0
$\begingroup$

Consider a permutation matrix $P$ and two vectors $x$, $v$ with 2-norm = 1 and all positive entries.

Are the optimal solutions $P^\ast$ of $\max_P \; (x^T \cdot Py)$ and $\min_P \; \|x-Py\|_2$ the same?

  • 0
    Minimizing $\|x-Py\|^2_2=\|x\|^2_2+\|y\|^2-2(x^T\cdot Py)$ over $P$ is equivalent to maximizing $x^T\cdot Py$.2011-10-26

1 Answers 1

1

Hint:

$$ \|x-Py\|_2^2=\|x\|_2^2+\|Py\|_2^2-2x^T\cdot Py=2(1-x^T\cdot Py) $$

  • 0
    Thanks!! One more question, if both of the vectors are able to permute freely, is the best strategy equals to sort both vectors in either ascending or descending order?2011-10-26
  • 0
    @Benjamin That approach works. In fact, you can fix one of the permutations arbitrarily, and optimize over the second permutation. (Note that $\| Px - Qy \| = \| x - P^{-1} Q y \|$.)2011-10-26
  • 0
    @SrivatsanNarayanan Thanks:)2011-10-26
  • 0
    @Benjamin That is the statement of the [Rearrangement inequality](http://en.wikipedia.org/wiki/Rearrangement_inequality). (It's not neccessary to assume x, y are unit norm and positive, by the way.)2011-11-09