Are the span of these vectors the same?

Question

Let's say we have to solve the equation $2x+3y+z=0$. If we let $x=s, y=t,$ then $z=-2s-3t$, and the solution set is \begin{bmatrix}s\\t\\-2s-3t\end{bmatrix} which is the span of the vectors \begin{bmatrix}1\\0\\-2\end{bmatrix} and \begin{bmatrix}0\\1\\-3\end{bmatrix} However, if instead we had let $y=t$ and $z=s$, we would have gotten \begin{bmatrix}-1.5s-.5t\\t\\s\end{bmatrix} or the span of the vectors \begin{bmatrix}-1\\0\\2\end{bmatrix} and \begin{bmatrix}-3\\2\\0\end{bmatrix}Are these results equivalent?

Answer 1

Yes they are. Notice that a plane has infinitely many vectors in it. So as long as we don't choose two vectors that are parallel to each other or none of them is $\vec{0}$, those two vectors will span the plane from which they are chosen (Since plane is 2-dimensional, we are chosing $2$ vectors). In other words, you can find infinitely many vector pairs on the same plane, which span that plane.

For example, in cartesian coordinate system, $$\bigg\{\begin{bmatrix}1\\0\end{bmatrix},{\begin{bmatrix}0\\1\end{bmatrix}}\bigg\}$$ is the most common base. But $$\bigg\{\begin{bmatrix}1\\1\end{bmatrix},{\begin{bmatrix}2\\1\end{bmatrix}}\bigg\}$$ also spans the same plane as well as $$\bigg\{\begin{bmatrix}1\\0\end{bmatrix},{\begin{bmatrix}4\\1\end{bmatrix}}\bigg\}$$ and so on.

In your case, notice that you can also combine the results that you have already found in order to have a different base as $$\begin{bmatrix}0\\1\\-3\end{bmatrix}, \begin{bmatrix}-3\\2\\0\end{bmatrix}$$ which also spans the given plane.

Answer 2

The second pair of vectors is obviously linearly independent so also spans a two-dimensional space. Therefore, it suffices to show that both are in the span of the first pair of vectors. This should be obvious for $[-1,0,2]^T$. For the second vector of the two, you need a linear combination of $[1,0,-2]^T$ and $[0,1,-3]^T$ that annihilates the last coordinate. The two possibilities are $3[1,0,-2]^T-2[0,1,-3]^T$ and $-3[1,0,-2]^T+2[0,1,-3]^T$. The latter is equal to $[-3,2,0]^T$, so you’re done.