Let the $n-1$ simplex be denoted as:
$$S_{n-1} = \{x = (x_1, \ldots, x_n) | \sum\limits_{i = 1}^nx_i = 1, 0\leq x_i \leq 1\}$$
Then the interior of a simplex is simply:
$$S_{n-1}^\circ = \{x = (x_1, \ldots, x_n) | \sum\limits_{i = 1}^nx_i = 1, 0<x_i < 1\}$$
I stumbled upon a post online that says the simplex is a vector space: https://golem.ph.utexas.edu/category/2016/06/how_the_simplex_is_a_vector_sp.html
I am finding it difficult to understand this claim, because I was taught that a vector space is anything that satisfies the axioms of being a vector space, one being that the zero vector is in the vector space. Clearly, the simplex doesn't have anything that looks like a zero vector.
In the post it claims that the barycenter $(\frac{1}{n}, \ldots, \frac{1}{n})$ is in effect the zero vector of the interior of the simplex.
But why is this the case? Because clearly adding another vector $v \in S^\circ_{n-1}$ with $0 = (\frac{1}{n}, \ldots, \frac{1}{n})$ produces another vector $w \neq v$...
Can someone please explain how this all works?
When constructing a vector space, you're allowed to define addition and scalar multiplication to be whatever operations you want (as long as they satisfy the axioms for a vector space). So the set $S^\circ_{n-1}$ is indeed naturally a vector space over $\mathbb{R}$ with zero element $(1/n,1/n,\dots,1/n)$, but the "addition" operation of this vector space is not just defined by componentwise addition (instead it is defined by componentwise multiplication, followed by rescaling to get an element of $S^\circ_{n-1}$). Similarly, scalar multiplication is not just coordinatewise multiplication (instead it is componentwise exponentiation, followed by rescaling). You can find all the details and the motivation for these definitions in the post you linked to.