Choose the false statement of given matrix? Let A be a square matrix such that $A^{3} =  0 $, but $A^{2} \neq 0$. Then which of
the following statements is not necessarily true?
(A) $A\neq  A^{2} $
(B) Eigenvalues of $ A^{2} $ are all zero
(C) $\text{rank}(A) > \text{rank}(A^{2})$
(D) $\text{rank}(A) > \text{trace}(A)$
My attempt : 
My correct answer is option 2), but I'm doubting little bits... is my answer is correct or incorrect pliz verified and tell me the solution, I would be more thankful.
 A: Your computation of the eigenvalues of $A^2$ with the given example is wrong.
Note that if $v$ is a non-zero eigenvector corresponding to a eigenvalue $\lambda$ we have  $Av=\lambda v$ and
$$A^nv= A^{n-1} (Av)= A^{n-1}(\lambda v)=\dots= \lambda^nv.$$ 
What may we conclude?
P.S. As regards option C, note that $\text{rank}(A^n)\geq \text{rank}(A^{n+1})$. Moreover if $\text{rank}(A^{n})=\text{rank}(A^{n+1})$, then $\text{rank}(A^{n})=\text{rank}(A^{n+k})$ for all $k\geq 0$. This implies that in our case also option C is true. 
A: $A\neq A^2$ must be true, otherwise $A^3 = A^2A = AA = A^2 \neq 0$.
The eigenvalues of $A^k$ are the eigenvalues of $A$ raised the the $k$th power, so for $A^3$ to be zero, the eigenvalues of $A$ must all be zero. Hence also the eigenvalues of $A^2$ are all zero and B must be true.
D must also be true. Trace is the sum of the eigenvalues, and since they are all zero (see the argument before), $\text{trace}(A) = 0$. The rank of $A$ must be positive since $A\neq 0$.
Only thing left is C. But can it be false either?
[EDIT]:
C must also be true: Of course $\text{rank}(A) \geq \text{rank}(A^2)$, since when you think $A$ as a linear mapping, the image of $A^2$ is a subset of the image of $A$. If $\text{rank}(A) = \text{rank}(A^2)$ this would mean that the images are equal. Hence they will be equal from the power $2$ onward and that means $\text{rank}(A^3) = \text{rank}(A^2) > 0$. But $A^3=0$, a contradiction.
(By the way, in your example you are mistaken about the rank of $A$. It can't be $3$ since otherwise $A$ would be invertible and its power would never be the zero matrix. Also the eigenvalues must be all zero.)
