How do I prove this matrices question? Let $$
A =
 \begin{pmatrix}
  1 & 1 & 1 \\
  1 & \omega^2 & \omega\\
  1 & \omega & \omega^2
 \end{pmatrix}$$
where $\omega \ne 1$ is a cube root of unity. If $\lambda_1,\lambda_2,\lambda_3$ denote the eigen values of $A^2, $ show that $|\lambda_1|+|\lambda_2|+|\lambda_3|\le9$
In my attempt, I found out the matrix $A^2.$ I know that sum of the eigen values is equal to the trace of the matrix. So, that gives $\lambda_1+\lambda_2+\lambda_3=3$. However, it doesn't seem the right way to reach the answer. Need some help.
 A: Assume $A$ is diagonalizable (disclaimer: I don't prove it here). Eigenvalues of conjungate matrix $A^*$ will be conjungates of eigenvalues of $A$ with same eigenvectors, so eigenvalues of their product will be squared modules of eigenvalues of $A$. It's easy to see that
$$
AA^* =
 \begin{pmatrix}
  1 & 1 & 1 \\
  1 & \omega^2 & \omega\\
  1 & \omega & \omega^2
 \end{pmatrix}
\begin{pmatrix}
  1 & 1 & 1 \\
  1 & \omega & \omega^2\\
  1 & \omega^2 & \omega
 \end{pmatrix}=
\begin{pmatrix}
  3 & * & * \\
  * & 3 & *\\
  * & * & 3
 \end{pmatrix}
$$
(no need to compute starred terms). Hence, if we denote eigenvalues of $A$ by $\tau_n$, we have
$$
\left|\tau_1\right|^2+\left|\tau_1\right|^2+\left|\tau_1\right|^2=9
$$
Since eigenvalues of $A^2$ are $\tau_n^2$, and $\left|\tau_n^2\right|=\left|\tau_n\right|^2$, we have proved the desired inequality.
A: $A^2 =\begin{bmatrix} 1 & 1 & 1 \\ 1 & \omega^2 & \omega\\ 1 & \omega & \omega^2
 \end{bmatrix}\cdot \begin{bmatrix}
  1 & 1 & 1 \\
  1 & \omega^2 & \omega\\
  1 & \omega & \omega^2
 \end{bmatrix}=\begin{bmatrix} 3 & 0 & 0 \\
0 &0 &3\\ 0 & 3 &0\end{bmatrix}$
$\mod \begin{bmatrix} 3-\lambda & 0 & 0 \\
0 &-\lambda &3\\ 0 & 3 &-\lambda\end{bmatrix}=0\implies (\lambda-3)(\lambda^2-9)=0$
Can you conclude now?
