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 $B$.)
 A: 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 ${\rm Im}(B)$. 
A: The null space of $M=\left( \begin{matrix}0&A\\ B&I\end{matrix}\right)$ will never be the same as the null space of $A$, because $M$ and $A$ have different numbers of columns. 
What I think you meant to say was that if $\vec v$ is in the null space of $A$, then $\left( \begin{matrix}0\\ \vec v\end{matrix}\right)$ is in the null space of $M$. In order for this to happen, the columns of $B$ must be linearly independent. However, it might not be sufficient.
It seems that the $RREF(M)$ should be of the form $\left( \begin{matrix}RREF(B) & C\\ 0&RREF(A)\end{matrix}\right)$; the proof of this is to use row operations to reduce the first few rows of $M$ to get $\left(\begin{matrix}0&RREF(A)\\ B&I\end{matrix}\right)$, then use some row operations to get $\left( \begin{matrix}0&RREF(A)\\ RREF(B) & C\end{matrix}\right)$, then swap rows to get the form I showed at the beginning, and then turn any entries above $A$'s pivots into 0s.
The matrix $C$ will be invertible until the last part; after changing entries into 0s, you'll get a formerly-invertible matrix, some of whose columns have been replaced with zero vectors.
I'm doing all of this off the top of my head. At this point, you might want to introduce some notation (indicating which columns have pivots in them), and then see whether you do get the null space of $A$. But this ought to start things off in a productive direction.
