Method for finding intersection between two basis What is the general way to find a basis for the intersection of two sub spaces? 
There's the method that use the fact that  if we take some vector $v\in V$ and $v\in U$ then every linear combination of one subspaces will be equal to the other and we can do: $a_1v_1+a_2v_2... = x_1u_1+x_2u_2... \Rightarrow a_1v_1+a_2v_2... - ( x_1u_1+x_2u_2...) =0$ and find all the $a_i$ in terms of $x_i$ but this method doesn't always work. 
Like in this example: Losing a dimension when finding intersection between subspaces
Or this example: $B_U=\{(-1,1,0,0),(-1,0,1,0),(-1,0,0,1)\} \\ B_V=\{(0,0,1,2),(0,1,0,-1),(-1,0,0,1)$
Where we have $(-c,b,a,2a-b+c)=(-x-y-z,x,y, z)$ so $b=x \\a=y\\ c=x+y+z \\ 2a-b+c=z$
Then $2y=z+x-y \Rightarrow z = -x+3y \Rightarrow c=4y \Rightarrow c=4a$ and now what?
Is there an even more general method to find the intersection?
 A: A method that works in general is as follows: (1) Write both subspaces as kernels of linear mappings, that is - for example - $U = \{x \in k^n \mid A_U x = 0\}$  for some matrix $A_U \in {\rm Mat}_{n}(k)$ (where $k$ is the ground field). 
(2) Then 
$$ U \cap V = \{x \in k^n \mid A_U x = 0, A_V x = 0 \}$$
Now solve this system of linear equations, giving a basis for $U \cap V$. 
For (1): As $U$ is a three-dimensional of $\def\R{\mathbf R}\R^4$, a matrix with kernel $U$ has rank $4-3 = 1$, so we are looking for an $A_U \in\def\Mat#1#2{{\rm Mat}_{#1,#2}(\R)}\Mat14$ with $U = \ker A_U$. Writing $A_U = (a \, b\, c\, d)$, we have the equations 
\begin{align*}
   -a + b &= 0\\
   -a + c &= 0\\ 
   -a + d &= 0
\end{align*}
That is $a = b = c= d$, choosing $d = 1$, we have $A_U = (1\, 1\, 1\, 1)$.
For $V$, we do the same, as $\dim V = 3$ also, we choose the ansatz $A_V = (a\, b\, c\, d)$, giving
\begin{align*}
   c + 2d &= 0\\
  b  - d &= 0\\
   a - d &= 0
\end{align*}
Again, choosing $d = 1$, we get $a = b = 1$ and $c = -2$, that is $A_V = (1\, 1 \, -2\, 1)$.
For (2): We have 
$$ U \cap V = \{x \in \R^4 \mid A_U x = A_V x = 0 \} $$
that is, $U \cap V$ is the solution set of the system 
\begin{align*}
  x_1 + x_2 + x_3 + x_4 &= 0\\
  x_1 + x_2 -2x_3 + x_4 &= 0
\end{align*}
Subtracting both equations, and dividing by 3, we are left with 
\begin{align*}
  x_1 + x_2 + x_3 + x_4 &= 0\\
  x_3 &= 0
\end{align*}
For $x_4 = 1$, $x_2 = 0$, we get $x = (-1, 0, 0, 1)$, for $x_4 = 0$, $x_2 = 1$, we get $x = (-1, 1, 0, 0)$. These form a basis of the kernel, i. e. $U \cap V$, so 
$$ B_{U \cap V} = \{ (-1,0,0, 1), (-1,1,0,0) \}$$ 
