We are giving a Polyhedron:
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The vertices or the extreme points of P is: (1,1,-3), (1,-3,1), (-3,-1,1)
We know from Linear Algebra that: Since the points are linearly independent; is the Dim(p)=3 ? - But is there no other linearly independent points in P?
In my book: Undergraduate Convexity It follows that if a set is affinely independent with m points (vectors) is the dim(P) = m-1 So by go a dimension up is the vertices of P affinely independent: (1,1,-3,1), (1,-3,1,1), (1,-3,1,1), (-3,-1,1,1) But we do need a 4th point that is affinely independent with these vectors, right? But isn't there infinitely many points in R^4 for P that is affinely independent?: Ex. the point (-1,-3,-1,1) which is in P, and is affinely independent with the 3 vertices.
So the dim(P) = m-1 1 = 4 -1 = 3 as the number of linearly independent points in P
Whats is the best way to determine the dim(P)?
You have three points, not collinear. Their convex hull is a triangle, in some plane. So it has dimension $2$.
In general with points $P_1,\ldots,P_n$, the dimension of their convex hull is the dimension of the vector space spanned by $\vec{P_1P_2},\vec{P_1P_3},\ldots,\vec{P_1P_n}$.