I'd be thankful if some could explain to me why the second equality is true…
I just can't figure it out. Maybe it's something really simple I am missing?
$\displaystyle\lim_{\epsilon\to0}\frac{\det(Id+\epsilon H)-\det(Id)}{\epsilon}=\displaystyle\lim_{\epsilon\to0}\frac{1}{\epsilon}\left[\det \begin{pmatrix}
1+\epsilon h_{11} & \epsilon h_{12} &\cdots & \epsilon h_{1n} \\
\epsilon h_{21} & 1+\epsilon h_{22} &\cdots \\
\vdots & & \ddots \\
\epsilon h_{n1} & & &1+\epsilon h_{nn}
\end{pmatrix}-1\right]$$\qquad\qquad\qquad\qquad\qquad\qquad=\displaystyle\sum_{i=1}^nh_{ii}=\text{trace}(H)$
Best Answer
Here is a conceptual proof that avoids expanding a complicated determinant:
The determinant of a linear operator (or of a square matrix) is the product of the eigenvalues, counting multiplicity. The trace of a linear operator (or of a square matrix) is the sum of the eigenvalues, counting multiplicity.
Now suppose that $\lambda_1, \dots, \lambda_n$ are the eigenvalues of $H$, counting multiplicity. Then the eigenvalues of $I + \epsilon H$ are $$ 1 + \epsilon \lambda_1, \dots, 1 + \epsilon \lambda_n, $$ counting multiplicity. Thus $$ \det(1 + \epsilon H) = (1 + \epsilon \lambda_1) \cdots (1 + \epsilon \lambda_n). $$ It is now clear that \begin{align*} \lim_{\epsilon\to 0} \frac{ \det(1 + \epsilon H) - 1}{\epsilon} &= \lambda_1 + \dots + \lambda_n\\ &= \text{trace } H, \end{align*} as desired.