$V = \operatorname{Im} T \oplus \operatorname{Im}T^t \oplus H$ where $T^2 = 0$ and $H = \ker T \cap \ker T^t$ 
Let $T$ be a square matrix regarded as a linear operator on a finite dimensional vector space $V$ such that $T^2 = 0$. Let $H =  \ker T \cap \ker T^t$. Show that $V = \operatorname{Im} T \oplus \operatorname{Im}T^t \oplus H$.


[my proof] 
If $T = 0$, $V = \operatorname{Im} T \oplus \operatorname{Im}T^t \oplus H = \{0\} \oplus \{0\} \oplus V = V$
If $T \neq 0$, $V = \ker T\cup \ker T^t$ by rank-nullity theorem (* But I am still confusing how the rank-nullity theorem proves this)
and we know $\operatorname{Im} T\subseteq \ker T$ and $\operatorname{Im} T^t\subseteq \ker T^t$ from $T^2 =0$.
Now we need to prove that $\ker T \setminus \operatorname{Im} T = \emptyset$ and $\ker T^t\setminus \operatorname{Im} T^t = \emptyset$ to finish the proof, but can't know how to prove it.
Any hint?
 A: Note that $\operatorname{Im} T^t \,\dot+\, \ker T = V$. Firstly, the sum is direct (i.e. intersection is trivial):
$$T^tx \in \operatorname{Im}T^t \cap \ker T \implies TT^tx = 0 \implies (T^tx)^t(T^tx) = x^tTT^tx = 0 \implies T^tx = 0$$
and secondly, $\dim\operatorname{Im} T^t + \dim\ker T = \dim\operatorname{Im} T + \dim\ker T = \dim V$, by the rank-nullity theorem.
Now, as you stated, $T^2 = 0$ implies that $\operatorname{Im} T \subseteq \ker T$ and $\operatorname{Im} T^t \subseteq \ker T^t$
$$V = \operatorname{Im} T^t \,\dot+\, \ker T \subseteq \ker T^t \,\dot+\, \ker T$$
so $\ker T^t \,\dot+\, \ker T = V$.
Now we claim that $V = \operatorname{Im} T \,\dot+\, \operatorname{Im} T^t \,\dot+ \,H$. 
Firstly, the sum is direct. Assume that $0 = Tx + T^ty + z$ with $x,y \in V, z \in H$. Applying $T$ and $T^t$ gives
$$0 = T^2x + TT^t + Tz = TT^ty \implies (T^ty)^tT^ty = y^tTT^ty = 0 \implies T^ty = 0$$
$$0 = T^tTx + (T^t)^2z + Tz = TT^ty \implies (Tx)^t(Tx) = x^tT^tTx = 0 \implies Tx =0$$
and thus also $z = 0$.
Secondly, we have:
\begin{align}
\dim H &= \dim (\ker T \cap \ker T^t) \\
&= \dim\ker T + \dim\ker T^t - \dim(\ker T\,\dot+\, \ker T^t) \\
&= \dim\ker T + \dim\ker T^t - \dim V
\end{align}
so 
\begin{align}
\dim\operatorname{Im} T  + \dim \operatorname{Im} T^t + \dim H  &= \dim\operatorname{Im} T  + \dim \operatorname{Im} T^t +
 \dim\ker T + \dim\ker T^t - \dim V\\
&= 2\dim V - \dim V\\
&= \dim V
\end{align}
by the rank-nullity theorem.
Hence we conclude
$$V = \operatorname{Im} T \,\dot+\, \operatorname{Im} T^t \,\dot+ \,H$$
