# When can a polytope in $\mathbb{R}^4$ be represented as a polyhedron in $\mathbb{R}^3$ (if ever)?

Can the convex hull of a set of vectors $x_1,x_2,...,x_n$ such that $x_i\in\mathbb{R}^4$ for each $i\in N$ be a polyhedron in $\mathbb{R}^3$? I think that the answer is affirmative, but I am not sure. Other than that, let me provide particular example. Consider the convex hull given by the following vectors: $$Co(X)=Co\{(1,0,1,1),(0,0,2,1),(0,0,1,2),(0,1,1,1),(1,1,1,0),(1,1,0,1),(1,2,0,0)\}$$

Then, my specific questions are:

1. Can the convex hull $Co(X)$ be represented as a polyhedron in $\mathbb{R}^3$?
2. If the answer to the previous question is affirmative, how do we do so? In other words, how do we find equivalent vertices in $\mathbb{R}^3$ that allow us to graphically represent $Co(X)$?

PS: If possible, provide some calculations on the required steps or else try to provide some useful resources.

If what you're asking is whether that convex hull is three dimensional (lives in a three dimensional affine subset of four space) then the way to find out is to subtract one point from all the rest and determine whether the resulting set of $n-1$ points is independent. If it is, the original polytope is four dimensional. If not, its dimension is less: the dimension of that span.
• The number of independent rows is the dimension of that polytope. If it's less than $4$ you're polytope has dimension less than $4$. You should be able to do the computation by hand for this small example. Commented Feb 6, 2018 at 15:37
• Your original matrix has $4$ independent rows but the translation when you subtract the first from the rest has just $3$. Think through this example: what is the dimension of the convex hull of $(1,0),(0,1)$ in the plane? That's a one dimensional line segment even though the two points are independent vectors. If this conversation continues in comments we're likely to get a message saying "continue in chat". You can send me email (easy to find) if you need more. Commented Feb 6, 2018 at 15:59