Interpret the matrix entries

The given matrix is $P=[p_{ij}]$ of order $n\times n$ with $p_{ij}=\omega^{i+j},$ where $\omega$ is a complex cube root of unity and $\omega\ne 1$.

For cube roots of unity, $\omega^3=1, \qquad 1+\omega+\omega^2=0.$

Write the matrix in separable form

Since $p_{ij}=\omega^{i+j}=\omega^i\omega^j,$ define the column vector $u=\begin{bmatrix} \omega^1 \\ \omega^2 \\ \vdots \\ \omega^n \end{bmatrix}.$ Then $P=uu^T.$

So $P$ is a rank-$1$ type matrix.

Compute $P^2$

Using $P=uu^T$, $P^2=(uu^T)(uu^T)=u(u^Tu)u^T.$ Hence $P^2=(u^Tu)\,P.$

Therefore, $P^2=0$ iff $u^Tu=0.$

Evaluate $u^Tu$

We have $u^Tu=\sum_{i=1}^n \omega^{2i}.$

Since $\omega^3=1$, the powers repeat with period $3$. Also $\omega^2$ is itself a non-real cube root of unity, so $1+\omega^2+\omega^4=1+\omega^2+\omega=0.$

Thus the geometric sum $\sum_{i=1}^n \omega^{2i}$ is zero exactly when $n$ is divisible by $3$.

Check the options

• A: $57$ $57\equiv 0 \pmod 3$ So $P^2=0$.

• B: $55$ $55\equiv 1 \pmod 3$ So $P^2\ne 0$.

• C: $58$ $58\equiv 1 \pmod 3$ So $P^2\ne 0$.

• D: $56$ $56\equiv 2 \pmod 3$ So $P^2\ne 0$.

Conclusion

$P^2=0$ only when $n$ is a multiple of $3$. Among the given options, only $\boxed{57}$ satisfies this.

Comparison with stored answer

Stored correct answer is: B, C, D.

But $55,58,56$ are not divisible by $3$, so for these values the sum $\sum_{i=1}^n \omega^{2i}\ne 0$, hence $P^2\ne 0$.

Therefore, the stored answer appears to be incorrect.