Discussion Forum

Let ω be a complex cube root of unity with ω ≠ 1 and P=[p_ij] be a n x n matrix with p_ij  =ω ^i+j. Then, P²≠ 0, when n is equal to 

Answer:

Explanation:

Here , P=[pij]_nxn with p_ij=w^i+j

 $\therefore$ when n=1

 $P=[p_{ij}]_{1\times1}=[w^{2}]\Rightarrow p^{2}=[w^{4}]\neq0$

 $\therefore$ when n=2

$P=[p_{ij]_{2\times2}}=\begin{bmatrix}p_{11} & p_{12} \\p_{21} & p_{22} \end{bmatrix}=\begin{bmatrix}w^{2} & w^{3} \\w^{3} & w^{4} \end{bmatrix}=\begin{bmatrix}w^{2} & 1 \\1 & w \end{bmatrix}$

$p^{2}=\begin{bmatrix}w^{2} & 1 \\1 & w \end{bmatrix}\begin{bmatrix}w^{2} & 1 \\1 & w \end{bmatrix}$

$p^{2}=\begin{bmatrix}w^{4}+1 & w^{2}+w \\w^{2}+w & 1+w^{2} \end{bmatrix}\neq0$

when n=3

$p=[p_{ij}]_{3\times3}=\begin{bmatrix}w^{2} & w^{3}&w^{4} \\w^{3} & w^{4}&w^{5}\\ w^{4}&w^{5}&w^{6} \end{bmatrix}$

    $=\begin{bmatrix}w^{2} & 1&w \\1& w&w^{2}\\ w&w^{2}&1 \end{bmatrix}$

 $p^{2}=\begin{bmatrix}w^{2} & 1&w \\1& w&w^{2}\\ w&w^{2}&1 \end{bmatrix}\begin{bmatrix}w^{2} & 1&w \\1& w&w^{2}\\ w&w^{2}&1 \end{bmatrix}$

$=\begin{bmatrix}0 & 0&0 \\0& 0&0\\ 0&0&0 \end{bmatrix}=0$

$\therefore$  $p^{2}$=0, when n  is mutilple of 3

$p^{2}\neq0$  , when n is not a multiple of 3

 $\Rightarrow$ n=57 is not possible

 $\therefore$   n=55,58,56 is possible