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: \begin{equation} 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)\} \end{equation}

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 $Co(X)$?

Thank you all very much in advanced for your time.

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


1 Answer 1


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.

I haven't worked out your particular example.

  • $\begingroup$ Thank you very much. This is certainly a useful answer. However, given my applied background, I am not really familiar with the techniques and concepts your are mentioning and that would get my problem solved. Would you be so kind to point me to some useful (and hopefully simple) resources that could teach me how to do what I need? $\endgroup$
    – EoDmnFOr3q
    Commented Feb 6, 2018 at 15:21
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    $\begingroup$ This is a standard technique in elementary linear algebra. You will probably find a link you can learn from by searching for find rank of a matrix : google.com/… For a Master's degree in economics you will probably have to learn some linear algebra. $\endgroup$ Commented Feb 6, 2018 at 15:23
  • $\begingroup$ Thank you for your comment. I can use Mathematica to determine the number of linearly independent rows; but I don't see how that helps me to answer my original question. PS: Sure I had to learn some matrix algebra. A whole different story is if I actually managed to learn all my teachers expected me to. $\endgroup$
    – EoDmnFOr3q
    Commented Feb 6, 2018 at 15:29
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    $\begingroup$ 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. $\endgroup$ Commented Feb 6, 2018 at 15:37
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    $\begingroup$ 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. $\endgroup$ Commented Feb 6, 2018 at 15:59

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