Showing the positivity of the determinant of $\mathfrak{sp}(n)$ without making use of diagonalization Let $\mathfrak{sp}(n)$ be the lie algebra of compact symplectic group $\mathrm{SP}(n)$, regarded as a compact form of $\mathfrak{sp}(2n,\mathbb{C})$, so we can talk about its (complex) determinant.
Let $M\in \mathfrak{sp}(n)$, then $M$ has purely imaginary eigenvalues $(ix_1,ix_2,\dots,ix_n,-ix_1,-ix_2,\dots,-ix_n)$, so
$$\det(M)=(x_1x_2\cdots x_n)^2\geq 0.$$
My question is

Is there a coordinate independent way to show that every element of $\mathfrak{sp}(n)$ has nonnegative determinant?

I would want an argument without using eigenvalues, nor anything that cannot be expressed as a function of the matrix entries.



*

*I will like to know if there is some geometric arguments.


I also want to understand the algebra behind. For $n=2$, I have tried expanding $\det(M)$, but I cannot find a way to express it as a sum of non-negative terms.
A useful way to show the positivity of an algebraic expression is to write it as a sum of terms, and each term is either a norm square, or can be shown to be non-negative by a direct application of the Cauchy-Schwarz inequality.  For example, we know $\mathrm{tr}(A^4)\geq 0$ because $\mathrm{tr}(A^4)=||A^2||^2$.
Of course we have $\det(M)=(x_1x_2\cdots x_n)^2$, but the problem $(x_1x_2\cdots x_n)$ is not expressible by $M$.




*Can $\det(M)$ be expressed a sum of such non-negative terms? If yes, what are they? If not, what are the extra ingredients we need to show the positivity apart from Cauchy Schwarz or completing squares?

Thanks in advanced!
 A: Here is a proof of $\det M\geq 0$ for $M\in \mathfrak{sp}(n)$ based on the lemma that every complex matrix is consimilar to a real matrix.
Acknowledgment: In what follows I was helped by feedback I received at MSE.
By construction, the $2n\times 2n$ complex matrix $M\in \mathfrak{sp}(n)$ is skew-Hermitian and Hamiltonian, which means that it has the $n\times n$ block decomposition
$$M=\begin{pmatrix}
A&B\\ C&-A^T\end{pmatrix},\;\;\text{with}\;\;A=-A^\ast,\;\;B=B^T=-C^\ast=-\bar{C}.$$
Here $M^T$ denotes the transpose, $\bar{M}$ the complex conjugate, and $M^\ast$ the conjugate transpose.
By continuity of the determinant it is sufficient to consider $\det A\neq 0$. Then Schur's determinant identity gives
$$\det M=\det(-AA^T-ACA^{-1}B)=\det(A\bar{A}+A\bar{B}A^{-1}B)$$
$$\qquad=\det(A\bar{A})\det(1+\bar{A}^{-1}\bar{B}A^{-1}B)$$
$$\qquad=|\det A|^2\det(1+\bar{X}X),\;\;\text{with}\;\;X=A^{-1}B.$$
Now I apply the consimilarity lemma, to write $X=SR\bar{S}^{-1}$ with $R$ a real matrix. This gives
$$\det M=|\det A|^2\det(1+\bar{S}R^2\bar{S}^{-1})=|\det A|^2\det(1+R^2)$$
$$\qquad=|\det A|^2\det(1+iR)\det(1-iR)$$
$$\quad=|\det A|^2|\det(1+iR)|^2\geq 0.$$
