Can such a set of vectors exist? Is it possible, given $n\geq 3,$ to find $\{\mathbf{x}_1,\;\mathbf{x}_2,\;\dots,\;\mathbf{x}_n\}\subset\mathbb{R}^n$ satisfying:


*

*$\mathbf{x}_i,\mathbf{x}_j$ linearly independent for any distinct $i,j$

*$\mathbf{x}_i,\mathbf{x}_j,\mathbf{x}_k$ linearly dependent for any distinct $i,j,k$


I can't see why this wouldn't be possible, but I am having trouble finding an explicit example. Any ideas would be appreciated.
 A: That $\mathbf{x}_i$ and $\mathbf{x}_j$ are linearly independent means they span a plane $P_{ij}$. That $\{\mathbf{x}_i,\mathbf{x}_j,\mathbf{x}_k\}$ is linearly dependent means that $\mathbf{x}_k \in P_{ij}$. For all distinct $i,j,k$. That means $P_{ij} = P_{km}$ for all $i\neq j$ and $k\neq m$. So it remains to find $n$ pairwise linearly independent vectors in a plane. Take for example $n$ points on the unit circle in the first quadrant.
A: This is absolutely possible.
For instance, let
$$
\mathbf{x}_i=\langle1,i,0,0,\ldots,0\rangle.
$$
Then if $\alpha_i\mathbf{x}_i+\alpha_j\mathbf{x}_j=\vec{0}$, it must be the case that $\alpha_j=-\alpha_i$ to get the correct first coordinate... but this doesn't work out for the second coordinate.
Now, note that the span of any two vectors is the two-dimensional subspace $\{\langle x,y,0,0,\ldots,0\mid x,y\in\mathbb{R}\}$, which clearly contains any third vector.
A: if $n=3$, we have : $x_3 = x_1 + x_2 $
It is easy to see that $(x_1,x_2)$,$(x_1,x_2)$,$(x_1,x_2)$ are linearly independent but $(x_1,x_2,x_3)$ is linearly dependent.
