Alternative way of finding a basis for a vector space The problem:
Let $U$ be a subspace of $\mathbb{R}^4$. $U$ is given by the span of the vectors $\begin{Bmatrix}  \begin{pmatrix}2\\0\\0\\-2\end{pmatrix},\begin{pmatrix}1\\-3\\3\\-1\end{pmatrix},\begin{pmatrix}-2\\0\\-4\\-2\end{pmatrix},\begin{pmatrix}-1\\-3\\-1\\-3\end{pmatrix},\begin{pmatrix}1\\-3\\1\\-3\end{pmatrix}\end{Bmatrix}$.
I need to find a basis of $U$.
My attempt:
I wrote the vectors into a matrix horizontally like this:
$\begin{pmatrix}2&0&0&-2\\1&-3&3&-1\\-2&0&-4&-2\\-1&-3&-1&-3\\1&-3&1&-3\end{pmatrix}$
Here, I tried to find linear combinations and was able to reduce the vectors down to the form:
$\begin{pmatrix}1&1&0&0\\0&1&0&1\\0&0&1&1\\0&0&0&0\\0&0&0&0\end{pmatrix}$
This leaves me with a basis for $U$ as follows: $B = \begin{Bmatrix}  \begin{pmatrix}1\\1\\0\\0\end{pmatrix},\begin{pmatrix}0\\1\\0\\1\end{pmatrix},\begin{pmatrix}0\\0\\1\\1\end{pmatrix}\end{Bmatrix}$.
By definition, a basis of a vector space is a set of linearly independent vectors, which span the vector space (in this example $U$). Since this is the case, $B$ is a basis for $U$.
Is my reasoning correct? If so, please let me know if there is a more standard way of writing down this procedure.

The standard way of solving this problem is to leave the five vectors listed from top to bottom, that is, as columns of $4 \times 5$ matrix.  Then use Gauss-Jordan elimination in the standard way.  At the end, the independent vectors (from the original set) are the ones that correspond to leading $1$'s in the (reduced) row echelon from.
Applying this, the reduce row echelon form is
$\begin{bmatrix} 1 && 0 && 0 && 0 && 0.5 \\ 0 && 1 && 0 && 1 && 1 \\ 0 && 0 && 1 && 1 && 0.5 \\ 0 && 0 && 0 && 0 && 0 \end{bmatrix} $
Therefore, a basis for the space spanned by the $5$ vectors is the set of the first $3$ vectors.
