Representing the determinant of a Hermitian matrix as an integral

Let $$M=\left (\omega\mathbb{I}-A\right )\left(\omega^{*}\mathbb{I}-A^{\dagger}\right)$$ be a Hermitian matrix of size $$n\times n$$ where $$A$$ is a real non symmetric matrix and $$\omega=a+\mathrm{i}b$$. $$A^{\dagger}$$ represents the conjugate transpose of $$A$$.

I want to compute $$\det[M]^{-\frac{1}{2}}$$.

I know that for a real symmetric matrix $$\Sigma$$ we can represent its determinant as a gaussian integral with real variables $$x_i$$: $$\frac{1}{|\Sigma|^{1 / 2}}=\int \frac{1}{(2 \pi)^{n / 2}} \exp \left(-\frac{1}{2}\mathbf{x}^{T} \Sigma\mathbf{x}\right)\mathrm{d}\mathbf{x}.$$

However in my case $$M$$ has complex values. I was wondering if we could extend this integral representation to Hermitian matrices. Among the feedback I got, these are the candidates: $$$$\det[M]^{-\frac{1}{2}}=\int \left ( \prod_{i} \frac{\mathrm{d} x_i}{\sqrt{2 \pi / i}}\right ) \exp \left\{-\frac{\mathrm{i}}{2} \sum_{i j }x_i\left (\sum_k\left(\omega \delta_{i k}-A_{i k}\right)\left(\omega^* \delta_{k j}-A_{k j}^T\right)\right ) x_j\right\}.$$$$ $$$$\det[M]^{-\frac{1}{2}}=\int\left(\prod_i \frac{d^{2} z_{i}}{\pi}\right) \exp \left\{-\sum_{i, j, k} z_{i}^{*}\left(\omega^{*} \delta_{i k}-J_{i k}^{T}\right)\left(\omega \delta_{k j}-J_{k j}\right) z_{j}\right\}$$$$ The second one involving complex variables seems intuitively the best suited. However I do not know whether this is correct, and I could use a simpler integral then I would prefer very much so.

Why would not this work: $$\det[M]^{-\frac{1}{2}}=\int \left ( \prod_{i} \frac{\mathrm{d} x_i}{\sqrt{2 \pi }}\right ) \exp \left\{-\frac{1}{2} \sum_{i j }x_i\left (\sum_k\left(\omega \delta_{i k}-A_{i k}\right)\left(\omega^* \delta_{k j}-A_{k j}^T\right)\right ) x_j\right\}.$$

I am very curious on what the correct way would be. Any remark or advice would be greatly appreciated!

edit: I consider the case where $$A$$ is real, and does not have complex entries anymore.

Second edit: I was told that I had to integrate over complex $$z_i$$ rather than real $$x_i$$. If this is true I would like to know why I can't use real integration.

• Your identity for the determinant of $\Sigma$ only seems to work if $\Sigma$ has all eigenvalues positive. Real and symmetric is not enough. Otherwise the integral will diverge, unless you interpret it somehow creatively. A similar-flavored condition should be made on $M$ (or $a$ and $b$). – Joonas Ilmavirta Jul 27 at 13:59
• I remember reading somewhere that if $A_{ij}$ is real then $M=(\omega \mathbb{l}-\boldsymbol{A})\left(\omega^{*} \mathbb{l}-\boldsymbol{A}^{T}\right)$ is positive semi definite, which makes the integral not diverge. However I don't see how I could prove it? Admitting it is true, the only concern would be when the eigenvalue is zero, (when $\omega$ is an eigenvalue of $A$) which makes the determinant not defined. This could be avoided by adding a small $\epsilon$ in the diagonal. Thus my identity could work and I don't need to have recourse to this $\mathrm{i}$ factor in my integral? – Matt Jul 27 at 18:36
• You have $M=TT^\dagger$ with $T=\omega I -A$, so it is positive semidefinite. – daw Jul 31 at 6:15
• Therefore I can simply use a real gaussian integral to represent the determinant of $M$? (and adding a small $\epsilon$ in the diagonal in order to prevent $0$ eigenvalues?) – Matt Jul 31 at 9:42

The following 2 Gaussian integral representations seem relevant to OP's problem:

1. Given a symmetric complex matrix $$A\in{\rm Mat}_{n\times n}(\mathbb{C})$$

• (i) such that the matrix $${\rm Re}A$$ is positive definite,

• (ii) or such that $$A$$ is an invertible imaginary matrix,

then the Gaussian integral is well-defined and is given by $$\int_{\mathbb{R}^n} \! d^n x ~e^{-\frac{1}{2} x^T A x} ~=~ \sqrt{\frac{(2\pi)^n}{\det A}}.\tag{1}$$ See e.g. this related Phys.SE post.

2. Given an invertible real matrix $$A\in{\rm Mat}_{n\times n}(\mathbb{R})$$, then $$\int_{\mathbb{R}^{2n}} \! d^n x~d^ny ~e^{ix^T A y} ~=~ \frac{(2\pi)^n}{|\det A|}.\tag{2}$$

Case (2) reduces to the case (1.ii) if we build the symmetric imaginary matrix $$\begin{pmatrix} 0 & -iA \cr -iA^T & 0 \end{pmatrix}$$ of twice size.

• Where could I find a proof of the second integral? When I try to compute it seems not to converge. First integrating $\vec{x}$ I get something in the lines of $\left[-\frac{\mathrm{i}e^{\mathrm{i}Axy}}{Ay}\right]_{x=-\infty}^{x=\infty}$ which does not seem to converge. – Matt Aug 20 at 10:07
• I mean, contrary to the previous integrals, none of the integrating variables seem coupled, we can do all integrals separately which I feel is a bit surprising – Matt Aug 20 at 10:16