No invertible matrix I'm trying to show:
If $A$ is a $n\times n$ matrix which is not invertible, then exists a matrix $n\times n$, $B$ such that: $AB=0$ with $B\neq 0$.
Thanks for your help. 
 A: If $A$ is not invertible then there exists a non-zero column vector $v$ such that $Av=0$. Take $B$ to be formed of $n$ columns all equal to the vector $v$.
If $A$ is invertible then $AB=0$ implies $B=0$ because you can multiply the equality by $A^{-1}$.
A: One short (but advanced) method to show that non-invertible matrices are zero divisors is to remember that $A$ is invertible iff its characteristic polynomial has nonzero constant term, and that matrices satisfy their minimal polynomials.
Then for a non-invertible matrix $A$ with minimal polynomial $p(x)$, $p(A)=A*q(A)=0$, where it is possible to factor out an $A$ because there is no constant term. Thus $q(A)$ is a nonzero matrix multiplying with $A$ to make zero.
A: Since the most obvious answers are taken, let me show here another method.
Suppose that $AB\ne0$ for all $B\ne0$. Then the map $X\mapsto AX$, is linear and injective (because if $AX=0$ then $X=0$ by the assumption). Now, since the set of $n\times n$ matrices is a finite dimensional vector space, an injective map of the space into itself is necessarily surjective. This means in particular that there exists $B$ such that $AB=I$, i.e. $A$ is invertible. 
The contrapositive of the previous paragraph then holds, and this is the required fact "$A$ non-invertible implies that there exists $B\ne0$ with $AB=0$. 
