I'm attempting to find the eigenvalues and eigenvectors from the following matrix :
\begin{pmatrix} -3\cos\theta&\sqrt{3}\sin\theta e^{iφ}&0&0\\\sqrt{3}\sin\theta e^{-iφ}&-\cos\theta&\sqrt{3}\sin\theta e^{iφ}&0\\0&\sqrt{3}\sin\theta e^{-iφ}&\cos\theta&\sqrt{3}\sin\theta e^{iφ}\\0&0&\sqrt{3}\sin\theta e^{-iφ}&3\cos\theta \end{pmatrix}
I can get as far as :
\begin{eqnarray} -\left(3\cos\theta+λ\right)\left(\cos\theta+λ\right)\left[\left(\cos\theta-λ\right)\left(3\cos\theta-λ\right)-3\sin^2\theta\right]+\\ -4\sin^2\theta\left(3\cos\theta-λ\right)\left(3\cos\theta+λ\right)+\sqrt{3}\sin\theta e^{-iφ}\left[\left(\cos\theta-λ\right)\left(3\cos\theta -λ\right)-3\sin^2\theta\right]=0 \end{eqnarray} However, the solution I'm provided with states that after some tedious algebraic manipulation I should arrive at : \begin{eqnarray} \left(\left(3\cos\theta-λ\right)\left(\cos\theta-λ\right)-3\sin^2\theta \right)\left(\left(3\cos\theta+λ\right)\left(\cos\left(\theta \right)+λ\right)-3\sin^2\theta \right)+ \\+4\sin^2\theta\left(3\cos\theta+λ\right)\left(3\cos\theta-λ\right)=0 \end{eqnarray} I'm having difficulty figuring out the necessary algebra between these two steps, where I assume I need to eliminate $e^{-i\phi}$ with the help of Euler's formula, but I still hit a brick wall.
Any suggestions?
UPDATE
Figured it out. I didn't multiply by my initial coefficients, as shown below, which end up cancelling $e^{-i\phi}$ and with a bit of rearranging I get the desired result.
\begin{eqnarray} \left(3cos\left(\theta\right)+λ\right)\{-\left(cos\left(\theta \right)-λ\right)\left[\left(3cos\left(\theta\right)-λ\right)\left(cos\left(\theta\right)-λ\right)-3sin^2\left(\theta \right)\right]-4sin^2\left(\theta\right)\left(3cos\left(\theta \right)-λ\right)\}+\sqrt{3}sin\left(\theta \right)e^{iϕ}\{\sqrt{3}sin\left(\theta \right)e^{-iϕ}\left[\left(3cos\left(\theta\right)-λ\right)\left(cos\left(\theta \right)-λ\right)-3sin^2\left(λ\right)\right]\}=0 \end{eqnarray}
Thanks for the assistance and pointing out my errors.
Your first equation is not correct. If you plug $\theta = 0$, the matrix becomes $\operatorname{diag}(-3,-1,1,3)$ and so the characteristic equation should be $(\lambda - 3)(\lambda + 3)(\lambda - 1)(\lambda + 1) = 0$. This is indeed consistent with the second equation but from the first equation you get
$$ -(1 + \lambda)(1 - \lambda)(3 - \lambda) = 0 $$
so you have made an error somewhere in your calculation of the determinant.