if $AB\neq 0$ for any non zero matrix $B$ then $A$ is invertible Question is to check that :
If $A$ is an $n\times n$ matrix over a field $F$ and $AB\neq 0$ for any non zero matrix $B_{n\times n}$ over $F$ then, $A$ is invertible.
This does make some sense to me but i am not sure how to prove this.
As $AB\neq 0$ for any $B$ , in particular, we have $A.A\neq 0$ i.e., $A^2\neq 0$
for similar reasons we see that $A^n\neq 0$ for any positive integer $n$
So, $A$ is not nilpotent... I see that this is just nilpotent...
I am stuck to prove that $A$ is invertible.
I do have some thoughts inbetween but nothing gives me simple way to conclude final result.
please help me to see this by giving some hints (I am sure this must be very easy)
Thank you
 A: Remember that $A$ is not invertible iff the equation $Ax=0$ has non-trivial solutions. Can you now construct a non-zero $B$ with $BA=0$?
A: Suppose $A$ isn't invertible. It follows that $(0,v)$ is an eigenpair of $A$, for some non-null $n\times 1$ vector $v$, that is, $Av=0_{n\times 1}$. Now consider the matrix $B$ whose columns are all equal to $v$.
A: Yet another proof. Consider the linear map $L_A : M_n(F)\to M_n(F)$ defined by $$\forall B\in M_n(F)\;:\; L_A(B)=AB\, .$$ The assumption means that $\ker (L_A)=\{ 0\}$, i.e. $L_A$ is 1-1. Since $M_n(F)$ is finite-dimensional, it follows that $L_A$ is invertible. In particular, one can find $B\in M_n(F)$ such that $AB=Id$; so $A$ is invertible.
A: Suppose $A=(a_{ij})_{n \times n}$ is not invertible. Then the vectors $A_{i}=(a_{i1}, \cdots, a_{in})(1 \le i \le n)$ are not linearly independent. This means that there is a $k$ -dimensional vector space $L$ with $k<n$ such that  all $A_i (1 \le i \le n)$ are  in $L$.
Let $b=(b_1,b_2, \cdots, b_n)$ be a non-zero vector being orthogonal to $L$. Let consider matrix $B$ whose each  column coincides with $b$. Then $A \times B=0$ and we get a contradiction.  
A: Let V be a vector space of all square matrices of size n then we can define a operator T on V such that T(B) = AB and if AB not equal to zero for any non-zero matrix B this implies that the Operator  T(B) = AB is non singular i.e. one - one . Here T is on finite dimensional Vector space so it must be onto hence identity matrix I must have pre-image in V i.e. there exist a non zero matrix M in V such that T(M) = AM = I.
