Are these two equalities equivalent: $ABA=0$ and $BA=0$? There are two equalities as follows:

$ABA=0$ , $BA=0$
where $A$ and $B$ are two $n\times n$ matrices.

Are the mentioned equalities the same? if not, why?
Thanks in advance.
 A: Not equivalent. Let $n = 2$. Take $A$ as the matrix that kills the unit vector $e_1$, say, and $B$ as the rotation of $90^\text{o}$. 
Then $A B A$ kills all vectors.
But $B A$ only kills those along $e_1$.
A: If these two equations would be the same then also for $B=A$ the equations $A^3=0$ and $A^2=0$ should be equivalent what certainly is not the case for all matrices,    
consider for example  the case of matrix   
$A= \begin{bmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0  \end{bmatrix}$,
$A^2= \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 0 & 0  \end{bmatrix}$,
$A^3= \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0  \end{bmatrix}$.
A: The equality $BA=0$ obviously implies $ABA=0$. But the reverse implication does not hold. For instance let 
$$
A=\begin{bmatrix} 1&0\\0&0\end{bmatrix},\ \ B=\begin{bmatrix} 0&0\\1&0\end{bmatrix} . 
$$
Then $ABA=0$, but $BA\ne0$. 
By changing $B$ to $$ B=\begin{bmatrix} 0&1\\1&0\end{bmatrix} ,$$one gets an example where $AB\ne0$, $BA\ne0$, and $ABA=0$. 
Edit: as mentioned by the user Improve, it is worth noting that the double implication does hold when $A$ is invertible. 
A: Consider $A = \begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix}, B = I$. Then $BA = AB = A \neq 0$ but $ABA = A^2 = 0$
