Consider a positive Hermitian $N \times N$ matrix $A$ with complex valued coefficients. We list its eigenvalues in increasing order and with their multiplicities, $\mu\_{1} \leq \mu\_{2} \leq \cdots \leq \mu\_{N}$ and consider the one parameter family of matrices $A+\lambda$. How can I verify that for any $\lambda>-\mu\_{1}$, the following
$$
\frac{d}{d \lambda} \log (\operatorname{det}(A+\lambda))=\operatorname{trace}(A+\lambda)^{-1}
$$
holds?
(This formula motivates the definition of a relative determinant.)



Computation to differentiate a determinant
The eigenvalues of $A+\lambda$ are $\{\mu\_j+\lambda\}$ which are positive by assumption. So


$$\frac{d}{d\lambda} \log\det(A+\lambda)
 = \frac{d}{d\lambda} \sum\_j \log (\lambda+\mu\_j)
 = \sum\_j (\lambda+\mu\_j)^{-1},
$$
which is the sum of the eigenvalues (i.e. the trace) of
$(A+\lambda)^{-1}$.


