My question is regarding spreading $m$ vectors in an $n$ dimensional space such that the vectors are maximally distant from each other. For example, let us say I have a 2-D space, and 3 vectors, the maximally distant spread would look as follows:
If I had 4 vectors in a 3D space it would look like this:
I'd like to know if there is a generic method to calculate a spread of vectors like this. I understand that one of the vectors would have to be given in advance so that all other $m-1$ would be aligned in relation to it.
So basically, given a vector $[v_1, v_2, ..., v_n]$ in an $n$-dimensional space I'd like to know of a way to determine what is the spread of another $m-1$ vectors such that all $m$ vectors have maximum distance between them.
This is too large to be a comment.
First of all you need to define a notion of angle in your vector space. This is done in the euclidean case with the dot product, for $x,y\in\mathbb{R}^n$, this is done as: $$ \cos \theta = \frac{x\cdot y}{\|x\| \|y\|}. $$ In an arbitrary inner product vector space you could define the angle between two vectors as: $$ \cos \theta = \frac{\langle x, y\rangle}{\|x\| \|y\|}. $$ Once defined, I think what you want is to maximize the sum of the angles, i.e, given a set of $m$ unitary vectors, $\{v_i\}$, in a $n$-dimensional inner product vector space, you want to find the maximum of the function: $$ f(v_1,\dots,v_m) = \sum_{0\leq i<j\leq m} \arccos\langle v_i,v_j\rangle. $$ I think here is where your problem gets a little bit complicated, since you'll need to get some calculus involved. Write each $v_i$ in terms of a given basis (for example $v_i= a_{1i}w_1 +\cdots + a_{ni}w_n$) and maximize the above function for all such scalars.