Are the vertices of a non-degenerate $n$-simplex in $\mathbb{R}^{n+1}$ linearly independent?

Question

By non-degenerate, I mean the $n+1$ vertices form the convex hull of the simplex and none of them lie within or on the $n-1$ simplex formed from the other points. I know that the vertices are affinely independent, but that doesn't necessarily imply that they're linearly independent as well

Answer 1

$\newcommand{\Reals}{\mathbf{R}}$The $n + 1$ vertices of an $n$-simplex (degenerate or not) in $\Reals^{n+1}$ may be viewed as displacement vectors from the origin, and therefore constitute a linearly independent set if and only if:

• They span $\Reals^{n+1}$; if and only if
• No hyperplane ($n$-dimensional linear subspace) contains the set of vertices; if and only if
• No hyperplane contains the simplex; if and only if
• No two vertices lie on a line through the origin, no three are coplanar with the origin, etc.

Answer 2

No. Any $n$-dimensional subspace of $R^{n+1}$ contains lots of nondegenerate $n$-simplices. For example, a segment of a line through the origin of $R^2$.

Answer 3

Since $0$ can be one of the vertices, they obviously don't need to be linearly independent. (Note that affine independence is translation invariant, but linear independence isn't.)