# determinant identity for invertible finite rank operators

I am currently reading a paper where the following identity, valid for an invertible finite - rank operator $T \colon \mathscr{H} \to \mathscr{H}$ on a separable Hilbert space, is given: $$\log \det T = - \frac{d}{dz} \{\text{Tr} \,T^{-z}\}|_{z = 0}$$ Here, $\text{Tr}$ stands for the trace defined on trace class operators, and $z$ is a complex variable. I would like to understand this identity better, in particular I would like to know whether the definition of the complex power on the left hand side is based on a functional integral in general, or whether there are alternative definitions for this. Is there a good source where I could read up on this?

Many thanks!

• If $T$ really is invertible with finite rank, your $H$ is finite-dimensional. I assume the assumption is rather $T-I$ is trace-class. To make sense of $\det T$ as the Fredholm determinant in the first place. Then this boils down to $\log \det (I+S)=\mathrm{Tr} \log (I+S)$ for every trace-class $S$. Commented Dec 15, 2013 at 0:53
• @julien yes the Hilbert space is finite dimensional in the first case. How is the complex power then defined in the case of a trace class operator, is there a source you'd recommend to read up on this? Many thanks! Commented Dec 15, 2013 at 8:38
• In the first case? I can only see one case in what you wrote. And if this is finite-dimensional, it does not make sense to mention the trace class. I'm confused. Commented Dec 15, 2013 at 14:17

In general you would use the holomorphic functional calculus $$f(T)=\frac1{2\pi i}\,\int_\Gamma \,f(z)\,(z-T)^{-1}\,dz,$$ where $\Gamma$ is a curve in the complex plane with the spectrum of $T$ inside the region it delimits, and $f$ is an analytic function on that region.
Now, if $H$ is finite-dimensional, things are way easier. Then you have $T=SJS^{-1}$ with $J$ the Jordan form, so that it has the eigenvalues $\lambda_1,\ldots,\lambda_n$ in the diagonal. Then $J^{-z}$ has $\lambda_1^{-z},\ldots,\lambda_n^{-z}$ in the diagonal, so $$\mbox{Tr}(T^{-z})=\mbox{Tr}(J^{-z})=\sum_{j=1}^n\lambda_j^{-z},$$ and $$\left.\frac{d}{dz}\right|_{z=0}\,\mbox{Tr}(T^{-z})=\left.\frac{d}{dz}\right|_{z=0}\sum_{j=1}^n\lambda_j^{-z}=-\left.\sum_{j=1}^n\lambda_j^{-z}\log\lambda_j\right|_{z=0}\, =-\sum_{j=1}^n\log\lambda_j\\ =-\log\left(\prod_{j=1}^n\lambda_j\right) =-\log\det T$$