# Prove that $\operatorname{trace}(A^{-1} \frac{dA}{dt}) = \sum_i^n \frac{\lambda_i'}{\lambda_i}$.

When proving Jacobi's formula for an invertible differentiable matrix $$A(t)$$ since $$\det A(t) = \prod_i^n \lambda_i(t)$$ where $$\lambda_i$$ are the generalized eigenvalues, we get \begin{aligned} \frac{d}{dt} \det(A(t)) & = \lambda_1' \lambda_2 \cdots \lambda_n + \cdots \lambda_1 \cdots \lambda_{n-1} \lambda_n' \\ & = (\lambda_1 \cdots \lambda_n) \left( \frac{\lambda_1'}{\lambda_1} + \cdots + \frac{\lambda_n'}{\lambda_n} \right) \\ & = \det(A) \operatorname{tr}(A^{-1} A') \end{aligned}

How does the last line follow?

If we use eigenvalue decomposition $$A(t) = P(t)D(t)P^{-1}(t)$$ where $$D(t)$$ contains the eigenvalues and $$P(t)$$ contains eigenvectors. Using chain rule and $$\frac{\delta P^{-1}}{\delta t} = -P^{-1}\frac{\delta P}{\delta t}P^{-1}$$, we get $$A'(t) = P'(t)D(t)P^{-1}(t) + P(t)D'(t)P^{-1}(t) - P(t)D(t)P^{-1}(t)P'(t)P^{-1}(t)$$ Note that the matrix product $$A^{-1}A'$$ is now expressed as $$A^{-1}A' = P(t)D^{-1}(t)P^{-1}(t)\Big(P'(t)D(t)P^{-1}(t) + P(t)D'(t)P^{-1}(t) - P(t)D(t)P^{-1}(t)P'(t)P^{-1}(t)\Big)$$ i.e. $$\text{trace} ( A^{-1}A' ) = T_1 + T_2 - T_3$$ where using linearity and cyclic properties of the trace we arrive at, $$T_1 = \text{trace}\Big(P(t)D^{-1}(t)P^{-1}(t)P'(t)D(t)P^{-1}(t) \Big) = \text{trace}\Big(P^{-1}(t)P'(t) \Big)$$ $$T_2 = \text{trace}\Big(P(t)D^{-1}(t)P^{-1}(t)P(t)D'(t)P^{-1}(t) \Big) = \text{trace}\Big(D^{-1}(t)D'(t) \Big) = \sum_k \frac{\lambda_k'}{\lambda_k}$$ $$T_3 = \text{trace}\Big(P(t)D^{-1}(t)P^{-1}(t)P(t)D(t)P^{-1}(t)P'(t)P^{-1}(t) \Big)= \text{trace}\Big(P^{-1}(t)P'(t) \Big)$$ So finally $$\text{trace} ( A^{-1}A' ) = T_2= \sum_k \frac{\lambda_k'}{\lambda_k}$$
• Ahmad, that does not work. Because $P$ depends on $t$ and because the $\lambda_i$ are not necessarily differentiable (except if the eigenvalues are pairwise distinct).
• Yes, that works when the eigenvalues are distinct. Then the matrices $P,D$ are differentiable.