For an $n \times n$ matrix, we can only have at most $n$ different eigenvalues.
Suppose I have $A = \begin{bmatrix}0 & \frac{1}{2}\\\frac{1}{2} & 0\end{bmatrix}$.
Using characteristic equation $0 = \det(A - \lambda I) = \lambda^2 - \frac{1}{4}$, I get $\lambda_1 = \lambda_2 = \pm \frac{1}{2}$.
This means I have either $\lambda_1 = \lambda_2 = \frac{1}{2}$ or $\lambda_1 = \lambda_2 = -\frac{1}{2}$. However, this gives me a total of $4 > n = 2$ eigenvalues.
Why am I getting so many eigenvalues? Shouldn't I get only 2?
Your eigenvalues are $\lambda_1 = \frac{1}{2}$ and $\lambda_2 = -\frac{1}{2}$. Note that if $\lambda$ is a root of $\det(A-\lambda I)$, the matrix $A - \lambda I $ is not invertible, hence there exists $v \in \mathbb{R}^n$ such that $(A- \lambda I)v = 0$, that is $Av = \lambda v$. So $\lambda$ is an eigenvalue. This is why we only count distinct eigenvalues, and we then talk about the dimension of their eigenspaces or their algebraic/geometric multiplicity.