Adding vector to span a space 
Let $A=\left\{\left(\begin{array}{c} 18 \\ 6 \\ -4 \\ 12 \end{array}\right),\left(\begin{array}{c} 6 \\ 2 \\ 2 \\ -6 \end{array}\right)\right\}$
   find the vectos to be added so A will span $\mathbb{R}^4$?

So what I did is :
\begin{pmatrix}
    18 & 6 & -4 & 12  \\
    6       & 2 & 2 & -6 \\
\end{pmatrix}
the row reduced echelon form is:
\begin{pmatrix}
    1 & \frac{1}{3} & 0 & 0  \\
    0       & 0 & 1 & -10 \\
\end{pmatrix}
What should I do next? I need to write the vectors that should be added with 1 and 0 only?

Linear independence
First, we must check to see if our initial two vectors are linearly independent. In this instance, the signs provide essential clues. The sign patterns reveal they aren't proportional:
$$
 \begin{array}{cccc}
   + & + & - & + \\
   + & + & + & - \\
 \end{array}
$$
Gram-Schmidt orthogonalization
The Gram-Schmidt orthogonalization of the vector set
$$
\left\{ \,
% 1
\left[
\begin{array}{r}
 18 \\ 6 \\ -4 \\ 12 \\
\end{array}
\right], 
% 2 
\left[
\begin{array}{r}
 6 \\ 2 \\ 2 \\ -6 \\
\end{array}
\right],
% 3 
\left[
\begin{array}{c}
 1 \\ 0 \\ 0 \\ 0 \\
\end{array}
\right],
% 4 
\left[
\begin{array}{r}
 0 \\ 0 \\ 0 \\ 1 \\
\end{array}
\right]
%
\, \right\}
$$
provides one solution
$$
\mathbb{R}^{4} = \text{span }
\left\{ \,
% 1
\frac{1}{2\sqrt{130}}
\left[
\begin{array}{r}
 9 \\ 3 \\ -2 \\ 6 \\
\end{array}
\right], 
% 2 
\frac{1}{\sqrt{130}}
\left[
\begin{array}{r}
 6 \\ 2 \\ 3 \\ -9 \\
\end{array}
\right],
% 3 
\frac{1}{\sqrt{10}}
\left[
\begin{array}{r}
 1 \\ -3 \\ 0 \\ 0 \\
\end{array}
\right],
% 4 
\frac{1}{\sqrt{10}}
\left[
\begin{array}{r}
 0 \\ 0 \\ 1 \\ -3 \\
\end{array}
\right]
%
\, \right\}
$$
