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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) $
Jacobis full formula

This comes from Jacobi's formula:

${\mathrm d} \log\left(\det (F(t))\right) = \mathrm{tr} (F(t)^{-1} {\mathrm d}F(t))$

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

Parents

Special case of

Context

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