Showing the positivity of the determinant of $\mathfrak{sp}(n)$ without making use of diagonalization

Question

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.

1. 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 $M$.

2. 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!

Answer 1

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 $$\det M=|\det A|^2\det(1+\bar{S}R^2\bar{S}^{-1})=|\det A|^2\det(1+R^2)$$ $$\qquad=|\det A|^2\det(1+iR)\det(1-iR)$$ $$\quad=|\det A|^2|\det(1+iR)|^2\geq 0.$$