Is there any relation between $Rank(A^2)$ and $Rank(A^3)$ if $Rank(A)=Rank(A^2)$? It is a question from my textbook :

$A$ is a square matrix of order $n\times n$. If $Rank(A)=Rank(A^2)$ then verify whether $Rank(A^2)=Rank(A^3)$ or not.

It is definite that $Rank(A^3)\leq Rank(A^2)$ but after that I cannot proceed.
Please anyone help me solve it. Thanks in advance. 
 A: Hint: In general, if $V$ is a vector space and $U$ is a subspace of $V$ such that $dim(U)\geq dim(V)$, then $U=V$ (the proof is simply based on the fact that any basis of $U$ will necessarily be linearly independent in $V$). 
$rank(A)$ is the dimension of $range(A)$. Can you show that $range(A^2)$ is a subspace of $range(A)$? How then can you strengthen the relationship between $range(A)$ and $range(A^2)$? Finally, what does this allow you to conclude about the relationship between $range(A^2)$ and $range(A^3)$?
A: If you are able to establish the following preparatory result: 
$$rank(A)=rank(A^2) \ \iff \ \exists B \ \text{invertible s.t.} \ A^2=BA$$
then, it will suffice to right-multiply the last relationship of (1) by $A$ to prove that 
$$rank(A)=rank(A^2) \ \implies \ rank(A^2)=rank(A^3)$$
A: In terms of triangularisation of $A$ which does not change the rank you may suppose $A$ upper triangular with the last eigenvalues the zero ones, if algebraic multiplicity and geometric multiplicity of $0$ as eigenvalue are equal, $\textrm{rank}(A^n)$ is constant for all $n$, now if not $A$ as upper triangular has its last $m\times m$  block (lower right) of the form $\begin{pmatrix}0&a\\0&0\end{pmatrix}$ $a$ is $1\times m-1$  not zero, elevating for $A^2$ this block turns  to all zero diminishing the rank a contradiction.
