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I have an optimization problem as below. I am having a hard time with the last constraint.

$\max \eta$

subject to

${\bf U}(:,m)^T{\bf A}{\bf U}(:,m)=0,m=1,2,\cdots,M$

here

$\bf{A}$ is a Binary Matrix of size $N\times N$ (given, known)

$\bf{U}$ is an optimization variable matrix $\bf U$ of size $N\times M$ (Binary matrix)

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    $\begingroup$ Welcome to OR.SE! It's a little unclear what you are asking. You said you are "having a hard time with the constraint," but what sort of trouble are you having exactly, what have you tried already, and what are you asking for help with? $\endgroup$ Jul 9, 2019 at 0:10
  • $\begingroup$ It might also help if you explain where this problem arises from (provide some context) and whether it is a homework-type problem, or part of a research project or something like that -- in other words, do you know for sure that it is possible to linearize/convexify (if that is indeed what you are asking), or are you trying to figure out whether it is possible? $\endgroup$ Jul 9, 2019 at 0:11
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    $\begingroup$ Also, if the question is only about the last constraints, maybe you can remove the other constraints and simplify the notation. For example, $x^\top A x = 0$. If you modify the notation, I will edit my answer accordingly. $\endgroup$ Jul 9, 2019 at 7:06

2 Answers 2

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The constraints $${\bf U}(:,m)^T{\bf A}{\bf U}(:,m)=0,m=1,2,\cdots,M$$ can be rewritten as $$\sum_{i=1}^N \sum_{j=1}^N A(i,j) U(i,m)U(j,m)=0,m=1,2,\cdots,M.$$

Next, you can linearize each of the $U(i,m)U(j,m)$ terms as explained here.

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    $\begingroup$ Thanks for your answer. So, we can linearize each of the $U(i,m)U(j,m)$ terms by introducing $Z(i,j,m)=U(i,m)U(j,m)$ and adding the following constraints $Z(i,:,m) \leq U(i,m)\\Z(:,j,m) \leq U(j,m)\\Z(i,j,m) \geq U(i,m) + U(j,m) - 1$ $\endgroup$ Jul 9, 2019 at 10:57
  • $\begingroup$ @dipaknarayanan That is correct! $\endgroup$ Jul 9, 2019 at 11:00
  • $\begingroup$ however, I am having difficulty in expressing the main constraint: $\sum_{i=1}^N\sum_{j=1}^NA(i,j)Z(i,j,m)=0$ $\endgroup$ Jul 9, 2019 at 11:06
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    $\begingroup$ What is the difficulty? If A(i,j) is given, this is simply a linear constraint. $\endgroup$ Jul 9, 2019 at 11:11
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Kevin Dalmeijer's answer is correct for the general case. Since $A$ is symmetric, there may be a method that involves fewer constraints. As suggested by Kevin's comment, I'm going to represent a typical equation with the simpler notation $x^T A x = 0$ (mostly to save typing).

A square matrix $A$ may have a square root $B$, such that $BB=A$. In some cases, such as when $A$ is positive semidefinite (implying symmetric), the square root is guaranteed to exist and will be symmetric ($B^T=B$). (If $A$ is positive definite, $x^TAx=0\implies x=0$ and there's not much to solve.) If your $A$ is such a matrix, you can compute the square root $B$ before solving the problem, rewrite $x^T Ax=0$ as $x^TB^TBx=0$, and observe that this is equivalent to $Bx=0$.

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  • $\begingroup$ In this problem, $\bf A$ is a binary matrix, and it is symmetric. $\endgroup$ Jul 9, 2019 at 20:45
  • $\begingroup$ @dipak narayanan Is A psd? $\endgroup$ Jul 9, 2019 at 21:00
  • $\begingroup$ @MarkL.Stone, no, it is not. eig(A) gives both positive and negative values. $\endgroup$ Jul 9, 2019 at 21:08
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    $\begingroup$ Then +1 for @prubins 's idea, but no dice in this case. $\endgroup$ Jul 9, 2019 at 21:14

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