Prove $A^{2}=0$ iff $C\left(A\right)=R^{0}\left(A\right) $ I'm learning some Linear Algebra through a University Textbook and I've come across this question which I have a hard time solving:
Let There be a square Matrix A. 
Prove that $A^{2}=0$ iff $C\left(A\right)=R^{0}\left(A\right) $
where $R^{0}\left(A\right)$ is the nullspace (kernel).
I just don't understand how to approach this question. I know how to find the four fundamental subspaces given a certain array, but not too much apart from that.
**EDIT: Apparently the term C(A) is not standard, in the textbook it refers to the subspace of the columns of A
Can anyone help? Thanks
 A: Saying $A^2=0$ means that every column of $A$ is in the kernel $R^\circ(A)$, since the columns of $A^2$ are obtained by applying $A$ to the columns of$~A$. In other words it is equivalent to $C(A)\subseteq R^\circ(A)$.
But you cannot conclude from $A^2=0$ that $C(A)=R^\circ(A)$. The rank nullity theorem says that always $\dim(C(A))+\dim(R^\circ(A))=n$ (the size of the square matrix$~A$), and this can even prevent the two subspaces from ever being equal, if $n$ is odd. In any case, one can always (if $n>0$) choose $A$ such that $C(A)\subsetneq R^\circ(A)$; for instance $A=0$, or when $n>2$ any matrix with just one nonzero entry, not situated on the main diagonal.
A: It might not be as formal as a mathematician would put it, but this is how I would put it:


*

*$Ker(A) =$ $\{x \in R^n: Ax=0\}$

*$C(A) =$ $\langle a_i \rangle,$ where $A=[a_1 a_2 ... a_n]$  and thus, $C(A) =$ $\sum_i\alpha_ia_i$ for $\alpha_i \in \mathbb{R}$. 


If:
$Ker(A) = C(A)$ $\leftrightarrow$ $\{x = \sum_i\alpha_ia_i \in R^n: Ax=0\}$
$\leftrightarrow$ $\{A (\sum_i\alpha_ia_i)=0\}$
$\leftrightarrow$ $\{A (A\alpha^T)=0\}$
$\leftrightarrow$ $\{A^2\alpha^T=0\}$
$\leftrightarrow$ $A^2=0$, assuming that $\alpha^T \in \mathbb{R}^n$ is a vector of arbitrary constants.  
Hope it helps you!
