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Suppose we have $m$ square matrices $M_1, M_2, \dots, M_m$ and a new matrix $M$ is created by a convex combination of $M_1, M_2, \dots, M_m$

$$M := \sum_{k=1}^m w_k M_k$$

where $\sum_{k=1}^m w_k = 1$ and $w_k \geq 0$.

How to find the weights $w_1,w_2,\dots, w_m$ for which as many main diagonal elements in $M$ as possible will be the smallest in their corresponding columns?

Could there be some theoretical results for this optimization problem?

  • 1
    For a specific column, the set of $w$ for which the main diagonal element is the smallest in that column is polyhedral. Now you want to find a point that is in as many polyhedrons as possible. My hunch is that this is NP-hard.2017-01-21
  • 0
    All elements in $M_1, M_2,..M_m$ are positive values.2017-01-22

1 Answers 1

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Given $n$ matrices $\mathrm A_1, \mathrm A_2, \dots, \mathrm A_n \in \mathbb R^{m \times m}$, we would like to find the real weight vector $\mathrm x \geq 0_n$ with $1_n^{\top} \mathrm x = 1$ such that the convex combination of these $n$ matrices

$$\mathrm A (\mathrm x) := \sum_{k=1}^n x_k \mathrm A_k$$

has the maximal number of columns where the entry on the main diagonal is the minimum of that column. If the $k$-th column of $\mathrm A (\mathrm x)$ has this property, then

$$(\mathrm e_i - \mathrm e_k)^{\top} \mathrm A (\mathrm x) \, \mathrm e_k \geq 0$$

for all $i \in \{1, 2, \dots, m\} \setminus \{k\}$, where $\mathrm e_i$ is the $i$-th vector of the standard basis. Rewriting the linear inequality above,

$$\left( (\mathrm e_i - \mathrm e_k)^{\top} \mathrm A_1 \, \mathrm e_k \right) x_1 + \left( (\mathrm e_i - \mathrm e_k)^{\top} \mathrm A_2 \, \mathrm e_k \right) x_2 + \cdots + \left( (\mathrm e_i - \mathrm e_k)^{\top} \mathrm A_n \, \mathrm e_k \right) x_n \geq 0$$

Writing the $m-1$ linear inequalities corresponding to column $k$ in matrix form, we obtain

$$\mathrm C_k \mathrm x \geq 0_{m-1}$$

Note that these $m-1$ linear inequalities can be "deactivated" by putting large negative numbers in the right-hand side. Let $\epsilon > 0$ be a very small positive real. Deactivating the linear inequalities,

$$\mathrm C_k \mathrm x \geq - \epsilon^{-1} 1_{m-1}$$

We now introduce a vector of binary decision variables, $\mathrm y \in \{0,1\}^m$, where $y_k = 1$ means that the $m-1$ linear equality constraints corresponding to column $k$ have been deactivated. As we would like to deactivate the least number of columns, we have the mixed-integer program (MIP)

$$\begin{array}{ll} \text{minimize} & \mathrm 1_m^{\top} \mathrm y\\ \text{subject to} & \mathrm C_1 \mathrm x + \epsilon^{-1} 1_{m-1} \mathrm e_1^{\top} \mathrm y \geq 0_{m-1}\\ & \mathrm C_2 \mathrm x + \epsilon^{-1} 1_{m-1} \mathrm e_2^{\top} \mathrm y \geq 0_{m-1}\\ & \qquad\qquad\vdots\\ & \mathrm C_m \mathrm x + \epsilon^{-1} 1_{m-1} \mathrm e_m^{\top} \mathrm y \geq 0_{m-1}\\ & 1_n^{\top} \mathrm x = 1\\ & \mathrm x \geq 0_n\\ & \mathrm y \in \{0,1\}^m\end{array}$$

  • 0
    One can solve MIPs using, say, [PuLP](https://www.coin-or.org/PuLP/).2017-02-02