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.
Best Answer
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 $W$.