Lets say we have an $n \times m$ matrix $A$, whose column vectors are $(\vec{\mathbf{0}},a_1,a_2,...a_j)$ are points in $\mathbb{R}^n$ and the non-zero vectors have unit length. Let $S:=\mathrm{span}\{a_i\}$. Now, we are given a trial vector, $v \in \mathbb{R}^n$. I want to determine if $v_{||S}$resides within the convex polytope formed from the column vectors of $A$.
Here's what I've tried so far:
Lets say that $A$ has a single non-zero column vector, $a_1$. Then the "polytope" will just be a line segment of length $1$ in $\mathbb{R}^n$. The projection of $v$ onto $S$ in this case can be found by solving the below equation:
$$v_{||a_1} = \lambda a_1: (v-\lambda a_1)\cdot a_1=0, \lambda \in \mathbb{R}$$
We can easily solve for $\lambda$:
$$\lambda = \frac{v \cdot a_1}{a_1 \cdot a_1}$$
Therefore, $v_{||a_1}$ is on the line segment formed by $a_1$ only if $\lambda \in [0,1]$
It seems that we can easily extend this to cases where there are more vectors in $A$ simply by verifying the above condition for each non-zero column vector of $A$. If the condition holds, then $v_{||S}$ must be inside the convex polytope formed by all the column vectors of $A$.
I was wondering if there was a rigorous proof of this or a counterexample?
UPDATED
The condition you give is necessary but not sufficient. Say a point is in the convex polytope so it can be written as ( here think of $v$ as projection of trial vector onto your subspace ) $v=\sum \lambda_j a_j$ where $\lambda_j$ are between 0 and 1 and sum upto 1. Now $v.a_k=\sum_j \lambda_j a_ja_k$ this will always be between 0 and 1 as all the $a_j$ are unit vectors for all $k$. But this is not a sufficient condtion. Say we have only two non zero column vector ( 2-D case ) $a_1=(1,0)$ and $a_2=(0,1)$ now if our trial vector is $v=(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$ ( projection of this vector onto your subspace is the vector itself ) then its component along $a_1,a_2$ is $\frac{1}{\sqrt{2}}$ but it does not lie inside the polytope.