Finding the basis of the intersection of a subspace and span I need help with determining the basis of $U_1 \cap U_2$ in the following problem:
Let $V=\mathbb{R}^4$. ${U_1} = \left\{ {\left( {\begin{array}{*{20}{c}}
  {{x_1}} \\ 
  {{x_2}} \\ 
  {{x_3}} \\ 
  {{x_4}} 
\end{array}} \right)\left| {{x_1} - {x_2} + {x_3} - 3{x_4} = 0} \right.} \right\}$ and $U_2=\left\langle {\left( {\begin{array}{*{20}{c}}
  1 \\ 
  1 \\ 
  0 \\ 
  3 
\end{array}} \right),\left( {\begin{array}{*{20}{c}}
  0 \\ 
  { - 1} \\ 
  0 \\ 
  1 
\end{array}} \right)} \right\rangle$. 
If $U_1$ is a subspace of $V$, determine a basis of $U_1 \cap U_2$.
My attempt:
I know that ${U_2} = \left\{ {\left( {\begin{array}{*{20}{c}}
  \lambda  \\ 
  {\lambda  - \mu } \\ 
  0 \\ 
  {3\lambda  + \mu } 
\end{array}} \right)\left| {\lambda ,\mu  \in \mathbb{R}} \right.} \right\}$, and that the next step is that I should choose an element in $U_1$ and in $U_2$, e.g. Let $w \in {U_1}$ and let $w \in {U_2}$. Then we know that $w$ is of the form $w = \left( {\begin{array}{*{20}{c}}
  \lambda  \\ 
  {\lambda  - \mu } \\ 
  0 \\ 
  {3\lambda  + \mu } 
\end{array}} \right)$, but I'm not sure what the procedure is from there.
 A: The next step is to note that\begin{align}U_1\cap U_2&=\left\{\begin{pmatrix}\lambda\\\lambda-\mu\\0\\3\lambda+\mu\end{pmatrix}\,\middle|\,\lambda-(\lambda-\mu)-3(3\lambda+\mu)=0\right\}\\&=\left\{\begin{pmatrix}\lambda\\\lambda-\mu\\0\\3\lambda+\mu\end{pmatrix}\,\middle|\,9\lambda+2\mu=0\right\}\\&=\left\{\begin{pmatrix}\lambda\\\frac{11}2\lambda\\0\\-\frac32\lambda\end{pmatrix}\,\middle|\,\lambda\in\mathbb{R}\right\}.\end{align}Can you take it from here?
A: $U_1$ is the set of vector such that $\begin{bmatrix} 1 &-1 &1 &-3\end{bmatrix}\begin{bmatrix} x_1 \\x_2 \\x_3 \\x_4\end{bmatrix} = 0$
$\begin{bmatrix} 1 &-1 &1 &-3\end{bmatrix}\begin{bmatrix} 1 \\1 \\0 \\3\end{bmatrix} = -9\\
\begin{bmatrix} 1 &-1 &1 &-3\end{bmatrix}\begin{bmatrix} 0 \\-1 \\0 \\1\end{bmatrix} = -2$
$2\begin{bmatrix} 1 \\1 \\0 \\3\end{bmatrix} - 9\begin{bmatrix} 0 \\-1 \\0 \\1\end{bmatrix} = \begin{bmatrix} 2 \\11 \\0 \\-3\end{bmatrix}$
A: If $(1,1,0,3)$ and $(0,-1,0,1)$ span $U_2$ then if $(x_1,x_2,x_3,x_4) \in U_2$ there exist $a,b$ such that:
$$ a(1,1,0,3)+b(0,-1,0,1)=(x_1,x_2,x_3,x_4)$$
We face,
$$ a=x_1, a-b = x_2, x_3=0, 3a+b=x_4 $$
Ok, so,
$$ b = x_1-x_2 \qquad \& \qquad b = x_4-3x_1 \ \ \Rightarrow \ \ x_1-x_2 = x_4-3x_1$$
Ok, in summary, $(x_1,x_2,x_3,x_4) \in U_2$ if we have
$$ 4x_1-x_2-x_4 = 0 \ \ \& \ \ x_3=0. $$
If $(x_1,x_2,x_3,x_4) \in U_1$ then we know $x_1-x_2+x_3-3x_4 = 0$. Consequently, to find $(x_1,x_2,x_3,x_4) \in U_1 \cap U_2$ we need to solve equations for both subspaces simultaneously:
$$ \left[ \begin{array}{cccc|c} 1 & -1 & 1 & -3 & 0 \\ 4 & -1 & 0 & -1 & 0 \\ 0 & 0 & 1 & 0 & 0 \end{array}\right] \sim 
\left[ \begin{array}{cccc|c} 1 & -1 & 0 & -9/3 & 0 \\ 0 & 1 & 0 & 11/3 & 0 \\ 0 & 0 & 1 & 0 & 0 \end{array}\right] \sim \left[ \begin{array}{cccc|c} 1 & 0 & 0 & 2/3 & 0 \\ 0 & 1 & 0 & 11/3 & 0 \\ 0 & 0 & 1 & 0 & 0 \end{array}\right] $$
Thus $x_1 = -2x_4/3$ and $x_2 = -11x_4/3$ and $x_3=0$ with $x_4$ free. In short,
$$ U_1 \cap U_2 = \text{span}(-2,-11,0,3). $$
