Let $\omega$ be a complex cube root of unity with $\omega\ne1$ and P=$[p_{ij}]$ be a $n$ x $n$ matrix with $p_{ij}$=$\omega^{i+j}$. Then $P^2\ne$O (null matrix), when $n=$

a) $57$

b) $55$

c) $58$

d) $56$

In this question, one thing I know is the determinant of $P$ will be zero if a row or a column repeats, for which the number of columns should be a multiple of $3$. However, is this condition sufficient for $P^2\ne0$? This is indeed giving me the answer as b,c,d, which is correct however my solution doesn't seems convincing enough.

Before answering (if you know the correct solution), please give a hint in the comments first. I will like to try it myself, and reach out to you if I was able to do it.

  • $\begingroup$ No power of $P$ will ever be zero. Do you mean $\det P^2 \neq 0$? $\endgroup$ Apr 24, 2017 at 13:19
  • $\begingroup$ @MatthewLeingang Sorry, I updated my question. $\endgroup$ Apr 24, 2017 at 13:21
  • $\begingroup$ For googling purposes, such a matrix (which has constant cross-diagonals) is known as a Hankel matrix. They arise in the context of orthogonal polynomials on the real line, so there's probably quite a bit known. (This is probably overkill, though.) $\endgroup$ Apr 24, 2017 at 13:54
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    $\begingroup$ The problem refers not to the square of the determinant, but to the square of the matrix itself, I believe it is one of the problems in the JEE (Hiss >,< ) 2013. jeeadv.iitr.ac.in/index.php/downloads $\endgroup$ Apr 24, 2017 at 14:28
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    $\begingroup$ As for a hint, try directly multiplying, factoring out a common term while evaluating the sum for each element, and exploit the fact that $1+ \omega + \omega ^2 = 0$ $\endgroup$ Apr 24, 2017 at 14:34


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