Let $$U=\operatorname{Span}\{v_1,v_2,v_3\}, V=\operatorname{Span}\{v_4,v_5,v_6\}$$
Where $$v_1=(1,28,2,39),v_2=(2,28,2,39),v_3=(-1,28,2,39)\\v_4=(0,8,0,11),v_5=(0,31,1,43),v_6=(0,-3,0,-4)$$
Find a basis for $U\cap V$
I have forgot the algorithm for finding intersection of subspaces, but in the exercise I was given an algorithm that I have yet met and would like to understand, how and why it works.
The first step is to find an homogeneous system s.t the subspace is the solution set (Null space).
To do so for $U$ we look at $$\left(\begin{array}{ccc|c}
1 & 2 & -1 &x\\
28 & 28& 28&y\\
2 & 2 & 2& z\\
39& 39&39& w \\
\end{array}\right)\sim \left(\begin{array}{ccc|c}
1 & 2 & -1 &x\\
1 & 1& 1& \frac{y}{28}\\
0 & 0 & 0& \frac{z}{2}-\frac{y}{28}\\
0& 0&0& \frac{w}{39}-\frac{y}{28} \\
\end{array}\right)$$
So the matrix that $U$ is her solution set is $$ \begin{pmatrix} 0&-\frac{1}{28}&\frac{1}{2}&0\\ 0&-\frac{1}{28}&0&\frac{1}{39} \end{pmatrix}$$
Doing the same with $V$ we get
$$\left(\begin{array}{ccc|c}
0 & 0 & 0 &x\\
8& 31& -3&y\\
0 & 1 & 0& z\\
11& 43&-4& w \\
\end{array}\right)\sim \left(\begin{array}{ccc|c}
8 & 31 & -3 &y\\
0 & 1 & 0& z\\
0 & 0& 1& 8w-11y-3z\\
0& 0&0& x \\
\end{array}\right)$$
So the matrix that $V$ is her solution set is $$ \begin{pmatrix} 1&0&0&0\end{pmatrix}$$
To find the intersection we solution for $$\left(\begin{array}{cccc|c}
0 & -\frac{1}{28} & \frac{1}{2} &0 &0\\
0 & -\frac{1}{28}& &\frac{1}{39}& 0\\
1 & 0 & 0& 0 &0\\
\end{array}\right) $$
Which is $$ \begin{pmatrix} 0&28&2&39\end{pmatrix}$$
a. Why does putting the vectors in columns and $x,y,z,w$ in the $b$ vector give us a system which the vectors are their solutions?
b. Why looking at the null space of the vectors (why we put the vectors in horizontal and not in vertical) that we found give us the basis of intersection?
C. Is there another way to find the basis of intersection?
Best Answer
You have written $U$ and $V$ explicitly as:
$$U = \left\{(x_1, x_2, x_3, x_4) \in \mathbb{R}^4 : -\frac{1}{28}x_2 + \frac12 x_3 = 0, -\frac{1}{28}x_2 + \frac1{39} x_4 = 0\right\}$$
$$V = \left\{(x_1, x_2, x_3, x_4) \in \mathbb{R}^4 : x_1 = 0\right\}$$
so
$$ U \cap V = \left\{(x_1, x_2, x_3, x_4) \in \mathbb{R}^4 : -\frac{1}{28}x_2 + \frac12 x_3 = 0, -\frac{1}{28}x_2 + \frac1{39} x_4 = 0, x_1 = 0\right\}$$ $$ = \left\{(x_1, x_2, x_3, x_4) \in \mathbb{R}^4 : \begin{pmatrix} 0 & -\frac{1}{28} & \frac{1}{2} &0 \\ 0 & -\frac{1}{28}& 0 &\frac{1}{39}\\ 1 & 0 & 0& 0\end{pmatrix}\begin{pmatrix}x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix} = 0\right\}$$
and now you just solve that linear system to obtain a solution $(x_1, x_2, x_3, x_4) = \left(0,28,2,39\right)$.
Here is another way:
First find bases for $U$ and $V$. We have:
$$3(1,28,2,39) -2(2,28,2,39) = (-1,28,2,39)$$
so $\{(1,28,2,39), (2,28,2,39)\}$ is a basis for $U$.
The set $\{(0,8,0,11), (0,31,1,43),(0,-3,0,-4)\}$ is linearly independent so it is a basis for $V$.
To find a basis for $U \cap V$, consider the set
$$\{(1,28,2,39), (2,28,2,39), (0,8,0,11), (0,31,1,43),(0,-3,0,-4)\}$$
and write every element as a linear combination of its predecessors, if possible.
The first four vectors are shown to be linearly independent. This gives $\dim (U + V) = 4$.
For $(0,-3,0,-4)$ we obtain:
$$(0,-3,0,-4) = \color{red}{-(1,28,2,39) + \frac12 (2,28,2,39)} - \frac52 (0,8,0,11) + (0,31,1,43)$$
The vector $$\color{red}{-(1,28,2,39) + \frac12 (2,28,2,39)} = {\left(0,-14,-1,- \frac{39}2\right)}$$
is therefore both in $U$ and $V$.
The dimension is
$$\dim(U \cap V) = \dim U + \dim V - \dim(U + V) = 1$$
so $U \cap V$ is in fact spanned by $\left(0,-14,-1,- \frac{39}2\right)$, or $\left(0,28,2,39\right)$ when multiplied by $-2$.
Therefore, $\{\left(0,28,2,39\right)\}$ is a basis for $U \cap V$.