Proving that $Ker A = Im A$ if and only if $2rank(A)=n$ and $A^2=0$ Let $A$ be an $n\times n$ matrix and we assume that $Ker A = Im A$.
We want to prove that $2rank(A)=n$ and that $\forall X \in R^n$, such that $A^2X=0$, it holds that $A^2=0$
How do I approach this question?
 A: The first part follows from the rank-nullity theorem: viewing $ A $ as a linear map, we have $ \dim (\ker A) + \dim (\textrm{Im} A) = n $. If $ \ker A = \textrm{Im} A $, then their dimensions are equal, and the result follows immediately.
For the second part, note that for any vector $ v $ we have $ Av \in \textrm{Im} A $ by definition of image. However, since the image and the kernel coincide, we must have that $ Av \in \ker A $ or $ A(Av) = A^2 v = 0 $ for any vector $ v $. This is only possible iff $ A^2 = 0 $.
A: If $x \in \mathbb{R}^n$, then $Ax \in \operatorname{Im}(A) = \operatorname{Ker}(A)$ and so $A(Ax) = (A^2)x = 0$. By taking $x = e_i$ (the standard basis vectors) or by the correspondence between matrices and linear maps, you can see that this implies that $A = 0$.
By the rank-nullity theorem,
$$ \dim \operatorname{Ker}(A) + \dim \operatorname{Im}(A) = 2 \dim \operatorname{Im}(A) = 2 \operatorname{rank}(A) = n. $$
A: Most easy answer should be:
Let $v\in V ,\ w\in Ker(A)$ thus $\forall w \quad Aw=0$
Since $Im(A)=Ker(A)$ and $Av\in Im(A)$ follows $A(Av)=A(w)=0$ for all $v \in V$.
Thus $A*A=A^2=0$
$\quad\square$
