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