# Dealing with determinants of $n\times n$ matrices!

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.

• No power of $P$ will ever be zero. Do you mean $\det P^2 \neq 0$? Apr 24, 2017 at 13:19
• @MatthewLeingang Sorry, I updated my question. Apr 24, 2017 at 13:21
• 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.) Apr 24, 2017 at 13:54
• 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 Apr 24, 2017 at 14:28
• As for a hint, try directly multiplying, factoring out a common term while evaluating the sum for each element, and exploit the fact that $1+ \omega + \omega ^2 = 0$ Apr 24, 2017 at 14:34