If $\operatorname{rank}(A)$ = $\operatorname{rank}(A^2)$, show that nullspace of $A$ = nullspace of $A^2$ Let $A$ be a square matrix.
If $\operatorname{rank}(A)$ = $\operatorname{rank}(A^2)$
Prove that nullspace of $A$ = nullspace of $A^2$
The first thing I notice is that this $\implies$ $\operatorname{nullity}(A)=\operatorname{nullity}(A^2)$
Then I am kinda stuck, any hints?
 A: You should show that the nullspace of $A$ is contained in the nullspace of $A^2$. By your observation, the dimension of the two nullspaces must necessarily be the same, so $\text{nullspace} A \subseteq \text{nullspace}A^2$ necessarily gives that they are equal.
A: Let $x$ $\in$ $R^n$, $A $$\in$ $R^{m\times n}$, then,
If $x$ satisfies $Ax = 0$, then  $x$ $\subset$ $N(A)$.
$A^2 x= A A x = A 0 = 0$  so  $x$ $\subset$ $N(A^2)$
Hope this helps
A: Hint: Can you see (or prove) that the nullspace of a matrix $A$ is a subspace of the nullspace of $A^2$?
A: For any matrix $C$,  let $\Psi(C)$ and $\mathscr{N} (C) $ denote the rank  and null space of $C$ respectively. 
Let $A$ be an $n\times n$ square matrix such that $\Psi(A) =\Psi(A^2)$. If $x\in\mathscr{N} (A) $, then one can show that $x\in\mathscr{N} (A^2)$. We thus have $\mathscr{N} (A)\subseteq\mathscr{N}(A^2)$. But since it is given that $\Psi(A) =\Psi(A^2)$, we have $\dim\mathscr{N} (A) =\dim\mathscr{N} (A^2)$. Therefore, we can conclude that $\mathscr{N} (A^2)\subseteq \mathscr{N} (A)$,  and hence $\mathscr{N} (A)=\mathscr{N} (A^2)$.
A: Hints
You can use the dimension theorem for linear transformations or matrices (as they are isomorphic). Try to think what is the range of $A^2$ in terms of the range of $A$. 


*

*What would happen if the ranges were not equal? 

*What conditions are necessary so that $range A$ = $range A^2$? 


As always thinking in terms of the rank is similar as thinking in terms of the null space as both are related by the dimension theorems.
