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I have been trying to prove this for a while but failed so far.

Let $A$ and $B$ are two positive, self-adjoint operators with compact resolvent on a Hilbert space $H$ defined on the same dense domain $D \subset H$ and suppose that $A^{-1}B$ extends from $D$ to a bounded operator on all of $H$. Suppose that both operators have trivial kernel. If $A^{-1}B - \mathrm{id}$ is trace-class, is then $$\det(A^{-1}B) = \prod_{j=1}^\infty \frac{\mu_j}{\lambda_j},$$ where $\lambda_1 \leq \lambda_2 \leq \dots$ are the eigenvalues of $A$ and $\mu_1 \leq \mu_2 \leq \mu_3 \leq \dots$ are the eigenvalues of $B$? In particular, do the assumptions imply that the product converges absolutely?

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  • $\begingroup$ Yes, if the operators commute. $\endgroup$
    – user1688
    Aug 25, 2015 at 16:14
  • $\begingroup$ that is obvious, yes $\endgroup$
    – Matthias Ludewig
    Aug 25, 2015 at 22:05
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    $\begingroup$ And also, you cannot expect the numeration of the eigenvallues being ascending for both operators. Say you have them ascending for one, then the other will preserve the eigenspaces and have its own eigenvalues there, so you can simultaneously diagonalize, but not both is ascending order. $\endgroup$
    – user1688
    Aug 26, 2015 at 6:10

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