Let $\omega$ be a complex cube root of unity with $\omega
eq 1$ and $P=\left[p_{i j} ight]$ be a

Question

Let $\omega$ be a complex cube root of unity with $\omega 
eq 1$ and $P=\left[p_{i j}ight]$ be a $n 	imes n$ matrix with $p_{i j}=\omega^{i+j}$. Then $P^{2} 
eq 0$, when $n=$\

Professor PC Thomas Classes · Answer

(b, c, d): Let $P^{2}=\left[m_{i j}ight]_{n 	imes n}$ Now $m_{i j}=\sum_{l=1}^{n} p_{i l} p_{l j}$ $$ =\sum_{l=1}^{n} \omega^{i+l} \omega^{l+j}=\omega^{i+j} \sum_{l=1}^{n} \omega^{2 l} $$ $=\omega^{i+j} \cdot \omega^{2}\left[1+\omega^{2}+\omega^{4}+\ldots+\omega 2^{n-2}ight]$\ $P^{2}$ is a null matrix when $n$ is a multiple of 3 .\ Hence, $n 
eq 3 k \Rightarrow P^{2} 
eq 0$\ Thus $n$ can be 55, 56 or 58 .\