Nullity of T vs. nullity of $T^2$ For a finite dimensional vector space, $V$, and  linear transformation, $T$, what is the relationship between the nullity of $T$ and the nullity of $T^2$? I read somwhere that if $nullity(T)=k$, then $nullity (T^2) \leq 2k$, but I would like to see a proof of this.
Here are somethings I know:

*

*$null (T) \subseteq null(T^2) \implies nullity(T) \leq nullity(T^2)$

*Possibility of splitting $null(T^2)$ into two sets, $A=\{\vec{v}\in V: \vec{v}\in null(T)\}$ and $B=\{\vec{v}\in V: \vec{v}\notin null(T)\}$

*If the above statement is valid, then $A$ and $B$ are disjoint, and $nullity(T^2)=dim(A+B)=dimA+dimB=k+dimB$ ?

*Then, I'd be proving that $dimA \leq k$ or $dimA \leq dimB$
I don't really know where to go from here! Any help would be appreciated!
 A: The kernel $\ker T^2 = T^{-1}(\ker T)$; consider the map $T : T^{-1}(\ker T) \to \ker T$.
A: In general, we have $nullity(BA) \leq nullity(A) + nullity(B)$.
To prove this, consider a basis $b_1, \ldots, b_n$ of $\ker B \cap im(A)$.
Then, take $a_1, \ldots, a_n$ such that for all $1 \leq i \leq n$, $A a_i = b_i$.
Finally, take $c_1, \ldots, c_m$ to be a basis of $\ker A$.
Then $a_1, \ldots, a_n, c_1, \ldots, c_m$ spans $\ker BA$ (and, in fact, is a basis - the proof it's a basis is left as an exercise).
For if we have $BAx = 0$, then $Ax \in \ker B \cap im(A)$. So we can write $Ax = \sum\limits_{i = 1}^n k_i b_i$. Then let $x' = \sum\limits_{i = 1}^n k_i a_i$.
Now $Ax = Ax'$, so $x - x' \in \ker A$. Therefore, $x - x'$ can be written as $\sum\limits_{i = 1}^m j_i c_i$. So we have $x = \sum\limits_{i = 1}^m j_i c_i + \sum\limits_{i = 1}^n k_i a_i$.
Note that since the $b_i$ are independent elements of $\ker B$, we must have $n \leq nullity(B)$. And of course we must have $m = nullity(A)$. Thus, we have $nullity(BA) = dim(\ker BA) \leq n + m \leq nullity(A) + nullity(B)$.
Of course, this immediately shows us that $nullity(T^2) \leq 2 nullity(T)$.
A: Let $U = \operatorname{null} T^2$ and $W = \operatorname{null} T$.
Note that $T(U) \subseteq W$. If we have $u \in U$, then $T^2 u = 0$, so $T(Tu) = 0$, which implies $Tu \in W$. We can therefore consider the restriction $T|_U : U \to W$.
It's also worth noting that $\operatorname{null} T|_U = \operatorname{null} T = W$. Certainly any $x \in \operatorname{null} T|_U$ satisfies $T x = 0$, as $T|_U$ and $T$ agree on all points in $U$ (where $T|_U$ is defined), thus $\operatorname{null} T|_U \subseteq \operatorname{null} T$. On the other hand, all elements of $\operatorname{null} T$ lie in $U = \operatorname{null} T^2$, as you pointed out in your question, so restricting $T$ to $U$ preserves the entire nullspace of $T$.
Using the rank-nullity theorem on $T|_U$, we see that
\begin{align*}
\operatorname{dim} U &= \operatorname{rank} T|_U + \operatorname{nullity} T_U \\
&= \dim \operatorname{range} T|_U + \dim \operatorname{null} T_U \\
&\le \dim W + \dim \operatorname{null} T_U \\
&= \dim W + \dim W = 2\dim W.
\end{align*}
That is, $\operatorname{nullity} T^2 \le 2\operatorname{nullity} T$.
