How to prove that a set spans a plane

Question

How do we prove that a set of vectors span a plane in $\mathbb{R}^3$?

(This is not the question I am asking for help with! This is an example of the method my teacher has given us to show that a set spans something!) For instance, to prove that the set $B = \{2, x-1, x^2+1\}$ spans $\mathbb{R}_2[x]$, my teacher has had us prove the following statement $$\exists w,y,z \in \mathbb{R}\ \mbox{such that}\ w*2+ y*(x-1) +z*(x^2+1) = ax^2 + bx + c.$$ Instead of $ax^2 + bx + c$, what would we use for a plane in $\mathbb{R}^3$?

Edit: To be more clear, the plane we have is $ax + by + cz=0$. How would I show that a set spans that?

Answer 1

The plane

$$ ax+by+cz=0$$

is the nullspace of the matrix

$$A=\left[\begin{matrix}a & b & c\end{matrix}\right]$$

Given an arbitrary set of vectors $S$, you can determine that they span the above plane by checking:

1. All of the vectors are in the nullspace of the matrix. That is $Av=0$ for all $v\in S$
2. $S$ contains at least two linearly independent vectors

Answer 2

David's already answered this in a neat way above, but let me try to fill some of the conceptual gaps:

The idea is: "using linear combinations of the elements in my set, can I reach every element of my space?". Your teacher has simply noted that every element of your space can be written uniquely as $ax^2 + bx + c$ for some $a, b, c$. In effect, your teacher is taking a nice basis you already know, $\{1, x, x^2\}$, and relating your new prospective spanning set to it.

If you happen to know a basis for your plane, e.g. $\left\{ \begin{pmatrix} 1\\3\\5 \end{pmatrix}, \begin{pmatrix} 2\\-1\\0 \end{pmatrix} \right\}$, then you can use that in the same way: every element of your plane can be written uniquely in the form $r \begin{pmatrix} 1\\3\\5 \end{pmatrix} + s\begin{pmatrix} 2\\-1\\0 \end{pmatrix}$, so it suffices to show that you can make everything of that form out of your prospective spanning set.

If you don't know a basis for your plane, you can always find one. I would wager that $\left\{ \begin{pmatrix} b\\-a\\0 \end{pmatrix}, \begin{pmatrix} 0\\c\\-b \end{pmatrix} \right\}$ is probably a basis for your plane - to show this, it suffices to check that these two vectors are linearly independent (and they will automatically be spanning by a dimension count). (N.B. of course, you need to be careful if $b = 0$ here, but I'm sure you can see how to fix it if this is the case.)

If you don't know a basis for your plane, and writing one down explicitly looks like a bit of a thankless task, this is where David's answer comes in. It's essentially a way of doing the same thing without the need to pick a basis.