A limit with determinants 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?

$\displaystyle\lim_{\epsilon\to0}\frac{\det(Id+\epsilon H)-\det(Id)}{\epsilon}=\displaystyle\lim_{\epsilon\to0}\frac{1}{\epsilon}\left[\det  \begin{pmatrix}
        1+\epsilon h_{11} & \epsilon h_{12} &\cdots & \epsilon h_{1n} \\
        \epsilon h_{21} & 1+\epsilon h_{22} &\cdots \\
        \vdots &  & \ddots \\
        \epsilon h_{n1} & & &1+\epsilon h_{nn}
        \end{pmatrix}-1\right]$
$\qquad\qquad\qquad\qquad\qquad\qquad=\displaystyle\sum_{i=1}^nh_{ii}=\text{trace}(H)$

 A: As I suggested in my comment, we proceed by expanding
$$|Id + \varepsilon H| = |A| = \left| \begin{array}{cccc}
1+\varepsilon h_{11} & \varepsilon h_{12} & \cdots & \varepsilon h_{1n}\\
\varepsilon h_{21} & 1+\varepsilon h_{22} & \cdots & \ \\
\vdots & \ & \ddots & \ &\\
\varepsilon h_{n1} & \ & \ & 1+\varepsilon h_{nn}
\end{array} \right|$$
in powers of $\varepsilon$. First, we have:
$$|A| = (1+\varepsilon h_{11}) \left| \begin{array}{cccc}
1+\varepsilon h_{22} & \varepsilon h_{23} & \cdots & \varepsilon h_{2n}\\
\varepsilon h_{32} & 1+\varepsilon h_{33} & \cdots & \ \\
\vdots & \ & \ddots & \ &\\
\varepsilon h_{n2} & \ & \ & 1+\varepsilon h_{nn}
\end{array} \right| + \varepsilon\sum_{j=2}^n (-1)^{1+j} h_{1j} \det(A_{1j})$$
where $\det(A_{1j}) = O(\varepsilon)$ (here, I am using the usual notation $A_{1j}$ to be the matrix obtained by deleting the first row and $j$th column from $A$).
Justification of this part: The minimal power of $\varepsilon$ would occur when the maximal number of diagonal terms is included. In this case (dealing with $(n-1)\times (n-1)$ minors, this would mean $n-2$ diagonal terms, since all terms of the cofactor expansion along the first row (with the exception of the first one, which I am separating from the rest of the computation) would exclude the $1+\varepsilon h_{11}$, as well as the diagonal entry that sits in the $j$th column. Finally, if there are $n-2$ diagonal terms multiplied together (in the minimal case), then there must be one off-diagonal term, introducing the (claimed) factor of $\varepsilon$.
We hence conclude that $$|A| = (1+\varepsilon h_{11}) \left| \begin{array}{cccc}
1+\varepsilon h_{22} & \varepsilon h_{23} & \cdots & \varepsilon h_{2n}\\
\varepsilon h_{32} & 1+\varepsilon h_{33} & \cdots & \ \\
\vdots & \ & \ddots & \ &\\
\varepsilon h_{n2} & \ & \ & 1+\varepsilon h_{nn}
\end{array} \right| + O(\varepsilon^2)$$
Continuing (inductively) to expand the determinant in this manner, we see that
\begin{align}
|A| &= \prod_{j=1}^n (1+\varepsilon h_{jj}) + O(\varepsilon^2)\\
&= 1+\varepsilon \sum_{j=1}^n h_{jj} + O(\varepsilon^2)
\end{align}
which exactly yields the desired equality, since this immediately says 
$$|A|-1 = \varepsilon \sum_{j=1}^n h_{jj} + O(\varepsilon^2),$$
hence 
\begin{align}
\lim_{\varepsilon \to 0} \frac{1}{\varepsilon}(|A|-1) &= \lim_{\varepsilon\to 0} \frac{1}{\varepsilon} \left( \varepsilon \sum_{j=1}^n h_{jj} + O(\varepsilon^2) \right)\\
&= \lim_{\varepsilon\to 0} \left( \sum_{j=1}^n h_{jj} + O(\varepsilon) \right)\\
&= \sum_{j=1}^n h_{jj} = \text{Trace}(H)
\end{align}
A: Hint:
$$\left|\begin{matrix}
1+\epsilon h_{11}&\epsilon h_{12}\\
\epsilon h_{21}&1+\epsilon h_{22}\\
\end{matrix}\right|=1+\epsilon h_{11}+\epsilon h_{22}+\epsilon^2\left(h_{11}h_{22}-h_{12}h_{21}\right).$$
Only the main diagonal generates terms in $\epsilon$. This generalizes to higher order, for instance using the expansion by minors.
A: 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.
A: actually, you're trying to calculate the differential $\phi$ of the function $\det : M_n(\mathbb{R}) \rightarrow \mathbb{R}$ at $I_n$, which is defined as
$$\forall H \in M_n(\mathbb{R}), \phi(H) = \lim\limits_{t \to 0} \dfrac{\det(I_n+tH)-\det(I_n)}{t}=
\dfrac{d}{dt}_{t=0} \det(I_n+tH)
$$
First, since $\det$ is a polynomial, it is differentiable at any point and $\phi$ is a linear form on $M_n(\mathbb{R})$. Therefore, it suffices to calculate its value on the canonical basis of $M_n(\mathbb{R})$.


*

*For $H=E_{i,i}$,  $\det(I_n+tE_{i,i})=1+t$, so $\phi(E_{i,i})=1$.

*For $H=E_{i,j},$ where $i \neq j, \det(I_n+tE_{i,j})=1$, so $\phi(E_{i,j})=0$.


So, $\phi=\textrm{Trace}$, since these two linear forms coincide on a basis of $M_n(\mathbb{R})$.
A: Late comments.


*

*This is a special case of Jacobi's formula.

*One way to prove the problem statement using Laplace/cofactor expansion, but easier to understand than other similar approaches and without creating a mess, is to use total derivative. Let
$$
g(t_{11},t_{12},\ldots,t_{nn})=\det\pmatrix{
1+t_{11}h_{11}&t_{12}h_{12}&\cdots&t_{1n}h_{1n}\\
t_{21}h_{21}&1+t_{22}h_{22}&\cdots&t_{2n}h_{2n}\\
\vdots&&\ddots&\vdots\\
t_{n1}h_{n1}&t_{n2}h_{n2}&\cdots&1+t_{nn}h_{nn}},
$$
where each $t_{ij}=t_{ij}(\epsilon)=\epsilon$ (yes, here each $t_{ij}$s is actually identical to $\epsilon$, but we'd like to view it as a function of $\epsilon$). Let also $f(\epsilon)=g\left(t_{11}(\epsilon),t_{12}(\epsilon),\ldots,t_{nn}(\epsilon)\right)$. Then
$$
f'(\epsilon)=\sum_{i,j}\frac{\partial g}{\partial t_{ij}}\frac{\partial t_{ij}}{\partial \epsilon}=\sum_{i,j}\frac{\partial g}{\partial t_{ij}}
$$
and in turn
$$
f'(0)=\sum_{i,j}\left.\frac{\partial g}{\partial t_{ij}}\right|_{t_{ij}=0}.
$$
However, to evaluate $\left.\frac{\partial g}{\partial t_{ij}}\right|_{t_{ij}=0}$, all other $t_{rs}$ are treated as zero and we are effectively finding the derivative of $g$ with respect to a single variable $t_{ij}$. By Laplace expansion along the $i$-row, we see that
$$
\left.\frac{\partial g}{\partial t_{ij}}\right|_{t_{ij}=0}=
\begin{cases}
h_{ii}&\text{ if } i=j,\\
0&\text{ otherwise}.
\end{cases}
$$
Therefore $f'(0)=\sum_i h_{ii}=\operatorname{trace}(H)$.

