Let $A,B$ be $n \times n$ matrices so that $AB = 0$ If $A,B \neq 0$ what do I know about $A$ and $B$? Let $A,B$ be $n \times n$ matrices so that $AB = 0$ If $A,B \neq 0$ what do I know about $A$ and $B$?
I want to expand my knowledge about matrices arithmetics and so. Supposing the above what do I know about both $A$ and $B$? I think I read somewhere something about the amount of $0$ columns in $A$ and $0$ rows in $B$, but I can't make the connection.
 A: First, you know that both matrices must be singular ($\det = 0$). If we had $A$ nonsingular ($\det A\ne 0$), then $A$ would have an inverse and we'd get $B=A^{-1}(AB)=A^{-1}0 = 0$. [Both Camerons commented this above, but the folks giving official "answers" missed this point.]
More specifically, you know that every vector $y=Bx$ satifies $Ay=0$, so $C(B)\subset N(A)$, where $C(B)$ is the column space (range) of $B$ and $N(A)$ is the nullspace (kernel) of $A$. [$(AB)x = A(Bx)$, so if $y=Bx$ for some $x\in\mathbb R^n$, then $Ay=0$. By definition, $C(B)$ is the set of all vectors that can be written as $Bx$ for some $x\in\mathbb R^n$.]
A: Image of $B$ is a subspace of kernel of $A$. Therefore, rank of $B$ is less or equal to $n-$ rank($A$).
A: You know that 
$$\det(AB)=\det(A)\times\det(B)=0$$
Therefor at least one of $A$ or $B$ has a $0$ determinant meaning that at least one of them is singular.
Addition (thanks to Ted):
In this case, we have $A,B\ne 0$, assume by way of contradiction that, WLOG,  $B$ is invertible, meaning that $|A| = 0$. For all $v\in V$ we have:
$$0=(AB)v = A(Bv)=Au=0\iff \ker A = V\iff A=0$$
contradiction.
A: The following is a complete solution to the problem. It shows that $A$ and $B$ are singular as others have pointed out, and it also shows that this is all there is to know. 
Theorem: Let $R$ be any non-zero commutative ring. Then $A\in \text{ Mat}_n(R)$ is a zero divisor iff $\det(A)=0.$ 
Proof: If $A$ is a zero divisor, there is a non-zero $B$ such that $AB=0$ so $A$ is not invertible. 
Now assume $A$ is not invertible, so $n\geq k:=\text{rank}(A)+1.$ We construct an $n \times k$ matrix $B$ which satisfies $AB=0,$ adding columns of zeros can make $B$ square. 
If $k=1$ any non-zero $B$ works since $A=0,$ so suppose $k\geq 2.$ Then some $k-1 \times k-1$ submatrix $C$ of $A$ has non-zero determinant. If $D$ is a $k\times k$ submatrix that contains $C$ then $\text{adj} D\neq 0$ since one of the entries is (up to sign) $\det(C).$ 
Suppose $D$ is the submatrix whose row numbers are $I=\{ i_1,\cdots, i_k\}$ and column numbers are $J=\{ j_1, \cdots, j_k \}$ which are both listed in increasing order. This means the $(s,t)$ entry of $D$ is the $(i_s, j_t)$ entry of $A.$ 
Define the $(j,s)$ entry of $B$ to be the $(t,s)$ entry of $\text{adj} D$ if $j=j_t$ for some $t$ and zero otherwise. That is, put $\text{adj} D$ in every row whose row number is in $J$ and zero elsewhere. 
For any $1\leq i\leq n, 1\leq s \leq k$ the $(i,s) $ entry of $AB$ is then
$$\sum_{j=1}^{n} a_{ij}b_{js} = \sum_{t=1}^k a_{ij_t} (\text{adj} D)_{(t,s)} =  \sum_{t=1}^k a_{ij_t} (-1)^{t+s} \det(D^{\hat{s}, \hat{t}}).$$
This is the determinant (via row expansion) of the $k\times k$ matrix $D'$ obtained by replacing the $s$-th row of $D$ with $(a_{ij_1}, a_{ij_2}, \cdots, a_{ij_k} ).$ There are three possible cases:


*

*If $i=i_s$ then $D'=D$ so $\det(D')=0.$ 

*If $i=i_t$ for $t\neq s$ then $D'$ repeats a row so $\det(D')=0$ again. 

*$D'$ is the result of permuting the rows of a $k\times k$
minor of $A$ so $\det(D')=0.$


Therefore every entry of $AB$ is zero.  $\Box$
Remark: This theorem can be used to give a proof that if $z_1,\cdots, z_n$ are linearly independent elements of $R^m$ (viewed as an $R$-module) then $n\leq m,$ which is a well known fact in the special case that $R$ is a P.I.D.
A: "something about the numbers of $0$ columns in $ A$ and $0$ rows in $B$" -- certainly, no. 
Take for example 
$$
A=\left(
  \begin{array}{cc}
    1 & -1 \\
    1 & -1 \\
  \end{array}
\right), \ \ \ \ 
\left(
  \begin{array}{cc}
   1  & 1 \\
   1  & 1 \\
  \end{array}
\right)
$$
But you can be sure that columns  of $A$ or of $B$ are linearly dependent, since  ${\rm det} A$ or ${\rm det} B$ equals $0$.
