How do I prove this matrices question?

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.

Answer 1

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.

Answer 2

$A^2 =\begin{bmatrix} 1 & 1 & 1 \\ 1 & \omega^2 & \omega\\ 1 & \omega & \omega^2 \end{bmatrix}\cdot \begin{bmatrix} 1 & 1 & 1 \\ 1 & \omega^2 & \omega\\ 1 & \omega & \omega^2 \end{bmatrix}=\begin{bmatrix} 3 & 0 & 0 \\ 0 &0 &3\\ 0 & 3 &0\end{bmatrix}$

$\mod \begin{bmatrix} 3-\lambda & 0 & 0 \\ 0 &-\lambda &3\\ 0 & 3 &-\lambda\end{bmatrix}=0\implies (\lambda-3)(\lambda^2-9)=0$

Can you conclude now?