Reasons why $AB$ is not full column rank

Question

Let $A$ be an $m \times n$ matrix with $Rank(A)=m$ and let $B$ be an $n \times k$ matrix with $Rank(B)=k$. Assume $k \leq m$. Under what conditions is $Rank(AB)<k$ and under what conditions is $Rank(AB) = k$?

I originally thought it had to do with $AB_i \neq 0$ but the solutions show that this is not correct.

edit: The question is now rewritten for a second time since I don't feel like the solutions answer the question I had in mind.

The original question was "Let $A$ be an $m \times n$ matrix with $Rank(A)=m$ and let $B$ be an $n \times k$ matrix with $Rank(B)=k$. Assume $k \leq m$. What are the reasons that $Rank(AB)<k$? i.e., why would $Rank(AB) \neq k$?"

After the first edit the question was: "Let $A$ be an $m \times n$ matrix with $Rank(A)=m$ and let $B$ be an $n \times k$ matrix with $Rank(B)=k$. Assume $k \leq m$. Let $B_1,\ldots,B_{k}$ be the columns of $B$ and assume that $AB_i \neq 0$, for $i=1,\ldots,k$. Is $Rank(AB)=k$?"

Answer 1

Here's an example:

$$\left[\begin{array}{ccc} 1&0&0\\ 0&1&0\end{array}\right] \left[\begin{array}{c} 0 \\ 0 \\1\end{array}\right]= \left[\begin{array}{c} 0 \\ 0 \end{array}\right]$$

Answer 2

Try $m=k=2,\; n=3$: $$A = \pmatrix{1 & 1 & 0\cr 1 & 1 & 1\cr},\ B = \pmatrix{1 & 0\cr 0 & 1\cr 0 & 0\cr},\ AB = \pmatrix{1 & 1\cr 1 & 1\cr} $$ $A$ and $B$ have rank $2$, and all $A B_i = \pmatrix{1\cr 1\cr} \ne 0$, but $AB$ has rank $1$.