Let $q$ be a power of an odd prime number $p$. Let us fix $\overline{\mathbb F}$ an algebraic closure of $\mathbb F_q$ and let $\sigma$ denote the Frobenius relative to $q$. Consider the group of unitary matrices defined as $$\mathrm{U}_n := \{M\in \mathrm{GL}_n(\overline{\mathbb F})\,|\, MM^* = I_n\}$$ where the conjugation $M^* := \,^t M ^{\sigma}$ is considered with respect to the $\sigma$-hermitian form $$(X,Y) := \sum_{i=1}^n X_iY_i^q$$ over $\overline{\mathbb F}^n$. Note that $\mathrm U_n \subset \mathrm{GL}_n(\mathbb F_{q^2})$.
Given a matrix $M\in \mathrm U_n$, its determinant must be a $q+1$-th root of unity.
Could the same be said of every eigenvalue of $M$ ?
As I consider an eigenvalue $\lambda \in \overline{\mathbb F}$ together with an eigenvector $X\in \overline{\mathbb F}^n$, I may write that $$\lambda^{q+1}(X,X) = (\lambda X,\lambda X) = (MX, MX) = (X,X)$$ And sure enough, the eigenvalue would be a $q+1$-th root of unity if $(X,X) \not = 0$.
But what if $(X,X) = \sum_{i=1}^n X_i^{q+1} = 0$ ?
If the eigenspace is of dimension at least $2$, we could expect that $X$ could be modified within this eigenspace so that $(X,X) \not = 0$. However, if the eigenspace is just a line, this can't be done.