Computation to differentiate a determinant [closed]

Question

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.)

Answer 1

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 $(A+\lambda)^{-1}$.

Answer 2

The condition $\lambda+\mu_1>0$ ensures that $M(\lambda)=A+\lambda $ is invertible, and then one can use Jacobi's formula $$\frac{d}{d\lambda} \det M(\lambda) = \det M(\lambda) \operatorname{tr} \left(M(\lambda)^{-1} \frac{d}{dt} M(\lambda)\right),$$ to find that $$\frac{d}{d \lambda} \ln \operatorname{det}(A+\lambda)=\frac{1}{\det(A+\lambda)}\frac{d}{d\lambda}\det(A+\lambda)=\text{tr}\,(A+\lambda)^{-1}.$$ This holds even if $A$ is not Hermitian, and cannot be diagonalized, as long as $A+\lambda$ is invertible.