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Perhaps the question is rather vague, but if we are given a non-convex set S, can we construct some invertible mapping f so that f(x) becomes a convex set?

In my problem , I am given a set of matrix that are all nonnegative and have the same nonnegative rank. I suppose this is a non convex set, but i want to come up with a invertible function f so that when do the entrywise mapping on each matrix, we get a convex set of matrix. Any suggestion ?

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  • $\begingroup$ Perhaps you left out a negation: "if we are given a non-convex set $S$..."? $\endgroup$
    – Joseph O'Rourke
    Feb 16, 2019 at 2:14
  • $\begingroup$ You need to require something about f. "An invertible function" is too general to make this a nontrivial question. $\endgroup$
    – Michael Renardy
    Feb 16, 2019 at 2:17
  • $\begingroup$ Perhaps this could help? Bertsekas, D. P. "Convexification procedures and decomposition methods for nonconvex optimization problems." Journal of Optimization Theory and Applications 29, no. 2 (1979): 169-197. PDF download. $\endgroup$
    – Joseph O'Rourke
    Feb 16, 2019 at 2:20
  • $\begingroup$ Assume $S$ has cardinality continuum, then take a bijection to $[0,1]$. Clearly you mean continuous and invertible? If $S$ is countable and has more than one point then I don't think it can be done :-) $\endgroup$
    – David Roberts ♦
    Feb 16, 2019 at 2:34
  • $\begingroup$ @DavidRoberts but does the bijection guarantee the resulting set of matrices convex? Just an emphasis, im doing a entrywise map via this invertible function. $\endgroup$
    – 蔣聞哲
    Feb 19, 2019 at 16:36

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For instance, the set of positive rank 1 $n\times n$ matrices is homeomorphic to $\mathbb{R}^n\setminus \{0\}$ modulo identification of $x$ with $-x$, or $\mathbb{C}^n\setminus \{0\}$ modulo the action of the circle group. For $n>1$ these spaces have nontrivial homology so cannot be homeomorphic to a convex set.

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  • $\begingroup$ hi, im not familiar with this homeomorphic topic. can you elaborate more on the impossibility of homeomorphic to a convex set? thanks ! $\endgroup$
    – 蔣聞哲
    Feb 19, 2019 at 16:33
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    $\begingroup$ I suspect you will need to learn a lot more math before this will make sense to you. $\endgroup$
    – Nik Weaver
    Feb 19, 2019 at 19:18
  • $\begingroup$ 'the set of positive rank 1 𝑛×𝑛 matrices is homeomorphic to ℝ𝑛∖{0} modulo identification of 𝑥 with −𝑥', can I understand this by seeing that any rank 1postive matrix can be written as the outer product of some vector in ℝ𝑛, so the double continuous bijection map would be a map from matrix in outer product form to that vector? And since we're considering positive matrices, we modulo out the identification of 𝑥 with −𝑥. What about rank-1 matrices that is n by m ? Thanks a lot ! $\endgroup$
    – 蔣聞哲
    Feb 20, 2019 at 18:00

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