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...

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 $U$.
