How do I determine the intersection of span A and span B? Consider the following 2 sets of vectors in $\mathbb R^4$: $A = \{v_1, v_2, v_3\}, B = \{w_1, w_2, w_3\}$. You are given that $A$ is a set of linearly independent vectors and that $B$ is a set of linearly independent vectors.
Let
$v_1 = (3,1,4,1), v_2 = (5,9,2,6), v_3 = (5,3,5,8), w_1 = (9,7,9,3), w_2 = (2,3,6,4), w_3 = (6,2,8,4)$.
(Wrote them sideways when it should be top to bottom in brackets but left to right shown)
Determine the intersection of span$\,A$ and span$\,B$ and write your answer as the span of a set of linearly independent vectors.
I'm really lost in class. Please show steps and answers that I can learn. Please help...
Thank you
 A: Ultimately you need to solve a system of equations:
$$ \alpha_1 v_1 + \alpha_2 v_2 + \alpha_3 v_3 = \beta_1 w_1 + \beta_2 w_2 + \beta_3 w_3$$
This will give you $3$ equations in $6$ variables,  you can then solve for possible values of $\alpha_i$ and $\beta_i$.  Look at a basis for the solutions to $\alpha_i$ and $\beta_i$ and use these to generate a spanning set of $V \cap W$.  Then reduce this spanning set to a basis.
A: The intersection will have dimension 2 or 3 so we can control our results. As your vectors doesn't look very nice we will use the gauss algorithm to make them sweeter. Sweet vectors are vectors with a lot of zeroes. When we calculate $$a_1=v_2-v_3=\begin{pmatrix} 5-5 \\ 9-3 \\ 2-5 \\ 6-8\\ \end{pmatrix} = \begin{pmatrix} 0 \\ 6 \\ -3 \\ -2 \\ \end{pmatrix}$$
Calculating 
$$a_2=3 v_2 -5 v_1= \begin{pmatrix} 15-15 \\ 27-5 \\ 6-20 \\ 18-5\\ \end{pmatrix}=
\begin{pmatrix} 0 \\ 22 \\-14\\ 13 \end{pmatrix}$$
When we calculate now 
$$a_3=3a_2-11 a_1= \begin{pmatrix} 0 \\ 66-66\\ -42+33\\ 39+22\\ \end{pmatrix}= \begin{pmatrix} 
0 \\ 0 \\-9 \\ 61 
\end{pmatrix}$$
Now we calculate 
$$a_4=3a_1 -a_3=\begin{pmatrix} 0 \\ 18\\ -9+9\\ -2-61 \\ \end{pmatrix}=\begin{pmatrix} 0 \\18 \\ 0 \\ 63\\ \end{pmatrix}=9\cdot \begin{pmatrix} 0\\ 2 \\ 0 \\ 7 \\\end{pmatrix}$$
Simplify $v_1$ like this too and the other set, and it will give you an easy (easier) system of equations.
Mathematica gives me those reduced echolons forms 
$$A=\left(
\begin{array}{cccc}
 1 & 0 & 0 & \frac{191}{18} \\
 0 & 1 & 0 & -\frac{67}{18} \\
 0 & 0 & 1 & -\frac{61}{9} \\
\end{array}
\right)\qquad B=
\left(
\begin{array}{cccc}
 1 & 0 & 0 & -\frac{28}{61} \\
 0 & 1 & 0 & -\frac{6}{61} \\
 0 & 0 & 1 & \frac{53}{61} \\
\end{array}
\right)
$$
where every row is a vector.
