Nullity and rank bounds for a nilpotent matrix 
Let $A=\mathbb R^{11}\to \mathbb R^{11}$ be a linear transformation such that $A^5=0$ and $A^4\neq 0$. Which of the following is true?
  a) $\operatorname{null}A\le7$
  b) $2\le\operatorname{null}A$
  c) $2\le\operatorname{rk}A\le9$


I don't know how i think about $A$ from this information. I think construct such as example where this hold but I can't construct any such of type example which satisfies $A^5=0$ but $A^4\ne0$.
I have an idea now: consider $T:\mathbb R^{11}\to\mathbb R^{11}$ such that $$T(x_1,x_2,...,x_{11})=(x_2,x_3,x_4,x_5,0,0,..,0)$$
$\Rightarrow T^5(x_1,x_2,...,x_{11})=(0,0,...,0)$. So we can say $\operatorname{null}T\le11-\operatorname{rk}T=7$

Only the answer a) is true. In fact, more precisely, the following inequalities are true :

$4\leq \text{Rank}(A)\leq 9 $  and $3\leq\text{Nullity}(A)\leq 7$ 

To prove it you need to know the Jordan decomposition for a nilpotent linear transformation:

A nilpotent linear transformation of degree $u$ (i.e. $A^u=0$ and $A^{u-1}\neq 0$) is similar to a block diagonal matrix
  :$$J = \begin{bmatrix}
J_{p_1} & \;     & \; \\
\;  & \ddots & \; \\ 
\;  & \;     & J_{p_k}\end{bmatrix}$$
  where each block  $J_{p_i}$ is a square matrix of size $p_i$ and of the form
  :$$J_{p_i} = 
\begin{bmatrix}
 0 & 1            & \;     & \;  \\
\;        & 0   & \ddots & \;  \\
\;        & \;           & \ddots & 1   \\
\;        & \;           & \;     & 0       
\end{bmatrix}.$$
  Where for all $i$, $0\leq p_i \leq u$ and at least one $p_i$ is such that $p_i=u$ ; moreover it is easy to see that $Nullity(A)=k$ (the number of blocs).

Here, with $u=5$ and the condition $p_i\leq u$, the minimum number of blocs is $3$ (since you need at least one bloc of five and a bloc of six is not possible to complete the bloc decomposition, so at least two bloc more are necessary to fulfill the 11 dimensions) : the $Nullity(A)\geq 3$ and, using the theorem $Rank(A) +Nullity(A)=11$ you will have $Rank(A)\leq 9$.
Now, you have at least one bloc of size 5, that is a bloc of rank 4 : the rank of the linear transformation is at least higher than 4, so $Rank(A)\geq 4$, which leads, again from $Rank(A) +Nullity(A)=11$, to $Nullity(A)\leq 9$.
