Why does a $n$-dimensional convex polytope require at least $n+1$ vertices?

Question

Given a set of $k$ vectors $\{\mathbf{v}_i\}$ one can define the convex polytope $$ \left\{\sum_{i=1}^k \lambda_i\mathbf{v}_i\,\middle|\,\forall i: 0 \leq \lambda_i \leq 1 \text{ and } \sum_{i=1}^k \lambda_i = 1\right\}\,. $$ But why does $k$ have to be of size at least $n+1$ in order for this polytope to be $n$-dimensional? I think I could probably figure out a reason, but none of the directions along which I'm thinking are as straightforward as expected.

Answer 1

Let $\bar{v}=\frac{1}{n}\sum_{k=1}^{n}v_{k}.$ This is clearly in the convex polytope generated by the $\{v_{i}\}$, and we observe that $\{v_{i}-\bar{v}\}$ contains $n$ vectors; but $$v_{n}-\bar{v}=(n\bar{v}-\sum_{k=1}^{n-1}v_{k})-\bar{v}=(n-1)\bar{v}-\sum_{k=1}^{n-1}v_{k}=-\sum_{k=1}^{n-1}(v_{k}-\bar{v}).$$

Then the whole polytope lies in the affine plane $\bar{v}+\mathrm{span}\{v_{i}-\bar{v}\}_{i=1}^{n-1},$ which is $(n-1)$-dimensional.