Give bases for col(A) and null(A) I have
A= $\begin{bmatrix}1&1&-3\\0&2&1\\1&-1&-4\end{bmatrix}$
I row reduce it to 
$\begin{bmatrix}1 &0& -3.5\\0&1&.5\\0&0&0\end{bmatrix}$
How do I find col(A) from the above info? Is it the pivot points correspond to the columns?
So col(A) would be $\begin{bmatrix}1\\0\\1\end{bmatrix}$and $\begin{bmatrix}1\\2\\-1\end{bmatrix}$
and for null(A) I got 
$\begin{bmatrix}3.5\\-.5\\1\end{bmatrix}$

Normally, you should column reduce to find a basis for the column space, or what amounts to the same, row-reduce the transpose matrix:
$$\begin{bmatrix}
1&0&1\\1&2&-1\\-3&1&-4\end{bmatrix}\rightsquigarrow\begin{bmatrix}
1&0&1\\0&2&-2\\0&1&-1\end{bmatrix}\rightsquigarrow\begin{bmatrix}
1&0&1\\0&1&-1\\0&0&0\end{bmatrix}$$
This proves the third column is a linear combination of the first two. Hence the first two column vectors are a basis of the column space.
Your method  is also valid, but it is a indirect method, as it uses the fact that the row rank and the column rank of a matrix are equal.
