# 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.

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This would be a lot easier if you said what $W$ is. – Chris Eagle Mar 8 '13 at 16:47
Note you actually have 4 vectors, not 3. A plane has dimension two. If two linearly independent vectors belong to it, they span it. – 1015 Mar 8 '13 at 16:50
i assume the subspace generated by the basis. Just choose two arbitrary points (they are a linear combination of the basis elements) out of $W$ and show the equation – Quickbeam2k1 Mar 8 '13 at 16:50
Tipos fixed. Sorry I'm typing on an iPad and it's pretty hard to do it right. – Susana Mar 8 '13 at 16:52
@Susana: You can find some good starting points on how to format mathematics on the site here. This AMS reference is very useful. – Zev Chonoles Mar 8 '13 at 17:26

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 $W$.

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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

$$\vec x =\left(\begin{array}{c} x \\ y \\ z \end{array}\right)= a \left(\begin{array}{r} 2 \\ 4 \\ -2 \end{array}\right)+b\left(\begin{array}{c} 2 \\ 1 \\ 1 \end{array}\right)$$

where $a$ and $b$ are numbers.

So your equations for $x,y$ and $z$ would be

$$\begin{array}{ccc} x &=& 2a+2b \\ y &=& 4a+b \\ z &=& -2a+b \end{array}$$

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.

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