Let $A\in M_3(\mathbb R)$ s.t. $\det A=1$ and one eigenvalue of $A$ is $\frac{-1+i\sqrt{3}}{2}$, find the rest of the eigenvalues.
Let $k_A(\lambda)\in \mathbb F[\lambda]$ denote the characteristic polynomial. Then,
$$k_A(\lambda)=\det(A-\lambda I),\deg k_A\le n=3$$ $A$ has at most three eigenvalues,i.e., $k_A(\lambda)$ has at most three roots. $$k_A\left(\frac{-1+i\sqrt{3}}{2}\right)=0\implies k_A\left(\frac{-1-i\sqrt{3}}{2}\right)=0$$
I first considered: $$k_A(\lambda)=\begin{vmatrix}a_{11}-\lambda&a_{12}&a_{a13}\\a_{21}&a_{22}-\lambda&a_{33}\\a_{31}&a_{32}&a_{33}-\lambda\end{vmatrix}$$ but decided to factorise:
$k_A(\lambda)=\displaystyle\sum_{i=0}^3\alpha_i\lambda_i=(-1)^3\left(\lambda-\frac{-1+i\sqrt{3}}{2}\right)\left(\lambda-\frac{-1-i\sqrt{3}}{2}\right)(\lambda-\lambda_0)\\=-(\lambda^2+\lambda+1)(\lambda-\lambda_0)=-\lambda^3+(\lambda_0-1)\lambda^2+(\lambda_0-1)\lambda+\lambda_0$,
where $\lambda_0$ denotes the third unknown eigenvalue. It seems $k_A(0)=\alpha_0=\det A=1=\lambda_0$
So, my answer is: $\frac{-1-i\sqrt{3}}{2}$ and $1$ are the remaining two eigenvalues.
Is it correct?
Alternately, the characteristic polynomial of $A$ has degree $3$ and real coefficients. Since $\frac{-1+i\sqrt{3}}{2}$ is a root, so is its complex conjugate, $\frac{-1-i\sqrt{3}}{2}$. The product of these two roots is $1$. The product of all three roots is $1$. Therefore, the third root is $1$.