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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: \begin{equation} \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\}. \end{equation} \begin{equation} \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\} \end{equation} 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.

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    $\begingroup$ 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 $b$). $\endgroup$ Jul 27, 2019 at 13:59
  • $\begingroup$ 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? $\endgroup$
    – Matt
    Jul 27, 2019 at 18:36
  • $\begingroup$ You have $M=TT^\dagger$ with $T=\omega I -A$, so it is positive semidefinite. $\endgroup$
    – daw
    Jul 31, 2019 at 6:15
  • $\begingroup$ 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?) $\endgroup$
    – Matt
    Jul 31, 2019 at 9:42

1 Answer 1

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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.

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  • $\begingroup$ 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. $\endgroup$
    – Matt
    Aug 20, 2019 at 10:07
  • $\begingroup$ 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 $\endgroup$
    – Matt
    Aug 20, 2019 at 10:16

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