If $A^2 =0$ then possible rank of $A$ 
Let, $A$ be a non zero matrix of order $8$ with $A^2 =0.$ Then one of the possible value for rank of $A$ is
(a) $5$ (b) $4$ (c) $6$  (d) $8$.

Attempt :
As , $A^2=0$ , so $A$ is a nilpotent matrix of order $2$. So , characteristic polynomial of $A$ is of the form $x^2f(x)$ , where $f(x)$ is a polynomial of degree $6$. So , possible values of $rank(A)$ are $4,5,6$.
Am I correct ?
 A: $Range(A)=\{Ax : x\in \mathbb{R}^8\}$
Claim : $Range(A)\subset Ker(A)$
Suppose $y=Ax\in Range(A)$
Then $Ay=A^2(x)=0$
$\implies y\in \ker (A)$
$\implies Range(A)\subset Ker(A)$
$\implies Rank(A)\leq Nullity (A)$
Therefore from Rank Nullity theorem We have 
$dim (\mathbb{R}^8)=8=Rank(A)+Nullity(A)$
$\implies 8\geq Rank(A)+Rank(A) $
$\implies 4\geq Rank(A)$
And Therefore possible value of Rank(A) is $4$.
A: Hint: Since $A^2 = 0$, what can you say about the null space and range of $A$?
A: As the minimal polynomial of $A$, i.e. $m(A)/x^2$, it should be $x$ or $x^2$, but $x$ is not possible as $A$ is non zero. So $m(A)=x^2$
Its Jordan canonical form has 4 Jordan blocks of order 2, 1 block of order 2 and other 6 blocks of order 1, 2 block of order 2 and 4 block of order 1, 3 block of order 2 and 2 block of order 1, so rank must be 4.
And also i.e. 3,2,1,0
A: Some more hints.


*

*If $A$ has a nullspace of dimension $N$, then at most N dimensions vanish if you apply $A$ once. 

*Then you have the rank-nullity theorem.

A: Apply formula rank (A^k) > equal k rank(A)- (k-1).n
0> equal 2×rank(A)-(2-1).8 hence rank is less than 4 hence maximum possible rank is 4
