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 
$\begin{bmatrix}3.5\\-.5\\1\end{bmatrix}$

Your answer and process seem correct.  That is, the vectors $(1,0,1)$ and $(1,2,-1)$ form a basis of the column space, while the vector $(3.5,-.5,1)$ forms a basis of the kernel.
