How to prove that a set spans a plane

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?

• $ax^2+bx+c$ is the generic form of a second degree polynomial. Why would you want a "different" generic form? – David Peterson Jan 10 '15 at 4:04
• Note: $\{1,x,x^{2}\}$ forms a basis for the space. So any element in the space can be written as some linear combination: $ax^{2} + bx + c$. – nchar Jan 10 '15 at 4:06
• I don't want a different generic form of a second degree polynomial. That was just an example for how my teacher is having us prove that a set spans something. I would like to know what form of a plane I could use there? For instance, the plane we have is ax + by + cz=0. How would I show that a set spans that? – Paula Jan 10 '15 at 4:07
• What does $\mathbb{R}_2[x]$ mean? How is that related to showing a set of vectors spans a plane in $\mathbb{R}^3$? – copper.hat Jan 10 '15 at 4:09
• Generally, a plane is of the form $\{x | \langle n, x \rangle = \alpha \}$, where $n$ is a normal to the plane. – copper.hat Jan 10 '15 at 4:10

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
• Bingo.There you go. – Mathemagician1234 Jan 10 '15 at 4:23

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.

• Thank you for your helpful explanation! The reason I'm actually trying to prove that a set spans that plane is because I'm trying to prove that a set (the set {(b,-a,0),(0,c,-b)} actually haha) is a basis of the plane and thus prove that the dimension of the plane is 2! – Paula Jan 10 '15 at 4:36
• @Pow: aha, okay - take an element (x,y,z) on your plane. Supposing $c \neq 0$, we can more helpfully write this as (x,y,$\frac{-ax-by}{c}$). Now it suffices for you to find r and s (in terms of a, b, c, x and y) such that r(b,-a,0) + s(0,c,-b) = (x,y,$\frac{-ax-by}{c}$) - shouldn't be too hard. (Then of course you have to deal with the case $c = 0$...) :) – Billy Jan 10 '15 at 4:41
• Awesome, thanks! – Paula Jan 10 '15 at 4:43