If $ker(R+T) = 0$ then $kerR \subseteq ImT$ I need to prove that if I have $2$ linear transformations: $R$ and $T$ that go from $W$ to $W$ (with finite dimension),If $R\circ T =0, ker(R+T) = 0$, then $ker(R) \subseteq  Im(T)$

To start, I came to some conclusions (some not very useful for the problem, but they may help).
$(R+T)(v)=R(v)+T(v) = 0$ if and only if $v = 0$. It is easy to see that if $v$ (nonzero) belongs to the kernel of $R$, then it does not belong to the kernel from $T$. My first idea was: if $v$ belongs to $ker(R)$ $(R(v) =0)$ then I should prove that $v$ belongs to the image of $T$ (i.e, there is a $x$ that belongs to W that $T(x)=v$). (But honestly I have a hard time thinking how to use hypotheses). I was thinking that maybe with the null rank theorem, (and with something related to the isomorphism of $R+T$) it could work. I would appreciate a hint.
 A: If I understood the question correctly. This is incorrect.
Let
$$
R=
\begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
\end{pmatrix},\
T=
\begin{pmatrix}
0 & 1 & 1 & 1 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 
\end{pmatrix}.
$$
The matrix $R+T$ is invertible, so $\operatorname{Ker}(R+T)=0$, $\operatorname{Ker} R=\langle e_2,e_3,e_4\rangle$, $\operatorname{Im} T=\langle e_1+e_2,e_1+e_3,e_1+e_4\rangle$.
We see that $e_2\notin\operatorname{Im}T$, hence $\operatorname{Ker} R\not\subset \operatorname{Im}T$.
Edit.
With the additional requirement that $RT=0$ this statement is true. Moreover, the equality $\operatorname{Ker}R=\operatorname{Im}T$ is satisfied.
Indeed, from equality $RT=0$ it follows that $\operatorname{Im}T\subset\operatorname{Ker}R$ means $\dim(\operatorname{Im} T)\leq \dim(\operatorname{Ker}R)$.
On the other hand since $R+T$ is invertible we have
$$
(R+T)(\operatorname{Ker}R)=T(\operatorname{Ker}R)\subset\operatorname{Im} T\Rightarrow \dim(\operatorname{Ker}R)\leq\dim(\operatorname{Im}T)
$$
So $\operatorname{Im}T\subset\operatorname{Ker}T$ and $\dim(\operatorname{Ker}R)=\dim(\operatorname{Im}T)$ hence it follows that $\operatorname{Im}T=\operatorname{Ker}R$.
