Let $A$ be a $2n \times 2n$ real symplectic matrix, i.e., $A$ satisfies $$A^TJA = J$$ where $J = \begin{pmatrix} 0 & I_n\\ -I_n & 0 \end{pmatrix}$. It is a well-known fact that $\det(A) = 1$. There are several proofs of this, including elementary ones such as outlined here and here. However, I would like to know if there is a direct way of showing that $\det(A) = 1$ from its eigenvalues, and none of the methods I found employed such an argument.
If $\lambda$ is an eigenvalue of $A$, then it can be shown that $\bar{\lambda}$, $\lambda^{-1}$, and $\bar{\lambda}^{-1}$ are also eigenvalues of $A$ (they need not be distinct from each other). At first glance, this observation seems to suggest that the eigenvalues cancel each other out when multiplied together, so that the determinant of $A$ is $1$. Nevertheless, I have yet to find a rigorous formulation of this argument.
Besides, this property of eigenvalues alone seems to be insufficient to prove our claim. As an illustration of this inadequacy, suppose $A$ is a $2 \times 2$ real symplectic matrix with eigenvalues $1$ and $-1$. Then they satisfy the property outlined above, but this implies that $\det(A) = -1$ (of course $A$ is not actually a symplectic matrix in this case). I think there are other properties of $A$ that need to be considered so that we can use the property of eigenvalues to prove the claim that $\det(A) = 1$.
In conclusion, my question is: Can we prove that $\det(A)$ by utilizing the observation that the eigenvalues of $A$ come in ''quadruplets'' of $\lambda$, $\bar{\lambda}$, $\lambda^{-1}$, and $\bar{\lambda}^{-1}$? An elementary argument with linear algebra is preferred (without Pfaffians or manifold theory), but I welcome any insight.