Given that $Av_1=Av_2$,
where $v_1$ and $v_2$ are two distinct nonzero vectors.
Subtracting, we get
$Av_1-Av_2=0$
which gives
$A(v_1-v_2)=0$.
Since $v_1\neq v_2$, we have
$v_1-v_2\neq 0$.
Thus, the homogeneous system $Ax=0$ has a nontrivial solution.
Hence, the null space of $A$ contains a nonzero vector, so $A$ is not invertible.
Therefore, $A$ is not full rank.
So, statement B is true.
Also, for an $n\times n$ matrix, if $A$ is not invertible, then
$\det(A)=0$.
So, statement D is true.
Further, if $\det(A)=0$, then $0$ is an eigenvalue of $A$.
So, statement A is true.
Now if $A$ is not invertible, its columns cannot be linearly independent.
So, statement C is false.
Therefore, the correct statements are $\boxed{\text{A, B and D are true}}$.