Describe the span of the given vectors geometrically and algebraically:
$\pmatrix{1\\0\\-1}$, $\pmatrix{-1\\1\\0}$, $\pmatrix{0\\-1\\1}$.
I have figured out that these vectors are linearly dependent since the system with the following augmented matrix
$$\begin{bmatrix}1 & -1 &0 & 0\\0 & 1 & -1 & 0\\-1 & 0 & 1 & 0\end{bmatrix}$$
led me to a parametric solution
$$t\pmatrix{1\\0\\-1} + t\pmatrix{-1\\1\\0} + t\pmatrix{0\\-1\\1} = 0$$ (so another solutions exist besides the trivial one).
How do I now determine what kind of a geometric object it represents? It can't be a 3D space since there are not 3 linearly independent vectors in the set.
I assume that it is an equation of a plane since it does not look like that any two vectors are a linear combination of third one (such a case would result in a line in 3D). How do I choose which two vectors from the three to use as directional vectors defining the plane? Thanks.
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$\begingroup$Your three vectors $\vec a,\vec b,\vec c$ are linearly dependent but the first two are linearly independent. Therefore $\operatorname{span}(\vec a,\vec b,\vec c)=\operatorname{span}(\vec a,\vec b)$. The span is a 2D plane. You could use any two vectors to do this, since any of your three vectors can be written as a linear combination of the other two.
$\endgroup$ 2 $\begingroup$Perhaps it is worth noting also that since your vectors transform to each other under permutations of the 3 axes, the plane which you seek is normal to the vector $(1,1,1)$. You can verify this by computing the inner product of this vector with your 3 vectors. This gives a nice way to visualize the plane. It is a plane which is tangent to the origin such that the 3 positive axes are all on one side, equi-angular from that plane.
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