Finding the basis and dimension of a vector space Find the basis and dimension of vector space $ L_{1}$ spanned by 
vectors $ a_{1} ,a_{2},a_{3} $, the basis and dimension of  vector space $ L_{2}$ spanned by 
vectors $ b_{1} ,b_{2},b_{3} $ 
and also $ L_{1} + L_{2} ,  L_{1} \cap
  L_{2}$
 $ a_{1} = (-10,10,-6,-2) ,a_{2} = (-6,4,-4,-4) ,a_{3} = (-2,-12,-4,-20) $
 $ b_{1} = (6,3,-6,4) ,b_{2} = (-4,0,0,-3) ,b_{3} = (-13,-7,9,-12) $
I've tried this:
\begin{pmatrix}-10 & 10 & -6 & -2\\
-6 & 4 & -4 & -4\\
-2 & -12 & -4 & -20
\end{pmatrix}
\begin{pmatrix}-2 & -12 & -4 & -20\\
0 & 70 & 14 & 98\\
0 & 0 & 0 & 0
\end{pmatrix}
So basis for  $ L_{1} $ is  $ (-2,-12,-4,-20), (0,70,14,98) $ ?
 A: For the first, form a matrix $A$ whose columns are the vectors $a_i$ and row reduce it:
$$\begin{bmatrix}
 -10 & -6 & -2 \\
 10 & 4 & -12 \\
 -6 & -4 & -4 \\
 -2 & -4 & -20
\end{bmatrix}
\rightarrow
\begin{bmatrix}
1 & 0 & -4 \\
 0 & 1 & 7 \\
 0 & 0 & 0 \\
 0 & 0 & 0
\end{bmatrix}
$$
There are leading ones in the first and second column of the row reduced matrix, so that means the first and second columns of $A$ (i.e., the vectors $a_1$ and $a_2$) are linearly independent and span the space $\text{span}\{a_1,a_2,a_3\}$. Thus $L_1$ is two dimensional and a basis for $L_1$ is $\{a_1,a_2\}$.
Adapt this technique to answer your other questions.
A: Your method is certainly a correct way of obtaining a basis for $L_1$. You can then do the same for $L_2$.  Another method is that outlined by JohnD in his answer.
Here's a neat way to do the rest, analogous to this second method: suppose that $u_1,u_2$ is a basis of $L_1$, and that $v_1,v_2,v_3$ (there may be no $v_3$) is a basis of $L_2$.  Row reduce the matrix with columns
$$
\pmatrix{u_1&u_2&v_1&v_2&v_3}
$$
The columns corresponding to the resulting pivots form a basis of $L_1 + L_2$.  The other columns correspond to a basis of $L_1 \cap L_2$.
