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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?

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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 $$\text{trace} ( A^{-1}A' ) = T_2= \sum_k \frac{\lambda_k'}{\lambda_k}$$

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  • $\begingroup$ 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). $\endgroup$
    – user91684
    Commented Mar 18, 2020 at 12:52
  • $\begingroup$ @loupblanc edited $\endgroup$ Commented Mar 18, 2020 at 13:02
  • $\begingroup$ Yes, that works when the eigenvalues are distinct. Then the matrices $P,D$ are differentiable. $\endgroup$
    – user91684
    Commented Mar 18, 2020 at 13:25
  • $\begingroup$ Very good computation. Many thanks for this nice answer @AhmadBazzi. $\endgroup$
    – pitonist
    Commented Mar 18, 2020 at 14:57
  • $\begingroup$ I did not know that there was an issue with differentiability. So the differentiability of the eigenvalues is not guaranteed. Hmm I will check this as well @loupblanc. $\endgroup$
    – pitonist
    Commented Mar 18, 2020 at 14:59

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