$U = \{ u \in \mathbb{R}^4 | 2u_1 + 3u_3 - 2u_4 = 0 \}$. That is, the set of four dimensional vectors with real components such that the constraint is satisfied.
Since $\dim \mathbb{R}^4 = 4$ and there is one constraint, we intuitively expect that $\dim U = 3$.
Some guesswork/experimentation can find a basis.
It should be clear that $b_1 = (0,1,0,0) \in U$.
Suppose $u_1 = 1, u_2, = u_3 =0$, then if we choose $u_4 = 1$, we see that
$b_2=(1,0,0,1) \in U$.
Now suppose $u_1=0, u_2 = 0, u_3 = 2$, then if we choose $u_4 = 3$, we have $b_3=(0,0,2,3) \in U$.
It is easy to check explicitly that if $u \in U$, then we can write $u$ as a combination of the $b_k$.
Explicitly, $u = u_1 b_2 + u_2 b_1 + {1 \over 2 } u_3 b_3$. (Unfortunate numbering of basis elements.)
It is also straightforward to check that the $b_k$ are
linearly independent.
If $\sum \alpha_k b_k = (\alpha_2, \alpha_1, 2 \alpha_3, \alpha_2+3 \alpha_3) = 0$, then we must have $\alpha_1 = \alpha_2 = \alpha_3 = 0$.
Hence $\dim U = 3$.
Inputs
$$
\alpha =
\left(
\begin{array}{rr}
1 & -5 \\ -4 & 2
\end{array}
\right),
\qquad
\beta =
\left(
\begin{array}{rr}
1 & 1 \\ -1 & 5
\end{array}
\right),
\qquad
\gamma =
\left(
\begin{array}{rr}
2 & -4 \\ -5 & 7
\end{array}
\right),
\qquad
\delta =
\left(
\begin{array}{rr}
1 & -7 \\ -5 & 1
\end{array}
\right)
$$
Find the basis for these matrices.
Matrix of row vectors
As noted by @Bernard, compose a matrix of row vectors. Flatten the matrices in this manner
$$
\left(
\begin{array}{rr}
1 & -5 \\ -4 & 2
\end{array}
\right)
\quad \Rightarrow \quad
\left(
\begin{array}{crrc}
1 & -5 & -4 & 2
\end{array}
\right)
$$
Compose the matrix
$$
\mathbf{A} =
\left(
\begin{array}{crrr}
1 & -5 & -4 & 2 \\\hline
1 & 1 & -1 & 5 \\\hline
2 & -4 & -5 & 7 \\\hline
1 & -7 & -5 & 1
\end{array}
\right)
$$
Row reduction
Column 1
$$
\left(
\begin{array}{rccc}
1 & 0 & 0 & 0 \\
-1 & 1 & 0 & 0 \\
-2 & 0 & 1 & 0 \\
-1 & 0 & 0 & 1 \\
\end{array}
\right)
%
\left(
\begin{array}{crrc}
1 & -5 & -4 & 2 \\
1 & 1 & -1 & 5 \\
2 & -4 & -5 & 7 \\
1 & -7 & -5 & 1 \\
\end{array}
\right)
%
=
%
\left(
\begin{array}{crrr}
\boxed{1} & -5 & -4 & 2 \\
0 & 6 & 3 & 3 \\
0 & 6 & 3 & 3 \\
0 & -2 & -1 & -1 \\
\end{array}
\right)
%
$$
Column 2
$$
\left(
\begin{array}{rccc}
1 & 0 & 0 & 0 \\
-1 & 1 & 0 & 0 \\
-2 & 0 & 1 & 0 \\
-1 & 0 & 0 & 1 \\
\end{array}
\right)
%
\left(
\begin{array}{crrr}
\boxed{1} & -5 & -4 & 2 \\
0 & 6 & 3 & 3 \\
0 & 6 & 3 & 3 \\
0 & -2 & -1 & -1 \\
\end{array}
\right)
%
=
%
\left(
\begin{array}{cccc}
\boxed{1} & 0 & -\frac{3}{2} & \frac{9}{2} \\
0 & \boxed{1} & \frac{1}{2} & \frac{1}{2} \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
\end{array}
\right)
%
$$
The fundamental rows are marked by the unit pivots.
Solution
The basis is
$$
\mathcal{B} = \left\{
\alpha, \, \beta \right\}
= \left\{
\left(
\begin{array}{rr}
1 & -5 \\ -4 & 2
\end{array}
\right),
\
\left(
\begin{array}{rr}
1 & 1 \\ -1 & 5
\end{array}
\right)
\right\}
$$
Best Answer
For the first, you wish to find vectors $(a_1, a_2, a_3)$ such that $a_1 - 3a_2 + 2a_3 = 0.$ If this equation came up in a row reduction process, we would identify $a_2$ and $a_3$ as free variables, since the pivot is in the first column. Thus we may write $a_1 = 3a_2 - 2a_3.$ From this we see that there are two free variables, and thus the dimension of the corresponding space is $2.$
Now for the basis. Any vector in this space has the form $(3a_2 - 2a_3, a_2, a_3)$. Decomposing shows that this is a sum of two vectors $(3a_2, a_2, 0)$ and $(-2a_3, 0, a_3)$, where $a_2$ and $a_3$ are free to be any real number. Thus we see that $(3, 1, 0)$ and $(-2, 0, 1)$ form a basis for this space. Two basis vectors, dimension two.
Let's now consider the collection of functions of the form $a\cos x + b\sin x.$ The zero function is in this collection, and it is easy to show that it is closed under linear combinations. A basis for this function space is the set $\{\sin x, \cos x\},$ with dimension 2.