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Let $A \in \mathbb{R}^{n\times n}$ be symmetric and positive definite. What is the following maximum?

$$\max_{x\in\{\pm1\}^n}x^T A x$$

My attempt:

if all $a_{ij}\geq 0$, then

\begin{equation} \max_{x\in\{\pm1\}^n}x^TAx=\sum a_{ij} \end{equation}

and in this case $x=[1\quad 1\quad\cdots\quad 1]'$ or $x=[-1\quad -1\quad\cdots\quad -1]'$.

if some $a_{ij}<0$, then situation is not clear, but we know that $x^TAx\leq\sum{|a_{ij}|}$.

For example:

\begin{equation} A= \begin{bmatrix} 2 & 3 & 1 \\ 3 & 10 & -8 \\ 1 & -8 & 17 \end{bmatrix}. \end{equation} \begin{equation} \max_{x\in\{\pm1\}^n}x^TAx=49<\sum |a_{ij}|=53 \end{equation} and in this case $x=[1\quad1\quad-1]'$ or $x=[-1\quad-1\quad1]'$. So idea is to sacrifice $2a_{13}$ since it is smaller than $2a_{23}$ and $2a_{12}$.

Is there any systematic way to tell the maximum of $x^TAx$ for any $n$ if $A$ is given? and if yes how to find $x$ that will give the maximum? Any suggestions are welcome :)

Rodrigo de Azevedo
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Lee
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  • in the example above, spectral radius of $A$ is 22.265 But maximum is 49. Please correct me if I am wrong – Lee Nov 15 '18 at 07:39
  • so $|x_i|$=1 and $||x||_{\infty}=1$ are similar? – Lee Nov 15 '18 at 07:49
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    what I wanted to say is that all elements of $x$ are equal to 1 or -1, so I think I will better stick to $|x_i|=1$ – Lee Nov 15 '18 at 07:54
  • @Lee and Rodrigo : Both of you are right. Sorry it is my mistake still, precising that $i$ goes from $1$ to $n$ would not hurt. An alternative way to write is simply $x\in{-1,1}^n$ – Surb Nov 15 '18 at 07:56
  • ok, let me change then – Lee Nov 15 '18 at 07:57
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    Related: https://math.stackexchange.com/q/2126489/339790 – Rodrigo de Azevedo Nov 15 '18 at 07:59
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    This is known as a Binary Quadratic Problem and is hard. There are some solvers for it, for instance http://www-lipn.univ-paris13.fr/BiqCrunch/ and you can have some success solving a sequence of semidefinite relaxations. – Michal Adamaszek Nov 15 '18 at 10:00
  • @RodrigodeAzevedo Using the link you have sent, I reformulate my problem to dual problem: \begin{array}{ll} \text{minimize} & \mbox{tr} (\mathrm X)\ \text{subject to} & \mathrm X - \mathrm M \succeq \mathrm O_n\end{array} where $X$ and $M$ are positive definite matrices and $X$ is diagonal. This link provides some similar problem (https://math.stackexchange.com/questions/2250901/minimize-trace-of-a-given-that-a-n-is-positive-semi-definite), but since $X$ is diagonal in my case $tr(X)>tr(M)$, if $M$ is non-diagonal. Any suggestions please? – Lee Nov 17 '18 at 04:27
  • @MichalAdamaszek thanks for suggestion. I need to get analytical solution. I already have solution using matlab – Lee Nov 17 '18 at 04:42

1 Answers1

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I found an answer, maybe will be useful for someone.

For details go to this paper: http://users.isy.liu.se/en/rt/claal20/Publications/bipartite_consensus.pdf

Here $A$ represents an undirected graph. If $A$ is structurally balanced (condition of structurally balanced given in the paper), then $\max x^TAx = \sum |a_{ij}|$ and we can find at which corner objective function achieves maximum. However, if $A$ is not structurally balanced, $\sum |a_{ij}|$ can never be achieved. In this case, we have to go through all corners numerically to find the maximum.

For example, $\begin{equation} A= \begin{bmatrix} 2 & 3 & 1 \\ 3 & 10 & -8 \\ 1 & -8 & 17 \end{bmatrix} \end{equation}$ is structurally unbalanced so we can never achieve 53. However, $\begin{equation} A= \begin{bmatrix} 2 & 3 & -1 \\ 3 & 10 & -8 \\ -1 & -8 & 17 \end{bmatrix} \end{equation}$ is structurally balanced and for $x=[1\quad 1\quad -1]'$ or $x=[-1\quad -1\quad 1]'$ we can achieve 53.

Lee
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