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Determinant differentiation
Theorem
context | $ A,\Omega\in\mathrm{Matrix}(n,\mathbb C) $ |
context | $ A $ … invertible |
postulate | $\frac{\partial}{\partial t}|_{t=0}\ \mathrm{det}(A+t\,\Omega) = \mathrm{tr}(A^{-1}\Omega)\cdot\mathrm{det}(A) $ |
Thus
$\dfrac{\partial}{\partial t}\left|_{t=0}\right. \dfrac{\mathrm{det}(A+t\,\Omega)}{\mathrm{det}(A)} = \mathrm{tr}(A^{-1}\Omega)$
and also implies
$ \mathrm{tr}(\Omega) = \dfrac{\partial}{\partial t}\left|_{t=0}\right. \mathrm{det}(I_n+t\,\Omega) $ |
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Jacobis full formula
This comes from Jacobi's formula:
$\mathrm{det}(F(t))^{-1}{\mathrm d} \det (F(t)) = \mathrm{tr} (F(t)^{-1} {\mathrm d}F(t))$
where $F(t)$ is a parameter dependent matrix
This is a special case of the product rule and generalizes
${\mathrm d}\left(u\cdot v\right) = u\,{\mathrm d}v+v\,{\mathrm d}u = u\cdot v\left(\dfrac{1}{u}{\mathrm d}u+\dfrac{1}{v}{\mathrm d}v\right)$.
which you get for
$F(t) := \mathrm{diag}(u(t),v(t))$
The expression $\dfrac{1}{u}{\mathrm d}u$ is the so called logarithmic derivative of $u$.
Perspective
The function of $t$ on the right acts like a generating function. Of course
$ \mathrm{tr}(\Omega) = \dfrac{\partial}{\partial t}\left|_{t=0}\right. \mathrm{tr}(K+t\,\Omega+{\mathcal O}(t^2)) $
but the det-formula involves a non-linear function.
Reference
Wikipedia: Jacobi's formula