Finding a Basis for this subspace Set $V=\mathbb{R}^{2x3}$ and let $U$ be a subspace of $V$ defined by:
\begin{equation*}
U=\{B=(b_{ij})\in V\mid b_{11} + b_{12} + b_{13} = -4(b_{21} + b_{22} + b_{23})\}.
\end{equation*}
I would greatly appreciate it if someone could help me understand how to find a basis for $U$ and the dimension of $U$.

We have $b_{11} + b_{12} + b_{13} = -4(b_{21} + b_{22} + b_{23})$ that is equivalent to $$b_{11} = - b_{12} - b_{13} - 4b_{21} - 4b_{22} - 4b_{23}.$$
Here $b_{12},b_{13},b_{21},b_{22},b_{23}$ are declared to be free variables and the general solution is
$$
  (- b_{12} - b_{13} - 4b_{21} - 4b_{22} -   
  4b_{23},b_{12},b_{13},b_{21},b_{22},b_{23}),
$$
where $b_{12},b_{13},b_{21},b_{22},b_{23} \in \mathbb{R}$.
Now we can find $5$ base vectors by substituting $(1,0,0,0,0)$, $(0,1,0,0,0)$, ..., $(0,0,0,0,1)$ for $b_{12},b_{13},b_{21},b_{22},b_{23}$ in the expression above. In matrix notation it should look as follows:
$$
  \begin{pmatrix}
    -1 & 1 & 0 \\
     0 & 0 & 0
  \end{pmatrix},
  \begin{pmatrix}
    -1 & 0 & 1 \\
     0 & 0 & 0
  \end{pmatrix},
  \begin{pmatrix}
    -4 & 0 & 0 \\
     1 & 0 & 0
  \end{pmatrix},
  \begin{pmatrix}
    -4 & 0 & 0 \\
     0 & 1 & 0
  \end{pmatrix},
  \begin{pmatrix}
    -4 & 0 & 0 \\
     0 & 0 & 1
  \end{pmatrix}.
$$
