Linear Transformation $T^{2} = 0$ what can we say about the relationship between $ker(T)$ and Im$(T)?$ I hope you can help me, I'm very new to linear algebra, I am given the linear transformation $T:V\rightarrow V$ that fullfills $T^2 = 0$,
what can be said of the relationship that exists between Im$(V)$ and $\ker(V)$? 
What I did so far is the following:
$T(Tx) = 0 \implies Tx \in \ker(T) \wedge Tx \in $Im$(T) \implies Tx \in \ker(T)\bigcap $ Im$(T)$
but $Tx$ should be equal to $0$ according to transformation's definition
because $Tx \in $ Im$(T)$ also belongs to $V$ and given the transformation's definition $T(Tx) = 0$ and picking $w=Tx$ whe could say $T(w) = 0$ $ \forall$ $w \in V$ so, can we conclude that $\ker(T)\bigcap$ Im$(T) = 0?$, so $\ker(T)$ and Im$(T)$ are disjoint sets? can we conclude anything else? am I right?
thanks for any replies, cheers.
 A: This one's a little difficult to answer without giving away the solution, but I'll try. 
Your logic is good until we get to the sentence that starts "but $Tx$ should be equal to $0$..." This does not follow from the fact that $TTx = 0$. Yes, it's true for particular values of $x$, but not for $x$ arbitrary.
For example, consider the map $\mathbb{R}^3 \to \mathbb{R}^3$ which sends $(x, y, z)$ to $(y, 0, 0)$. Clearly, the square of this map is $0$. The image of this map is the $x$-axis. The kernel of this map is the $(x, z)$-plane. What is the relation between them? (It falls right out of the implication you have already proven.)
A: We have the inclusion: $\text{Im } T \subseteq \ker T$. 
Proof: let $w \in \text{Im } T$. There exists $v \in V$, such that $w = Tv$. Hence, $Tw = T^2 v = 0$.
A: quite easy
Look, $T^{2}\left(X\right)=0$ $\Longrightarrow $T
$\left(X\right)\in
{Ker}\left(T\right)$
We know that $T\left(X\right)$ is range of $T$
it implies $R\left(T\right)$ $\subseteq$ $N\left(T\right)$
That means Range of T $\subseteq$ Nullity Of T
