Rank of matrix products Suppose $A_1$, $A_2$ and $A_3$ are $3$ by $3$ matrices of rank $2$ such that their kernels are linearly independent. Is the following true?
Define: $V_1=A_2A_1$, $V_2=A_3A_2$ and $V_3=A_1A_3$. Then each $V_i$ is rank $1$ and $V_iV_j=0$ for $i\neq j$.
Thanks for your help!
Stan

If I interpret the question correctly the assumption is wrong, take 
\begin{align*}
A_1&= \begin{pmatrix} 0& 1 & 0\\ 0&0 &1\\ 0 & 0 & 0\\ \end{pmatrix} \qquad \operatorname{kernel} \begin{pmatrix} \alpha \\0 \\ 0 \end{pmatrix} \\
A_2&= \begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 &0 \\ 0 & 0 &0 \end{pmatrix}
\qquad \operatorname{kernel} \begin{pmatrix} 0 \\ 0 \\ \beta \\ \end{pmatrix}\\
A_3&= \begin{pmatrix} 0 & 0& 1\\ 0& 0 & 0 \\ 1 & 0 & 0\end{pmatrix} \qquad \operatorname{kernel}\begin{pmatrix} 0 \\\gamma \\ 0\\ \end{pmatrix}
\end{align*}
Than $V_1 V_2= A_2 A_1^2 A_3$ which is 
\begin{align*}
V_1V_2&= \begin{pmatrix} 1 & 0 &0\\ 1 & 1 &0 \\ 0 & 0 & 0 \\ \end{pmatrix} \cdot 
\begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0& 0 & 0\\ \end{pmatrix}\cdot 
\begin{pmatrix} 0 & 0 & 1 \\0 & 0 & 0 \\ 1 & 0 & 0 \\ \end{pmatrix} \\
&= \begin{pmatrix} 1 & 0 & 0 \\1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}
\end{align*}
checked with mathematica, so we have shown that $V_i V_j=0$ for $i \neq j$ is wrong, we will show what else won't work with this example. 
$$V_1=A_2 A_1 = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 1  & 1 \\ 0 & 0 & 0 \\ \end{pmatrix} $$ 
So rank of $V_1$ is 2 not 1. 
For $V_2$ we have 
$$A_3 A_2 = \begin{pmatrix} 0 & 0 & 0 \\0 & 0 & 0\\ 1 & 0 & 0 \end{pmatrix}
$$ 
Here it works, and even for $V_3$ we have 
$$A_1  A_3 = \begin{pmatrix}  0  & 0 & 0 \\ 1 & 0 & 0 \\0 & 0 & 0 \\ \end{pmatrix} $$
