| 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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This comes from Jacobi's formula:
${\mathrm d} \det (F(t)) = \det (F(t)) \mathrm{tr} (F(t)^{-1} {\mathrm d}F(t))$
where $F(t)$ is a parameter dependent matrix
This is a special case of the product rule and generalizes
${\mathrm d}\left(a\cdot b\right) = a\,{\mathrm d}b+b\,{\mathrm d}a = a\cdot b\left(\dfrac{1}{a}{\mathrm d}a+\dfrac{1}{b}{\mathrm d}b\right)$.
which you get for
$F(t) := \mathrm{diag}(a(t),b(t))$,
which can be seen $a(t)\cdot b(t)=\det\,F(t)$ representing the changing area of a rectangle.
The expression $\dfrac{1}{a}{\mathrm d}a$ is the so called logarithmic derivative of $a$ and scale invariant.
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.
Wikipedia: Jacobi's formula