If $A^2=0$ then what can we say about the column space of $A$ . I came across a question in one of the entrance examinations, which was Let $T$ be a linear transformation from $\Bbb{R}^n$ to $\Bbb{R}^n$, such that $T^2=0$. Then show that $r(\operatorname{rank} \text{of }T)\le n/2$
So I know that every linear transformation can be represented by a matrix and if $T^2=0$ then we know that $T$ is singular. But that only gives me that $r< n$.
I was thinking of going ahead with induction but I'm a little confused that would induction be a good way to prove it or not.
 A: $A^2=0\iff \operatorname{ran}A\subseteq\ker A$. Thus, $\operatorname{rk}A\le\dim\ker A$. But $$\operatorname{rk}A=n-\dim\ker A$$ Thus $\operatorname{rk}A\le n-\operatorname{rk}A$, q.e.d.
A: The correct line of reasoning is as follows:
Because $T^2 = 0$, we see that $T(T(x)) = 0$ for all $x$.  
Recall: the range of $T$ (i.e. the "column space of $A$") is defined to be $\operatorname{ran}(T) = \{T(x) : x \in \Bbb R^n\}$. The kernel of $T$ (i.e. the "null space of $A$") is defined to be $\ker(T) = \{y: T(y) = 0\}$.  By definition, we have $r = \dim \operatorname{ran}(T)$.  By the rank-nullity theorem, we have
$$
\dim \operatorname{ran}(T) + \dim \ker(T) = n \implies \dim \ker(T) = n-r
$$
Now, because $T(T(x)) = 0$, we know that $\operatorname{ran}(T) \subset \ker(T)$.  This implies that $\dim \operatorname{ran}(T) \leq \dim \ker(T)$.  In other words, 
$$
r \leq n - r
$$
But this implies that $r \leq n/2$, as desired.
A: $$\operatorname{dim} \ker T  +\operatorname{dim} \operatorname{Im} T = n$$
For a vector $v\in \operatorname{Im} T$ , by definition $ v= Tu $ and so $Tv = T^2 u = 0$ meaning that
$$\operatorname{Im} T \subseteq \ker T $$
Meaning 
$$ \operatorname{dim} \operatorname{Im} T \leq \operatorname{dim} \ker T $$
Substituting that in the first equation we get
$$n = \operatorname{dim} \ker T  +\operatorname{dim} \operatorname{Im} T \geq \operatorname{dim} \operatorname{Im} T + \operatorname{dim} \operatorname{Im} T= 2 \operatorname{dim} \operatorname{Im} T $$
$$\operatorname{dim} \operatorname{Im} T \leq \frac{n}{2} $$
QED
