Finding basis of subspace from homogenous equations (checking) The question;
$U = \{x |Ax = 0\}$ If $ A = \begin{bmatrix}1 & 2 & 1 & 0 & -2\\ 2 & 1 & 2 & 1 & 2\\1 & 1 & 0 & -1 & -2\\ 0 & 0 & 2 & 0 & 4\end{bmatrix}$
Find a basis for $U$.

To make it linearly independent, I reduce the rows of $A$;

$\begin{bmatrix}1 & 0 & 0 & 0 & 0\\0 & 1 & 0 & 0 & -2\\0 & 0 & 1 & 0 & 2\\0 & 0 & 0 & 1 & 0\end{bmatrix}\begin{bmatrix}x_1 & x_2 & x_3 & x_4\end{bmatrix} = 0$
So I figure I just have to list the rows out as vectors;
Basis vectors = $\begin{pmatrix}1\\0\\0\\0\\0\end{pmatrix} , \begin{pmatrix}0\\1\\0\\0\\-2\end{pmatrix} , \begin{pmatrix}0\\0\\1\\0\\2\end{pmatrix} , \begin{pmatrix}0\\0\\0\\1\\0\end{pmatrix}$
(can't figure out how to surround it all with curly brackets in MathJax).

Am I doing this properly? It feels too simple...
 A: Observe that row operations doesn't change the solution for the system of linear equations.  From the last column of the reduced row echelon form (RREF) of $A$,
$$\begin{bmatrix}1 & 0 & 0 & 0 & 0\\0 & 1 & 0 & 0 & -2\\0 & 0 & 1 & 0 & 2\\0 & 0 & 0 & 1 & 0\end{bmatrix}\begin{bmatrix}x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5\end{bmatrix} = 0,\tag{*}\label{*}$$
we see that if $x = (x_1,\dots,x_5)^T \in U$, $Ax = 0$, so $\sum\limits_{j=1}^5 a_i x_i = 0$, where $a_i$ denotes the $i$-th column of $A$.  From \eqref{*}, we observe that $(x_1,x_2,x_3,x_4,x_5) = (0,-2,2,0,-1)$ is a possible solution.  By the rank-nullity theorem, since $\mathop{\mathrm{rank}}(A) = 4$, so $\dim U = \dim\mathsf{N}(A) = 5 - 4 = 1$.  Hence, we conclude that $x = (0,-2,2,0,-1)^T$ is a basis for $U$.
A: $U$ is the set of vectors $\vec x$, whose image under $f_A$ is $\vec 0$.
Since your matrix $A$ is rank 4 and $f_A:K^5 \rightarrow K^4$, it must have a $1$ dimensional Kernel.
So your answer should be subspace of $K^5$ spanned by 1 vector. What vector could that be from your reduced matrix?
