Prove that a subspace is a plane So I have four vectors $v_1= (2,4,-2), v_2=(2,1,1), v_3=(3,3,0), v_4=(4,2,2)$. It's matrix $A$ is
$\begin{pmatrix}
2 & 2 & 3 & 4 \\
4 & 1 & 3 & 2 \\
-2 & 1 & 0 & 2 \\
 \end{pmatrix} \leadsto \begin{pmatrix}
2 & 0 & 1 & 0 \\
0 & 1 & 1 & 2 \\
0 & 0 & 0 & 0 \\
\end{pmatrix}$
So I know a basis is $b = \{(2,4,-2), (2,1,1)\}$ and it's nullspace is $$\mathrm{span}\{(-1/2, -1,1,0), (0,-2,0,1)\}$$ that's clear for me I understand perfectly.
Now, the book's question is


Prove that the subspace $W$ (generated by $v_1$, $v_2$, $v_3$ and $v_4$) is the plane $x-y-z=0$.


I'm self learning linear algebra (following MIT's Open Course Ware) but not on the videos nor on the book is this explained. Now, looking the basis I've found I can see that (2-4+2 = 0 and (2 - 1 -1 = 0) but, how is this precisely related to the plane, I mean, how can I solve the question. Thanks in advance.
 A: Checking that your two basis elements satisfy that equation suffices to check that $W$ is indeed that plane.  The reason is because, if we let $P$ be the plane, by checking that each of the basis elements is contained in $P$ you have shown $W \subseteq P$.  You also know that both $W$ and $P$ are $2$-dimensional, thus you know $W = P$.
Of course, this assumes that you have reached the point in the course where you know a little something about dimensions.  If this isn't the case you'll have to finish the proof by showing $P \subseteq W$.  Do this by finding a basis for $P$ and checking that these basis elements are contained in $W$.
A: The solution space $W$ of your system is a plane floating in a 3-dimensional space. You know that because your basis has two vectors and each of them has 3 components. A solution is then any vector
\begin{equation}
 \vec x =\left(\begin{array}{c} x \\ y \\ z \end{array}\right)= a \left(\begin{array}{r} 2 \\ 4 \\ -2 \end{array}\right)+b\left(\begin{array}{c} 2 \\ 1 \\ 1 \end{array}\right)
\end{equation}
where $a$ and $b$ are numbers.
So your equations for $x,y$ and $z$ would be
\begin{equation}
  \begin{array}{ccc}
    x &=& 2a+2b \\
    y &=& 4a+b \\
    z &=& -2a+b
  \end{array}
\end{equation}
You must show that this fullfills the plane equation $x-y-x=0$, so you just substitute your $x,y$ and $z$ inside the equation
$2a+2b-4a-b+2a-b=0$
and see it is true.
Finished.
