Relation of eigenvalues of a matrix $A^2$ , where $A$ is a matrix consisting cube roots of unity Matrix $A$ is
$$
        \begin{pmatrix}
        1 & 1 & 1 \\
        1 & w^2 & w \\
        1 & w & w^2 \\
        \end{pmatrix}
$$
Where $1,w,w^2$ are cube roots of unity.
My problem is to obtain the result $|\lambda_1| + |\lambda_2| +|\lambda_3| \le 9$, where, $\lambda_1 , \lambda_2, \lambda_3$ are eigenvalues of  matrix $A^2$.
My approach was, to use the fact that eigenvalues of $A^2$ are squares of eigenvalues of $A$. But I couldn't find a way to relate them through an in equation. 
 A: In general, fix $n\in\mathbb{N}$ and pick a primitive $n$-th root of unity $\omega\in \mathbb{C}$.  Define $$\mathbf{A}:=\left[\omega^{ij}\right]_{i,j\in\{0,1,2,\ldots,n-1\}}\,.$$
Then, we have
$$\mathbf{A}^2=\left[\sum_{\mu=0}^{n-1}\,\omega^{(i+j)\mu}\right]_{i,j\in\{0,1,2,\ldots,n-1\}}\,.$$
Ergo, $\mathbf{A}^2$ is given by the $n$-by-$n$ permutation matrix
$$\mathbf{A}^2=n\,\left[\begin{array}{c|cccccc}
1&0&0&\cdots&0&0&0\\
\hline
0&0&0&\cdots&0&0&1\\
0&0&0&\cdots&0&1&0\\
0&0&0&\cdots&1&0&0\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\
0&0&1&\cdots&0&0&0\\
0&1&0&\cdots&0&0&0
\end{array}\right]\,.$$
Therefore,
$$\mathbf{A}^4=n^2\,\mathbf{I}_n$$
where $\mathbf{I}_n$ is the $n$-by-$n$ identity matrix.  Hence, all eigenvalues of $\mathbf{A}$ must belong to the set $\big\{+\sqrt{n},-\sqrt{n},+\sqrt{n}\text{i},-\sqrt{n}\text{i}\}$.
This means the eigenvalues of $\mathbf{A}^2$ lie in the set $\{+n,-n\}$.  In fact, since 
$$\text{Trace}\left(\mathbf{A}^2\right)=\begin{cases}n\,,&\mbox{if }n\mbox{ is odd}\,,\\
2n\,,&\mbox{if }n\mbox{ is even}\,,\end{cases}$$
we see that the eigenvalue $+n$ of $\mathbf{A}^2$ has multiplicity $\left\lceil\frac{n+1}{2}\right\rceil$, whereas the eigenvalue $-n$ of $\mathbf{A}^2$ has multiplicity $\left\lfloor\frac{n-1}{2}\right\rfloor$.
