If $Col(A)$ and $Col(B)$ are linearly independent, then $AB = 0$ What does $Col(A)$ and $Col(B)$ are linearly independent mean? 
It is from an old linear algebra book and I don't see this used very often, does it mean a basis of $Col(A)$ and a basis of $Col(B)$ are linearly independent?
Assume the above definiton, then it means $Col(A)$ and $Col(B)$ are linearly independent iff  $Col(A) \cap Col(B) = \{0\}$. 
Just from this how could I show $AB = 0$?
Edit: I guess we need $A$ and $B$ to be symmetric, from the counter example $\begin{bmatrix} 1 & 2 \\ 1 &2 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 1 &1 \end{bmatrix}$.
 A: First, an example:
$$
\begin{align}
  \mathbf{A} \mathbf{B} &= \mathbf{0} \\[5pt]
\left(
\begin{array}{cccc}
 1 & 0 & 0 & 0 \\
 0 & 1 & 0 & 0 \\
 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 \\
\end{array}
\right)\,
\left(
\begin{array}{cccc}
 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 \\
 0 & 0 & 1 & 0 \\
 0 & 0 & 0 & 1 \\
\end{array}
\right)
&= 
\left(
\begin{array}{cccc}
 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 \\
 0 & 0 & 0 & 0 \\
\end{array}
\right)
\end{align}
$$
Certainly $\mathbf{A} \mathbf{B} = 0$.
Look at the matrix-vector multiplication
$$
  \mathbf{A} \mathbf{B} x \overset{?}{=} \mathbf{0}
$$
Define 
$$
  y = \mathbf{B} x
$$
When does
$$
\mathbf{A} y = \mathbf{0}?
$$
When the vector $y$ is not in the column space of $\mathbf{A}$. Yet the vector $y$ is in the column space of $\mathbf{B}$ by construction:
$$
  y = \mathbf{B} x = x_{1} \mathbf{B}_{1} + x_{2} \mathbf{B}_{2} + \dots 
$$
It is assembled using the column vectors of $\mathbf{B}$.
The proof demonstrates that if the matrix product is $\mathbf{0}$, the multiplicand matrices must be linearly independent. 
As noted by @Nan Li, it does not show the reverse direction, that the product of regular matrices is necessarily $\mathbf{0}$.
