<s>[INST] 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.
 [/INST] 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.
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