Basis for intersecting subspaces - is there a trick here?

Question

I'm doing this problem, which gives me these subspaces of $\mathbb{R}^4$ $$U=\text{span}\left\{\;\begin{pmatrix} 3\\ 2\\4 \\ -1\end{pmatrix},\;\begin{pmatrix} 1\\ 2\\1 \\ -2\end{pmatrix},\;\begin{pmatrix} -2\\ 3\\3 \\ 2\end{pmatrix}\;\right\}\qquad V=\text{span}\left\{\;\begin{pmatrix} 1\\ 4\\2 \\ 4\end{pmatrix},\;\begin{pmatrix} -1\\ 1\\-1 \\ -1\end{pmatrix},\;\begin{pmatrix} 3\\ 1\\2 \\ 0\end{pmatrix}\;\right\}$$ I'm meant to find a basis for $U\bigcap V$. From the dimension of subspace equation I know that it has dimension $2$.

I first tried playing around with the bases for $U$ to get something in $V$, and vice versa, but I just can't spot anything... The only other option I can think of is to write down a system of equations, but this would have $6$ unknowns and $4$ equations which doesn't feel like a good approach.

Is there another way of doing this? Or have I just missed a fairly obvious trick?

Answer 1

First, put the given sets into one set which will be system of generators for $U \cup V$, call it $B$ then reduce it to base.

Write the vectors you kicked out as a linear combination of the vectors from $B$, and put all vectors from $U$ on one side and all vectors from $V$ on the other side.

Say, your vector $v$ looks like $v = 5x + 2y$, where $v, x \in V$ and $y \in U$, write it as $v -5x = 2y$ and $2y$ will be one of the vectors for the base.

Answer 2

For a vector $x \in U \cap V$ there must exist solution vectors $\lambda^U$, $\lambda^V$ for the linear combinations regarding $U$ and $V$ each: $$ x = A_U \lambda^U = A_V \lambda^V \iff \\ A_U \lambda^U + (-A_V) \lambda^V = 0 \iff \\ (A_U \mid -A_V) (\lambda^U, \lambda^V)^t = 0 $$ This matrix is

>> A = [u1, u2, u3, -v1, -v2, -v3]
A =

   3   1  -2  -1   1  -3
   2   2   3  -4  -1  -1
   4   1   3  -2   1  -2
  -1  -2   2  -4   1  -0

We apply Gaussian elimination:

>> rref(A)
ans =

   1.00000   0.00000   0.00000   0.00000   0.55556  -0.69841
   0.00000   1.00000   0.00000   0.00000  -0.88889   0.17460
   0.00000   0.00000   1.00000   0.00000  -0.11111   0.39683
   0.00000   0.00000   0.00000   1.00000  -0.00000   0.28571

From the last row we take $$ \lambda^V_1 + 0.28571 \lambda^V_3 = 0 $$ and this means \begin{align} x &= \lambda^V_1 v_1 + \lambda^V_2 v_2 + \lambda^V_3 v_3 \\ &= -0.28571 \lambda^V_3 v_1 + \lambda^V_2 v_2 + \lambda^V_3 v_3 \\ &= \lambda^V_2 v_2 + \lambda^V_3 (-0.28571 v_1 + v_3) \\ &= \lambda^V_2 b_1 + \lambda^V_3 b_2 \end{align} So (modulo computation errors) a base of $U\cap V$ is $\{ b_1, b_2 \}$ with \begin{align} b_1 &= v_2 \\ b_2 &= -0.28571 v_1 + v_3 \end{align}