Gaussian integral formula for matrix product

Question

I am looking for a way to prove that $$ \det (M \cdot N) = \det(M)\det(N) \tag{0}$$ Where $M$ and $N$ are matrices with continuous indices, so that $\det$ is a functional determinant. A way to show that $(0)$ is wrong would also be welcomed.

This question is about the following formula, $$ \int\text{d}\vec{x} \exp(- \sum_{ij}x^i A_{ij}x^j) = \left (\det A_{ij}\right )^{-1/2}\left (2\pi\right )^{D/2}. \tag{1} $$ Now, we would like this identity to be compatible with, $$ \int\text{d}\vec{x} \exp(- \sum_{ijk}x^i A_{ik}B_{kj}x^j) = \left (\det A\cdot B\right )^{-1/2}\left (2\pi\right )^{D/2} = \left (\det A\right )^{-1/2}\left (\det B\right )^{-1/2}\left (2\pi\right )^{D/2}.\tag{2} $$ Any idea how to prove this? I am interested, eventually, in the generalisation of this formula to path integrals, namely, given the path integral
$$ \int\mathcal{D}\phi \exp\left[- \int\text{d}x\text{d}y \phi(x)M(x,y)\phi(y)\right] =C \left (\det M\right )^{-1/2}, \tag{3} $$ where now $\det M$ is a functional determinant, i ask the question whether it makes sense to write the generalised formula, $$\begin{align} \int\mathcal{D}\phi \exp\left[- \int\text{d}x\text{d}y \text{d}z\phi(x)M(x,y)N(y,z)\phi(z)\right] =& \left (\det M\cdot N\right )^{-1/2}\cr =& \left (\det M\right )^{-1/2} \left (\det N\right )^{-1/2}.\end{align} \tag{4} $$

[UPDATE]: I might have an answer now: let us just consider, $$\det M\cdot N = \prod_i \lambda_i[M\cdot N],\tag{5}$$ where $\lambda_i[M\cdot N]$ are the the eigenvalues of the matrix $M\cdot N$. This formula is valid even for continuous matrices, such as the laplacian operator $\partial^2 \delta(x-y)$. If the commutator $[M,N] = 0$, then the two matrices can be diagonalised in the same basis, and $\lambda_i[M\cdot N] = \lambda_i[M]\lambda_i[N]$, with no sum over $i$. Then formula (4) can be proven at least in the simple case in which the commutator vanishes. A trivial example of this is for $M = A$ and $N = A^{-1}$, for any invertible matrix $A$, which leads to $\det A\cdot A^{-1}=1$. Also, in case $M\cdot M^T = f(x) \delta(x-y)$, this would imply that $$\det M\cdot M^T = (\det M)^2 = \det f(x) \delta(x-y) = \prod_x f(x)\tag{6}$$ and so on. These seem trivial cases, but since we are talking of functional determinants they constitute a powerful computational tool. How much do you agree with this attempt of a solution? It is not very formal, but i don't see where it could go wrong.

Answer 1

The statement seems to be wrong even for an infinite number of discrete indices.

Consider for example the vector space of square integrable functions on the positive integers, i.e. sequences $\{f_1,f_2,\cdots\}$ s.t. $\sum_{i>0} |f_i|^2 < \infty$, and consider the shift operator $S:f \mapsto Sf$, where

$Sf = \{ f_2,f_3,\cdots \} \ . $

Consider furthermore the operator $S^\dagger: f \mapsto S^\dagger f$ with

$S^\dagger f = \{ 0, f_1,f_2, \cdots \} \ . $

Now

$SS^\dagger f = \{f_1,f_2,\cdots \} = f \ \ \ \text{ but } \ \ \ S^\dagger S f = \{ 0, f_2,f_3,\dots \} \ . $

That is, $SS^\dagger$ has all eigenvalues $1$, while $S^\dagger S$ has one eigenvalue $0$ and all other eigenvalues $1$. Hence

$ \det( SS^\dagger) = 1 \ \ \ \ \ \text{while} \ \ \ \ \det (S^\dagger S ) = 0 \ .$

Now if it were true that $\det(M N) = \det(M) \det(N)$, than we would have proven that $1 = 0$.

Answer 2

The following comments seem relevant to OP's problem:

1. For a matrix $A\in{\rm Mat}_{n\times n}(\mathbb{C})$, define the symmetrized matrix $$A_+~:=~ \frac{A+A^T}{2}.\tag{A}$$

2. Then the Gaussian integral reads $$ \int_{\mathbb{R}^n} \! d^n x ~e^{-\frac{1}{2} x^T A x} ~=~ \sqrt{\frac{(2\pi)^n}{\det A_+}}\tag{B}$$ if the matrix ${\rm Re}A_+$ is positive definite, cf. e.g. this math.SE post.

3. Similarly, $$ \int_{\mathbb{R}^n} \! d^n x ~e^{-\frac{1}{2} x^T AB x} ~=~ \sqrt{\frac{(2\pi)^n}{\det (AB)_+}}\tag{C}$$ if the matrix ${\rm Re}(AB)_+$ is positive definite, cf. OP's eq. (2).