Null space of a matrix with a particular block decomposition I am considering rectangular matrices that can be expressed in the following block form:
$$
M=\begin{pmatrix}
 0 & A \\
 B & I
\end{pmatrix},
$$
where the $I$ in the bottom right is square, but the zero matrix in the upper left need not be.
I am interested in the properties of the null spaces of such matrices. In my application the null space of the full matrix is sometimes just the null space of $B$, but sometimes not. I'm interested in sufficient and/or necessary conditions for this to be the case. Are any such conditions known?
(Note: when I say the null space of $M$ is the null space of $B$, I mean that  $\begin{pmatrix}v\\0\end{pmatrix}$ is in the null space of $M$ if and only if $v$ is in the null space of $B$.)

Suppose that 
$A$ has $p$ columns, and $q$ rows(and so $B$ has $q$ columns and $p$ rows). For for a vector $x$, write $x=(v,w)$, with $v\in \mathbb{R}^q$ and $w\in \mathbb{R}^p$. Then $Mx=(Aw,Bv+w)$. So $x\in {\rm Ker}(M)$ if and only if $Aw=0$, and $Bv=-w$. This imply that $ABv=0$, hence $v\in {\rm Ker(AB)}$. If now $v\in {\rm Ker}(AB)$, then $(v,-Bv)$ is in ${\rm Ker}(M)$, So ${\rm Ker}(M)=\{(v,-Bv), v\in {\rm Ker}(AB)\}$. Now to have ${\rm Ker}(M) =\{(v,0), v\in {\rm Ker}(B)\}$, we need to have $ {\rm Ker}(AB)={\rm Ker}(B)$. This is the case when $A$ is injective on ${\rm Im}(B)$. 
