I'm having troubles understanding following problem:
Find a basis for $U \cap V$ with:
\begin{equation*} U = \text{span}\big\{ \underbrace{\begin{pmatrix} 0 \\ 1 \\ 0 \\ 1 \end{pmatrix}}_{= \vec{u}_1},\underbrace{\begin{pmatrix} 1 \\ 0 \\ -1 \\ 0 \end{pmatrix}}_{= \vec{u}_2}\big\} \qquad\qquad V = \text{span}\big\{ \underbrace{\begin{pmatrix} -1 \\ 1 \\ 0 \\ 0 \end{pmatrix}}_{=\vec{v}_1},\underbrace{\begin{pmatrix} -2 \\ 0 \\ 1 \\ -1 \end{pmatrix}}_{= \vec{v}_2}\big\} \end{equation*}
I know that I have to find solutions to this equation: \begin{equation} \vec{0} = \alpha \vec{u}_1 + \beta \vec{u}_2 - \gamma \vec{v}_1 - \epsilon \vec{u}_2 = \begin{pmatrix} \vec{u}_1 & \vec{u}_2 & -\vec{v}_1 & -\vec{v}_2 \end{pmatrix} \begin{pmatrix} \alpha \\ \beta \\ \gamma \\ \epsilon \end{pmatrix} \end{equation}
\begin{align} U \cap V &= \text{ker}\begin{pmatrix} \vec{u}_1 & \vec{u}_2 & -\vec{v}_1 & -\vec{v}_2 \end{pmatrix} = \text{ker}\begin{pmatrix} 0 & 1 & 1 & 2 \\ 1 & 0 & -1 & 0 \\ 0 & -1 & 0 & -1 \\ 1 & 0 & 0 & 1 \end{pmatrix}\\ &=\text{ker}\begin{pmatrix} 0 & 0 & 1 & 1 \\ 0 & 0 & -1 & -1 \\ 0 & 1 & 0 & 1 \\ 1 & 0 & 0 & 1 \end{pmatrix} =\text{ker}\begin{pmatrix} 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 \\ 1 & 0 & 0 & 1 \end{pmatrix} \\ &= \text{ker}\begin{pmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix} = \text{span} \big\{ \begin{pmatrix} -1 \\ -1 \\ -1 \\ 1 \end{pmatrix}\big\} =\text{span}\big\{ \begin{pmatrix} 1 \\ 1 \\ 1 \\ -1 \end{pmatrix}\big\} \end{align}
Now I found a basis and therefore I'm done with the problem.
However, how can this vector be a basis of $U \cap V$ if it is clear to see that:
\begin{equation} \begin{pmatrix} 1 \\ 1 \\ 1 \\ -1 \end{pmatrix} \notin U \qquad\qquad \begin{pmatrix} 1 \\ 1 \\ 1 \\ -1 \end{pmatrix} \notin V \end{equation}
Note that when you solve the equation below:
\begin{equation} \vec{0} = \alpha \vec{u}_1 + \beta \vec{u}_2 - \gamma \vec{v}_1 - \epsilon \vec{u}_2 = \begin{pmatrix} \vec{u}_1 & \vec{u}_2 & -\vec{v}_1 & -\vec{v}_2 \end{pmatrix} \begin{pmatrix} \alpha \\ \beta \\ \gamma \\ \epsilon \end{pmatrix} \end{equation}
You find all the scalars which defines the vectors in the intersection between U and V but not the vectors. Hence, you should use the scalars you found to obtain the vectors, Choose: $\alpha = 1$, $\beta = 1$ for the vectors in U basis, and $\gamma = -1$, $\epsilon = 1$ for the vectors in V basis, and you will get: \begin{pmatrix} 1 \\ 1 \\-1 \\ 1 \end{pmatrix} which is indeed in U∩V.