1
$\begingroup$

Quoted from section 4.1 in the paper Aggregating Crowdsourced Binary Ratings:

... We will also denote the (scaled) top eigenvector of a matrix $M$ as $v_1(M) = \arg\min_x \|M - xx^T\|_2, x^{(1)}\ge0$.

I understand the concept of scaled leading eigenvector, but I don't understand the constraint $x^{(1)}\ge0$. Specifically, what is the meaning of the superscript $(1)$?

My first guess was that it means the first entry of $x$, but in the rest of the paper, they used the standard notation $x_i$ to represent vector entries, so I don't think this is the case.


Edit: Rodrigo de Azevedo suggests in the comments that different co-authors of the paper might use different notations, and perhaps it does mean $x_1\ge0$. If this is the case, I have the following questions:

  • Have you seen any paper (in applied mathematics and related fields) that uses this weird notation?
  • What is the reason behind the $x_1\ge0$ requirement? Can we somehow relate the $v_1(M)$ in this definition with the leading eigenvector in the "standard" definition $\arg\max_x \frac{x^TMx}{\|x\|}$?
$\endgroup$
  • $\begingroup$ Have you tried contacting the authors? $\endgroup$ – Rodrigo de Azevedo Jan 4 at 9:17
  • $\begingroup$ @RodrigodeAzevedo Not yet, but I've found a few typos and unrigorous proofs in the paper, so I guess it's some kind of mistake :( $\endgroup$ – hklel Jan 4 at 21:09
  • $\begingroup$ Four authors. Perhaps one started using $x^{(1)}$ to denote the first entry of vector $x$, and then some other decided to use $x_1$ instead. And they forgot to change the notation in one instance. Perhaps. $\endgroup$ – Rodrigo de Azevedo Jan 4 at 21:23
  • $\begingroup$ @RodrigodeAzevedo Thanks for the reply! If that's the case, can you take a look at the updated question? $\endgroup$ – hklel Jan 4 at 22:03
1
$\begingroup$

My guess is that it is the first component of the vector. This is so that the problem has a more unique solution since $xx^T = (-x)(-x)^T$. This can still fail if $x^{(1)}=0$ but depending on the matrices involved this could be very unlikely

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.