How do I verify that a set of vectors is a basis for the given plane

Question

I have a set of 2 vectors: $\{ (1,2,0), (0,2,-1) \}$. I have to show that this set is a basis for the plane with equation: $2x_1 - x_2 -2x_3 = 0$. I know that the normal vector of the plane is $\langle 2, -1, 2\rangle$. I also know that I have to check for linear dependence.

But do I check linear dependence only with the 2 vectors in the set? Because when I use Gauss-Jordan elimination to check for linear dependence for all 3 vectors (the 2 in the set and the normal vector of the plane), I get an inconsistent solution. What do I do?

Answer 1

both vectors $(1,2,0)$ and $(0,2,-1)$ are on the plane $2x_1-x_2-2x_3 = 0.$ you don't need the normal to the plane to show that these vectors are linearly independent. in fact, if you have only two nonzero vectors, then they are linearly dependent if one is not the multiple of the other as the case her.

Answer 2

The equation of the plane directly gives a normal: $$ 0 = 2 x_1 - x_2 - 2 x_3 = (2,-1,-2)^T \cdot (x_1,x_2,x_3)^T = n \cdot x $$ Now examining the scalar product for every vector in the span: $$ n \cdot x = n \cdot \left( c_1 a_1 + c_2 a_2 \right) = c_1 \left(n \cdot a_1 \right) + c_2 \left(n \cdot a_2\right) = 0 $$ as the $a_i$ are normal to $n$. Thus the span is part of the plane.

Can the plane have more vectors than the span? Its defining normal vector is base of a 1D orthogonal space, so it must be 2D, as expected from a plane.

The span might be of dimension less than 2. It is at least 1D as the generating $a_i$ are not null vectors. That would only leave linear dependence as cause for a 1D span, thus we test this: $$ c_1 a_1 + c_2 a_2 = 0 \iff \\ \left[ \begin{array}{rr|r} 1 & 0 & 0 \\ 2 & 2 & 0 \\ 0 & -1 & 0 \end{array} \right] \iff \left[ \begin{array}{rr|r} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & -1 & 0 \end{array} \right] \iff \left[ \begin{array}{rr|r} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{array} \right] \iff \\ \left( \begin{array}{r} c_1 \\ c_2 \end{array} \right) = \left( \begin{array}{r} 0 \\ 0 \end{array} \right) $$ There is only 0 as solution, so they are linear independent. Thus the span is 2D and equal to the plane. The $a_i$ form a basis for the plane.