\gamma)$ would be a nonzero element in $F\cap G$ contradicting the hypothesis), and hence $dim((f+g)(\Gamma))=dim(\Gamma)=rk(g)$.
Then
$$
rk(f)+rk(g)=rk(f+g) \leq dim((f+g)(\Gamma))+dim((f+g)(Nuc(g))) = rk(g)+dim(f(Nuc(g)))
$$
So $rk(f) \leq dim(f(Nuc(g)))$, but on the other hand we trivially have that
$f(Nuc(g)) \subseteq Im(f)$, so those two subpaces must be equal, and this is the second part of (ii). The last part of (ii) is deduced similarly, reversing the roles of $f$ and $g$.
Proof of $(ii) \Rightarrow (i)$ : suppose (ii) is true.
I claim that the subspaces $A=(f+g)(Nuc(g))=f(Nuc(g))$ and $B=(f+g)(\Gamma)$ have trivial intersection, i.e. $A\cap B=\lbrace 0 \rbrace$. Otherwise, we would have a nonzero $z\in Nuc(g)$ and a nonzero $\gamma \in \Gamma$ such that
$f(z)=(f+g)(\gamma)$, and then $f(z-\gamma)=g(\gamma)$ would be a nonzero element of $Im(f)\cap Im(g)$, contradicting the hypothesis.
From the fact that $Im(f+g)=A+B$, we may then deduce
$$rk(f+g)=dim(A)+dim(B)=dim(f(Nuc(g)))+dim(B)=rk(f)+dim((f+g)(\Gamma))=rk(f)+rk(g)$$
where in the last step we use the same argument as in the proof of $(i) \Rightarrow (ii)$. This finishes the proof.
