Nilpotent matrix in $3$ dimensional vector space This is a part of a long problem and I'm stuck in two questions of it.
Let $E$ a $3$ dimensional $\mathbb R$- vector space and $g\in\mathcal{L}(E)$ such that $g^3=0$. So the first question is to prove that
$$\dim\left(\ker g\cap \mathrm{Im}\ g\right)\leq 1$$
and the second is: assume that $g^2\ne 0$, prove that if $\ker g=\ker g^2$ then $E=\ker g\oplus \mathrm{Im}\ g$ and deduce a contradiction, hence determinate $\dim\ker g$. Thanks for any help.
 A: For the first part dim(im$\,g$)$ + $ dim $(\ker\,g) = 3$.
The possible combinations are $0 + 3$, $1 + 2$, $2 + 1$, $3 + 0$ and you can see that any intersection cannot have dimension greater than 1.
For the second part:
\begin{align*}
g^3(x) &= 0,\, \forall x \in E \\
g^2(g(x)) &= 0 ,\, \forall x \in E \\
&\Rightarrow \text{ im } g \subset \ker g^2 - (1)
\end{align*}
Assume $\ker g = \ker g^2$ then
\begin{align*}
(1) &\Rightarrow \text{ im } g \subset \ker g \\
&\Rightarrow g(g(x)) = 0,\, \forall x \in E \\
&\Rightarrow g^2 = 0
\end{align*}
This is a contradiction as $g^2 \neq 0 - (2)$ 
\begin{align*}
g^2(x) &= g(g(x)) = 0,\, \forall x \in \ker g \\
&\Rightarrow \ker g \subset \ker g^2 \\
&\Rightarrow \dim (\ker g) \leq \dim (\ker g^2) \\
(2) &\Rightarrow \dim (\ker g) < \dim (\ker g^2) - (3)
\end{align*}
Clearly, as $g^2 \neq 0$, we have with (1) and (3):
\begin{align*}
1 &\leq \dim (\ker g) < \dim (\ker g^2) \leq 2 \\
&\Rightarrow \dim (\ker g) = 1
\end{align*}
Addendum 1: Sometimes playing with an example helps to see what's going on. Find the kernel and image of $A$ and $A^2$ in the following example in $\mathbb R^3$.
$A = \begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}$
