TITLE: Finding a basis for a given subspace of $\Bbb R^4$
QUESTION [6 upvotes]: Find a basis for the subspace $ W = \{(x, y, z, w) \in\Bbb R^4
: y − 2z + w = 0\}$. What is $\dim(W)$?
I don't seem to understand how to solve this problem. I just don't know where to start I am not given an A matrix to work with. possibly the b matrix is $[0, 1, -2,1]$ but is that enough to find the basis?

REPLY [6 votes]: Well, every vector in $W$ may be written as
$$
\begin{bmatrix}
x\\ y\\ z\\ 2z-y
\end{bmatrix}
=x
\begin{bmatrix}
1\\0\\0\\0
\end{bmatrix}
+
y
\begin{bmatrix}
0\\ 1\\ 0\\ -1
\end{bmatrix}
+
z
\begin{bmatrix}
0\\0\\1\\ 2
\end{bmatrix}
$$
One easily checks that the collection
$$
\mathcal B=
\left\{
\begin{bmatrix}
1\\0\\0\\0
\end{bmatrix}
,
\begin{bmatrix}
0\\ 1\\ 0\\ -1
\end{bmatrix}
,
\begin{bmatrix}
0\\0\\1\\ 2
\end{bmatrix}
\right\}
$$
is linearly independent. Hence $\mathcal B$ is a basis for $W$ which implies $\dim(W)=3$.

REPLY [4 votes]: You could write every vector in the subspace as:
$$(x,y,z,w)=(x,2z-w,z,w)=x(1,0,0,0)+z(0,2,1,0)+w(0,-1,0,1)$$
Then being the vectors $(1,0,0,0),(0,2,1,0),(0,-1,0,1)$ independents the dimension of the subspace is 3 (because every vector is linear combination of that 3 vectors)