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 
$\begin{bmatrix}3.5\\-.5\\1\end{bmatrix}$
 A: 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.
A: 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.
A: To clear up confusion, work out the steps.


*

*Form the augmented matrix
$$
% A I
\left[
\begin{array}{c|c}
 \mathbf{A} & \mathbf{I}_{3} \\
\end{array}
\right]
%
=
% 
\left[
\begin{array}{rcr|ccc}
 1 & 0 & 1 & 1 & 0 & 0 \\
 1 & 2 & -1 & 0 & 1 & 0 \\
 -3 & 1 & -4 & 0 & 0 & 1 \\
\end{array}
\right]
$$

*Clear column 1.
$$
% E
\left[
\begin{array}{rcc}
 1 & 0 & 0 \\
 -1 & 1 & 0 \\
 3 & 0 & 1 \\
\end{array}
\right]
% in
\left[
\begin{array}{rcr|ccc}
 1 & 0 & 1 & 1 & 0 & 0 \\
 1 & 2 & -1 & 0 & 1 & 0 \\
 -3 & 1 & -4 & 0 & 0 & 1 \\
\end{array}
\right]
=
% out
\left[
\begin{array}{ccr|rcc}
 1 & 0 & 1 & 1 & 0 & 0 \\
 0 & 2 & -2 & -1 & 1 & 0 \\
 0 & 1 & -1 & 3 & 0 & 1 \\
\end{array}
\right]
%
$$

*Clear column 2.
$$
% E
\left[
\begin{array}{crc}
 1 & 0 & 0 \\
 0 & \frac{1}{2} & 0 \\
 0 & -\frac{1}{2} & 1 \\
\end{array}
\right]
% in
\left[
\begin{array}{ccr|rcc}
 1 & 0 & 1 & 1 & 0 & 0 \\
 0 & 2 & -2 & -1 & 1 & 0 \\
 0 & 1 & -1 & 3 & 0 & 1 \\
\end{array}
\right]
=
% out
\left[
\begin{array}{ccr|rrc}
 1 & 0 & 1 & 1 & 0 & 0 \\
 0 & 1 & -1 & -\frac{1}{2} & \frac{1}{2} & 0 \\
 0 & 0 & 0 & \frac{7}{2} & -\frac{1}{2} & 1 \\
\end{array}
\right]
%
$$


The 
$$
\begin{align}
%
\left[
\begin{array}{c|c}
 \mathbf{A} & \mathbf{I}_{3} \\
\end{array}
\right]
&=
%
\left[
\begin{array}{rcr|ccc}
 1 & 0 & 1 & 1 & 0 & 0 \\
 1 & 2 & -1 & 0 & 1 & 0 \\
 -3 & 1 & -4 & 0 & 0 & 1 \\
\end{array}
\right] \\
%
&\qquad \qquad \qquad
\Downarrow \\
%
\left[
\begin{array}{c|c}
 \mathbf{E_{A}} & \mathbf{R} \\
\end{array}
\right]
&=
%
\left[
\begin{array}{ccr|rrc}
 \boxed{1} & 0 & 1 & 1 & 0 & 0 \\
 0 & \boxed{1} & -1 & -\frac{1}{2} & \frac{1}{2} & 0 \\\hline
 0 & 0 & 0 & \color{red}{\frac{7}{2}} & \color{red}{-\frac{1}{2}} & \color{red}{1} \\
\end{array}
\right]
%
\end{align}
$$
The unit pivots (boxed) in the matrix $\mathbf{E_{A}}$ identifies the fundamental columns of the images. The red vector in $\mathbf{R}$ is the span of the null space:resolution
$$
\boxed{
 \color{blue}{\text{col }  \mathbf{A}} \oplus 
 \color{red} {\text{null } \mathbf{A}} =
 \color{blue}{\mathcal{R}\left(  \mathbf{A}\right)} \oplus 
 \color{red} {\mathcal{N}\left(  \mathbf{A}^{*}\right)} =
%
\color{blue} {
\text{span } \left\{ \,
\left[
\begin{array}{r}
  1 \\
  1 \\
 -3 \\
\end{array}
\right],
%
\left[
\begin{array}{r}
  0 \\
  2 \\
  1 \\
\end{array}
\right]
\, \right\}}
% % %
\oplus
% % %
%
\color{red} {
\text{span } \left\{ \,
\left[
\begin{array}{r}
  \frac{7}{2} \\
 -\frac{1}{2} \\
  1 \\
\end{array}
\right]
\, \right\}}}
$$
