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 $U$.

Note that $B\in U$ if and only if
\begin{align*}
B
&=
\begin{bmatrix}
b_{11} & b_{12} & b_{13} \\
b_{21} & b_{22} & b_{23}
\end{bmatrix} \\
&=
\begin{bmatrix}
-b_{12}-b_{13}-4\,b_{21} -4\,b_{22}-4\,b_{23} & b_{12} & b_{13} \\
b_{21} & b_{22} & b_{23}
\end{bmatrix} \\
&= b_{12}\underbrace{\begin{bmatrix}
-1 & 1 & 0 \\
0 & 0 & 0
\end{bmatrix}}_{=u_1}
+
b_{13}
\underbrace{
\begin{bmatrix}
-1 & 0 & 1 \\
0&0&0
\end{bmatrix}}_{=u_2}
+b_{21}
\underbrace{
\begin{bmatrix}
-4 & 0 & 0 \\
1&0&0
\end{bmatrix}}_{=u_3} \\
&\qquad\qquad\qquad\qquad\qquad\qquad+
b_{22}
\underbrace{
\begin{bmatrix}
-4 & 0 & 0 \\
0&1&0
\end{bmatrix}}_{=u_4}
+
b_{23}
\underbrace{
\begin{bmatrix}
-4 & 0 & 0 \\
0&0&1
\end{bmatrix}}_{=u_5}
\end{align*}
This proves that $\{u_1,u_2,u_3,u_4,u_5\}$ is a basis for $U$. In particular, $\dim U=5$.
