Nonzero Square of Matrix with Complex Cube Roots of Unity

To determine the values of $n$ for which $P^2 \neq 0$, let us first calculate the entries of the matrix $P^2$.

The given matrix is $P = [p_{ij}]_{n \times n}$, where: $p_{ij} = \omega^{i+j}$ Here, $\omega$ is a complex cube root of unity with $\omega \neq 1$, which implies: $\omega^3 = 1 \quad \text{and} \quad 1 + \omega + \omega^2 = 0$

Let the product matrix be $P^2 = [q_{ik}]_{n \times n}$. The element in the $i$-th row and $k$-th column of $P^2$ is given by: $q_{ik} = \sum_{j=1}^{n} p_{ij} p_{jk} = \sum_{j=1}^{n} \omega^{i+j} \omega^{j+k} = \omega^{i+k} \sum_{j=1}^{n} \omega^{2j}$

Since $\omega^{i+k} \neq 0$ for all $i, k$, the matrix $P^2$ will be the zero matrix if and only if the scalar sum: $S = \sum_{j=1}^{n} \omega^{2j} = \omega^2 + \omega^4 + \omega^6 + \dots + \omega^{2n}$ is equal to zero.

Using the property $\omega^3 = 1$, the powers of $\omega^2$ are periodic with period 3:

• For $j = 1$: $\omega^2$
• For $j = 2$: $\omega^4 = \omega$
• For $j = 3$: $\omega^6 = 1$

The sum of any three consecutive terms is: $\omega^2 + \omega + 1 = 0$

Therefore, the sum $S$ depends on the value of $n \pmod 3$:

1. If $n = 3k$ ($n$ is a multiple of $3$): $S = k(\omega^2 + \omega + 1) = k(0) = 0 \implies P^2 = 0$

2. If $n = 3k + 1$: $S = 0 + \omega^2 = \omega^2 \neq 0 \implies P^2 \neq 0$

3. If $n = 3k + 2$: $S = 0 + \omega^2 + \omega = -1 \neq 0 \implies P^2 \neq 0$

Hence, $P^2 \neq 0$ if and only if $n$ is not a multiple of $3$ ($n \not\equiv 0 \pmod 3$).

Now, evaluating the given options:

• (A) $n = 57 = 3 \times 19 \implies n \equiv 0 \pmod 3 \implies P^2 = 0$
• (B) $n = 55 = 3 \times 18 + 1 \implies n \equiv 1 \pmod 3 \implies P^2 \neq 0$
• (C) $n = 58 = 3 \times 19 + 1 \implies n \equiv 1 \pmod 3 \implies P^2 \neq 0$
• (D) $n = 56 = 3 \times 18 + 2 \implies n \equiv 2 \pmod 3 \implies P^2 \neq 0$

Thus, $P^2 \neq 0$ when $n$ equals $55$, $58$, or $56$.

Correct Options: B, C, D