Let $ S : \mathbb{C}^3 \to \mathbb{C}^3$ be an invertible linear transformation such tha $ S^{-1} + S + 2I = 0 $. I need to find all possible eigenvalues of $ S$ and prove whether $ S $ is unitary or not. Is my proof correct?
Let $ \lambda $ be an eigenvalue of $ S $. Therefore, there exists a $ v \neq 0 $ such that $ Sv = \lambda v$.
$ S = -S^{-1} - 2I \Rightarrow (-S^{-1} - 2I)v = \lambda v \Rightarrow S^{-1}v = (-\lambda - 2)v \Rightarrow v = \lambda(-\lambda - 2)v \Rightarrow 1 = -\lambda² - 2\lambda \Rightarrow \lambda = -1$.
Let $ S $ be the linear transformation defined by $ Sv = \begin{pmatrix} -1 & 0 & 0 \\ 0 & -1 & 1 \\ 0 & 0 & -1 \end{pmatrix}v $.
Let $ E $ be the standard basis. Therefore, $ [S]_E = \begin{pmatrix} -1 & 0 & 0 \\ 0 & -1 & 1 \\ 0 & 0 & -1 \end{pmatrix}$.
Therefore, $ [S^{-1}]_E = [S]_E^{-1} = \begin{pmatrix} -1 & 0 & 0 \\ 0 & -1 & -1 \\ 0 & 0 & -1 \end{pmatrix} \Rightarrow [S]_E + [S]_E^{-1} + 2I = 0 \Rightarrow S + S^{-1} + 2I = 0$.
$ E $ is an orthonormal basis with respect to the standard inner product, and therefore, if $ S $ were unitary, then $ [S]_E $ would be unitary too, but the norm of the third column is $ \sqrt{2} $, which means it is not.
Therefore, the transformation is not necessarily unitary.